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https://ru.scribd.com/document/296540847/Statics-an | [
"Вы находитесь на странице: 1из 13\n\nPROBLEM 4.\n\n3\nA T-shaped bracket supports the four loads shown. Determine the\nreactions at A and B (a) if a = 10 in., (b) if a = 7 in.\n\nSOLUTION\nFree-Body Diagram:\n\nFx = 0: Bx = 0\nM B = 0: (40 lb)(6 in.) (30 lb)a (10 lb)(a + 8 in.) + (12 in.) A = 0\n\n(40a 160)\n12\n\nA=\n\n(1)\n\nM A = 0: (40 lb)(6 in.) (50 lb)(12 in.) (30 lb)(a + 12 in.)\n\n(10 lb)(a + 20 in.) + (12 in.) B y = 0\nBy =\n\nBx = 0, B =\n\nSince\n(a)\n\n(b)\n\n(1400 + 40a)\n12\n(1400 + 40a )\n12\n\n(2)\n\nFor a = 10 in.,\n\nEq. (1):\n\nA=\n\n(40 10 160)\n= +20.0 lb\n12\n\nEq. (2):\n\nB=\n\n(1400 + 40 10)\n= +150.0 lb\n12\n\nB = 150.0 lb W\n\nEq. (1):\n\nA=\n\n(40 7 160)\n= +10.00 lb\n12\n\nA = 10.00 lb W\n\nEq. (2):\n\nB=\n\n(1400 + 40 7)\n= +140.0 lb\n12\n\nB = 140.0 lb W\n\nA = 20.0 lb W\n\nFor a = 7 in.,\n\nPROPRIETARY MATERIAL. 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,\nreproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited\ndistribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,\nyou are using it without permission.\n347\n\nPROBLEM 4.15\nThe bracket BCD is hinged at C and attached to a control\ncable at B. For the loading shown, determine (a) the tension\nin the cable, (b) the reaction at C.\n\nSOLUTION\n\nAt B:\n\nTy\nTx\n\n0.18 m\n0.24 m\n\n3\nTy = Tx\n4\n\n(a)\n\n(1)\n\nTx = +1600 N\n\nFrom Eq. (1):\n\nTy =\n\n3\n(1600 N) = 1200 N\n4\n\nT = Tx2 + Ty2 = 16002 + 12002 = 2000 N\n\n(b)\n\nT = 2.00 kN W\n\nFx = 0: Cx Tx = 0\nCx 1600 N = 0 C x = +1600 N\n\nC x = 1600 N\n\nFy = 0: C y Ty 240 N 240 N = 0\nC y 1200 N 480 N = 0\nC y = +1680 N\n\nC y = 1680 N\n\n= 46.4\nC = 2320 N\n\nC = 2.32 kN\n\n46.4 W\n\nPROPRIETARY MATERIAL. 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,\nreproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited\ndistribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,\nyou are using it without permission.\n359\n\nPROBLEM 4.31\nNeglecting friction, determine the tension in cable ABD and the\nreaction at support C.\n\nSOLUTION\nFree-Body Diagram:\n\nM C = 0: T (0.25 m) T (0.1 m) (120 N)(0.1 m) = 0\n\nFx = 0: C x 80 N = 0\n\nC x = +80 N\n\nFy = 0: C y 120 N + 80 N = 0\n\nC y = +40 N\n\nT = 80.0 N W\nC x = 80.0 N\nC y = 40.0 N\nC = 89.4 N\n\n26.6 W\n\nPROPRIETARY MATERIAL. 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,\nreproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited\ndistribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,\nyou are using it without permission.\n379\n\nPROBLEM 4.38\nA light rod AD is supported by frictionless pegs at B and C and\nrests against a frictionless wall at A. A vertical 120-lb force is\napplied at D. Determine the reactions at A, B, and C.\n\nSOLUTION\nFree-Body Diagram:\n\nA = 69.28 lb\n\nA = 69.3 lb\n\nM B = 0: C (8 in.) (120 lb)(16 in.) cos 30\n\n+ (69.28 lb)(8 in.)sin 30 = 0\nC = 173.2 lb\n\nC = 173.2 lb\n\n60.0 W\n\nB = 34.6 lb\n\n60.0 W\n\nM C = 0: B(8 in.) (120 lb)(8 in.) cos 30\n\n+ (69.28 lb)(16 in.) sin 30 = 0\nB = 34.6 lb\n\nCheck:\n\nFy = 0: 173.2 34.6 (69.28)sin 30 (120)sin 60 = 0\n\n0 = 0 (check) \u0003\n\nPROPRIETARY MATERIAL. 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,\nreproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited\ndistribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,\nyou are using it without permission.\n386\n\nPROBLEM 4.1\nTwo crates, each of mass 350 kg, are placed as shown in the\nbed of a 1400-kg pickup truck. Determine the reactions at each\nof the two (a) rear wheels A, (b) front wheels B.\n\nSOLUTION\nFree-Body Diagram:\n\nW = (350 kg)(9.81 m/s 2 ) = 3.4335 kN\n\nWt = (1400 kg)(9.81 m/s 2 ) = 13.7340 kN\n\n(a)\n\nRear wheels:\n\n(3.4335 kN)(3.75 m) + (3.4335 kN)(2.05 m)\n\n+ (13.7340 kN)(1.2 m) 2 A(3 m) = 0\n\nA = +6.0659 kN\n\n(b)\n\nFront wheels:\n\nA = 6.07 kN W\n\nFy = 0: W W Wt + 2 A + 2 B = 0\n3.4335 kN 3.4335 kN 13.7340 kN + 2(6.0659 kN) + 2B = 0\nB = +4.2346 kN\n\nB = 4.23 kN W\n\nPROPRIETARY MATERIAL. 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,\nreproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited\ndistribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,\nyou are using it without permission.\n345\n\nPROBLEM 4.5\nA hand truck is used to move two kegs, each of mass 40 kg.\nNeglecting the mass of the hand truck, determine (a) the vertical\nforce P that should be applied to the handle to maintain\nequilibrium when = 35, (b) the corresponding reaction at each\nof the two wheels.\n\nSOLUTION\nFree-Body Diagram:\nW = mg = (40 kg)(9.81 m/s 2 ) = 392.40 N\na1 = (300 mm)sin (80 mm)cos\na2 = (430 mm)cos (300 mm)sin\nb = (930 mm)cos\n\nFrom free-body diagram of hand truck,\n\nDimensions in mm\n\nM B = 0: P(b) W ( a2 ) + W (a1 ) = 0\n\n(1)\n\nFy = 0: P 2W + 2 B = 0\n\n(2)\n\n= 35\n\nFor\n\na1 = 300sin 35 80 cos 35 = 106.541 mm\n\na2 = 430 cos 35 300sin 35 = 180.162 mm\nb = 930cos 35 = 761.81 mm\n\n(a)\n\nFrom Equation (1):\n\nP(761.81 mm) 392.40 N(180.162 mm) + 392.40 N(106.54 mm) = 0\nP = 37.921 N\n\n(b)\n\nor P = 37.9 N W\n\nFrom Equation (2):\n\n37.921 N 2(392.40 N) + 2 B = 0\n\nor B = 373 N W\u0003\n\nPROPRIETARY MATERIAL. 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,\nreproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited\ndistribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,\nyou are using it without permission.\n349\n\nPROBLEM 4.11\nThree loads are applied as shown to a light beam supported by\ncables attached at B and D. Neglecting the weight of the beam,\ndetermine the range of values of Q for which neither cable\nbecomes slack when P = 0.\n\nSOLUTION\n\nM B = 0: (3.00 kN)(0.500 m) + TD (2.25 m) Q (3.00 m) = 0\n\nQ = 0.500 kN + (0.750) TD\n\n(1)\n\nM D = 0: (3.00 kN)(2.75 m) TB (2.25 m) Q(0.750 m) = 0\n\nQ = 11.00 kN (3.00) TB\n\n(2)\n\nQ 11.00 kN\n\nQ 0.500 kN\n\nFor neither cable to be slack,\n\n0.500 kN Q 11.00 kN W\n\nPROPRIETARY MATERIAL. 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,\nreproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited\ndistribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,\nyou are using it without permission.\n355\n\nPROBLEM 4.23\nDetermine the reactions at A and B when (a) h = 0,\n(b) h = 200 mm.\n\nSOLUTION\nFree-Body Diagram:\n\n37.5\nB=\n0.25 0.866h\n\n(a)\n\n(1)\n\nWhen h = 0,\nB=\n\nFrom Eq. (1):\n\n37.5\n= 150 N\n0.25\n\nB = 150.0 N\n\n30.0 W\n\nFy = 0: Ax B sin 60 = 0\nAx = (150)sin 60 = 129.9 N\n\nA x = 129.9 N\n\nFy = 0: Ay 150 + B cos 60 = 0\nAy = 150 (150) cos 60 = 75 N\n\nA y = 75 N\n\n= 30\nA = 150.0 N\n\n(b)\n\nA = 150.0 N\n\n30.0 W\n\nWhen h = 200 mm = 0.2 m,\n\nFrom Eq. (1):\n\nB=\n\n37.5\n= 488.3 N\n0.25 0.866(0.2)\n\nB = 488 N\n\n30.0 W\n\nFx = 0: Ax B sin 60 = 0\nAx = (488.3) sin 60 = 422.88 N\n\nA x = 422.88 N\n\nFy = 0: Ay 150 + B cos 60 = 0\nAy = 150 (488.3) cos 60 = 94.15 N\n\nA y = 94.15 N\n\n= 12.55\nA = 433.2 N\n\nA = 433 N\n\n12.55 W\n\nPROPRIETARY MATERIAL. 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,\nreproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited\ndistribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,\nyou are using it without permission.\n369\n\nPROBLEM 4.35\nA movable bracket is held at rest by a cable attached at C and by\nfrictionless rollers at A and B. For the loading shown, determine (a) the\ntension in the cable, (b) the reactions at A and B.\n\nSOLUTION\nFree-Body Diagram:\n\n(a)\n\nFy = 0: T 600 N = 0\nT = 600 N W\n\n(b)\n\nFx = 0: B A = 0\n\nB=A\n\nNote that the forces shown form two couples.\n\nM = 0: (600 N)(600 mm) A(90 mm) = 0\nA = 4000 N\nB = 4000 N\nA = 4.00 kN\n\n; B = 4.00 kN\n\nW\u0003\n\nPROPRIETARY MATERIAL. 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,\nreproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited\ndistribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,\nyou are using it without permission.\n383\n\nPROBLEM 4.46\nA tension of 20 N is maintained in a tape as it passes through the\nsupport system shown. Knowing that the radius of each pulley is\n10 mm, determine the reaction at C.\n\nSOLUTION\nFree-Body Diagram:\n\nFx = 0: C x + (20 N) = 0\n\nC x = 20 N\n\nFy = 0: C y (20 N) = 0\n\nC y = +20 N\n\nC = 28.3 N\n\n45.0 W\u0003\n\nM C = 0: M C + (20 N)(0.160 m) + (20 N) (0.055 m) = 0\n\nM C = 4.30 N m\n\nM C = 4.30 N m\n\nPROPRIETARY MATERIAL. 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,\nreproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited\ndistribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,\nyou are using it without permission.\n394\n\nPROBLEM 4.54\nRod AB is acted upon by a couple M and two forces, each of magnitude P.\n(a) Derive an equation in , P, M, and l that must be satisfied when the rod\nis in equilibrium. (b) Determine the value of corresponding to equilibrium\nwhen M = 150 N m, P = 200 N, and l = 600 mm.\n\nSOLUTION\nFree-Body Diagram:\n\n(a)\n\nFrom free-body diagram of rod AB:\n\nM C = 0: P(l cos ) + P(l sin ) M = 0\nor sin + cos =\n\n(b)\n\nM\nW\nPl\n\nFor M = 150 lb in., P = 20 lb, and l = 6 in.,\n\nsin + cos =\n\n150 lb in.\n5\n= = 1.25\n(20 lb)(6 in.) 4\nsin 2 + cos 2 = 1\n\nUsing identity\n\nsin + (1 sin 2 )1/2 = 1.25\n\n(1 sin 2 )1/2 = 1.25 sin\n1 sin 2 = 1.5625 2.5sin + sin 2\n2sin 2 2.5sin + 0.5625 = 0\n\nsin =\n=\n\nor\n\n2(2)\n2.5 1.75\n4\n\nsin = 0.95572 and sin = 0.29428\n\n= 72.886 and = 17.1144\nor = 17.11 and = 72.9 W\n\nPROPRIETARY MATERIAL. 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,\nreproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited\ndistribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,\nyou are using it without permission.\n404\n\nPROBLEM 4.56\nA slender rod AB, of weight W, is attached to blocks A and B that\nmove freely in the guides shown. The constant of the spring is k,\nand the spring is unstretched when = 0. (a) Neglecting the weight\nof the blocks, derive an equation in W, k, l, and that must be\nsatisfied when the rod is in equilibrium. (b) Determine the value\nof when W = 75 lb, l = 30 in., and k = 3 lb/in.\n\nSOLUTION\nFree-Body Diagram:\n\nSpring force:\n\nFs = ks = k (l l cos ) = kl (1 cos )\n\nM D = 0: Fs (l sin ) W cos = 0\n2\n\n(a)\n\nkl (1 cos )l sin\nkl (1 cos ) tan\n\n(b)\n\nFor given values of\n\nW\nl cos = 0\n2\n\nW\n=0\n2\n\nor (1 cos ) tan =\n\nW\nW\n2kl\n\nW = 75 lb\nl = 30 in.\nk = 3 lb/in.\n(1 cos ) tan = tan sin\n75 lb\n=\n2(3 lb/in.)(30 in.)\n= 0.41667\n\nSolving numerically,\n\n= 49.710\n\nor\n\n= 49.7 W\n\nPROPRIETARY MATERIAL. 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,\nreproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited\ndistribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,\nyou are using it without permission.\n406\n\nPROBLEM 4.58\nA collar B of weight W can move freely along the vertical rod shown.\nThe constant of the spring is k, and the spring is unstretched when\n= 0. (a) Derive an equation in , W, k, and l that must be satisfied\nwhen the collar is in equilibrium. (b) Knowing that W = 300 N,\nl = 500 mm, and k = 800 N/m, determine the value of corresponding\nto equilibrium.\n\nSOLUTION\nFirst note:\n\nT = ks\n\nwhere\n\nk = spring constant\ns = elongation of spring\nl\n=\nl\ncos\nl\n(1 cos )\n=\ncos\nkl\nT=\n(1 cos )\ncos\n\n(a)\n\nFrom F.B.D. of collar B:\n\nor\n\n(b)\n\nFor\n\nFy = 0: T sin W = 0\nkl\n(1 cos )sin W = 0\ncos\n\nor tan sin =\n\nW\nW\nkl\n\nW = 3 lb\nl = 6 in.\nk = 8 lb/ft\n6 in.\nl=\n= 0.5 ft\n12 in./ft\ntan sin =\n\nSolving numerically,\n\n3 lb\n= 0.75\n(8 lb/ft)(0.5 ft)\n\n= 57.957\n\nor\n\n= 58.0 W\n\nPROPRIETARY MATERIAL. 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,\nreproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited\ndistribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,\nyou are using it without permission.\n409"
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https://zims-en.kiwix.campusafrica.gos.orange.com/wikipedia_en_all_nopic/A/Source_field | [
"# Source field\n\nIn theoretical physics, a source field is a field $J$",
null,
"whose multiple\n\n$S_{source}=J\\Phi$",
null,
"appears in the action, multiplied by the original field $\\Phi$",
null,
". Consequently, the source field appears on the right-hand side of the equations of motion (usually second-order partial differential equations) for $\\Phi$",
null,
". When the field $\\Phi$",
null,
"is the electromagnetic potential or the metric tensor, the source field is the electric current or the stress–energy tensor, respectively.\n\nAll Green's functions (correlators) may be formally found via Taylor expansion of the partition sum considered as a function of the source fields. This method is commonly used in the path integral formulation of quantum field theory. The general method by which such source fields can be utilized to obtain propagators in both quantum, statistical-mechanics and other systems is outlined in the article on the partition function."
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"https://zims-en.kiwix.campusafrica.gos.orange.com/wikipedia_en_all_nopic/I/m/359e4f407b49910e02c27c2f52e87a36cd74c053.svg",
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"https://zims-en.kiwix.campusafrica.gos.orange.com/wikipedia_en_all_nopic/I/m/a0c9fd1256246d138a799b0e346a17dc8481facd.svg",
null,
"https://zims-en.kiwix.campusafrica.gos.orange.com/wikipedia_en_all_nopic/I/m/aed80a2011a3912b028ba32a52dfa57165455f24.svg",
null,
"https://zims-en.kiwix.campusafrica.gos.orange.com/wikipedia_en_all_nopic/I/m/aed80a2011a3912b028ba32a52dfa57165455f24.svg",
null,
"https://zims-en.kiwix.campusafrica.gos.orange.com/wikipedia_en_all_nopic/I/m/aed80a2011a3912b028ba32a52dfa57165455f24.svg",
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http://cloud.originlab.com/doc/Origin-Help/SLogistic1-FitFunc | [
"# 30.1.163 SLogistic1\n\n## Function",
null,
"$y=\\frac a{1+e^{-k\\left( x-x_c\\right) }}$\n\n## Brief Description\n\nSigmoidal Logistic function, type 1.\n\n## Sample Curve",
null,
"## Parameters\n\nNumber: 3\n\nNames: a, xc, k\n\nMeanings: a = amplitude, xc = center, k = coefficient\n\nLower Bounds: xc > 0\n\nUpper Bounds: none\n\n## Script Access\n\n nlf_slogistic1(x,a,xc,k)\n\n## Function File\n\nFITFUNC\\SLOGIST1.FDF\n\nGrowth/Sigmoidal"
]
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"http://d2mvzyuse3lwjc.cloudfront.net/doc/en/UserGuide/images/SLogistic1/math-09c8aa107c06170f79bafd4f69456d5e.png",
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"http://d2mvzyuse3lwjc.cloudfront.net/doc/en/UserGuide/images/SLogistic1/CFF_Image464.jpg",
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https://mathhelpboards.com/threads/mutualism-autonomous-system.448/ | [
"# [SOLVED]mutualism autonomous system\n\n#### dwsmith\n\n##### Well-known member\nConsider two species whose survival depend on their mutual cooperation. An example would be a species of the bee that feeds primarily on the nectar of one plant species and simultaneously pollinates that plant. One simple model of this mutualism is given by the autonomous system:\n\\begin{align}\n\\frac{dx}{dt} =& -ax + bxy\\notag\\\\\n\\frac{dy}{dt} =& -my + nxy\\notag\n\\end{align}\n\nInterpret the constants a, b, m, and n in terms of the physical problem.\n\nIn the absence of cooperation, the system would be\n\\begin{align}\n\\frac{dx}{dt} =& -ax\\\\\n\\frac{dy}{dt} =& -my\\notag\n\\end{align}\n\nSo a and m are non-negative. So a and m are the deaths proportional to the given population?\n\nI don't know what to say about b and n though.\n\n#### springfan25\n\n##### New member\nfor interpretation purposes is often handy to look at the rate of change per unit of existing population:\n\n$\\frac{\\frac{dx}{dy}}{x}=-a +by$\n\nso i would describe b as the increase in (proportional) growth in x per unit of y population.\n\nsimilarly for n.\n\n•",
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"dwsmith"
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https://analystnotes.com/cfa-study-notes-hypothesis-tests-concerning-differences-between-means.html | [
"### Why should I choose AnalystNotes?\n\nAnalystNotes specializes in helping candidates pass. Period.\n\n##### Subject 9. Hypothesis Tests Concerning Differences between Means\nIn practice, analysts often want to know whether the means of two populations are equal or whether one is larger than the other.\n\nIf it is reasonable to believe that the samples are from populations at least approximately normally distributed and that the samples are also independent of each other, whether a mean value differs between the two populations can be tested. The test procedure is the same as before. There are just a couple of modifications that need to be made.\n\nAs mentioned previously, the null hypothesis involves an equal sign. So, in this situation, the null hypothesis would be that the two unknown population means are equal. The alternative hypothesis would involve one of >, < or ≠.\n\nThe rest of the testing procedure is the same, but the test statistic is different. It's now time to look at what formula should be used. Be warned, though, that the formulae in this section are horrific.\n\nThe test statistic to be used in this section is a t-value, but it varies based on the assumptions. The assumption has been made throughout that the population means are normally distributed.\n\n1. Test statistic for a test of the difference between two population means (normally distributed populations, population variances unknown but assumed equal):",
null,
"where sp2 = [(n1 - 1)s12 + (n2 - 1)s22] / (n1 + n2 - 2) is a pooled estimator of the common variance. The number of degrees of freedom is n1 + n2 - 2. Normally, the degrees of freedom are given by n - 1, but here there are two samples. Combine the sample sizes and then subtract 1 for each sample, or 2 in total. This gives n1 + n2 - 2 as the degrees of freedom.\n\n2. Test statistic for a test of the difference between two population means (normally distributed populations, unequal and unknown population variances):",
null,
"where tables are used to show the t-distribution using \"modified\" degrees of freedom computed with the formula:",
null,
"A practical tip is to compute the t-statistic before computing the degrees of freedom. Whether the t-statistic is significant will sometimes be obvious.\n\nExample\n\nFrom a class of Science students, a sample of 36 is drawn; the mean grade is found to be 62% with a standard deviation of 10. From a class of Arts students, a sample of 49 is drawn; the mean grade is found to be 59% with a standard deviation of 9.6. Assuming that the grades in both classes have a normal distribution, and that the population variances are equal, test at the 5% level for a statistically significant difference in the mean grades of the two classes.\n\nNote that since the sample standard deviations are 10 and 9.6 respectively, the assumption that the population variances are equal seems valid. Had there been a big discrepancy in the sample values, the assumption of equality of population variances would have carried less weight.\n\n\"1\" will be used to represent the Science students and \"2\" to represent the Arts students.\n\nStep 1: State the hypotheses.\n\nYou are testing for differences in the population means μ1 and μ2. Since the question does not specify a direction, a two-sided test is appropriate. The hypotheses are therefore:\n\nH0: μ1 - μ2 = 0\nHa: μ1 - μ2 ≠ 0\n\nStep 2: Identify the test statistic and its probability distribution.\n\nThe appropriate test statistic is the one that assumes equal variances. It is the t-value (discussed earlier).\n\nStep 3: Specify the significance level.\n\nYou are told to test at the 5% level, so α = 0.05.\n\nStep 4: State the decision rule.\n\nThis is a two-sided test, so you need to split your area equally between both tails. You thus have 2.5% of area in each. The total degrees of freedom are: n1 + n2 - 2 = 36 + 49 - 2 = 83. Your tables don't have 83 degrees of freedom, so use 80, which is the closest value to 83. From the t-table, the critical values are therefore -1.99 and 1.99.",
null,
"The value above determines your decision. If your test statistic lies to the left of -1.99 or to the right of 1.99, you will reject H0; otherwise, you will not reject H0.\n\nYou might notice that your critical values of -1.99 and 1.99 are very close to the corresponding z-values of -1.96 and 1.96. This is because, as explained earlier, as the degrees of freedom increase, the t-values approach the z-values. Since 83 degrees of freedom is a large number, the t-graph here closely resembles a z-graph.\n\nStep 5: Collect the data in the sample and calculate the necessary value(s) using the sample data.\n\nThe question gives you the necessary sample values: x-bar1 = 62, x-bar2= 59, s1 = 10, s2 = 9.6, n1 = 36 and n2 = 49.\n\nRecall the formula from earlier: s2p = [35 x 100 + 48 x 92.16] / 83 = 95.466. Now you can substitute into the test statistic: = (62 - 59 - 0) /",
null,
"= 1.3988.\n\nThe t-value of 1.3988 is now compared with your critical values of -1.99 and 1.99.\n\nStep 6: Make a decision regarding the hypotheses.\n\nSince the value of the test statistic is less extreme (i.e., closer to zero) than the positive critical value, the test statistic falls in the acceptance region. You would thus not reject H0 at the 5% significance level.",
null,
"Step 7: Make a decision based on the test results.\n\nYou can now conclude that the difference in the average marks of Science and Arts students is not significant when testing at the 5% level.\n\nNote:\n\n• Had you used a p-value approach here, you would have obtained a p-value between 0.1 and 0.2 (closer to 0.2). You can check this for yourself as an exercise. You cannot obtain the p-value exactly from t-tables. Because the value is larger than 0.05, you would not reject H0, so your results are consistent with those above.\n• Had you made Science population 2 and Arts population 1, your test statistic would have worked out to be -1.3988, but this would have made no difference in your conclusion, since the test is two-sided. It is better, however, to be guided by the question in this regard.\n\nLearning Outcome Statements\n\nh. identify the appropriate test statistic and interpret the results for a hypothesis test concerning the equality of the population means of two at least approximately normally distributed populations, based on independent random samples with 1) equal or 2) unequal assumed variances;\n\nCFA® Level I Curriculum, 2020, Volume 1, Reading 11\n\nUser Comment\nStoner 16 observations. 3 independent variables on a multiple regression. When testing for significance using the output from the regression, how many d/f? n-k-1? or n-1? or something else? Please advise.\nroli79 I just went through this stuff last night. For multiple regression: df = n-(k+1) n = sample size k = number of parameters being estimated the + 1 accounts for the intercept. df = 16-(3+1) df = 12 I am pretty confident in that answer. If anyone else disagrees please post. This is the wrong los to ask this question, though.\nwhiteknight why should we use t-statistics in case both the population variance is known ? any comments...\nachu population variance is unknown here.\nepizi Yea but here you are interested in the equality of the population variance to know which steps to follow.If it wasn't equal, you were to calculated the degree of freedom using the most complex fomular above\ncschulz316 what's the deal with that super complicated \"modified\" degrees of freedom?",
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https://socratic.org/questions/if-the-sum-of-three-consecutive-integers-is-69-what-are-the-integers | [
"If the sum of three consecutive integers is 69, what are the integers?\n\nMar 3, 2018\n\n$22 , 23 , 24$\n\nExplanation:\n\nAlgebra method:\n\nLet $x$ be the middle number.\n\nThen the three numbers are: $x - 1 , x , x + 1$\n\nThe sum of these numbers: $x - 1 + x + x + 1 = 3 x$\n\nThe sum of these numbers equal $69$: $3 x = 69$\n\nSolve for $x$\n\n$x = 23$\n\nSubstitute $x = 23$ into $x - 1$ and $x + 1$: $23 - 1 = 22 , 23 + 1 = 24$\n\nCheck: $22 + 23 + 24 = 69$ and they are consecutive.\n\nNormally though, my thinking process would be like this:\n\n1. Find the mean\n2. Are they consecutive integers, odd integers or even integers?\n3. If they're consecutive, +1 and -1 to the mean. If they're odd/even, +2 and -2 to the mean. (Of course it would be different if there are 5 consecutive numbers, 7, etc)"
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https://www.gradesaver.com/textbooks/math/algebra/intermediate-algebra-connecting-concepts-through-application/chapter-1-linear-functions-1-4-intercepts-and-graphing-1-4-exercises-page-62/18 | [
"## Intermediate Algebra: Connecting Concepts through Application\n\n(a) The $P$-intercept is $4934$, which means that the population in Minnesota in the year $2000$ was around $4,934,000$ (b) The $t$-intercept is $-\\frac{4934}{41}\\approx -120$, which means that the population in Minnesota $121$ years before the year $2000$, which was in $1880$, was $0$.\n(a) The $P$-intercept can be found by setting $t=0$. Thus, the $P$-intercept is: \\begin{align*} P&=41t+4934\\\\ P&=41(0)+4934\\\\ P&=0+4934\\\\ P&=4934 \\end{align*} The $P$-intercept is $4394$. This means that in the year $2000$, there were $4,934,000$ people in Minnesota. (b) The $t$-intercept can be found by setting $P=0$. Thus, the $t$-intercept is: \\begin{align*} P&=41t+4934\\\\ 0&=41(t)+4934\\\\ -4934&=41t\\\\ -\\frac{4934}{41}&=t\\\\ t&\\approx -120 \\end{align*} The $t$-intercept is $-\\frac{4934}{41}$ or approximately $-120$. This means that the population in Minnesota was $0$ approximately $121$ years ago or in the year $1880$."
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.92151093,"math_prob":0.999824,"size":939,"snap":"2019-51-2020-05","text_gpt3_token_len":348,"char_repetition_ratio":0.142246,"word_repetition_ratio":0.07874016,"special_character_ratio":0.44728434,"punctuation_ratio":0.104166664,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9999105,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-12-15T13:05:20Z\",\"WARC-Record-ID\":\"<urn:uuid:e818f994-439e-4a33-a34f-d52800d2329d>\",\"Content-Length\":\"78185\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:a53c4357-394f-4c2f-9f31-e7dcd77779aa>\",\"WARC-Concurrent-To\":\"<urn:uuid:67df3d88-e27c-485d-9c5e-c627e69419e7>\",\"WARC-IP-Address\":\"54.210.73.90\",\"WARC-Target-URI\":\"https://www.gradesaver.com/textbooks/math/algebra/intermediate-algebra-connecting-concepts-through-application/chapter-1-linear-functions-1-4-intercepts-and-graphing-1-4-exercises-page-62/18\",\"WARC-Payload-Digest\":\"sha1:UU7AN64FN6LWHNFUO6H7LBR23Z4PHRNC\",\"WARC-Block-Digest\":\"sha1:XAM64RHWQWFJIUXIKKM5NBYMEC7IKCL6\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-51/CC-MAIN-2019-51_segments_1575541308149.76_warc_CC-MAIN-20191215122056-20191215150056-00517.warc.gz\"}"} |
https://zh.m.wikipedia.org/wiki/%E8%83%A1%E5%85%8B%E5%AE%9A%E5%BE%8B | [
"# 胡克定律\n\n$\\sigma =E\\varepsilon$",
null,
"$\\Delta L={\\frac {1}{E}}\\times L\\times {\\frac {F}{A}}={\\frac {1}{E}}\\times L\\times \\sigma$",
null,
"$\\Delta L$",
null,
"：總伸長（縮減）量。胡克定律用17世纪英国物理学家罗伯特·胡克的名字命名。胡克提出该定律的过程颇有趣味，他于1676年发表了一句拉丁语字谜，谜面是：ceiiinosssttuv。两年后他公布了谜底是：ut tensio sic vis，意思是“力如伸长（那样变化）”（见参考文献），这正是胡克定律的中心内容。\n\n## 弹簧方程\n\n$F=-kx$\n\n$k$ ：弹簧的劲度系数(彈力係數)，由材料性质、几何外形决定，负号：弹簧产生的弹力与其伸长(压縮)的方向相反，这种弹力称为回復力，表示它有使系统回復平衡的趋势。满足上式的弹簧称为线性弹簧\n\n$U={1 \\over 2}kx^{2}$\n\n$\\omega _{n}={\\sqrt {k \\over m}}$\n\n1. 最大强度\n2. 屈服强度\n3. 破坏点\n4. 加工硬化英语Work hardening\n5. 颈缩\n\n$\\sigma _{ij}=\\sum _{kl}c_{ijkl}\\cdot \\varepsilon _{kl}$\n\n## 胡克定律的张量形式\n\n（在牛顿流体中的类比参见粘性词条。）\n\n$\\varepsilon _{ij}=\\left({\\frac {1}{3}}\\varepsilon _{kk}\\delta _{ij}\\right)+\\left(\\varepsilon _{ij}-{\\frac {1}{3}}\\varepsilon _{kk}\\delta _{ij}\\right)$\n\n$\\sigma _{ij}=3K\\left({\\frac {1}{3}}\\varepsilon _{kk}\\delta _{ij}\\right)+2G\\left(\\varepsilon _{ij}-{\\frac {1}{3}}\\varepsilon _{kk}\\delta _{ij}\\right)$\n\n${\\begin{cases}\\varepsilon _{11}={\\cfrac {1}{Y}}\\left(\\sigma _{11}-\\nu (\\sigma _{22}+\\sigma _{33})\\right)\\\\\\varepsilon _{22}={\\cfrac {1}{Y}}\\left(\\sigma _{22}-\\nu (\\sigma _{11}+\\sigma _{33})\\right)\\\\\\varepsilon _{33}={\\cfrac {1}{Y}}\\left(\\sigma _{33}-\\nu (\\sigma _{11}+\\sigma _{22})\\right)\\\\\\varepsilon _{12}={\\cfrac {\\sigma _{12}}{2G}}\\\\\\varepsilon _{13}={\\cfrac {\\sigma _{13}}{2G}}\\\\\\varepsilon _{23}={\\cfrac {\\sigma _{23}}{2G}}\\end{cases}}$\n\n### 正交各向异性材料\n\n${\\begin{pmatrix}\\sigma _{11}\\\\\\sigma _{22}\\\\\\sigma _{33}\\\\\\sigma _{12}\\\\\\sigma _{23}\\\\\\sigma _{31}\\\\\\end{pmatrix}}={\\begin{pmatrix}C_{11}&C_{12}&C_{13}&0&0&0\\\\C_{12}&C_{22}&C_{23}&0&0&0\\\\C_{13}&C_{23}&C_{33}&0&0&0\\\\0&0&0&C_{44}&0&0\\\\0&0&0&0&C_{55}&0\\\\0&0&0&0&0&C_{66}\\end{pmatrix}}{\\begin{pmatrix}\\varepsilon _{11}\\\\\\varepsilon _{22}\\\\\\varepsilon _{33}\\\\\\varepsilon _{12}\\\\\\varepsilon _{23}\\\\\\varepsilon _{31}\\\\\\end{pmatrix}}$\n\n## 参考文献\n\n• Y. C. Fung （冯元桢）, Foundations of Solid Mechanics, Prentice-Hall Inc., Englewood Cliffs, New Jersey, 1965\n• A.C. Ugural, S.K. Fenster, Advanced Strength and Applied Elasticity, 4th ed"
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| {"ft_lang_label":"__label__zh","ft_lang_prob":0.9706621,"math_prob":0.9999646,"size":2323,"snap":"2022-27-2022-33","text_gpt3_token_len":2461,"char_repetition_ratio":0.04484692,"word_repetition_ratio":0.0,"special_character_ratio":0.16185966,"punctuation_ratio":0.0625,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9998534,"pos_list":[0,1,2,3,4,5,6],"im_url_duplicate_count":[null,null,null,10,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-07-05T02:07:37Z\",\"WARC-Record-ID\":\"<urn:uuid:fa6dad43-a442-429c-bf89-157139cbd77b>\",\"Content-Length\":\"121778\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:244c8dd9-2587-41e8-ad60-078c6978a18c>\",\"WARC-Concurrent-To\":\"<urn:uuid:c778f957-f9b6-4e29-9876-ae1a4c79520f>\",\"WARC-IP-Address\":\"208.80.154.224\",\"WARC-Target-URI\":\"https://zh.m.wikipedia.org/wiki/%E8%83%A1%E5%85%8B%E5%AE%9A%E5%BE%8B\",\"WARC-Payload-Digest\":\"sha1:PHQBACF6I4JQKKZZ2RTIOYWZ4BAYJAQM\",\"WARC-Block-Digest\":\"sha1:ML7PY47JRPYFHOZJ3R34GQSNIJQ2YBOY\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-27/CC-MAIN-2022-27_segments_1656104506762.79_warc_CC-MAIN-20220704232527-20220705022527-00164.warc.gz\"}"} |
https://uk.mathworks.com/help/stats/parallelcoords.html | [
"Documentation\n\n# parallelcoords\n\nParallel coordinates plot\n\n## Syntax\n\n``parallelcoords(x)``\n``parallelcoords(x,Name,Value)``\n``parallelcoords(ax,___)``\n``h = parallelcoords(___)``\n\n## Description\n\n````parallelcoords(x)` creates a parallel coordinates plot of the multivariate data in the matrix `x`. Use a parallel coordinates plot to visualize high dimensional data, where each observation is represented by the sequence of its coordinate values plotted against their coordinate indices.```\n\nexample\n\n````parallelcoords(x,Name,Value)` creates a parallel coordinates plot with additional options specified by one or more `Name,Value` pair arguments. For example, you can standardize the data in `x` or label the coordinate tick marks along the horizontal axis of the plot.```\n````parallelcoords(ax,___)` creates a parallel coordinates plot using the axes specified by the axes graphic object `ax`, using any of the previous syntaxes.```\n\nexample\n\n````h = parallelcoords(___)` returns a column vector of handles to the `Line` objects created by `parallelcoords`, with one handle for each row of `x`.```\n\n## Examples\n\ncollapse all\n\nLoad the Fisher iris sample data.\n\n`load fisheriris`\n\nThe data contains four measurements (sepal length, sepal width, petal length, and petal width) from three species of iris flowers. The matrix `meas` contains all four measurements for each of 150 flowers. The cell array `species` contains the species name for each of the 150 flowers.\n\nCreate a cell array that contains the name of each measurement variable in the sample data.\n\n`labels = {'Sepal Length','Sepal Width','Petal Length','Petal Width'};`\n\nCreate a parallel coordinate plot using the measurement data in `meas`. Use a different color for each group as identified in `species`, and label the horizontal axis using the variable names.\n\n`parallelcoords(meas,'Group',species,'Labels',labels)`",
null,
"The resulting plot contains one line for each observation (flower). The color of each line indicates the flower species.\n\nLoad the Fisher iris sample data.\n\n`load fisheriris`\n\nThe data contains four measurements (sepal length, sepal width, petal length, and petal width) from three species of iris flowers. The matrix `meas` contains all four measurements for each of 150 flowers. The cell array `species` contains the species name for each of the 150 flowers.\n\nCreate a cell array that contains the name of each measurement variable in the sample data.\n\n`labels = {'Sepal Length','Sepal Width','Petal Length','Petal Width'};`\n\nCreate a parallel coordinates plot using the measurement data in `meas`. Plot only the median, 25 percent, and 75 percent quartile values for each group identified in `species`. Label the horizontal axis using the variable names.\n\n```parallelcoords(meas,'group',species,'labels',labels,... 'quantile',.25)```",
null,
"The plot shows the median values for each group as a solid line and the quartile values as dotted lines of the same color. For example, the solid blue line shows the median value measured for each variable on `setosa` irises. The dotted blue line below the solid blue line shows the 25th percentile of measurements for each variable on `setosa` irises. The dotted blue line above the solid blue line shows the 75th percentile of measurements for each variable on `setosa` irises.\n\nLoad the Fisher iris sample data.\n\n`load fisheriris`\n\nThe data contains four measurements (sepal length, sepal width, petal length, and petal width) from three species of iris flowers. The matrix `meas` contains all four measurements for each of 150 flowers. The cell array `species` contains the species name for each of the 150 flowers.\n\nCreate a cell array that contains the name of each measurement variable in the sample data.\n\n`labels = {'Sepal Length','Sepal Width','Petal Length','Petal Width'};`\n\nCreate a parallel coordinates plot using the measurement data in `meas`. Plot only the median, 25 percent, and 75 percent quartile values for each group identified in `species`. Label the horizontal axis using the variable names. Set the line width to 2.\n\n```parallelcoords(meas,'group',species,'labels',labels,... 'quantile',.25,'LineWidth',2)```",
null,
"Specifying `'LineWidth'` in this way sets the width of every line in the plot to 2.\n\nRecreate the parallel coordinates plot, but this time, use handles to increase the width of only the line representing the median value for each measurement made on irises in the `setosa` group.\n\n```h = parallelcoords(meas,'group',species,'labels',labels,... 'quantile',.25)```",
null,
"```h = 9x1 Line array: Line (median) Line (lower quantile) Line (upper quantile) Line (median) Line (lower quantile) Line (upper quantile) Line (median) Line (lower quantile) Line (upper quantile) ```\n\nThe returned column vector `h` contains handles that correspond to each line object created by `parallelcoords`. For example, h(1) corresponds to the median line for the first grouping variable (`setosa`).\n\nUse dot notation to increase the width of the line showing the median value for each measurement made on irises in the `setosa` group.\n\n`h(1).LineWidth = 2;`",
null,
"## Input Arguments\n\ncollapse all\n\nMultivariate input data, specified as an n-by-p matrix of numeric values. n is the number of rows of `x`, and each row corresponds to an observation in `x`. p is the number of columns in `x`, and each column corresponds to a variable in `x`.\n\n`parallelcoords` treats `NaN` values in `x` as missing values and does not plot those coordinate values.\n\nData Types: `single` | `double`\n\nAxes for plot, specified as an axes graphic object. If you do not specify `ax`, then `parallelcoords` creates the plot using the current axis. For more information on creating an axes graphic object, see `axes` and Axes Properties.\n\n### Name-Value Pair Arguments\n\nSpecify optional comma-separated pairs of `Name,Value` arguments. `Name` is the argument name and `Value` is the corresponding value. `Name` must appear inside quotes. You can specify several name and value pair arguments in any order as `Name1,Value1,...,NameN,ValueN`.\n\nExample: `'Group',species,'Quantile',.25` plots the median, 25 percent, and 75 percent quartile values for the input data, using a different color for each group identified in the variable `species`.\n\nGrouping variable for input data, specified as the comma-separated pair consisting of `'Group'` and a numeric array containing a group index for each observation. Alternatively, the array can be a categorical variable, character matrix, string array, or cell array containing a group name for each observation.\n\nData Types: `single` | `double` | `categorical` | `char` | `string` | `cell`\n\nHorizontal axis labels, specified as the comma-separated pair consisting of `'Labels'` and a character array, string array, or cell array containing the label names.\n\nExample: `'Labels',{'Sepal Width','Sepal Length'}`\n\nData Types: `char` | `string` | `cell`\n\nQuantiles of input data to plot, specified as the comma-separated pair consisting of `'Quantile'` and a numeric value in the range (0,1). If you specify a value alpha for `'Quantile'`, then `parallelcoords` plots only the median, alpha, and 1 – alpha quantiles for each of the variables (columns) in `x`.\n\nThe quantile plot option provides a useful summary of the data when `x` contains many observations.\n\nExample: `'Quantile',.25`\n\nData Types: `single` | `double`\n\nMethod to standardize input data, specified as the comma-separated pair consisting of `'Standardize'` and one of the following.\n\n `'on'` Scale each column of `x` to have a mean equal to 0 and a standard deviation equal to 1 before plotting. `'PCA'` Create plot from the principal component scores of `x`, in order of decreasing eigenvalues. `parallelcoords` removes rows of `x` containing missing values (`NaN`) for PCA standardization. `'PCAStd'` Create plot using the standardized principal component scores.\n\nExample: `'Standardize','on'`\n\n#### Tips\n\n• You can modify certain aspects of the plot lines by specifying a property name and value for any of the properties listed in Line Properties. However, this approach applies the modification to all the lines in the plot. To modify only certain plot lines, use the syntax that returns graphics handles and use dot notation to adjust each line property individually. For an illustration, see Adjust Line Properties in Parallel Coordinates Plot.\n\n## Output Arguments\n\ncollapse all\n\nGraphic handles for line objects, returned as a vector of `Line` graphic handles. Graphic handles are unique identifiers that you can use to query and modify the properties of a specific line on the plot. To view and set properties of line objects, use dot notation. For information on using dot notation, seeAccess Property Values (MATLAB). For information on the `Line` properties that you can set, see Line Properties.\n\nIf you use the `'Quantile'` name-value pair argument, then `h` contains one handle for each of the three lines objects created. If you use both the `'Quantile'` and the `'Group'` name-value pair arguments, then `h` contains three handles for each group.\n\n## Alternative Functionality\n\nAlternatively, you can create a `ParallelCoordinatesPlot` object by using the `parallelplot` function.\n\n• Unlike the `parallelcoords` function, `parallelplot` allows you to plot tabular data that includes categorical variables.\n\n• `parallelplot` does not support the plotting of quantiles for numeric data. However, the `ParallelCoordinatesPlot` object contains the `DataNormalization` property, which provides several data normalization methods for coordinates with numeric values.\n\nTo control the appearance and behavior of the object, change the ParallelCoordinatesPlot Properties.\n\nDownload ebook"
]
| [
null,
"https://uk.mathworks.com/help/examples/stats/win64/ParallelCoordinatesPlotForGroupedDataExample_01.png",
null,
"https://uk.mathworks.com/help/examples/stats/win64/ParallelCoordinatesPlotWithQuantileValuesExample_01.png",
null,
"https://uk.mathworks.com/help/examples/stats/win64/AdjustLinePropertiesInParallelCoordinatesPlotExample_01.png",
null,
"https://uk.mathworks.com/help/examples/stats/win64/AdjustLinePropertiesInParallelCoordinatesPlotExample_02.png",
null,
"https://uk.mathworks.com/help/examples/stats/win64/AdjustLinePropertiesInParallelCoordinatesPlotExample_03.png",
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.6065703,"math_prob":0.93014735,"size":1422,"snap":"2019-35-2019-39","text_gpt3_token_len":291,"char_repetition_ratio":0.21086037,"word_repetition_ratio":0.016304348,"special_character_ratio":0.1673699,"punctuation_ratio":0.11261261,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99163866,"pos_list":[0,1,2,3,4,5,6,7,8,9,10],"im_url_duplicate_count":[null,2,null,2,null,2,null,2,null,2,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-08-17T22:35:58Z\",\"WARC-Record-ID\":\"<urn:uuid:c0958c70-3908-492a-9339-c8bb0b27706f>\",\"Content-Length\":\"101126\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:44e93ec5-b80d-4fbb-88fa-8c2cc73f0ebb>\",\"WARC-Concurrent-To\":\"<urn:uuid:5fd76aff-fe66-478e-8b36-518ae339801c>\",\"WARC-IP-Address\":\"104.118.179.86\",\"WARC-Target-URI\":\"https://uk.mathworks.com/help/stats/parallelcoords.html\",\"WARC-Payload-Digest\":\"sha1:P55ERMI3XLIXYHPXGGGNTV7DRYL3D77Z\",\"WARC-Block-Digest\":\"sha1:GUM5KXBP4LJIANIRTNIO3VO6X7NPEMZH\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-35/CC-MAIN-2019-35_segments_1566027313501.0_warc_CC-MAIN-20190817222907-20190818004907-00528.warc.gz\"}"} |
http://www.sxglgf.com/article/37239.htm | [
" 解析php二分法查找数组是否包含某一元素_php技巧_澳门金沙网上娱乐 - 澳门金沙国际_澳门金沙娱乐注册_澳门金沙娱乐场极速入口\n\n# 解析php二分法查找数组是否包含某一元素\n\n<?php\n\n\\$searchValue = (int)\\$_GET['key'];\n\nfunction search(array \\$array, \\$value)\n{\n\\$max = count(\\$array)-1;\n\\$min = 0;\n\\$isAscSort = \\$array[\\$min] < \\$array[\\$max];\n\nwhile (TRUE) {\n\\$sum = \\$min+\\$max;\n\\$midKey = (int)(\\$sum%2 == 1 ? ceil(\\$sum/2) : \\$sum/2);\n\nif (\\$max < \\$min) {\nreturn -1;\n} else if (\\$value == \\$array[\\$midKey]) {\nreturn 1;\n} else if (\\$value > \\$array[\\$midKey]) {\n\\$isAscSort ? \\$min = \\$midKey+1 : \\$max = \\$midKey-1;\n} else if (\\$value < \\$array[\\$midKey]) {\n\\$isAscSort ? \\$max = \\$midKey-1 : \\$min = \\$midKey+1;\n}\n}\n}\n\n\\$array = array(\n'4', '5', '7', '8', '9', '10', '11', '12'\n);\n// 正序\necho search(\\$array, \\$searchValue);\n\n// 逆序\nrsort(\\$array);\necho search(\\$array, \\$searchValue);"
]
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| {"ft_lang_label":"__label__zh","ft_lang_prob":0.65004444,"math_prob":0.9892308,"size":2163,"snap":"2019-43-2019-47","text_gpt3_token_len":1172,"char_repetition_ratio":0.1454377,"word_repetition_ratio":0.8734694,"special_character_ratio":0.36107257,"punctuation_ratio":0.21556886,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9941467,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-11-22T07:38:15Z\",\"WARC-Record-ID\":\"<urn:uuid:790c2282-f8b0-4bed-83f0-eb3a9dc691bc>\",\"Content-Length\":\"33830\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:1bb5e6ca-9ff5-4497-951b-bbabd0c836bd>\",\"WARC-Concurrent-To\":\"<urn:uuid:c84b87ba-3f3c-4662-a2d0-7319336eee18>\",\"WARC-IP-Address\":\"166.88.211.179\",\"WARC-Target-URI\":\"http://www.sxglgf.com/article/37239.htm\",\"WARC-Payload-Digest\":\"sha1:EVEWU2BDIXMDVOTMIZTO3NKXK2URDJ4C\",\"WARC-Block-Digest\":\"sha1:ITXXV5KFAZLJTERLIA62GQSOA4ZXKKOK\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-47/CC-MAIN-2019-47_segments_1573496671245.92_warc_CC-MAIN-20191122065327-20191122093327-00387.warc.gz\"}"} |
https://www.procustoms.org/2020/10/nptel14-sep-04-dec-week-3-program-1and.html | [
"# “HAPPINESS SHOULD BE A FUNCTION WITHOUT ANY PARAMETERS”\n\n##",
null,
"Programming in JAVA.\n\nJava Week 3: Q1\n\nThis program is related to the generation of Fibonacci numbers.\n\nFor example: 0, 1, 1, 2, 3, 5, 8, 13,… is a Fibonacci sequence where 13 is the 8thFibonacci number. A partial code is given and you have to complete the code as per the instruction given as comments.\n\nPROGRAM:\n\n```if(n==1)\nreturn 0;\nelse if(n==2)\nreturn 1;\nelse\nreturn fib(n-1)+fib(n-2);```\n\nJava Week 3: Q2\n\nDefine a class Point with two fields x and y each of type double. Also , define a method distance(Point p1, Point p2) to calculate the distance between points p1 and p2 and return the value in double. Complete the code segment given below. Use Math.sqrt( ) to calculate the square root.\n\nPROGRAM:\n\n```class Point{\n\ndouble x,y;\n\nvoid distance(Point p1,Point p2){\ndouble diff,xx,yy;\nxx=p2.x-p1.x;\nyy=p2.y-p1.y;\n\ndiff=Math.sqrt(xx*xx+yy*yy); // sqrt((x2-x2)2 +(y2-y1)2) simply use distance formula\nSystem.out.print(diff);\n\n}\n\n}```\n\nWEEK-3 :Programming in JAVA"
]
| [
null,
"https://1.bp.blogspot.com/-YwwOSztsSxQ/XgM6NJzunHI/AAAAAAAAEOY/YQOut3MdrIwN9-ZPzcDzKUHk7ppUgNUEACLcBGAsYHQ/s320/nptel.PNG",
null
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.7482654,"math_prob":0.99643344,"size":1447,"snap":"2020-45-2020-50","text_gpt3_token_len":392,"char_repetition_ratio":0.08939709,"word_repetition_ratio":0.0,"special_character_ratio":0.2639945,"punctuation_ratio":0.16819572,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9996563,"pos_list":[0,1,2],"im_url_duplicate_count":[null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-10-24T12:20:26Z\",\"WARC-Record-ID\":\"<urn:uuid:e06fb76b-9b0f-4108-8df5-1207000b2d96>\",\"Content-Length\":\"222983\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:29e235af-1353-4302-a758-ddea6f3390fd>\",\"WARC-Concurrent-To\":\"<urn:uuid:52efe01e-ac94-4dc8-ab05-102b377597f2>\",\"WARC-IP-Address\":\"172.217.7.147\",\"WARC-Target-URI\":\"https://www.procustoms.org/2020/10/nptel14-sep-04-dec-week-3-program-1and.html\",\"WARC-Payload-Digest\":\"sha1:GR7GFFMS37XFQGO567Y4IWINCUFOR2PH\",\"WARC-Block-Digest\":\"sha1:CWAHGMYZPXD6JRU3CQKDEA6E4YJ3UPCD\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-45/CC-MAIN-2020-45_segments_1603107882581.13_warc_CC-MAIN-20201024110118-20201024140118-00705.warc.gz\"}"} |
https://kw2hp.co.uk/189-hp-equals-to-how-many-kw-kilowatts/ | [
"# 189 HP equals to how many kW (kilowatts)?\n\n189 HP after calculating to kW (kilowatts) is 138,97 kW and equals 186,41 BHP.\n\nHow to calculate that 189 HP is 138,97 kW (kilowatts)?\n\nIt’s very simple – just multiply 189 HP by 0,74. It gives 138,97 kW (kilowatts).\n\nHow useful was this post?\n\nClick on a star to rate it!\n\nAverage rating 0 / 5. Vote count: 0\n\nNo votes so far! Be the first to rate this post.\n\nWe are sorry that this post was not useful for you!\n\nLet us improve this post!\n\nTell us how we can improve this post?"
]
| [
null
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.57814646,"math_prob":0.96273136,"size":429,"snap":"2021-21-2021-25","text_gpt3_token_len":144,"char_repetition_ratio":0.16941176,"word_repetition_ratio":0.9459459,"special_character_ratio":0.4079254,"punctuation_ratio":0.16981132,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9907423,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-06-24T16:28:13Z\",\"WARC-Record-ID\":\"<urn:uuid:e1ef6c93-d409-453d-aac6-e4542fb25b1c>\",\"Content-Length\":\"42679\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:0c808f00-dc0c-4c90-b1e2-a4d47d94f965>\",\"WARC-Concurrent-To\":\"<urn:uuid:0abc17e8-73a9-4f15-97d8-f89cf63fe02d>\",\"WARC-IP-Address\":\"195.78.66.124\",\"WARC-Target-URI\":\"https://kw2hp.co.uk/189-hp-equals-to-how-many-kw-kilowatts/\",\"WARC-Payload-Digest\":\"sha1:MSXI5L4E2XEFPVQG7OCNAZP6QQDX5TT6\",\"WARC-Block-Digest\":\"sha1:2YRUDAB2JJJNO4UHFVH2MWGSFILUKC3T\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-25/CC-MAIN-2021-25_segments_1623488556133.92_warc_CC-MAIN-20210624141035-20210624171035-00478.warc.gz\"}"} |
https://au.mathworks.com/help/gads/patternsearch-climbs-mt-washington.html | [
"# Pattern Search Climbs Mount Washington\n\nThis example shows visually how pattern search optimizes a function. The function is the height of the terrain near Mount Washington, as a function of the x-y location. In order to find the top of Mount Washington, we minimize the objective function that is the negative of the height. (The Mount Washington in this example is the highest peak in the northeastern United States.)\n\nThe US Geological Survey provides data on the height of the terrain as a function of the x-y location on a grid. In order to be able to evaluate the height at an arbitrary point, the objective function interpolates the height from nearby grid points.\n\nIt would be faster, of course, to simply find the maximum value of the height as specified on the grid, using the `max` function. The point of this example is to show how the pattern search algorithm operates; it works on functions defined over continuous regions, not just grid points. Also, if it is computationally expensive to evaluate the objective function, then performing this evaluation on a complete grid, as required by the `max` function, will be much less efficient than using the pattern search algorithm, which samples a small subset of grid points.\n\n### How Pattern Search Works\n\nPattern search finds a local minimum of an objective function by the following method, called polling. In this description, words describing pattern search quantities are in bold. The search starts at an initial point, which is taken as the current point in the first step:\n\n1. Generate a pattern of points, typically plus and minus the coordinate directions, times a mesh size, and center this pattern on the current point.\n\n2. Evaluate the objective function at every point in the pattern.\n\n3. If the minimum objective in the pattern is lower than the value at the current point, then the poll is successful, and the following happens:\n\n3a. The minimum point found becomes the current point.\n\n3b. The mesh size is doubled.\n\n3c. The algorithm proceeds to Step 1.\n\n4. If the poll is not successful, then the following happens:\n\n4a. The mesh size is halved.\n\n4b. If the mesh size is below a threshold, the iterations stop.\n\n4c. Otherwise, the current point is retained, and the algorithm proceeds at Step 1.\n\nThis simple algorithm, with some minor modifications, provides a robust and straightforward method for optimization. It requires no gradients of the objective function. It lends itself to constraints, too, but this example and description deal only with unconstrained problems.\n\n### Preparing the Pattern Search\n\nTo prepare the pattern search, load the data in `mtWashington.mat`, which contains the USGS data on a 472-by-345 grid. The elevation, Z, is given in feet. The vectors x and y contain the base values of the grid spacing in the east and north directions respectively. The data file also contains the starting point for the search, X0.\n\n`load mtWashington`\n\nThere are three MATLAB® files that perform the calculation of the objective function, and the plotting routines. They are:\n\n1. `terrainfun`, which evaluates the negative of height at any x-y position. `terrainfun` uses the MATLAB function `interp2` to perform two-dimensional linear interpolation. It takes the Z data and enables evaluation of the negative of the height at all x-y points.\n\n2. `psoutputwashington`, which draws a 3-d rendering of Mt. Washington. In addition, as the run progresses, it draws spheres around each point that is better (higher) than previously-visited points.\n\n3. `psplotwashington`, which draws a contour map of Mt. Washington, and monitors a slider that controls the speed of the run. It shows where the pattern search algorithm looks for optima by drawing + signs at those points. It also draws spheres around each point that is better than previously-visited points.\n\nIn the example, `patternsearch` uses `terrainfun` as its objective function, `psoutputwashington` as an output function, and `psplotwashington` as a plot function. We prepare the functions to be passed to `patternsearch` in anonymous function syntax:\n\n```mtWashObjectiveFcn = @(xx) terrainfun(xx, x, y, Z); mtWashOutputFcn = @(xx,arg1,arg2) psoutputwashington(xx,arg1,arg2, x, y, Z); mtWashPlotFcn = @(xx,arg1) psplotwashington(xx,arg1, x, y, Z);```\n\n### Pattern Search Options Settings\n\nNext, we create options for pattern search. This set of options has the algorithm halt when the mesh size shrinks below 1, keeps the mesh unscaled (the same size in each direction), sets the initial mesh size to 10, and sets the output function and plot function:\n\n```options = optimoptions(@patternsearch,'MeshTolerance',1,'ScaleMesh',false, ... 'InitialMeshSize',10,'UseCompletePoll',true,'PlotFcn',mtWashPlotFcn, ... 'OutputFcn',mtWashOutputFcn,'UseVectorized',true);```\n\n### Observing the Progress of Pattern Search\n\nWhen you run this example you see two windows. One shows the points the pattern search algorithm chooses on a two-dimensional contour map of Mount Washington. This window has a slider that controls the delay between iterations of the algorithm (when it returns to Step 1 in the description of how pattern search works). Set the slider to a low position to speed the run, or to a high position to slow the run.\n\nThe other window shows a three-dimensional plot of Mount Washington, along with the steps the pattern search algorithm makes. You can rotate this plot while the run progresses to get different views.\n\n```[xfinal ffinal] = patternsearch(mtWashObjectiveFcn,X0,[],[],[],[],[], ... [],[],options)```\n```Optimization terminated: mesh size less than options.MeshTolerance. ```",
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"",
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"```xfinal = 1×2 316130 4904295 ```\n```ffinal = -6280 ```\n\nThe final point, `xfinal`, shows where the pattern search algorithm finished; this is the x-y location of the top of Mount Washington. The final objective function, `ffinal`, is the negative of the height of Mount Washington, 6280 feet. (This should be 6288 feet according to the Mount Washington Observatory).\n\nExamine the files `terrainfun.m`, `psoutputwashington.m`, and `psplotwashington.m` to see how the interpolation and graphics work.\n\nThere are many options available for the pattern search algorithm. For example, the algorithm can take the first point it finds that is an improvement, rather than polling all the points in the pattern. It can poll the points in various orders. And it can use different patterns for the poll, both deterministic and random. Consult the Global Optimization Toolbox User's Guide for details."
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"https://au.mathworks.com/help/examples/globaloptim/win64/mtwashdemo_01.png",
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"https://au.mathworks.com/help/examples/globaloptim/win64/mtwashdemo_02.png",
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.8388927,"math_prob":0.96666366,"size":6401,"snap":"2022-27-2022-33","text_gpt3_token_len":1381,"char_repetition_ratio":0.16507737,"word_repetition_ratio":0.024072217,"special_character_ratio":0.20965475,"punctuation_ratio":0.15673469,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.966694,"pos_list":[0,1,2,3,4],"im_url_duplicate_count":[null,2,null,2,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-06-28T22:22:23Z\",\"WARC-Record-ID\":\"<urn:uuid:3de41cc4-d19d-409c-8f73-ccb2304268eb>\",\"Content-Length\":\"79906\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:37f1f750-6a4f-4749-83fe-7537ed18febe>\",\"WARC-Concurrent-To\":\"<urn:uuid:c9bebb75-31fb-4b7a-b3b6-eb9196601289>\",\"WARC-IP-Address\":\"23.218.145.211\",\"WARC-Target-URI\":\"https://au.mathworks.com/help/gads/patternsearch-climbs-mt-washington.html\",\"WARC-Payload-Digest\":\"sha1:PXZMGUPLO3FSN6N3OA7BSSWI3QXGZB4O\",\"WARC-Block-Digest\":\"sha1:CZU7WLRKEDCLLLXW7QHU3J4C6B6JDZK4\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-27/CC-MAIN-2022-27_segments_1656103617931.31_warc_CC-MAIN-20220628203615-20220628233615-00249.warc.gz\"}"} |
https://iitutor.com/volumes-using-integration/ | [
"# Volumes using Integration\n\n## Volume of Revolution\n\nWe can use integration to find volumes of revolution between $x=a$ and $x=b$.\nWhen the region enclosed by $y=f(x)$, the $x$-axis, and the vertical lines $x=a$ and $x=b$ is revolved through $2 \\pi$ or $360^{\\circ}$about the $x$-axis to generate a solid, the volume of the solid is given by:\n\\begin{align} \\displaystyle V &= \\lim_{h \\rightarrow 0} \\sum_{x=a}^{x=b}{\\pi \\big[f(x)\\big]^2 h} \\\\ &= \\int_{a}^{b}{\\pi \\big[f(x)\\big]^2}dx \\\\ &= \\pi \\int_{a}^{b}{y^2}dx \\end{align}\n\n### Example 1\n\nFind the volume of the solid generated when the line $y=x$ for $1 \\le x \\le 3$ is rovolved through $2 \\pi$ or $360^{\\circ}$ around the $x$-axis.\n\n\\begin{align} \\displaystyle V &= \\pi \\int_{1}^{3}{y^2}dx \\\\ &= \\pi \\int_{1}^{3}{x^2}dx \\\\ &= \\pi \\Big[\\dfrac{x^3}{3}\\Big]_{1}^{3} \\\\ &= \\dfrac{\\pi}{3} \\big[x^3\\big]_{1}^{3} \\\\ &= \\dfrac{\\pi}{3} \\big[3^3 – 1^3\\big] \\\\ &= \\dfrac{\\pi}{3} \\times 26 \\\\ &= \\dfrac{26 \\pi}{3} \\text{ units}^3 \\end{align}\n\n### Example 2\n\nFind the volume of the solid generated when the line $y=\\sqrt{x}$ for $0 \\le x \\le 2$ is rovolved through $2 \\pi$ or $360^{\\circ}$ around the $x$-axis.\n\n\\begin{align} \\displaystyle V &= \\pi \\int_{0}^{2}{y^2}dx \\\\ &= \\pi \\int_{0}^{2}{\\sqrt{x}^2}dx \\\\ &= \\pi \\int_{0}^{2}{x}dx \\\\ &= \\pi \\Big[\\dfrac{x^2}{2}\\Big]_{0}^{2} \\\\ &= \\dfrac{\\pi}{2}\\pi \\big[x^2\\big]_{0}^{2} \\\\ &= \\dfrac{\\pi}{2}\\pi \\big[2^2 – 0^2\\big]_{0}^{2} \\\\ &= \\dfrac{\\pi}{2} \\times 4 \\\\ &= 2 \\pi \\text{ units}^3 \\end{align}",
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https://www.classcentral.com/course/introduction-to-mathematics-for-finance-and-busin-37579 | [
"Class Central is learner-supported. When you buy through links on our site, we may earn an affiliate commission.\n\n# Introduction to Mathematics for Finance and Business\n\n## Overview\n\nThis course will allow you to use mathematical equations to describe and analyze certain problems that appear in the areas of business and finance. For example, it analyzes how the performance of an asset or distributing a product is modeled, how an optimization process is done in a portfolio or how aversion to risk can be described to an investor. Additionally, it will help you understand how certain algorithms are solved to analyze large amounts of data, make forecasts and describe trends. The course uses programming and simulation tools to convey the main concepts.\n\nThe course starts from the concept of function, that is the basis of mathematical modeling and allows identifying the relationship between a group of variables of interest. Different types of functions are analyzed , such as linear, polynomial and exponential functions, providing specific examples to the areas of finance and business. The concepts are supported with the help of packages such as R / Python or Matlab. In this way, interactive and visual learning is motivated, which in addition to transmitting mathematical concepts, establishes the bases to develop computational skills.\n\n## Syllabus\n\nTopic 1: Linear Functions\n\n1.1 Geometric presentation of a linear equation. Linear interpolation. Slope concept.\n\n1.2 First order difference equation. Introduction to simulation.\n\nTopic 2: Linear Algebra\n\n2.1 Solution of simultaneous equations. Concept of a vector and a matrix.. Geometric representation.\n\n2.2 Matrix operations. Characterization of the number of solutions in a system. Linear Transformation Concept.\n\nTopic 3: Nonlinear functions\n\n3.1 Polynomials and root determination. Concepts of Concavity and Convexity.\n\n3.2 Exponential and Logarithmic Equations .\n\nTopic 4: Infinitesimal calculus.\n\n4.1 Concept of limit and derivation rules. Interpretation of the derivative as a measure of sensitivity.\n\n4.2 Approach to nonlinear functions through Taylor series. Non-linear optimization and First Order Conditions.\n\nTopic 5 : Financial Mathematics.\n\n5 .1 Yield and Discount Rates, Geometric Series, Present Value Formulas,\n\n5 .2 Calculation of Perpetuities, Annuities and IRR of a bond.\n\n### Taught by\n\nLuis A. Hernández Arámburo\n\n## Reviews\n\nStart your review of Introduction to Mathematics for Finance and Business\n\n###",
null,
"Never Stop Learning!\n\nGet personalized course recommendations, track subjects and courses with reminders, and more."
]
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null,
"https://ccweb.imgix.net/https%3A%2F%2Fwww.classcentral.com%2Fimages%2Fsignup-illus-mobile.png",
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.8463532,"math_prob":0.9026892,"size":2712,"snap":"2021-04-2021-17","text_gpt3_token_len":556,"char_repetition_ratio":0.11816839,"word_repetition_ratio":0.0,"special_character_ratio":0.18989676,"punctuation_ratio":0.14767933,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9854604,"pos_list":[0,1,2],"im_url_duplicate_count":[null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-04-11T21:41:08Z\",\"WARC-Record-ID\":\"<urn:uuid:3126095b-3d3c-4f1f-953b-44386bdddd3e>\",\"Content-Length\":\"480809\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:05cfbf7e-7ea0-47d9-97fa-b3ccfe9136d7>\",\"WARC-Concurrent-To\":\"<urn:uuid:94310d42-38ef-492e-bfbd-32cbbc259978>\",\"WARC-IP-Address\":\"34.68.4.21\",\"WARC-Target-URI\":\"https://www.classcentral.com/course/introduction-to-mathematics-for-finance-and-busin-37579\",\"WARC-Payload-Digest\":\"sha1:YL3QECBDOBEZBBPLLIBUMWSLB6P2DXQK\",\"WARC-Block-Digest\":\"sha1:IDGA3SKL5YRTFIUF3L3SS7FFXBX4626R\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-17/CC-MAIN-2021-17_segments_1618038065492.15_warc_CC-MAIN-20210411204008-20210411234008-00620.warc.gz\"}"} |
https://www.livmathssoc.org.uk/cgi-bin/sews.py?DotProduct | [
"The \"Dot Product\" of two vectors is a scalar number that represents how much of one is in the direction of the other.\n\nThe dot product is linear in the lengths of the vectors, so for any scalar $c$ and vectors ${\\bf~a}$ and ${\\bf~b}$ we have $(c{\\bf~a}){.}{\\bf~b}=c({\\bf~a}{.}{\\bf~b})$\n\nThe dot product is commutative.\n\nThe geometric interpretation is that ${\\bf~a}{.}{\\bf~b}$ is the product of the lengths of the vectors, times the cosine of the angle between them. This implies that if the vectors are orthogonal then the dot product is zero.\n\nIf the vectors are finite dimensional then the dot product is obtained by summing the point-wise products of the coordinates.\n\nSo if ${\\bf~a}=\\left[\\begin{matrix}a_1\\\\a_2\\\\a_3\\end{matrix}\\right]$ and ${\\bf~b}=\\left[\\begin{matrix}b_1\\\\b_2\\\\b_3\\end{matrix}\\right]$ then ${\\bf~a}{\\bf.b}=a_1b_1+a_2b_2+a_3c_3.$"
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.8486955,"math_prob":0.9999691,"size":1079,"snap":"2022-27-2022-33","text_gpt3_token_len":326,"char_repetition_ratio":0.15534884,"word_repetition_ratio":0.025974026,"special_character_ratio":0.28822985,"punctuation_ratio":0.062801935,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99997354,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-07-02T20:57:45Z\",\"WARC-Record-ID\":\"<urn:uuid:2319c30d-f412-4ca7-808d-47d29eaf72ae>\",\"Content-Length\":\"5559\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:0e89a8d0-37be-4363-83c8-319028d0724b>\",\"WARC-Concurrent-To\":\"<urn:uuid:43cc167e-17ab-4141-96ff-0bbccd10a81c>\",\"WARC-IP-Address\":\"193.35.57.242\",\"WARC-Target-URI\":\"https://www.livmathssoc.org.uk/cgi-bin/sews.py?DotProduct\",\"WARC-Payload-Digest\":\"sha1:245XVC3JKGCJL2GOM6H67OIBRQOQE325\",\"WARC-Block-Digest\":\"sha1:VNONSV4UYOSMD3L5VYVKS7I5DQCUWFBV\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-27/CC-MAIN-2022-27_segments_1656104204514.62_warc_CC-MAIN-20220702192528-20220702222528-00795.warc.gz\"}"} |
http://adicodes.com/powershell-in-sharepoint-2010-basics-part-2/?shared=email&msg=fail | [
"# PowerShell in Sharepoint 2010 – Basics Part 2\n\nPowerShell in Sharepoint 2010 – Basics Part 2\n0 votes, 0.00 avg. rating (0% score)\n\n## Introduction\n\nWe learned how to use variable, arrays and hashtables in Powershell basics part 1. We will check how we can use arithmetic operators, assignment operators and comparision operators in powershell\n\n### Arithmetic Operators\n\nFollowing is the list of arithmetic operators and their usage\n\n#### Usage\n\n+ Adds two values Ex: 10+20\nOuput: 30Ex: “Hi” + “There”\nOutput: HiThere\n\nEx: [string]30 + “items”\nOutput: 30items\n\nSubtracts values Ex: 10 – 3\nOutput: 7\n* Multiplies two values\n\n(Also multiplys string value with numerics)\n\nEx: 10 * 3\nOutput: 30\n\nEx: “Hi” * 3\nOutput: HiHiHi\n\n/ Divide one value by another Ex: 10 / 3\nOutput: 3\n% Returns the remainder from devision Ex: 10 % 3\nOutput: 1\n\n### Assignment Operators\n\nFollowing is the list of\nassignment operators and their usage\n\n#### Usage\n\n= Sets the value of a variable to the specified value Ex: \\$myVariable = 1\n+= Increases the value of a variable by the specified value or appends to the\nexisting value\nEx:\n\n\\$myVariable = “Microsoft”\n\\$myVariable += ” ”\n\\$myVariable += “Sharepoint2010”\n\n\\$myVariable output:\n\nMicrosoft Sharepoint2010\n\n−= Decreases the value of a variable by the specified value Ex:\n\n\\$myVariable = 10\n\\$myVariable -= 3\n\n\\$myVariable output:7\n\n*= Multiplies the value of a variable by the specified value or appends the\nspecified value to the existing value\nEx:\n\n\\$myVariable = 10\n\\$myVariable *= 3\n\\$myVariable output: 30\n\n/= Divides the value of a variable by the specified value Ex:\n\n\\$myVariable = 10\n\\$myVariable /= 3\n\n\\$myVariable output: 3\n\n%= Divides the value of a variable by the specified value and assigns the remainder\nto the variable\nEx:\n\n\\$myVariable = 10\n\\$myVariable %= 3\n\n\\$myVariable output: 1\n\n++ Increases the value by one Ex:\n\n\\$myVariable = 10\n\\$myVariable ++\n\n\\$myVariable output: 11\n\nDecreases the value by one Ex:\n\n\\$myVariable = 10\n\\$myVariable —\n\\$myVariable output: 9\n\n### Comparision Operators\n\nFollowing is the list of comparison operators and their usage. This is one of the\nfrequently used operators with different syntax not similar to c#.\nSo, most of the developers tend to look for the reference how to use comparison operators\nNote that powershell is not case sensitive. Comparision operators will resemble operators in c# but we have\nto exclude the case sensitivity.\n\n#### Usage\n\n-eq equal to -eq operator is same as “==” in c#\n\nIf either side of the ‘-eq’ operator values are equal, it returns true\n\nEx:\nif(“variable1Value” -eq “variable1Value”)\n{\nwrite-host “equal”\n}\n\nOutput: equal\n\n-ne Not equal to -ne operator is same as “!=” in c#.\n\nIf either side of the ‘-ne’ operator values are not equal, it returns true\n\nEx:\nif(“variable1Value” -ne “variable2Value”)\n{\nwrite-host “notequal”\n}\n\nOutput: notequal\n\n-gt greater than -ne operator is same as “>” in c#.\n\nIf the left value of ‘-gt’ operator is greater than right value,it returns true.\n\nEx:\nif(100 -gt 50)\n{\nwrite-host “value is greater”\n}\n\nOutput: value is greater\n\n-lt less than -ne operator is same as “!=” in c#\nIf the left value of ‘-lt’ operator is less than right value,it returns true.\n\nEx:\nif(50 -lt 100)\n{\nwrite-host “value is less”\n}Output: value is less\n\n-le less than or equal to -le operator is same as “<=” in c#\nIf the left value of ‘-le’ operator is less than or equal to right value,it returns true.\n\nEx:\nif(100 -le 100)\n{\nwrite-host “either less than or equal”\n}\n\nOutput: either less than or equal\n\n-ge greater than or equal to -ge operator is same as “>=” in c#.\n\nIf the left value of ‘-ge’ operator is greater than or equal to right value,it returns true.\n\nEx:\nif(100 -ge 100)\n{\nwrite-host “either greater than or equal”\n}\n\nOutput: either greater than or equal\n\n-like Match with wildcard character (*) The –like and –notlike operators are similar to the –eq and –ne operators, but instead of matching exact values, they allow wildcards to be used.\n\nFor case sensitivity comparison, use ‘-clike’ operator\nEx:\n“This is test” -like “This is test”\nOutput : TrueEx:\n“This is test” -like “This is*”\nOutput : True\n\n-notlike Does not Match with wildcard character (*) The –like and –notlike operators are similar to the –eq and –ne operators, but instead of matching exact values, they allow wildcards to be used.\n\nEx:\n“This is test” -notlike “This is test”\nOutput : FalseEx:\n“This is test” -notlike “Is it test*”\nOutput : True\n\n-match Evaluates a regular expression against the operand on the left;\nreturns True if the match is successful. Operator ‘-cmatch’ is nfor case sensitive\nThe –match and –notmatch operators try to match one or more of the set of string values on the left side of the operation using regular expressions.\nEx:\n“This is test” -match “This is test”\nOutput : TrueEx: regular expression that checks if the string starts with “This”\n“This is test” -match “^This”\nOutput : True\n-notmatch Evaluates a regular expression against the operand on the left;\nreturns True if the match is not successful\nThe –match and –notmatch operators try to match one or more of the set of string values on the left side of the operation using regular expressions.\nEx:\n“This is test” -notmatch “This is test”\nOutput : FalseEx for regular expression that checks if the string starts with “This”\n“This is test” -notmatch “^This”\nOutput : False\n-replace Replaces all or part of a value with the specified value using regular expressions You can use the -replace operator to search for and replace a specific pattern.\n\nEx: Example that replace the word\n\n“This is test” -replace “test” “done”\nOutput : This is done\n\nWe use also use regular expression.\nUsing ^ character to match the beginning of the original string.\n\nEx:\n“This is test, this is good” -replace “^This” “That”\nOutput : That is test, this is good\n\n## Conclusion\n\nHope we got some good understanding on using operators.\nWe will further check some more basics of powershell to implement in sharepoint 2010.",
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"April 16, 2012 ·",
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"Adi ·",
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"No Comments",
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"Tags: , , , , , , , , · Posted in: Packaging and Deployment, Powershell, Sharepoint 2010"
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https://valenceelectrons.com/ | [
"# Valence Electrons\n\nThe total number of electrons in the last shell of an atom is called the valence electrons. That is, the total number of electrons in the last orbit of an element after electron configuration is called the valence electron.\n\nThe valence electrons of the element play an important role in that element. Valence electrons participate in the formation of bonds and compounds of one element with another. In this article, we will learn more about valence electrons.\n\n## What are valence electrons?\n\nThe nucleus is located at the center of the atom. Protons and neutrons are located in the nucleus. The electrons are located around the nucleus at a certain distance and in a certain circular path.\n\nThese circular pathways are called shell or energy levels. This shell or energy level is known as orbit. According to scientist Niels Bohr, energy levels are three-dimensional spherical. That is, the shells are somewhat like football-shaped spheres.\n\nThe shell closest to the nucleus is the first orbit. This first shell is the smallest. This shell can hold a maximum of two electrons. The second shell is larger than the first shell. This orbit(shell) can have a maximum of eight electrons.\n\nAs the shell gets bigger, the electron holding capacity in the shell increases. The total number of electrons in the last shell after arranging the electrons of the element is called the valence electron. Depending on the element, the valence electrons may be paired or unpaired.\n\n## Where are the valence electrons located in the atom?\n\nWe already know what the valence electron is about. This time we will know where the valence electron is located in the atom. To know the details about the valence electrons first you have to know about the electron configuration of the element.\n\nHowever, the answer to the question of where the valence electron is located in the atom is that the valence electron is located in the last orbit of the atom.\n\nThat is, the total number of electrons in the last orbit of an atom is the number of valence electrons in that element. However, in order to determine the number of valence electrons of a particular atom, one must know the electron configuration of that atom.\n\n## How to find valence electrons in an atom?\n\nThe electrons in the last orbit of the atom are the valence electrons. The valence electrons are in the last orbit of the atom. To determine the number of valence electrons of a particular atom, the electrons of that element have to be arranged.\n\nHowever, in addition to the electron configuration, valence electrons can be determined by group and block in the periodic table.\n\n## Determination of valence electrons by electron configuration\n\nThe simplest and most accurate way to determine the valence electron is to know in detail the electron configuration of that element. Valence electrons can not be accurately determined by block and group. But valence electrons can be easily identified by electron configuration.\n\nThis time we will see how to arrange the electrons of an element. Many rules have to be followed to arrange the electrons of the element. For example, Bohr Principal, Aufbau Principle, Hund Principle, Poly Exclusion Principle. However, it is possible to easily determine the valence electron if the Bohr principal knows.\n\nScientist Niels Bohr was the first to give an idea of the atom’s orbit. He provided a model of the atom in 1913. The complete idea of the orbit is given there. The electrons of the atom revolve around the nucleus in a certain circular path. These circular paths are called orbit. These orbits are expressed by n. [ n = 1,2 3 4 . . .]\n\nK is the name of the first orbit, L is the second, M is the third, N is the name of the fourth orbit. The electron holding capacity of each orbit is 2n2. [Where, n = 1, 2 3, 4. . .]\n\nNow,\n\n• n = 1 for K orbit. The electron holding capacity of K orbit is 2n2 = 2 × 12 = 2 electrons.\n• For L orbit, n = 2. The electron holding capacity of the L orbit is 2n2 = 2 × 22 = 8 electrons.\n• n=3 for M orbit. The maximum electron holding capacity in M orbit is 2n2 = 2 × 3= 18 electrons.\n• n=4 for N orbit. The maximum electron holding capacity in N orbit is 2n2 = 2 × 32 = 32 electrons.\n\nWe already know that the total number of electrons in the last orbit of an atom is the number of valence electrons in that element. For example, the atomic number of oxygen is eight. Therefore, the atom of the oxygen element has a total of eight electrons.\n\nThe electron configuration of oxygen shows that there are two electrons in the first orbit and a total of six electrons in the second orbit. The total of eight electrons of oxygen atom are in the first and second orbits.\n\nFrom the electron configuration of oxygen atoms, we see that the second orbit of oxygen is the last shell. And there are a total of 8 electrons in the last orbit. So, we can easily say that the oxygen atom has six valence electrons.\n\nSimilarly, the atomic number of sulfur is 16. Therefore, the total number of electrons in a sulfur atom is 16. The electron configuration of sulfur shows that there are a total of two electrons in the first orbit and a total of eight electrons in the second orbit.\n\nAnd the third orbit has a total of six electrons. From the electron configuration of sulfur, it is understood that the third orbit is the last orbit of sulfur and there are a total of six electrons. So, we can easily say that the number of valence electrons of sulfur is six.\n\n### Determination of valence electrons through the electron configuration of the Aufbau principle\n\nThe German physicist Aufbau first proposed an idea of electron configuration through sub-orbits. The Aufbau method is to do electron configuration through the sub-energy level. These sub-orbitals are expressed by ‘l’.\n\nThe Aufbau principle is that the electrons present in the atom will first complete the lowest energy orbital and then gradually continue to complete the higher energy orbital.\n\nThese orbitals are named s, p, d, f. The electron holding capacity of these orbitals is s = 2, p = 6, d = 10 and f = 14. The Aufbau electron configuration method is 1 s 2s 2p 3s 3p 4s 3d 4p 5s 4d 5p 6s 4f 5d 6p 7s 5f 6d.\n\nThe atomic number of nitrogen is seven. Therefore, the total number of electrons in a nitrogen atom is seven. The electron configuration of nitrogen is 1s2 2s2 2p3.\n\nThe electron configuration of nitrogen shows that there are two electrons in the first orbit and a total of five electrons in the second orbit. From the electron configuration of nitrogen, it is understood that the second orbit of nitrogen is the last orbit.\n\nThe electron configuration of nitrogen shows that there are two electrons in the first orbit and a total of five electrons in the second orbit. The electron configuration of nitrogen indicates that the second orbit of nitrogen is the last orbit and the last orbit has a total of 5 electrons. Therefore, we can easily say that the nitrogen atom has a total of 5 valence electrons.\n\nSimilarly, the atomic number of argon is 18. The total number of electrons in an argon atom is 18. The electron configuration of an argon atom is 1s2 2s2 2p6 3s2 3p6.\n\nThe electron configuration of argon shows that there are two electrons in the first shell, eight electrons in the second shell, and a total of eight electrons in the third orbit.\n\nFrom the electron configuration of argon, we understand that the third orbit of argon is the last orbit and the last shell has a total of eight electrons. Therefore, the number of valences electrons in argon are eight.\n\n## Determination of valence electrons by group\n\nValence electrons can be easily identified by groups and blocks. However, this method has several limitations in determining valence electrons. There are a total of 18 groups in the periodic table.\n\nGroup-1 elements are called alkali metals. The number of valence electrons in all the elements in this group is one. For example, the electron configuration of sodium is 1s2 2s2 2p6 3s1. The electron configuration of sodium shows that sodium is an element of group-1 and that sodium has one electron in its last orbit. So, the valence electron of sodium is one.\n\nGroup-2 elements are called alkaline earth metals. The valence electrons of all the elements in this group are two. For example, the electron configuration of magnesium shows that there are two electrons in the last orbit of magnesium.\n\nTherefore, the group of magnesium is two. The valence electrons of all the elements in this group, including magnesium are two.\n\nGroup-3 to 12 elements are called transition metal. The valence electrons of these elements vary by group. To determine the valence electrons of transition elements must know the electron configuration of that element. However, the valence electrons of transition elements range from 3 to 12.\n\nFor example, the electron configuration of vanadium is 1s2 2s2 2p6 3s2 3p6 3d3 4s2. The electron configuration of vanadium shows that there are a total of five electrons in the last orbit of vanadium. So, the valence electrons of vanadium are five. Vanadium is a transition element.\n\nThe number of valence electrons in all the elements of group-13 is three. Group-13 is also called Boron Group. The electron configuration of boron atom shows that there are three electrons in the last orbit of the boron. The valence electrons of all the elements of group-13 including boron are three.\n\nThe first element of group-14 is carbon. Therefore, group number fourteen is also called the carbon group. The electron configuration of carbon shows that carbon has a total of four electrons in its final orbit. So, the valence electrons of carbon are 4. From this, we can understand that the valence electron of all the elements of group-14 is 4.\n\nThe first element of group-15 is nitrogen. Therefore, this group is also called the nitrogen group. An element of this group is phosphorus. The electron configuration of phosphorus shows that It has a total of five electrons in its final orbit. So, the valence electrons of phosphorus are five. The valence electrons of all the elements in this group, including nitrogen, are five.\n\nThe valence electrons of all the elements of group-16 are six. Group number sixteen is called the chalcogens group. The first element in this group is oxygen. Selenium is an element of group-16. The atomic number of selenium is 34.\n\nThe electron configuration of selenium shows that the last shell has a total of six electrons. So, the valence electrons of selenium are 6. All the elements of group-16 have six valence electrons.\n\nGroup-17 is called the halogen group. The elements in this group are fluorine, chlorine, bromine, iodine, astatine. The valence electrons of the elements of group-17 are seven. The electron configuration of bromine shows that it has a total of seven electrons in its last shell.\n\nArranging the electrons of the other elements in this group shows that there are a total of seven electrons in the last shell of each element. Therefore, the valence electrons of all the elements in the halogen group are seven.\n\nThe elements of group-18 are inert elements. These elements do not easily participate in any chemical reaction. These elements remain in the gaseous state at normal temperatures. For these reasons, the elements of Group-18 are called Noble gas.\n\nThe octaves of these elements are full except for helium. That is, the last shell contains eight electrons. For example, the atomic number of argon is eighteen. That is, the argon atom has a total of eighteen electrons.\n\nThe electron configuration of argon shows that the last shell of an argon atom has a total of eight electrons. So, we can say that the number of valence electrons in the element of group-18 is eight. Only two valence electrons of helium.\n\n## Determination of valence electrons through block\n\nThe periodic table is divided into four blocks by electron configuration. The blocks are s, p, d, f. The specific valence electrons of any element cannot be determined by the block. However, it is possible to determine the maximum and minimum valence electrons of the elements in that block. For example, the s-block contains a total of fourteen elements. The valence electrons of these elements are limited to one or two.\n\nArranging the electrons of the s-block elements shows that the electron configuration ends in the s-orbital. For example, The electron configuration of calcium is 1s2 2s2 2p6 3s2 3p6 4s2. The electron configuration of calcium implies that the electron configuration of calcium ends in the s-orbital and the last shell has a total of two electrons. So, the valence electrons of calcium are two.\n\nThe p-block consists of a total of six groups from 13 to 18. The number of maximum valence electrons in the elements of this block is eight and the number of minimum valence electrons is three. The electron configuration of the p-block elements shows that the electron configuration ends in a p-orbital. Therefore, these elements are called p-block elements.\n\nFor example, the electron configuration of aluminum is 1s2 2s2 2p6 3s2 3p1. The electron configuration of aluminum implies that the electron configuration ends in a p-orbital and the last shell has a total of three electrons. So, aluminum has three valence electrons.\n\nThe d-block consists of all the elements from group-3 to 12. These elements are called transition elements. The valence electrons of these elements range from 3 to 12. From the electron configuration of the d-block elements, it is understood that the electron configuration ends in the d-orbital. Therefore, these elements are called d-block elements.\n\nFor example, the electron configuration of cobalt is 1s22s22p63s23p6 4s2 3d7. The electron configuration of cobalt shows that the electron configuration ends in a d-orbital and that the last shell has a total of nine electrons. Therefore, the valence electron of cobalt atom is 9.\n\nThe elements in the lanthanides and actinides series are called f-block elements. The valence electrons of these elements range from 3 to 16.\n\n## The importance of valence electrons in chemical reactions and bond formation\n\nThe electrons in the last shell of the atom participate in chemical reactions and bond formation. Valence electrons participate in both ionic or covalent bonds. The last shell of an element has 1, 2 or 3 electrons, those elements are called metals.\n\nThese elements leave electrons and turn into positive ions. Elements that have 5, 6, or 7 electrons in the last shell are called non-metals. All these elements receive electrons and turn into negative ions.\n\nThe bond formed by the exchange of electrons between metals and non-metals is called an ionic bond. For example, sodium and chlorine combine to form sodium chloride through ionic bonding.\n\nSodium has 1 valence electron and chlorine has 7 valence electrons. Sodium leaves 1 electron in its last shell and chlorine accepts that electron. As a result, sodium chloride bonds are formed.\n\nAgain, oxygen atoms have 6 valence electrons and hydrogen atoms have 1 valence electron. Oxygen and hydrogen are non-metallic elements. Hydrogen and oxygen together form covalent bonds through electron share.\n\nThe last shell of an oxygen atom has 6 electrons. The oxygen atom completes the octave by sharing electrons with the two hydrogen atoms.\n\nOn the other hand, hydrogen acquires the electron configuration of helium and comes to a stable state. In this way, hydrogen and oxygen make water(H2O) through electron sharing.\n\n## Determination of valence electrons of exceptional elements\n\nThe electron configuration of copper is somewhat exceptional. The atomic number of copper elements is 29. That is, the copper atom has a total of 29 electrons. Like other elements, the electron configuration of copper is 1s2 2s2 2p6 3s2 3p6 4s2 3d9.\n\nIn this case, the valence electrons of copper are two. But this electron configuration in copper is not correct. The d-orbital can hold a maximum of ten electrons. From the electron configuration of copper, we can see that it has nine electrons in its d-orbital.\n\nThe previous orbital(4s) has two electrons. 1 electron is transferred from the 4s orbital to the d-orbital. The d-orbital becomes stable as a result of the transfer of an electron. As a result, the electron configuration of copper changes to 1s2 2s2 2p6 3s2 3p6 4s1 3d10.\n\nFrom the correct electron configuration of copper, we can see that there is 1 electron in the last orbit. So, the valence electron of copper is 1. However, the valence electrons of copper can be easily determined by following the Bohr principle.\n\n## Characteristics of valence electron\n\n• The total number of electrons in the last shell of an element after electron configuration is called the valence electron.\n• The valence electron of the main group elements are in the last shell.\n• The valence electron of the transition element may be in the inner orbital.\n• Inert elements have eight valence electrons.\n• Characteristic determination of the element can be done through valence electron.\n• Group and block of elements can be determined by valence electrons.\n• The electrical conductivity of an element can be determined by valence electrons.\n• Valence electrons participate in chemical reactions and bond formation.\n• The valency of an element can be determined by valence electrons.\n\n## Conclusion\n\nThe number of electrons in the last shell is the valence electron of that element. To determine the valence electron must have a good idea about the electron configuration. Valence electrons participate in chemical reactions and bond formation. Characteristics of elements are determined by valence electrons.\n\n## What are valence electrons?\n\nThe total number of electrons in the last shell of an atom is called the valence electrons. That is, the total number of electrons in the last orbit of an element after electron configuration is called the valence electron.\n\n## Where are the valence electrons located in the atom?\n\nThe valence electron is located in the last orbit of the atom. That is, the total number of electrons in the last orbit of an atom is the number of valence electrons in that element.\n\n## How to find valence electrons in an atom?\n\nThe valence electrons are in the last orbit of the atom. To determine the number of valence electrons of a particular atom, the electrons of that element have to be arranged.\n\n## How many valence electrons of hydrogen(H) have?\n\nThe valence electron of hydrogen is 1. The valence electrons of hydrogen participates in bond formation. The hydrogen atom has only one electron. The symbol for hydrogen is ‘H’. Leaving 1 electron, hydrogen can turn into a positive ion.\n\n## How many valence electrons does helium(He) have?\n\nThe valence electrons of helium are two and the total number of electrons is two. The symbol for the helium element is ‘He’. Helium does not participate in chemical reactions and bond formation.\n\n## How many valence electrons does lithium(Li) have?\n\nThe valence electron of lithium is one. Lithium is called alkali metal. Lithium is the element of group-1 and the symbol is ‘Li’. Lithium atoms participate in the formation of bonds through valence electrons.\n\n## How many valence electrons does beryllium(Be) have?\n\nThe valence electrons of beryllium are two. Beryllium is the fourth element of the periodic table and the symbol is ‘Be’.\n\n## How many valence electrons does boron(B) have?\n\nThe valence electrons of boron are three. Boron is the fifth element of the periodic table and the symbol is ‘B’.\n\n## How many valence electrons does carbon(C) have?\n\nThe last shell of carbon has four electrons, so the valence electrons of carbon have four.\n\n## How many valence electrons does nitrogen(N) have?\n\nThe last shell of nitrogen has five electrons, so the valence electrons of nitrogen have five.\n\n## How many valence electrons does oxygen(O) have?\n\nThe valence electrons of oxygen are six. Oxygen is the eighth element of the periodic table and the symbol is ‘O’.\n\n## How many valence electrons does fluorine(F) have?\n\nThe valence electrons of fluorine has seven. Fluorine is a halogen element and its symbol is F.\n\n## How many valence electrons does neon(Ne) have?\n\nThe valence electrons of neon(Ne) has eight. Neon is an inert element and its symbol is ‘Ne’.\n\n## How many valence electrons does sodium(Na) have?\n\nThe last shell(orbit) of sodium has one electron. Therefore, the valence electron of sodium(Na) is one.\n\n## How many valence electrons does magnesium(Mg) have?\n\nThe last shell(orbit) of magnesium has two electrons. Therefore, the valence electrons of magnesium are two.\n\n## How many valence electrons does aluminum(Al) have?\n\nThe last shell(orbit) of aluminum has three electrons. Therefore, the valence electrons of aluminum(Al) are three."
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https://www.numberempire.com/2211169 | [
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"0 / 12\n\n# Number 2211169\n\ntwo million two hundred eleven thousand one hundred sixty nine\n\n### Properties of the number 2211169\n\n Factorization 1487 * 1487 Divisors 1, 1487, 2211169 Count of divisors 3 Sum of divisors 2212657 Previous integer 2211168 Next integer 2211170 Is prime? NO Previous prime 2211161 Next prime 2211179 2211169th prime 36119947 Is a Fibonacci number? NO Is a Bell number? NO Is a Catalan number? NO Is a factorial? NO Is a regular number? NO Is a perfect number? NO Polygonal number (s < 11)? square(1487) Binary 1000011011110101100001 Octal 10336541 Duodecimal 8a7741 Hexadecimal 21bd61 Square 4889268346561 Square root 1487 Natural logarithm 14.60903189292 Decimal logarithm 6.3446219370439 Sine 0.14797921593614 Cosine -0.98899047096063 Tangent -0.14962653360291\nNumber 2211169 is pronounced two million two hundred eleven thousand one hundred sixty nine. Number 2211169 is a composite number. Factors of 2211169 are 1487 * 1487. Number 2211169 has 3 divisors: 1, 1487, 2211169. Sum of the divisors is 2212657. Number 2211169 is not a Fibonacci number. It is not a Bell number. Number 2211169 is not a Catalan number. Number 2211169 is not a regular number (Hamming number). It is a not factorial of any number. Number 2211169 is a deficient number and therefore is not a perfect number. Number 2211169 is a square number with n=1487. Binary numeral for number 2211169 is 1000011011110101100001. Octal numeral is 10336541. Duodecimal value is 8a7741. Hexadecimal representation is 21bd61. Square of the number 2211169 is 4889268346561. Square root of the number 2211169 is 1487. Natural logarithm of 2211169 is 14.60903189292 Decimal logarithm of the number 2211169 is 6.3446219370439 Sine of 2211169 is 0.14797921593614. Cosine of the number 2211169 is -0.98899047096063. Tangent of the number 2211169 is -0.14962653360291\n\n### Number properties\n\n0 / 12\nExamples: 3628800, 9876543211, 12586269025"
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https://www.mathworks.com/matlabcentral/answers/8422-optimization | [
"# Optimization\n\n2 views (last 30 days)\nCharles on 30 May 2011\nClosed: MATLAB Answer Bot on 20 Aug 2021\n*Modified Post*\nI have a 3D matrix of this form:\ngauss = values(3,3,3)\ni.e\ngauss(:,:,1) =\n0.0155 0.0622 0.0155\n0.0622 0.2494 0.0622\n0.0155 0.0622 0.0155\ngauss(:,:,2) =\n0.0622 0.2494 0.0622\n0.2494 1.0000 0.2494\n0.0622 0.2494 0.0622\ngauss(:,:,3) =\n0.0155 0.0622 0.0155\n0.0622 0.2494 0.0622\n0.0155 0.0622 0.0155\nWhat I want to do is create another 3D matrix of larger dimension, i.e of size (10,3,3)\nI.e NewMatrix = zeros(10,3,3);\nWhat I want:\nNewMatrix(1:3,:,:) = NewMatrix(1:3,:,:) + gauss;\nNewMatrix(2:4,:,:) = NewMatrix(2:4,:,:) + gauss;\nNewMatrix(4:6,:,:) = NewMatrix(4:6,:,:) + gauss;\n:\n:\nNewMatrix(8:10,:,:) = NewMatrix(8:10,:,:) + gauss;\nSo essentially I want to use the 'gauss' matrix fill the values of 'NewMatrix'. However, the placement of the matrix 'gauss' in 'NewMatrix' is to be optimized so that the values of 'NewMatrix' are as uniform as possible. Let me know if more clarification is needed. Any suggestion will be appreciated.\nIvan van der Kroon on 5 Jun 2011\nJust put all weigths zero. I assume this is too trivial. I also noticed you opened a new topic, where you reformulated it for 1-D."
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https://matthewmumpower.com/publications/proceedings/2014/mumpower/impact-global-nuclear-mass-model-uncertainties-r-process-abundance-predictions | [
"## The impact of global nuclear mass model uncertainties on $r$-process abundance predictions\n\n### M. Mumpower, R. Surman, A. Aprahamian\n\nPublished CGS15 (2014)\n\nRapid neutron capture or \\'$r$-process\\' nucleosynthesis may be responsible for half the production of heavy elements above iron on the periodic table. Masses are one of the most important nuclear physics ingredients that go into calculations of $r$-process nucleosynthesis as they enter into the calculations of reaction rates, decay rates, branching ratios and Q-values. We explore the impact of uncertainties in three nuclear mass models on $r$-process abundances by performing global monte carlo simulations. We show that root-mean-square (rms) errors of current mass models are large so that current $r$-process predictions are insufficient in predicting features found in solar residuals and in $r$-process enhanced metal poor stars. We conclude that the reduction of global rms errors below $100$ keV will allow for more robust $r$-process predictions.\n\n## Mail\n\nMatthew Mumpower\nLos Alamos National Lab\nMS B283\nTA-3 Bldg 123\nLos Alamos, NM 87545"
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http://ixtrieve.fh-koeln.de/birds/litie/document/37471 | [
"# Document (#37471)\n\nAuthor\nRauber, A.\nTitle\nDigital preservation in data-driven science : on the importance of process capture, preservation and validation\nSource\nProceedings of the 2nd International Workshop on Semantic Digital Archives held in conjunction with the 16th Int. Conference on Theory and Practice of Digital Libraries (TPDL) on September 27, 2012 in Paphos, Cyprus [http://ceur-ws.org/Vol-912/proceedings.pdf]. Eds.: A. Mitschik et al\nYear\n2012\nPages\nS.7-17\nAbstract\nCurrent digital preservation is strongly biased towards data objects: digital files of document-style objects, or encapsulated and largely self-contained objects. To provide authenticity and provenance information, comprehensive metadata models are deployed to document information on an object's context. Yet, we claim that simply documenting an objects context may not be sufficient to ensure proper provenance and to fulfill the stated preservation goals. Specifically in e-Science and business settings, capturing, documenting and preserving entire processes may be necessary to meet the preservation goals. We thus present an approach for capturing, documenting and preserving processes, and means to assess their authenticity upon re-execution. We will discuss options as well as limitations and open challenges to achieve sound preservation, speci?cally within scientific processes.\nContent\nVgl. auch: http://sda2012.dke-research.de.\n\n## Similar documents (author)\n\n1. Rauch, C.; Rauber, A.: Anwendung der Nutzwertanalyse zur Bewertung von Strategien zur langfristigen Erhaltung digitaler Objekte (2005) 4.86\n```4.856803 = sum of:\n4.856803 = weight(author_txt:rauber in 4860) [ClassicSimilarity], result of:\n4.856803 = fieldWeight in 4860, product of:\n1.0 = tf(freq=1.0), with freq of:\n1.0 = termFreq=1.0\n9.713606 = idf(docFreq=6, maxDocs=42596)\n0.5 = fieldNorm(doc=4860)\n```\n2. Becker, C.; Rauber, A.: Decision criteria in digital preservation : what to measure and how (2011) 4.86\n```4.856803 = sum of:\n4.856803 = weight(author_txt:rauber in 457) [ClassicSimilarity], result of:\n4.856803 = fieldWeight in 457, product of:\n1.0 = tf(freq=1.0), with freq of:\n1.0 = termFreq=1.0\n9.713606 = idf(docFreq=6, maxDocs=42596)\n0.5 = fieldNorm(doc=457)\n```\n3. Rauber, K.; Nilges, A.: Was hieß noch mal schnell \"Unterbegriff\" auf Englisch? : Finden Sie die Antwort im Glossary to Terms of Information Literacy (2011) 4.86\n```4.856803 = sum of:\n4.856803 = weight(author_txt:rauber in 519) [ClassicSimilarity], result of:\n4.856803 = fieldWeight in 519, product of:\n1.0 = tf(freq=1.0), with freq of:\n1.0 = termFreq=1.0\n9.713606 = idf(docFreq=6, maxDocs=42596)\n0.5 = fieldNorm(doc=519)\n```\n4. Bashir, S.; Rauber, A.: On the relationship between query characteristics and IR functions retrieval bias (2011) 4.86\n```4.856803 = sum of:\n4.856803 = weight(author_txt:rauber in 629) [ClassicSimilarity], result of:\n4.856803 = fieldWeight in 629, product of:\n1.0 = tf(freq=1.0), with freq of:\n1.0 = termFreq=1.0\n9.713606 = idf(docFreq=6, maxDocs=42596)\n0.5 = fieldNorm(doc=629)\n```\n5. Klein, A.; Mitschang, J.; Nilges, A.; Oberhausen, B.; Rauber, K.; Weiß, A.: \"Aus der Praxis für die Praxis\" : ein Glossar zu Begriffen der Informationskompetenz (2008) 2.43\n```2.4284015 = sum of:\n2.4284015 = weight(author_txt:rauber in 2462) [ClassicSimilarity], result of:\n2.4284015 = fieldWeight in 2462, product of:\n1.0 = tf(freq=1.0), with freq of:\n1.0 = termFreq=1.0\n9.713606 = idf(docFreq=6, maxDocs=42596)\n0.25 = fieldNorm(doc=2462)\n```\n\n## Similar documents (content)\n\n1. 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Dobratz, S.; Neuroth, H.: nestor: Network of Expertise in long-term STOrage of digital Resources : a digital preservation initiative for Germany (2004) 0.22\n```0.22166362 = sum of:\n0.22166362 = product of:\n0.69269884 = sum of:\n0.004944102 = weight(abstract_txt:data in 2375) [ClassicSimilarity], result of:\n0.004944102 = score(doc=2375,freq=1.0), product of:\n0.04688777 = queryWeight, product of:\n1.0045604 = boost\n3.3742545 = idf(docFreq=3964, maxDocs=42596)\n0.01383266 = queryNorm\n0.10544545 = fieldWeight in 2375, product of:\n1.0 = tf(freq=1.0), with freq of:\n1.0 = termFreq=1.0\n3.3742545 = idf(docFreq=3964, maxDocs=42596)\n0.03125 = fieldNorm(doc=2375)\n0.0076863114 = weight(abstract_txt:science in 2375) [ClassicSimilarity], result of:\n0.0076863114 = score(doc=2375,freq=1.0), product of:\n0.06292356 = queryWeight, product of:\n1.1637317 = boost\n3.908901 = idf(docFreq=2322, maxDocs=42596)\n0.01383266 = queryNorm\n0.122153156 = fieldWeight in 2375, product of:\n1.0 = tf(freq=1.0), with freq of:\n1.0 = termFreq=1.0\n3.908901 = idf(docFreq=2322, maxDocs=42596)\n0.03125 = fieldNorm(doc=2375)\n0.014255587 = weight(abstract_txt:document in 2375) [ClassicSimilarity], result of:\n0.014255587 = score(doc=2375,freq=2.0), product of:\n0.07539017 = queryWeight, product of:\n1.2738069 = boost\n4.2786365 = idf(docFreq=1604, maxDocs=42596)\n0.01383266 = queryNorm\n0.1890908 = fieldWeight in 2375, product of:\n1.4142135 = tf(freq=2.0), with freq of:\n2.0 = termFreq=2.0\n4.2786365 = idf(docFreq=1604, maxDocs=42596)\n0.03125 = fieldNorm(doc=2375)\n0.010835684 = weight(abstract_txt:context in 2375) [ClassicSimilarity], result of:\n0.010835684 = score(doc=2375,freq=1.0), product of:\n0.07911136 = queryWeight, product of:\n1.3048652 = boost\n4.3829594 = idf(docFreq=1445, maxDocs=42596)\n0.01383266 = queryNorm\n0.13696748 = fieldWeight in 2375, product of:\n1.0 = tf(freq=1.0), with freq of:\n1.0 = termFreq=1.0\n4.3829594 = idf(docFreq=1445, maxDocs=42596)\n0.03125 = fieldNorm(doc=2375)\n0.06915431 = weight(abstract_txt:digital in 2375) [ClassicSimilarity], result of:\n0.06915431 = score(doc=2375,freq=18.0), product of:\n0.11889229 = queryWeight, product of:\n1.9591546 = boost\n4.3871174 = idf(docFreq=1439, maxDocs=42596)\n0.01383266 = queryNorm\n0.5816551 = fieldWeight in 2375, product of:\n4.2426405 = tf(freq=18.0), with freq of:\n18.0 = termFreq=18.0\n4.3871174 = idf(docFreq=1439, maxDocs=42596)\n0.03125 = fieldNorm(doc=2375)\n0.07529151 = weight(abstract_txt:authenticity in 2375) [ClassicSimilarity], result of:\n0.07529151 = score(doc=2375,freq=1.0), product of:\n0.28807038 = queryWeight, product of:\n2.4899783 = boost\n8.363679 = idf(docFreq=26, maxDocs=42596)\n0.01383266 = queryNorm\n0.26136497 = fieldWeight in 2375, product of:\n1.0 = tf(freq=1.0), with freq of:\n1.0 = termFreq=1.0\n8.363679 = idf(docFreq=26, maxDocs=42596)\n0.03125 = fieldNorm(doc=2375)\n0.112674944 = weight(abstract_txt:objects in 2375) [ClassicSimilarity], result of:\n0.112674944 = score(doc=2375,freq=7.0), product of:\n0.24823529 = queryWeight, product of:\n3.268835 = boost\n5.489905 = idf(docFreq=477, maxDocs=42596)\n0.01383266 = queryNorm\n0.45390382 = fieldWeight in 2375, product of:\n2.6457512 = tf(freq=7.0), with freq of:\n7.0 = termFreq=7.0\n5.489905 = idf(docFreq=477, maxDocs=42596)\n0.03125 = fieldNorm(doc=2375)\n0.39785635 = weight(abstract_txt:preservation in 2375) [ClassicSimilarity], result of:\n0.39785635 = score(doc=2375,freq=14.0), product of:\n0.52297777 = queryWeight, product of:\n5.810968 = boost\n6.506224 = idf(docFreq=172, maxDocs=42596)\n0.01383266 = queryNorm\n0.76075196 = fieldWeight in 2375, product of:\n3.7416575 = tf(freq=14.0), with freq of:\n14.0 = termFreq=14.0\n6.506224 = idf(docFreq=172, maxDocs=42596)\n0.03125 = fieldNorm(doc=2375)\n0.32 = coord(8/25)\n```\n3. Hendley, T.: ¬The preservation of digital material (1996) 0.19\n```0.18823154 = sum of:\n0.18823154 = product of:\n0.9411577 = sum of:\n0.08465208 = weight(abstract_txt:stated in 5154) [ClassicSimilarity], result of:\n0.08465208 = score(doc=5154,freq=1.0), product of:\n0.107243516 = queryWeight, product of:\n1.0742782 = boost\n7.2168646 = idf(docFreq=84, maxDocs=42596)\n0.01383266 = queryNorm\n0.78934455 = fieldWeight in 5154, product of:\n1.0 = tf(freq=1.0), with freq of:\n1.0 = termFreq=1.0\n7.2168646 = idf(docFreq=84, maxDocs=42596)\n0.109375 = fieldNorm(doc=5154)\n0.037924893 = weight(abstract_txt:context in 5154) [ClassicSimilarity], result of:\n0.037924893 = score(doc=5154,freq=1.0), product of:\n0.07911136 = queryWeight, product of:\n1.3048652 = boost\n4.3829594 = idf(docFreq=1445, maxDocs=42596)\n0.01383266 = queryNorm\n0.47938618 = fieldWeight in 5154, product of:\n1.0 = tf(freq=1.0), with freq of:\n1.0 = termFreq=1.0\n4.3829594 = idf(docFreq=1445, maxDocs=42596)\n0.109375 = fieldNorm(doc=5154)\n0.08068002 = weight(abstract_txt:digital in 5154) [ClassicSimilarity], result of:\n0.08068002 = score(doc=5154,freq=2.0), product of:\n0.11889229 = queryWeight, product of:\n1.9591546 = boost\n4.3871174 = idf(docFreq=1439, maxDocs=42596)\n0.01383266 = queryNorm\n0.67859757 = fieldWeight in 5154, product of:\n1.4142135 = tf(freq=2.0), with freq of:\n2.0 = termFreq=2.0\n4.3871174 = idf(docFreq=1439, maxDocs=42596)\n0.109375 = fieldNorm(doc=5154)\n0.09329974 = weight(abstract_txt:processes in 5154) [ClassicSimilarity], result of:\n0.09329974 = score(doc=5154,freq=1.0), product of:\n0.16503395 = queryWeight, product of:\n2.308227 = boost\n5.1687922 = idf(docFreq=658, maxDocs=42596)\n0.01383266 = queryNorm\n0.56533664 = fieldWeight in 5154, product of:\n1.0 = tf(freq=1.0), with freq of:\n1.0 = termFreq=1.0\n5.1687922 = idf(docFreq=658, maxDocs=42596)\n0.109375 = fieldNorm(doc=5154)\n0.6446009 = weight(abstract_txt:preservation in 5154) [ClassicSimilarity], result of:\n0.6446009 = score(doc=5154,freq=3.0), product of:\n0.52297777 = queryWeight, product of:\n5.810968 = boost\n6.506224 = idf(docFreq=172, maxDocs=42596)\n0.01383266 = queryNorm\n1.232559 = fieldWeight in 5154, product of:\n1.7320508 = tf(freq=3.0), with freq of:\n3.0 = termFreq=3.0\n6.506224 = idf(docFreq=172, maxDocs=42596)\n0.109375 = fieldNorm(doc=5154)\n0.2 = coord(5/25)\n```\n4. Dalkir, K.: Knowledge management (2009) 0.17\n```0.1707329 = sum of:\n0.1707329 = product of:\n0.7113871 = sum of:\n0.026626162 = weight(abstract_txt:science in 86) [ClassicSimilarity], result of:\n0.026626162 = score(doc=86,freq=3.0), product of:\n0.06292356 = queryWeight, product of:\n1.1637317 = boost\n3.908901 = idf(docFreq=2322, maxDocs=42596)\n0.01383266 = queryNorm\n0.42315093 = fieldWeight in 86, product of:\n1.7320508 = tf(freq=3.0), with freq of:\n3.0 = termFreq=3.0\n3.908901 = idf(docFreq=2322, maxDocs=42596)\n0.0625 = fieldNorm(doc=86)\n0.10542756 = weight(abstract_txt:encapsulated in 86) [ClassicSimilarity], result of:\n0.10542756 = score(doc=86,freq=1.0), product of:\n0.18027717 = queryWeight, product of:\n1.3928412 = boost\n9.356931 = idf(docFreq=9, maxDocs=42596)\n0.01383266 = queryNorm\n0.5848082 = fieldWeight in 86, product of:\n1.0 = tf(freq=1.0), with freq of:\n1.0 = termFreq=1.0\n9.356931 = idf(docFreq=9, maxDocs=42596)\n0.0625 = fieldNorm(doc=86)\n0.109996974 = weight(abstract_txt:capturing in 86) [ClassicSimilarity], result of:\n0.109996974 = score(doc=86,freq=1.0), product of:\n0.23365143 = queryWeight, product of:\n2.2424898 = boost\n7.532381 = idf(docFreq=61, maxDocs=42596)\n0.01383266 = queryNorm\n0.47077382 = fieldWeight in 86, product of:\n1.0 = tf(freq=1.0), with freq of:\n1.0 = termFreq=1.0\n7.532381 = idf(docFreq=61, maxDocs=42596)\n0.0625 = fieldNorm(doc=86)\n0.05331414 = weight(abstract_txt:processes in 86) [ClassicSimilarity], result of:\n0.05331414 = score(doc=86,freq=1.0), product of:\n0.16503395 = queryWeight, product of:\n2.308227 = boost\n5.1687922 = idf(docFreq=658, maxDocs=42596)\n0.01383266 = queryNorm\n0.32304952 = fieldWeight in 86, product of:\n1.0 = tf(freq=1.0), with freq of:\n1.0 = termFreq=1.0\n5.1687922 = idf(docFreq=658, maxDocs=42596)\n0.0625 = fieldNorm(doc=86)\n0.20335907 = weight(abstract_txt:documenting in 86) [ClassicSimilarity], result of:\n0.20335907 = score(doc=86,freq=1.0), product of:\n0.40289086 = queryWeight, product of:\n3.606497 = boost\n8.075996 = idf(docFreq=35, maxDocs=42596)\n0.01383266 = queryNorm\n0.5047498 = fieldWeight in 86, product of:\n1.0 = tf(freq=1.0), with freq of:\n1.0 = termFreq=1.0\n8.075996 = idf(docFreq=35, maxDocs=42596)\n0.0625 = fieldNorm(doc=86)\n0.21266316 = weight(abstract_txt:preservation in 86) [ClassicSimilarity], result of:\n0.21266316 = score(doc=86,freq=1.0), product of:\n0.52297777 = queryWeight, product of:\n5.810968 = boost\n6.506224 = idf(docFreq=172, maxDocs=42596)\n0.01383266 = queryNorm\n0.406639 = fieldWeight in 86, product of:\n1.0 = tf(freq=1.0), with freq of:\n1.0 = termFreq=1.0\n6.506224 = idf(docFreq=172, maxDocs=42596)\n0.0625 = fieldNorm(doc=86)\n0.24 = coord(6/25)\n```\n5. McCallum, S.H.: Preservation metadata standards for digital resources : what we have and what we need (2005) 0.14\n```0.14128833 = sum of:\n0.14128833 = product of:\n0.70644164 = sum of:\n0.012360256 = weight(abstract_txt:data in 5354) [ClassicSimilarity], result of:\n0.012360256 = score(doc=5354,freq=1.0), product of:\n0.04688777 = queryWeight, product of:\n1.0045604 = boost\n3.3742545 = idf(docFreq=3964, maxDocs=42596)\n0.01383266 = queryNorm\n0.26361364 = fieldWeight in 5354, product of:\n1.0 = tf(freq=1.0), with freq of:\n1.0 = termFreq=1.0\n3.3742545 = idf(docFreq=3964, maxDocs=42596)\n0.078125 = fieldNorm(doc=5354)\n0.025200557 = weight(abstract_txt:document in 5354) [ClassicSimilarity], result of:\n0.025200557 = score(doc=5354,freq=1.0), product of:\n0.07539017 = queryWeight, product of:\n1.2738069 = boost\n4.2786365 = idf(docFreq=1604, maxDocs=42596)\n0.01383266 = queryNorm\n0.33426848 = fieldWeight in 5354, product of:\n1.0 = tf(freq=1.0), with freq of:\n1.0 = termFreq=1.0\n4.2786365 = idf(docFreq=1604, maxDocs=42596)\n0.078125 = fieldNorm(doc=5354)\n0.07058031 = weight(abstract_txt:digital in 5354) [ClassicSimilarity], result of:\n0.07058031 = score(doc=5354,freq=3.0), product of:\n0.11889229 = queryWeight, product of:\n1.9591546 = boost\n4.3871174 = idf(docFreq=1439, maxDocs=42596)\n0.01383266 = queryNorm\n0.5936492 = fieldWeight in 5354, product of:\n1.7320508 = tf(freq=3.0), with freq of:\n3.0 = termFreq=3.0\n4.3871174 = idf(docFreq=1439, maxDocs=42596)\n0.078125 = fieldNorm(doc=5354)\n0.06664267 = weight(abstract_txt:processes in 5354) [ClassicSimilarity], result of:\n0.06664267 = score(doc=5354,freq=1.0), product of:\n0.16503395 = queryWeight, product of:\n2.308227 = boost\n5.1687922 = idf(docFreq=658, maxDocs=42596)\n0.01383266 = queryNorm\n0.4038119 = fieldWeight in 5354, product of:\n1.0 = tf(freq=1.0), with freq of:\n1.0 = termFreq=1.0\n5.1687922 = idf(docFreq=658, maxDocs=42596)\n0.078125 = fieldNorm(doc=5354)\n0.5316579 = weight(abstract_txt:preservation in 5354) [ClassicSimilarity], result of:\n0.5316579 = score(doc=5354,freq=4.0), product of:\n0.52297777 = queryWeight, product of:\n5.810968 = boost\n6.506224 = idf(docFreq=172, maxDocs=42596)\n0.01383266 = queryNorm\n1.0165975 = fieldWeight in 5354, product of:\n2.0 = tf(freq=4.0), with freq of:\n4.0 = termFreq=4.0\n6.506224 = idf(docFreq=172, maxDocs=42596)\n0.078125 = fieldNorm(doc=5354)\n0.2 = coord(5/25)\n```"
]
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.6775102,"math_prob":0.9974092,"size":15114,"snap":"2019-51-2020-05","text_gpt3_token_len":5842,"char_repetition_ratio":0.24698874,"word_repetition_ratio":0.42911878,"special_character_ratio":0.5346698,"punctuation_ratio":0.283424,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99974984,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-01-20T09:53:43Z\",\"WARC-Record-ID\":\"<urn:uuid:268ba38c-85aa-4021-bee5-f9e0063b3b35>\",\"Content-Length\":\"29934\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:209553c7-4e5a-48af-969e-47ddd96f3387>\",\"WARC-Concurrent-To\":\"<urn:uuid:8d3fa4dd-4e96-45e7-a18c-e4ef64bbc2d7>\",\"WARC-IP-Address\":\"139.6.160.6\",\"WARC-Target-URI\":\"http://ixtrieve.fh-koeln.de/birds/litie/document/37471\",\"WARC-Payload-Digest\":\"sha1:K5MCWCGLZQBVIV5EP3O2Q2WENY5B5WKG\",\"WARC-Block-Digest\":\"sha1:ETOZN5ZFHA3HSKFI7UCIOWBAF2KDBNO2\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-05/CC-MAIN-2020-05_segments_1579250598217.23_warc_CC-MAIN-20200120081337-20200120105337-00260.warc.gz\"}"} |
https://short-fact.com/how-do-you-calculate-cp-and-cpk/ | [
"# How do you calculate CP and Cpk?\n\n## How do you calculate CP and Cpk?\n\nA perfectly centered process will have Cp = Cpk. Both Cpk and Ppk relate the standard deviation and centering of the process about the midpoint to the allowable tolerance specifications. An estimate for Cpk = Cp(1-k). and since the maximum value for k is 1.0, then the value for Cpk is always equal to or less than Cp.\n\n### How do you calculate CP in Six Sigma?\n\nIt is calculated by finding the difference between the upper and lower specification. That difference is then divided by 6 standard deviations. The higher the value is, the better the process capability.\n\n#### How do you calculate CP in Excel?\n\nHow to Calculate CPK With Excel\n\n1. Launch Microsoft Excel and type “Data” in A1, “Upper Limit” in B1, “Average” in C1, “StDev” in D1, and “Cpk” in E1.\n2. Type “1” in A2, “2” in A3, “3” in A4, “4” in A5, “5” in A6, “6” in A7, “7” in A8, “8” in A9, “9” in A10, and “10” in A11.\n\nWhat is a CP value?\n\nCp is a ratio of the specification spread to the process spread. The process spread is often defined as the 6-sigma spread of the process (that is, 6 times the within-subgroup standard deviation). Higher Cp values indicate a more capable process.\n\nWhat is a good CP value?\n\nIn general, the higher the Cpk, the better. A Cpk value less than 1.0 is considered poor and the process is not capable. A value between 1.0 and 1.33 is considered barely capable, and a value greater than 1.33 is considered capable. But, you should aim for a Cpk value of 2.00 or higher where possible.\n\n## How many Sigma is 1.67 Cpk?\n\nSigma level table\n\nTwo sided table\nCpk Ppk Sigma level PPM out of tolerance\n1.33 4.0 63.342\n1.50 4.5 6.795\n1.67 5.0 0.573\n\n### What is CP Cpk?\n\nCp and Cpk, commonly referred to as process capability indices, are used to define the ability of a process to produce a product that meets requirements. In other words, they define what is expected from an item for it to be usable.\n\n#### What is minimum CP value?\n\nWhat does CP of 1.33 mean?\n\nIn simple words, it measures producer’s capability to produce a product within customer’s tolerance range. Cpk is used to estimate how close you are to a given target and how consistent you are to around your average performance. Cpk = or >1.33 indicates that the process is capable and meets specification limits.\n\nWhat does a Cp of 1.5 mean?\n\nSo a Cp of 1.5 means the process can fit inside the specification 1.5 times. A Cp greater than one is obviously desirable. However, the example has a Cp greater than one and yet it still has data outside the specification. This is due to the position of the overall average relative to the specification.\n\n## What is minimum Cpk value?\n\nA Cpk value less than 1.0 is considered poor and the process is not capable. A value between 1.0 and 1.33 is considered barely capable, and a value greater than 1.33 is considered capable.\n\n### What is CP and Cpk full form?\n\n#### Which is the best formula for the formula CPK?\n\nThe Cp can perform the best process if that process is centered on its midpoint. The minimum value of “k” is 0.0 and the maximum is 1.0. A perfect centered process will have Cp = Cpk. An estimate for Cpk = Cp (1-k).\n\nHow to calculate the CPK using CPL and CPU?\n\nDetermine the Cpk. Once you have the Cpl and Cpu calculated, you can put this into the Cpk formula. The Formula for Cpk = Min (Cpl, Cpu) The standard deviation is multiplied by three because six standard deviations (or six sigmas), account for just about every eventuality in a process using a normal distribution curve.\n\nHow is CPK calculated for off centered processes?\n\nTo calculate Cpk, compare the average of the data to both the upper and lower specification limit. An off-centered process will have a greater risk of fallout to the specification limit closest to the process mean. The reported Cpk will be the one that measures the highest risk.\n\n## What is the meaning of the CPK index?\n\nProcess Capability Index (Cpk) Definition: Process capability index (cpk) is the measure of process capability. It shows how closely a process is able to produce the output to its overall specifications."
]
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.93371016,"math_prob":0.9603977,"size":4430,"snap":"2022-40-2023-06","text_gpt3_token_len":1161,"char_repetition_ratio":0.15273385,"word_repetition_ratio":0.14563107,"special_character_ratio":0.26072234,"punctuation_ratio":0.12783505,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.985864,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-10-05T13:15:46Z\",\"WARC-Record-ID\":\"<urn:uuid:839fb2cb-4d2b-4382-b7b7-04840581cb84>\",\"Content-Length\":\"141517\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:fcf861a6-6129-4ba9-843f-6e7ba5006f85>\",\"WARC-Concurrent-To\":\"<urn:uuid:98226b8a-7f34-48ee-8439-17d2c6081f54>\",\"WARC-IP-Address\":\"104.21.73.15\",\"WARC-Target-URI\":\"https://short-fact.com/how-do-you-calculate-cp-and-cpk/\",\"WARC-Payload-Digest\":\"sha1:QMRTRX4E4AHMBDPU6H7VZUQE3RFRHLXI\",\"WARC-Block-Digest\":\"sha1:XU34HLKFHZLFYM4K3IH7SEJKGYREJALH\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-40/CC-MAIN-2022-40_segments_1664030337625.5_warc_CC-MAIN-20221005105356-20221005135356-00412.warc.gz\"}"} |
https://search.r-project.org/CRAN/refmans/BMS/html/coef.bma.html | [
"coef.bma {BMS} R Documentation\n\n## Posterior Inclusion Probabilities and Coefficients from a 'bma' Object\n\n### Description\n\nReturns a matrix with aggregate covariate-specific Bayesian model Averaging: posterior inclusion probabilites (PIP), post. expected values and standard deviations of coefficients, as well as sign probabilites\n\n### Usage\n\n ## S3 method for class 'bma'\ncoef(object, exact = FALSE, order.by.pip = TRUE, include.constant = FALSE,\nincl.possign = TRUE, std.coefs = FALSE, condi.coef = FALSE, ...)\n\n#equivalent:\nestimates.bma(object, exact = FALSE, order.by.pip = TRUE, include.constant = FALSE,\nincl.possign = TRUE, std.coefs = FALSE, condi.coef = FALSE)\n\n\n### Arguments\n\n object a 'bma' object (cf. bms) exact if exact=FALSE, then PIPs, coefficients, etc. will be based on aggregate information from the sampling chain with posterior model distributions based on MCMC frequencies (except in case of enumeration - cf. 'Details'); if exact=TRUE, estimates will be based on the nmodel best models encountered by the sampling chain, with the posterior model distribution based on their exact marginal likelihoods - cf. 'Details' below. order.by.pip order.by.pip=TRUE orders the resulting matrix according to posterior inclusion probabilites, order.by.pip=FALSE ranks them according to the original data (order of the covariates as in provided in X.data to bms), default TRUE include.constant If include.constant=TRUE then the resulting matrix includes the expected value of the constant in its last row. Default FALSE incl.possign If incl.possign=FALSE, then the sign probabilites column (cf. 'Values' below) is omitted from the result. Default TRUE std.coefs If std.coefs=TRUE then the expected values and standard deviations are returned in standardized form, i.e. as if the original data all had mean zero and variance 1. If std.coefs=FALSE (default) then both expected values and standard deviations are returned 'as is'. condi.coef If condi.coef=FALSE (default) then coefficients β_i and standard deviations are unconditional posterior expected values, as in standard model averaging; if condi.coef=FALSE then they are given as conditional on inclusion (equivalent to β_i / PIP_i). ... further arguments for other coef methods\n\n### Details\n\nMore on the argument exact:\nIn case the argument exact=TRUE, the PIPs, coefficient statistics and conditional sign probabilities are computed on the basis of the (500) best models the sampling chain encountered (cf. argument nmodel in bms). Here, the weights for Bayesian model averaging (BMA) are the posterior marginal likelihoods of these best models.\nIn case exact=FALSE, then these statistics are based on all accepted models (except burn-ins): If mcmc=\"enumerate\" then this are simply all models of the traversed model space, with their marginal likelihoods providing the weights for BMA.\nIf, however, the bma object bmao was based on an MCMC sampler (e.g. when bms argument mcmc=\"bd\"), then BMA statistics are computed differently: In contrast to above, the weights for BMA are MCMC frequencies, i.e. how often the respective models were encountered by the MCMC sampler. (cf. a comparison of MCMC frequencies and marginal likelihoods for the best models via the function pmp.bma).\n\n### Value\n\nA matrix with five columns (or four if incl.possign=FALSE)\n\n Column 'PIP' Posterior inclusion probabilities ∑ p(γ|i \\in γ, Y) / sum p(γ|Y) Column 'Post Mean' posterior expected value of coefficients, unconditional E(β|Y)=∑ p(γ|Y) E(β|γ,Y), where E(β_i|γ,i \\notin γ, Y)=0 if condi.coef=FALSE, or conditional on inclusion (E(β|Y) / ∑ p(γ|Y, i \\in γ) ) if condi.coef=TRUE Column 'Post SD' posterior standard deviation of coefficients, unconditional or conditional on inclusion, depending on condi.coef Column 'Cond.Pos.Sign' The ratio of how often the coefficients' expected values were positive conditional on inclusion. (over all visited models in case exact=FALSE, over the best models in case exact=TRUE) Column 'Idx' the original order of covariates as the were used for sampling. (if included, the constant has index 0)\n\n### Author(s)\n\nMartin Feldkircher and Stefan Zeugner\n\nbms for creating bma objects, pmp.bma for comparing MCMC frequencies and marginal likelihoods.\n\n### Examples\n\n#sample, with keeping the best 200 models:\ndata(datafls)\nmm=bms(datafls,burn=1000,iter=5000,nmodel=200)\n\n#standard BMA PIPs and coefficients from the MCMC sampling chain, based on\n# ...how frequently the models were drawn\ncoef(mm)\n\n#standardized coefficients, ordered by index\ncoef(mm,std.coefs=TRUE,order.by.pip=FALSE)\n\n#coefficients conditional on inclusion:\ncoef(mm,condi.coef=TRUE)\n\n#same as\nests=coef(mm,condi.coef=FALSE)\nests[,2]/ests[,1]\n\n#PIPs, coefficients, and signs based on the best 200 models\nestimates.bma(mm,exact=TRUE)\n\n#... and based on the 50 best models\ncoef(mm[1:50],exact=TRUE)\n\n\n\n[Package BMS version 0.3.4 Index]"
]
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.83166164,"math_prob":0.9605447,"size":3194,"snap":"2021-43-2021-49","text_gpt3_token_len":854,"char_repetition_ratio":0.11755486,"word_repetition_ratio":0.056155507,"special_character_ratio":0.24452098,"punctuation_ratio":0.1936508,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9971195,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-12-04T16:06:49Z\",\"WARC-Record-ID\":\"<urn:uuid:efc8dcc8-4d89-426a-9600-47161e1e930c>\",\"Content-Length\":\"8101\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:c6da3439-76b5-4e30-93b0-83e4df8a23c0>\",\"WARC-Concurrent-To\":\"<urn:uuid:49b1c95b-f591-4f36-b601-6c278f7a06f1>\",\"WARC-IP-Address\":\"137.208.57.46\",\"WARC-Target-URI\":\"https://search.r-project.org/CRAN/refmans/BMS/html/coef.bma.html\",\"WARC-Payload-Digest\":\"sha1:VAS75E6QZB2MQHGWYFKRO74GRBK52VKZ\",\"WARC-Block-Digest\":\"sha1:5JV3WJMOVJQW5BR5Q3U2O47IWMZ7C4KQ\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-49/CC-MAIN-2021-49_segments_1637964362999.66_warc_CC-MAIN-20211204154554-20211204184554-00076.warc.gz\"}"} |
https://cntofu.com/book/169/docs/1.0/nlp_word_embeddings_tutorial.md | [
"# 词嵌入：编码形式的词汇语义",
null,
"• The mathematician ran to the store.\n• The physicist ran to the store.\n• The mathematician solved the open problem.\n\n• The physicist solved the open problem.\n\n• 我们发现数学家和物理学家在句子里有相同的作用，所以在某种程度上，他们有语义的联系。\n• 当看见物理学家在新句子中的作用时，我们发现数学家也有起着相同的作用。\n\n## Getting Dense Word Embeddings（密集词嵌入）",
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"",
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"",
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"",
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"是两个向量的夹角。 这就意味着，完全相似的单词相似度为1。完全不相似的单词相似度为-1。\n\n## Pytorch中的词嵌入\n\n# 作者: Robert Guthrie\n\nimport torch\nimport torch.nn as nn\nimport torch.nn.functional as F\nimport torch.optim as optim\n\ntorch.manual_seed(1)\n\n\nword_to_ix = {\"hello\": 0, \"world\": 1}\nembeds = nn.Embedding(2, 5) # 2 words in vocab, 5 dimensional embeddings\nlookup_tensor = torch.tensor([word_to_ix[\"hello\"]], dtype=torch.long)\nhello_embed = embeds(lookup_tensor)\nprint(hello_embed)\n\n\n\ntensor([[ 0.6614, 0.2669, 0.0617, 0.6213, -0.4519]],\ngrad_fn=<EmbeddingBackward>)\n\n\n\n## 例子： N-Gram语言模型",
null,
"",
null,
"是单词序列的第i个单词。 在本例中，我们将在训练样例上计算损失函数，并且用反向传播算法更新参数。\n\nCONTEXT_SIZE = 2\nEMBEDDING_DIM = 10\n# 我们用莎士比亚的十四行诗 Sonnet 2\ntest_sentence = \"\"\"When forty winters shall besiege thy brow,\nAnd dig deep trenches in thy beauty's field,\nThy youth's proud livery so gazed on now,\nWill be a totter'd weed of small worth held:\nThen being asked, where all thy beauty lies,\nWhere all the treasure of thy lusty days;\nTo say, within thine own deep sunken eyes,\nWere an all-eating shame, and thriftless praise.\nHow much more praise deserv'd thy beauty's use,\nIf thou couldst answer 'This fair child of mine\nShall sum my count, and make my old excuse,'\nProving his beauty by succession thine!\nThis were to be new made when thou art old,\nAnd see thy blood warm when thou feel'st it cold.\"\"\".split()\n# 应该对输入变量进行标记，但暂时忽略。\n# 创建一系列的元组，每个元组都是([ word_i-2, word_i-1 ], target word)的形式。\ntrigrams = [([test_sentence[i], test_sentence[i + 1]], test_sentence[i + 2])\nfor i in range(len(test_sentence) - 2)]\n# 输出前3行，先看下是什么样子。\nprint(trigrams[:3])\n\nvocab = set(test_sentence)\nword_to_ix = {word: i for i, word in enumerate(vocab)}\n\nclass NGramLanguageModeler(nn.Module):\n\ndef __init__(self, vocab_size, embedding_dim, context_size):\nsuper(NGramLanguageModeler, self).__init__()\nself.embeddings = nn.Embedding(vocab_size, embedding_dim)\nself.linear1 = nn.Linear(context_size * embedding_dim, 128)\nself.linear2 = nn.Linear(128, vocab_size)\n\ndef forward(self, inputs):\nembeds = self.embeddings(inputs).view((1, -1))\nout = F.relu(self.linear1(embeds))\nout = self.linear2(out)\nlog_probs = F.log_softmax(out, dim=1)\nreturn log_probs\n\nlosses = []\nloss_function = nn.NLLLoss()\nmodel = NGramLanguageModeler(len(vocab), EMBEDDING_DIM, CONTEXT_SIZE)\noptimizer = optim.SGD(model.parameters(), lr=0.001)\n\nfor epoch in range(10):\ntotal_loss = 0\nfor context, target in trigrams:\n\n# 步骤 1\\. 准备好进入模型的数据 (例如将单词转换成整数索引,并将其封装在变量中)\ncontext_idxs = torch.tensor([word_to_ix[w] for w in context], dtype=torch.long)\n\n# 步骤 2\\. 回调torch累乘梯度\n# 在传入一个新实例之前，需要把旧实例的梯度置零。\nmodel.zero_grad()\n\n# 步骤 3\\. 继续运行代码，得到单词的log概率值。\nlog_probs = model(context_idxs)\n\n# 步骤 4\\. 计算损失函数（再次注意，Torch需要将目标单词封装在变量里）。\nloss = loss_function(log_probs, torch.tensor([word_to_ix[target]], dtype=torch.long))\n\n# 步骤 5\\. 反向传播更新梯度\nloss.backward()\noptimizer.step()\n\n# 通过调tensor.item()得到单个Python数值。\ntotal_loss += loss.item()\nlosses.append(total_loss)\nprint(losses) # 用训练数据每次迭代，损失函数都会下降。\n\n\n\n[(['When', 'forty'], 'winters'), (['forty', 'winters'], 'shall'), (['winters', 'shall'], 'besiege')]\n[518.6343855857849, 516.0739576816559, 513.5321269035339, 511.0085496902466, 508.5003893375397, 506.0077188014984, 503.52977323532104, 501.06553316116333, 498.6121823787689, 496.16915798187256]\n\n\n\n## 练习：计算连续词袋模型的词向量",
null,
"Pytorch中，通过填充下面的类来实现这个模型，有两条需要注意：\n\n• 考虑下你需要定义哪些参数。\n• 确保你知道每步操作后的结构，如果想重构，请使用.view()\nCONTEXT_SIZE = 2 # 左右各两个词\nraw_text = \"\"\"We are about to study the idea of a computational process.\nComputational processes are abstract beings that inhabit computers.\nAs they evolve, processes manipulate other abstract things called data.\nThe evolution of a process is directed by a pattern of rules\ncalled a program. People create programs to direct processes. In effect,\nwe conjure the spirits of the computer with our spells.\"\"\".split()\n\n# 通过对raw_text使用set()函数，我们进行去重操作\nvocab = set(raw_text)\nvocab_size = len(vocab)\n\nword_to_ix = {word: i for i, word in enumerate(vocab)}\ndata = []\nfor i in range(2, len(raw_text) - 2):\ncontext = [raw_text[i - 2], raw_text[i - 1],\nraw_text[i + 1], raw_text[i + 2]]\ntarget = raw_text[i]\ndata.append((context, target))\nprint(data[:5])\n\nclass CBOW(nn.Module):\n\ndef __init__(self):\npass\n\ndef forward(self, inputs):\npass\n\n# 创建模型并且训练。这里有些函数帮你在使用模块之前制作数据。\n\ndef make_context_vector(context, word_to_ix):\nidxs = [word_to_ix[w] for w in context]\nreturn torch.tensor(idxs, dtype=torch.long)\n\nmake_context_vector(data, word_to_ix) # example\n\n\n\n[(['We', 'are', 'to', 'study'], 'about'), (['are', 'about', 'study', 'the'], 'to'), (['about', 'to', 'the', 'idea'], 'study'), (['to', 'study', 'idea', 'of'], 'the'), (['study', 'the', 'of', 'a'], 'idea')]"
]
| [
null,
"https://img.cntofu.com/book/pytorch-doc-zh/docs/1.0/img/cf775cf1814914c00f5bf7ada7de4369.jpg",
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"https://img.cntofu.com/book/pytorch-doc-zh/docs/1.0/img/62ca51f900bf27324a2be4e6b8609f4b.jpg",
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"https://img.cntofu.com/book/pytorch-doc-zh/docs/1.0/img/a0be0498fd8216177330deffbfcb6ea2.jpg",
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"https://img.cntofu.com/book/pytorch-doc-zh/docs/1.0/img/8b4e6bfa073defa91d3f23cdec8f1f0e.jpg",
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"https://img.cntofu.com/book/pytorch-doc-zh/docs/1.0/img/320ccbbf907b47c4b407365b392e4367.jpg",
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"https://img.cntofu.com/book/pytorch-doc-zh/docs/1.0/img/1ed346930917426bc46d41e22cc525ec.jpg",
null,
"https://img.cntofu.com/book/pytorch-doc-zh/docs/1.0/img/3c3d846fb2913b4605e8d59bc5a14e6c.jpg",
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"https://img.cntofu.com/book/pytorch-doc-zh/docs/1.0/img/aa38f107289d4d73d516190581397349.jpg",
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"https://img.cntofu.com/book/pytorch-doc-zh/docs/1.0/img/56cc30cd24196bea5f078cb9b1f1d0dc.jpg",
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| {"ft_lang_label":"__label__zh","ft_lang_prob":0.7497033,"math_prob":0.9123802,"size":6632,"snap":"2021-21-2021-25","text_gpt3_token_len":3793,"char_repetition_ratio":0.07815329,"word_repetition_ratio":0.016474465,"special_character_ratio":0.27623641,"punctuation_ratio":0.19425173,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9590908,"pos_list":[0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18],"im_url_duplicate_count":[null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-06-15T09:51:55Z\",\"WARC-Record-ID\":\"<urn:uuid:b661b33d-5299-4fae-8350-4523f0fbdfb5>\",\"Content-Length\":\"58961\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:fca785bd-1d27-4c86-ae01-fa3e502fad79>\",\"WARC-Concurrent-To\":\"<urn:uuid:f5cd1a1c-f552-435b-9761-0a2365be6102>\",\"WARC-IP-Address\":\"47.75.243.192\",\"WARC-Target-URI\":\"https://cntofu.com/book/169/docs/1.0/nlp_word_embeddings_tutorial.md\",\"WARC-Payload-Digest\":\"sha1:EJMA4HB6WITRK3LJ2X5DK3FA6PF2NFWJ\",\"WARC-Block-Digest\":\"sha1:VPRA3KWIQ4CAKH3RIJ6MOSXXWZFM4OEJ\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-25/CC-MAIN-2021-25_segments_1623487620971.25_warc_CC-MAIN-20210615084235-20210615114235-00038.warc.gz\"}"} |
https://math.stackexchange.com/questions/tagged/finite-state-machine | [
"# Questions tagged [finite-state-machine]\n\nFor questions about finite state machine, which is a mathematical model of computation.\n\n59 questions\nFilter by\nSorted by\nTagged with\n39 views\n\n### Prove\\Disprove: If $L_1$ is not a CFL, and $L_2$ is finite, then $L_1 \\cup L_2$ is not a CFL.\n\nI am getting ready for finals, and encountered this question in a past assignment. I haven't proved this then and I don't understand how I can prove it now. Prove\\Disprove: If $L_1$ is not a Context-...\n23 views\n\n### How do I convert the following NFA to a DFA?\n\nFrom my textbook, I am trying to convert the above NFA to a DFA (no solution provided). I went through the steps to convert it and ended up with the following DFA solution: At first glance, I thought ...\n19 views\n\n### Finite State Machines: Can a state diagram have more than one self-loop?\n\nAccording to Schaum's Outlines, the state diagrams of Finite State Machines are labelled directed graphs. Now, it is entirely possible that the resultant state after transition returns to itself more ...\n22 views\n\n### If L is regular, then $L \\setminus \\{ \\epsilon \\}$ is regular.\n\nI need to show that if $L$ is regular language, then $L \\setminus \\{ \\epsilon \\}$. I was thinking: If $L$ contains $\\epsilon$, then in the FSA the start state is a final state. But by making the start ...\n50 views\n\n### Is Shuffling of a regular language regular too?\n\nIf $A$ be regular language, How we can prove that $A^{'}$ is regular too? $A^{'} = \\{a_{2}a_{1}a_{4}a_{3} ... a_{2n}a_{2n-1} \\mid a_{1}a_{2}a_{3}...a_{2n} \\in A\\}$ Is there any way to prove that even/...\n24 views\n\n### Pumping Lemma to prove language not regular, formatting $x$\n\nI need help with a question/verification I'm even thinking about it correctly. I'm trying to use the PL to prove $L$ is not regular. $L = \\{\\{a,b,c\\}^* \\mid |a| < |b| \\wedge |a| < |c|\\}$. This ...\n78 views\n\n15 views\n\n### Minimize the finite state machine given by the following state table\n\nMinimize the finite state machine given by the following state table\n24 views\n\n### Calculating a transition probability in FSM\n\nI have a Finite State Machine represented in following form: ...\n67 views\n\n### Finite State Automata\n\nit's been a while since I've done FSA's so I'm a little rusty, bear with me. I'm creating an FSA to parse Integer and Decimal tokens for a class. The two tokens have the regex Integer: $0|[1-9][0-9]^*$...\n48 views\n\n### Non-linear recurrence relation with Kronecker delta\n\nI am studying a game in number theory and I have come across some non-linear (coupled) recurrence relations which involve what I've been referring to as Kronecker deltas (or unit sample functions). An ...\n12 views\n\n### Are the following conversions from DFAs to GNFAs correct?\n\nThis machine (a) should accept all strings that start with a 1 and end with a 0. This machine (b) should accept all strings that contain three or more 1s. This machine (c) should accept all strings ...\n93 views\n\n### Floyd Invariant Principle on a deck of cards [closed]\n\nThe below problem has been taken from Mathematics for Computer Science (MIT Opencourseware https://ocw.mit.edu/courses/electrical-engineering-and-computer-science/6-042j-mathematics-for-computer-...\n81 views\n\n245 views\n\n### First Order Logic knowledge base problem\n\nI would like to present in the predicate logic the knowledge base and then check if the one provided formula is satisfied using the defined knowledgebase. I am trying to do this using SPASS prover, ...\n882 views\n\n### Creating a deterministic Finite state machine that accepts even 1s parity\n\nIve been working on this project for over a week now and its coming due soon, and im in no way going to finish it soon. the project is essentially where we have been assigned 3 \"Codes\" which form a ..."
]
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.89949864,"math_prob":0.838228,"size":14069,"snap":"2021-31-2021-39","text_gpt3_token_len":3758,"char_repetition_ratio":0.14816922,"word_repetition_ratio":0.041152265,"special_character_ratio":0.27052385,"punctuation_ratio":0.12338794,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9911298,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-08-04T22:13:38Z\",\"WARC-Record-ID\":\"<urn:uuid:ff37b2bd-d3d9-42b7-a0b0-6e29197a4326>\",\"Content-Length\":\"285602\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:7ce7040c-c72c-40ba-8b53-e1137cd89a92>\",\"WARC-Concurrent-To\":\"<urn:uuid:7b778adf-1fe8-4e04-8683-789f589f54b7>\",\"WARC-IP-Address\":\"151.101.65.69\",\"WARC-Target-URI\":\"https://math.stackexchange.com/questions/tagged/finite-state-machine\",\"WARC-Payload-Digest\":\"sha1:YJPHYEXTTXWHVX75GBRX3VVIQFSPGTUY\",\"WARC-Block-Digest\":\"sha1:QPVVZSCZET3VWIHYNZISL3442EIH33IL\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-31/CC-MAIN-2021-31_segments_1627046155188.79_warc_CC-MAIN-20210804205700-20210804235700-00535.warc.gz\"}"} |
https://stage.geogebra.org/m/hfP9Chhg | [
"# Construct a Triangle\n\nAuthor:\ndylan.busa\nTendopro\n\n## Construct a Triangle\n\n1. Use the Circle with Centre through Point to draw 2 circles that are the same size. In other words each circle passes through the centre point of the other. 2. Use the Intersect tool to mark the top intersection of the 2 circles. 3. Now, using the line tool, draw lines joining this point of intersection with the centres of the circles and join the centres to each other to create a triangle. 4. Use the Distance of Length tool to measure the length of each side of this triangle.\n\n## Question 1\n\nWhat kind of triangle have you constructed?\n\n## Question 2\n\nWhat is the size of each angle in this triangle? Use the Angle tool to measure each angle if you are not sure.\n\nCheck all that apply"
]
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.8711481,"math_prob":0.8738817,"size":734,"snap":"2021-31-2021-39","text_gpt3_token_len":167,"char_repetition_ratio":0.15479451,"word_repetition_ratio":0.0,"special_character_ratio":0.21934605,"punctuation_ratio":0.105960265,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9933235,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-09-19T13:21:26Z\",\"WARC-Record-ID\":\"<urn:uuid:6c8724a4-9323-4bd9-bb05-00956a137109>\",\"Content-Length\":\"45128\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:8ee9a6aa-3f46-41fb-ad1c-10a88395f3dc>\",\"WARC-Concurrent-To\":\"<urn:uuid:29d8ec86-0d2f-4f9e-8c5f-4453f6f91fbb>\",\"WARC-IP-Address\":\"13.32.150.11\",\"WARC-Target-URI\":\"https://stage.geogebra.org/m/hfP9Chhg\",\"WARC-Payload-Digest\":\"sha1:I2BC2UADXKK6O6TJYNRENDRKS5HL67EP\",\"WARC-Block-Digest\":\"sha1:ZSYQ2MB4DXADSGANPHKOWIHHJGMGNQXP\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-39/CC-MAIN-2021-39_segments_1631780056890.28_warc_CC-MAIN-20210919125659-20210919155659-00580.warc.gz\"}"} |
https://www.mathsisfun.com/fractions.html | [
"# Fractions\n\nHow many parts of a whole\n\n### Slice a pizza, and we get fractions:",
null,
"",
null,
"",
null,
"1/2 1/4 3/8 (One-Half) (One-Quarter) (Three-Eighths)\n\nThe top number says how many slices we have.\nThe bottom number says how many equal slices the whole pizza was cut into.\n\n## Equivalent Fractions\n\nSome fractions may look different, but are really the same, for example:\n\n 4/8 = 2/4 = 1/2 (Four-Eighths) (Two-Quarters) (One-Half)",
null,
"=",
null,
"=",
null,
"It is usually best to show an answer using the simplest fraction ( 1/2 in this case ). That is called Simplifying, or Reducing the Fraction\n\n## Numerator / Denominator\n\nWe call the top number the Numerator, it is the number of parts we have.\nWe call the bottom number the Denominator, it is the number of parts the whole is divided into.\n\nNumeratorDenominator\n\nYou just have to remember those names! (If you forget just think \"Down\"-ominator)\n\nIt is easy to add fractions with the same denominator (same bottom number):\n\n 1/4 + 1/4 = 2/4 = 1/2 (One-Quarter) (One-Quarter) (Two-Quarters) (One-Half)",
null,
"+",
null,
"=",
null,
"=",
null,
"One-quarter plus one-quarter equals two-quarters, equals one-half\n\nAnother example:\n\n 5/8 + 1/8 = 6/8 = 3/4",
null,
"+",
null,
"=",
null,
"=",
null,
"Five-eighths plus one-eighth equals six-eighths, equals three-quarters\n\n## Adding Fractions with Different Denominators\n\nBut what about when the denominators (the bottom numbers) are not the same?\n\n 3/8 + 1/4 = ?",
null,
"+",
null,
"=",
null,
"Three-eighths plus one-quarter equals ... what?\n\nWe must somehow make the denominators the same.\n\nIn this case it is easy, because we know that 1/4 is the same as 2/8 :\n\n 3/8 + 2/8 = 5/8",
null,
"+",
null,
"=",
null,
"Three-eighths plus two-eighths equals five-eighths\n\nThere are two popular methods to make the denominators the same:\n\n(They both work nicely, use the one you prefer.)\n\n## Other Things We Can Do With Fractions\n\nWe can also:\n\nVisit the Fractions Index to find out even more."
]
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"https://www.mathsisfun.com/images/fractions/pie-1-2.jpg",
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"https://www.mathsisfun.com/images/fractions/pie-1-4.jpg",
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"https://www.mathsisfun.com/images/fractions/pie-3-8.jpg",
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"https://www.mathsisfun.com/images/fractions/pie-4-8.jpg",
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"https://www.mathsisfun.com/images/fractions/pie-2-4.jpg",
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"https://www.mathsisfun.com/images/fractions/pie-1-2.jpg",
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"https://www.mathsisfun.com/images/fractions/pie-1-4.jpg",
null,
"https://www.mathsisfun.com/images/fractions/pie-1-4.jpg",
null,
"https://www.mathsisfun.com/images/fractions/pie-2-4.jpg",
null,
"https://www.mathsisfun.com/images/fractions/pie-1-2.jpg",
null,
"https://www.mathsisfun.com/images/fractions/pie-5-8.jpg",
null,
"https://www.mathsisfun.com/images/fractions/pie-1-8.jpg",
null,
"https://www.mathsisfun.com/images/fractions/pie-6-8.jpg",
null,
"https://www.mathsisfun.com/images/fractions/pie-3-4.jpg",
null,
"https://www.mathsisfun.com/images/fractions/pie-3-8.jpg",
null,
"https://www.mathsisfun.com/images/fractions/pie-1-4.jpg",
null,
"https://www.mathsisfun.com/images/fractions/pie-huh.jpg",
null,
"https://www.mathsisfun.com/images/fractions/pie-3-8.jpg",
null,
"https://www.mathsisfun.com/images/fractions/pie-2-8.jpg",
null,
"https://www.mathsisfun.com/images/fractions/pie-5-8.jpg",
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]
| {"ft_lang_label":"__label__en","ft_lang_prob":0.8505164,"math_prob":0.99745935,"size":1626,"snap":"2020-10-2020-16","text_gpt3_token_len":492,"char_repetition_ratio":0.15967941,"word_repetition_ratio":0.014134276,"special_character_ratio":0.30811808,"punctuation_ratio":0.08571429,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9971612,"pos_list":[0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28,29,30,31,32,33,34,35,36,37,38,39,40],"im_url_duplicate_count":[null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,9,null,6,null,null,null,null,null,null,null,null,null,null,null,6,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-02-29T13:39:47Z\",\"WARC-Record-ID\":\"<urn:uuid:72198012-b2c1-45df-80db-8f89efdd766c>\",\"Content-Length\":\"13628\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:ef6ce476-06e1-4113-b2c3-dc964876f673>\",\"WARC-Concurrent-To\":\"<urn:uuid:a0f23457-cebf-4645-85b3-b83f41985d24>\",\"WARC-IP-Address\":\"104.20.212.26\",\"WARC-Target-URI\":\"https://www.mathsisfun.com/fractions.html\",\"WARC-Payload-Digest\":\"sha1:BIEUAWDDOZ73SZGOYELUT5YF7NGZVXTH\",\"WARC-Block-Digest\":\"sha1:RWBDHUK37IWP23G4NAJESTDZAWZBCVWB\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-10/CC-MAIN-2020-10_segments_1581875149238.97_warc_CC-MAIN-20200229114448-20200229144448-00435.warc.gz\"}"} |
https://blog.coolight.cool/cvector%E7%9A%84%E5%A4%A9%E5%9D%91/ | [
"# [c++]记录使用vector遇到过的天坑问题\n\n/ 0评论 / 162阅读 / 0点赞\n\n## vector\n\n* vector是c++的一个模板容器，同时也是一个动态数组，是数组就意味着它支持随机访问。\n\n* 它几乎与arraylist一样，但vector是线程安全的，因此vector的性能比arraylist弱。\n\n* 使用时需要#include <vector>\n\n## 注意\n\n• 指向vector里面的元素的指针是迭代器iterator，而不是一般的变量的那个星号*。\n• back()：返回最后一个元素的引用。\n• end()： 返回的迭代器不可用，它返回的是指向最后一个元素的下一位置，因此想要访问最后一个元素，应该写end()-1，或使用back()。\n• 当vector内没有元素时，begin()和end()的返回值一样。\n• 注意对迭代器的运算顺序\n• 末尾两行如果执行会在运行时报错：\n• cannot seek vector iterator before begin\n• 即迭代器移动越界了\n``````vector<int> arr;\nfor (int i = 10; i--;)\t\t//向vector压入数据\narr.push_back(i);\n\nvector<int>::iterator it_p = arr.begin();//获取vector第一个的迭代器\nfor (; it_p < arr.end(); ++it_p)//依次顺序输出\ncout << *it_p << \" \";\n\ncout << arr.back() << endl;\t//输出最后一个元素的值\n\nit_p = arr.begin();\nit_p += 4;\t\t\t//把迭代器移动4个位置\ncout << *(it_p - 3) << endl;\t//不会报错\n//cout << *(it_p - 6 + 3) << endl;//如果编译器有优化提前计算出了结果-3，则不会报错\nint a = 6, b = 3;\n//cout << *(it_p - a + b) << endl;//会报错``````\n• -\n• 运行结果：\n• -\n• 在末尾的两行，it_p本来是指向第5个位置（下标为4），此时（it_p - 6 + 3）先算 it_p - 6 则会因为越界而在运行时报错。尽管它最终的结果等同于it_p - 3 是不会越界的。最后一句也是如此。\n• 解决方法当然也很简单，把后面的数值计算加个括号，注意符号可能需要改变\n• 如 （it_p - 6 + 3）改为（it_p - (6 - 3)）或（it_p + (3 - 6)）\n• 同理（it_p - a + b）改为 (it_p - (a - b)) 或 （it_p + (b - a)）\n• 当然vector是动态数组，是可以用[]随机访问的，即支持arr[-6 + 7]，这样写是没有问题的，它会先计算最终结果（-6 + 7）= 1。\n\n• 默认\n• 护眼\n• 夜晚\n• Serif\n• Sans"
]
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| {"ft_lang_label":"__label__zh","ft_lang_prob":0.8591137,"math_prob":0.99337775,"size":1205,"snap":"2022-27-2022-33","text_gpt3_token_len":769,"char_repetition_ratio":0.13738552,"word_repetition_ratio":0.0,"special_character_ratio":0.38257262,"punctuation_ratio":0.13068181,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9867478,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-08-08T17:00:42Z\",\"WARC-Record-ID\":\"<urn:uuid:ba44a47f-0e66-4125-9ff2-52e3103bec97>\",\"Content-Length\":\"32564\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:c392924c-0d36-45b5-9542-52300257f421>\",\"WARC-Concurrent-To\":\"<urn:uuid:d7306e94-876b-4db1-9d9a-e79c7d33048a>\",\"WARC-IP-Address\":\"101.33.245.234\",\"WARC-Target-URI\":\"https://blog.coolight.cool/cvector%E7%9A%84%E5%A4%A9%E5%9D%91/\",\"WARC-Payload-Digest\":\"sha1:RATXMF7KF36JFF3GL2OHDAT7SOSITV7G\",\"WARC-Block-Digest\":\"sha1:YJ3JRZ2VIPRIS47AJAB6CROWGFXSI5SV\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-33/CC-MAIN-2022-33_segments_1659882570868.47_warc_CC-MAIN-20220808152744-20220808182744-00091.warc.gz\"}"} |
http://mehap.cclal.org/rstudio-ggplot-bar-chart/ | [
"# Rstudio Ggplot Bar Chart\n\nbeginners guide to creating grouped and stacked bar charts .\n\nmake a bar plot with ggplot the practical r .\n\ncreate a percentage stacked bar chart tidyverse .\n\nr showing data values on stacked bar chart in ggplot2 .\n\ndetailed guide to the bar chart in r with ggplot r bloggers .\n\nggplot dodged vs faceted bar chart r bloggers .\n\nr ggplot2 how to add percentage or count labels above .\n\nr how to plot a stacked and grouped bar chart in ggplot .\n\ndetailed guide to the bar chart in r with ggplot r bloggers .\n\nr quick help creating a stacked bar chart ggplot2 .\n\nstacked bar chart in r ggplot free table bar chart .\n\nstacked barplot in r programming .\n\nr adding percentage labels to a bar chart in ggplot2 .\n\nr ggplot bar chart count free table bar chart .\n\nbar chart in r ggplot2 free table bar chart .\n\ncharts how to produce stacked bars within grouped .\n\nr bar graph ggplot2 free table bar chart .\n\nhow to put labels over geom bar for each bar in r with .\n\nr graph gallery rg 8 multiple arranged error bar plot .\n\nhow to create a ggplot stacked bar chart datanovia .\n\nbar chart in r ggplot2 godola .\n\nr generate paired stacked bar charts in ggplot using .\n\nr language grafico a barre verticale e orizzontale r .\n\nggplot2 error bars finished quick start guide r .\n\nfacet specific ordering for stacked bar chart tidyverse .\n\nmake a bar plot with ggplot r bloggers .\n\neasily plotting grouped bars with ggplot rstats r bloggers .\n\nr showing data values on stacked bar chart in ggplot2 ."
]
| [
null
]
| {"ft_lang_label":"__label__en","ft_lang_prob":0.53744376,"math_prob":0.76174474,"size":1481,"snap":"2020-45-2020-50","text_gpt3_token_len":345,"char_repetition_ratio":0.2620176,"word_repetition_ratio":0.1971831,"special_character_ratio":0.21944632,"punctuation_ratio":0.097222224,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9924656,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-12-01T12:37:34Z\",\"WARC-Record-ID\":\"<urn:uuid:c17816b3-520e-444d-8c11-7ac75f57ef30>\",\"Content-Length\":\"36222\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:e21871f4-4d01-494d-8ddb-64df2ec517f3>\",\"WARC-Concurrent-To\":\"<urn:uuid:28d06533-84a7-4139-8cc9-ff84cdbc2339>\",\"WARC-IP-Address\":\"213.202.241.219\",\"WARC-Target-URI\":\"http://mehap.cclal.org/rstudio-ggplot-bar-chart/\",\"WARC-Payload-Digest\":\"sha1:YRDPYZEHPRTFV7U42IPS5VNF3GHOIZHH\",\"WARC-Block-Digest\":\"sha1:JOGKSTDZVLFZK4CSWTIOQRQI73EBC3EF\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-50/CC-MAIN-2020-50_segments_1606141674082.61_warc_CC-MAIN-20201201104718-20201201134718-00224.warc.gz\"}"} |
https://docs.ray.io/en/latest/ray-air/examples/automl_with_ray_air.html | [
"# Simple AutoML for time series with Ray AIR#\n\nAutoML (Automatic Machine Learning) boils down to picking the best model for a given task and dataset. In this Ray Core example, we showed how to build an AutoML system which will chooses the best statsforecast model and its corresponding hyperparameters for a time series regression task on the M5 dataset.\n\nThe basic steps were:\n\n1. Define a set of autoregressive forecasting models to search over. For each model type, we also define a set of model parameters to search over.\n\n2. Perform temporal cross-validation on each model configuration in parallel.\n\n3. Pick the best performing model as the output of the AutoML system.\n\nWe see that these steps fit into the framework of a hyperparameter optimization problem that can be tackled with the Ray AIR Tuner!\n\nIn this notebook, we will show how to:\n\n1. Create an AutoML system with Ray AIR for time series forecasting.\n\n2. Leverage the higher-level Tuner API to define the model and hyperparameter search space, as well as parallelize cross-validation of different models.\n\n3. Analyze results to identify the best-performing model type and model parameters for the time-series dataset.\n\nSimilar to the Ray Core example, we will be using only one partition of the M5 dataset for this example.\n\n## Setup#\n\nLet’s first start by installing the statsforecast and ray[air] packages.\n\n!pip install statsforecast\n!pip install ray[air]\n\n\nNext, we’ll make the necessary imports, then initialize and connect to our Ray cluster!\n\nimport time\nimport itertools\nimport pandas as pd\nimport numpy as np\nfrom collections import defaultdict\nfrom statsforecast import StatsForecast\nfrom statsforecast.models import ETS, AutoARIMA, _TS\nfrom pyarrow import parquet as pq\nfrom sklearn.metrics import mean_squared_error, mean_absolute_error\n\nimport ray\nfrom ray import air, tune\n\nif ray.is_initialized():\nray.shutdown()\nray.init(runtime_env={\"pip\": [\"statsforecast\"]})\n\n\nNote\n\nWe may want to run on multiple nodes, and setting the runtime_env to include the statsforecast module will guarantee that we can access it on each worker, regardless of which node it lives on.\n\n## Read a partition of the M5 dataset from S3#\n\nWe first obtain the data from an S3 bucket and preprocess it to the format that statsforecast expects. As the dataset is quite large, we use PyArrow’s push-down predicate as a filter to obtain just the rows we care about without having to load them all into memory.\n\ndef get_m5_partition(unique_id: str) -> pd.DataFrame:\n\"s3://anonymous@m5-benchmarks/data/train/target.parquet\",\nfilters=[(\"item_id\", \"=\", unique_id)],\n)\nY_df = ds1.to_pandas()\n# StatsForecasts expects specific column names!\nY_df = Y_df.rename(\ncolumns={\"item_id\": \"unique_id\", \"timestamp\": \"ds\", \"demand\": \"y\"}\n)\nY_df[\"unique_id\"] = Y_df[\"unique_id\"].astype(str)\nY_df[\"ds\"] = pd.to_datetime(Y_df[\"ds\"])\nY_df = Y_df.dropna()\nconstant = 10\nY_df[\"y\"] += constant\nY_df = Y_df[Y_df.unique_id == unique_id]\nreturn Y_df\n\ntrain_df = get_m5_partition(\"FOODS_1_001_CA_1\")\ntrain_df\n\nunique_id ds y\n0 FOODS_1_001_CA_1 2011-01-29 13.0\n1 FOODS_1_001_CA_1 2011-01-30 10.0\n2 FOODS_1_001_CA_1 2011-01-31 10.0\n3 FOODS_1_001_CA_1 2011-02-01 11.0\n4 FOODS_1_001_CA_1 2011-02-02 14.0\n... ... ... ...\n1936 FOODS_1_001_CA_1 2016-05-18 10.0\n1937 FOODS_1_001_CA_1 2016-05-19 11.0\n1938 FOODS_1_001_CA_1 2016-05-20 10.0\n1939 FOODS_1_001_CA_1 2016-05-21 10.0\n1940 FOODS_1_001_CA_1 2016-05-22 10.0\n\n1941 rows × 3 columns\n\n## Create a function that performs cross-validation#\n\nNext, we will define two methods below:\n\n1. cross_validation performs temporal cross-validation on the dataset and reports the mean prediction error across cross-validation splits. See the visualizations in the analysis section below to see what the cross-validation splits look like and what we are averaging across. The n_splits and test_size parameters are used to configure the cross-validation splits, similar to TimeSeriesSplit from sklearn.\n\n2. compute_metrics_and_aggregate is a helper function used in cross_validation that calculates the aggregated metrics using the dataframe output of StatsForecast.cross_validation. For example, we will calculate the mean squared error between the model predictions and the actual observed data, averaged over all training splits. This metric gets reported to Tune as mse_mean, which is the metric we will use to define the best-performing model.\n\nWe will run this cross-validation function on all the model types and model parameters we are searching over, and the model that produces the lowest error metric will be the output of this AutoML example. Notice that model_cls_and_params is passed to the function via the config parameter. This is how Tune will set the corresponding model class and parameters for each trial.\n\nfrom ray.air import Checkpoint, session\n\ndef cross_validation(config, Y_train_df=None):\nassert Y_train_df is not None, \"Must pass in the dataset\"\n\n# Get the model class\nmodel_cls, model_params = config.get(\"model_cls_and_params\")\nfreq = config.get(\"freq\")\nmetrics = config.get(\"metrics\", {\"mse\": mean_squared_error})\n\n# CV params\ntest_size = config.get(\"test_size\", None)\nn_splits = config.get(\"n_splits\", 5)\n\nmodel = model_cls(**model_params)\n\n# Default the parallelism to the # of cross-validation splits\nparallelism_kwargs = {\"n_jobs\": n_splits}\n\n# Initialize statsforecast with the model\nstatsforecast = StatsForecast(\ndf=Y_train_df,\nmodels=[model],\nfreq=freq,\n**parallelism_kwargs,\n)\n\n# Perform temporal cross-validation (see sklearn.TimeSeriesSplit)\ntest_size = test_size or len(Y_train_df) // (n_splits + 1)\n\nstart_time = time.time()\nforecasts_cv = statsforecast.cross_validation(\nh=test_size,\nn_windows=n_splits,\nstep_size=test_size,\n)\ncv_time = time.time() - start_time\n\n# Compute metrics (according to metrics)\ncv_results = compute_metrics_and_aggregate(forecasts_cv, model, metrics)\n\n# Report metrics and save cross-validation output DataFrame\nresults = {\n**cv_results,\n\"cv_time\": cv_time,\n}\ncheckpoint_dict = {\n\"cross_validation_df\": forecasts_cv,\n}\ncheckpoint = Checkpoint.from_dict(checkpoint_dict)\nsession.report(results, checkpoint=checkpoint)\n\ndef compute_metrics_and_aggregate(\nforecasts_df: pd.DataFrame, model: _TS, metrics: dict\n):\nunique_ids = forecasts_df.index.unique()\nassert len(unique_ids) == 1, \"This example only expects a single time series.\"\n\ncutoff_values = forecasts_df[\"cutoff\"].unique()\n\n# Calculate metrics of the predictions of the models fit on\n# each training split\ncv_metrics = defaultdict(list)\nfor ct in cutoff_values:\n# Get CV metrics for a specific training split\n# All forecasts made with the same cutoff date\nsplit_df = forecasts_df[forecasts_df[\"cutoff\"] == ct]\nfor metric_name, metric_fn in metrics.items():\ncv_metrics[metric_name].append(\nmetric_fn(\nsplit_df[\"y\"], split_df[model.__class__.__name__]\n)\n)\n\n# Calculate aggregated metrics (mean, std) across training splits\ncv_aggregates = {}\nfor metric_name, metric_vals in cv_metrics.items():\ntry:\ncv_aggregates[f\"{metric_name}_mean\"] = np.nanmean(\nmetric_vals\n)\ncv_aggregates[f\"{metric_name}_std\"] = np.nanstd(\nmetric_vals\n)\nexcept Exception as e:\ncv_aggregates[f\"{metric_name}_mean\"] = np.nan\ncv_aggregates[f\"{metric_name}_std\"] = np.nan\n\nreturn {\n\"unique_ids\": list(unique_ids),\n**cv_aggregates,\n\"cutoff_values\": cutoff_values,\n}\n\n\n## Define the model search space#\n\nWe want to search over the following set of models and their corresponding parameters:\n\nsearch_space = {\nAutoARIMA: {},\nETS: {\n\"season_length\": [6, 7],\n\"model\": [\"ZNA\", \"ZZZ\"],\n}\n}\n\n\nThis translates to 5 possible (model_class, model_params) configurations, which we generate using the helper function below.\n\ndef generate_configurations(search_space):\nfor model, params in search_space.items():\nif not params:\nyield model, {}\nelse:\nconfigurations = itertools.product(*params.values())\nfor config in configurations:\nconfig_dict = {k: v for k, v in zip(params.keys(), config)}\nyield model, config_dict\n\nconfigs = list(generate_configurations(search_space))\nconfigs\n\n[(statsforecast.models.AutoARIMA, {}),\n(statsforecast.models.ETS, {'season_length': 6, 'model': 'ZNA'}),\n(statsforecast.models.ETS, {'season_length': 6, 'model': 'ZZZ'}),\n(statsforecast.models.ETS, {'season_length': 7, 'model': 'ZNA'}),\n(statsforecast.models.ETS, {'season_length': 7, 'model': 'ZZZ'})]\n\n\n## Create a Tuner to run a grid search over configurations#\n\nNow that we have defined the search space as well as the cross-validation function to apply to each configuration inside that search space, we can define our Ray AIR Tuner to launch the trials in parallel.\n\nHere’s a summary of what we are doing in the code below:\n\n• First, we include the training dataset using tune.with_parameters, which will put the dataset into the Ray object storage so that it can be retrieved as a common reference from every Tune trial.\n\n• Next, we define the Tuner param_space. We use Tune’s tune.grid_search to create one trial for each (model_class, model_params) tuple that we want to try. The rest of the parameters are constants that will be passed into the config parameter along with model_cls_and_params.\n\n• Finally, we specify that we want to minimize the reported mse_mean metric.\n\nWe can launch the trials by using Tuner.fit, which returns a ResultGrid that we can use for analysis.\n\ntuner = tune.Tuner(\ntune.with_parameters(cross_validation, Y_train_df=train_df),\nparam_space={\n\"model_cls_and_params\": tune.grid_search(configs),\n\"n_splits\": 5,\n\"test_size\": 1,\n\"freq\": \"D\",\n\"metrics\": {\"mse\": mean_squared_error, \"mae\": mean_absolute_error},\n},\ntune_config=tune.TuneConfig(\nmetric=\"mse_mean\",\nmode=\"min\",\n),\n)\n\nresult_grid = tuner.fit()\n\n\nGreat, we’ve computed cross-validation metrics for all the models! Let’s get the result of this AutoML system by selecting the best-performing trial using ResultGrid.get_best_result!\n\nbest_result = result_grid.get_best_result()\n\n\nWe can take a look at the hyperparameter config of the best result:\n\nbest_result.config\n\n{'model_cls_and_params': (statsforecast.models.ETS,\n{'season_length': 6, 'model': 'ZNA'}),\n'n_splits': 5,\n'test_size': 1,\n'freq': 'D',\n'metrics': {'mse': <function sklearn.metrics._regression.mean_squared_error(y_true, y_pred, *, sample_weight=None, multioutput='uniform_average', squared=True)>,\n'mae': <function sklearn.metrics._regression.mean_absolute_error(y_true, y_pred, *, sample_weight=None, multioutput='uniform_average')>}}\n\n\nWithin this config, we can pull out the model type and parameters that resulted in the lowest forecast error!\n\nbest_model_cls, best_model_params = best_result.config[\"model_cls_and_params\"]\nprint(\"Best model type:\", best_model_cls)\nprint(\"Best model params:\", best_model_params)\n\nBest model type: <class 'statsforecast.models.ETS'>\nBest model params: {'season_length': 6, 'model': 'ZNA'}\n\n\nWe can also inspect the reported metrics:\n\nprint(\"Best mse_mean:\", best_result.metrics[\"mse_mean\"])\nprint(\"Best mae_mean:\", best_result.metrics[\"mae_mean\"])\n\nBest mse_mean: 0.64205205\nBest mae_mean: 0.7200615\n\n\n## Analysis#\n\nFinally, let’s wrap up this AutoML example by performing some basic analysis and plotting.\n\n### Visualize Temporal Cross-validation Splits#\n\nLet’s first take a look at how cross-validation is being performed. This plot shows how our parameters of n_splits=5 and test_size=1 are being used to generate the cross-validation splits. Only the last 50 points in the dataset are shown for visualization purposes.\n\nFor each of the 5 splits, the blue ticks represent the data used to train the model. The orange tick is the index that the model is tying to predict, and it’s just a single point due to setting test_size=1. The metrics are calculated by comparing the predicted value to the actual data at the orange data point. The grey points represent data that is not considered for the split.\n\ncutoff_values_for_cv = best_result.metrics[\"cutoff_values\"]\ntest_size = best_result.config.get(\"test_size\")\nmse_per_split = best_result.metrics[\"mse_mean\"]\ncutoff_idxs = [np.where(train_df[\"ds\"] == ct) for ct in cutoff_values_for_cv]\ncolors = np.array([\"blue\", \"orange\", \"grey\"])\n\nimport matplotlib.pyplot as plt\n\nshow_last_n = 50\n\nplt.figure(figsize=(8, 3))\nfor i, cutoff_idx in enumerate(cutoff_idxs):\ndataset_idxs = np.arange(len(train_df))[-show_last_n:]\ncolor_idxs = np.zeros_like(dataset_idxs)\ncolor_idxs[dataset_idxs > cutoff_idx] = 1\ncolor_idxs[dataset_idxs > cutoff_idx + test_size] = 2\nplt.scatter(\nx=dataset_idxs,\ny=np.ones_like(dataset_idxs) * i,\nc=colors[color_idxs],\nmarker=\"_\",\nlw=8\n)\n\nplt.title(\nf\"Showing last {show_last_n} training samples of the {len(cutoff_idxs)} splits\\n\"\n\"Blue=Training, Orange=Test, Grey=Unused\"\n)\nplt.show()",
null,
"### Visualize model forecasts#\n\nEarlier, we saved the cross-validation output DataFrame inside a Ray AIR Checkpoint. We can use this to visualize some predictions of the best model! The predictions are pulled from the cross-validation results, where each step is predicted with horizon=1.\n\nAgain, we only show the last 50 timesteps for visualization purposes.\n\ndef plot_model_predictions(result, train_df):\nmodel_cls, model_params = result.config[\"model_cls_and_params\"]\n\n# Get the predictions from the data stored within this result's checkpoint\ncheckpoint_dict = result.checkpoint.to_dict()\nforecast_df = checkpoint_dict[\"cross_validation_df\"]\n\n# Only show the last 50 timesteps of the ground truth data\nmax_points_to_show = 50\nplt.figure(figsize=(10, 4))\nplt.plot(\ntrain_df[\"ds\"][-max_points_to_show:],\ntrain_df[\"y\"][-max_points_to_show:],\nlabel=\"Ground Truth\"\n)\nplt.plot(\nforecast_df[\"ds\"],\nforecast_df[model_cls.__name__],\nlabel=\"Forecast Predictions\"\n)\nplt.title(\nf\"{model_cls.__name__}({model_params}), \"\nf\"mae_mean={result.metrics['mse_mean']:.4f}\\n\"\nf\"Showing last {max_points_to_show} points\"\n)\nplt.legend()\n\nplt.show()\n\nplot_model_predictions(best_result, train_df)",
null,
"We can also visualize the predictions of the other models.\n\n# Plot for all results\nfor result in result_grid:\nplot_model_predictions(result, train_df)",
null,
"",
null,
"",
null,
"",
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""
]
| [
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"https://docs.ray.io/en/latest/_images/automl_with_ray_air_28_0.png",
null,
"https://docs.ray.io/en/latest/_images/automl_with_ray_air_30_0.png",
null,
"https://docs.ray.io/en/latest/_images/automl_with_ray_air_32_0.png",
null,
"https://docs.ray.io/en/latest/_images/automl_with_ray_air_32_1.png",
null,
"https://docs.ray.io/en/latest/_images/automl_with_ray_air_32_2.png",
null,
"https://docs.ray.io/en/latest/_images/automl_with_ray_air_32_3.png",
null,
"https://docs.ray.io/en/latest/_images/automl_with_ray_air_32_4.png",
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https://fr.mathworks.com/help/finance/frac2cur.html | [
"Documentation\n\n# frac2cur\n\nFractional currency value to decimal value\n\n## Syntax\n\n``Decimal = frac2cur(Fraction,Denominator)``\n\n## Description\n\nexample\n\n````Decimal = frac2cur(Fraction,Denominator)` converts a fractional currency value to a decimal value. `Fraction` is the fractional currency value input as a character vector, and `Denominator` is the denominator of the fraction.```\n\n## Examples\n\ncollapse all\n\nThis example shows how to convert a fractional currency value to a decimal value.\n\n`Decimal = frac2cur('12.1', 8)`\n```Decimal = 12.1250 ```\n\n## Input Arguments\n\ncollapse all\n\nFractional currency values, specified as a character vector or cell array of character vectors.\n\nData Types: `char` | `cell`\n\nDenominator of the fractions, specified as a scalar or vector using numeric values for the denominator.\n\nData Types: `double`\n\n## Output Arguments\n\ncollapse all\n\nDecimal currency value, returned as a scalar or vector with numeric decimal values.\n\nData Types: `double`"
]
| [
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.62081844,"math_prob":0.98275316,"size":264,"snap":"2019-35-2019-39","text_gpt3_token_len":60,"char_repetition_ratio":0.1923077,"word_repetition_ratio":0.0,"special_character_ratio":0.17045455,"punctuation_ratio":0.0952381,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99928445,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-09-22T09:00:37Z\",\"WARC-Record-ID\":\"<urn:uuid:9f1c298f-c683-4d3f-8965-ac17af7a2863>\",\"Content-Length\":\"71364\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:27b284eb-8fcf-4d68-b61e-91630c7e3fc8>\",\"WARC-Concurrent-To\":\"<urn:uuid:574b5614-26c3-4a42-8553-51a159290135>\",\"WARC-IP-Address\":\"104.110.193.39\",\"WARC-Target-URI\":\"https://fr.mathworks.com/help/finance/frac2cur.html\",\"WARC-Payload-Digest\":\"sha1:IQVDEGBGZSO646GPCV45Y2QUSXMKFVKA\",\"WARC-Block-Digest\":\"sha1:YNBKVIV4WN7FLT2BLEVGZM262HAHDZRW\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-39/CC-MAIN-2019-39_segments_1568514575402.81_warc_CC-MAIN-20190922073800-20190922095800-00270.warc.gz\"}"} |
https://www.colorhexa.com/0c9d8c | [
"# #0c9d8c Color Information\n\nIn a RGB color space, hex #0c9d8c is composed of 4.7% red, 61.6% green and 54.9% blue. Whereas in a CMYK color space, it is composed of 92.4% cyan, 0% magenta, 10.8% yellow and 38.4% black. It has a hue angle of 173 degrees, a saturation of 85.8% and a lightness of 33.1%. #0c9d8c color hex could be obtained by blending #18ffff with #003b19. Closest websafe color is: #009999.\n\n• R 5\n• G 62\n• B 55\nRGB color chart\n• C 92\n• M 0\n• Y 11\n• K 38\nCMYK color chart\n\n#0c9d8c color description : Dark cyan.\n\n# #0c9d8c Color Conversion\n\nThe hexadecimal color #0c9d8c has RGB values of R:12, G:157, B:140 and CMYK values of C:0.92, M:0, Y:0.11, K:0.38. Its decimal value is 826764.\n\nHex triplet RGB Decimal 0c9d8c `#0c9d8c` 12, 157, 140 `rgb(12,157,140)` 4.7, 61.6, 54.9 `rgb(4.7%,61.6%,54.9%)` 92, 0, 11, 38 173°, 85.8, 33.1 `hsl(173,85.8%,33.1%)` 173°, 92.4, 61.6 009999 `#009999`\nCIE-LAB 58.116, -38.083, -0.82 16.941, 26.084, 28.951 0.235, 0.362, 26.084 58.116, 38.092, 181.234 58.116, -46.057, 4.431 51.072, -30.168, 2.141 00001100, 10011101, 10001100\n\n# Color Schemes with #0c9d8c\n\n• #0c9d8c\n``#0c9d8c` `rgb(12,157,140)``\n• #9d0c1d\n``#9d0c1d` `rgb(157,12,29)``\nComplementary Color\n• #0c9d43\n``#0c9d43` `rgb(12,157,67)``\n• #0c9d8c\n``#0c9d8c` `rgb(12,157,140)``\n• #0c669d\n``#0c669d` `rgb(12,102,157)``\nAnalogous Color\n• #9d430c\n``#9d430c` `rgb(157,67,12)``\n• #0c9d8c\n``#0c9d8c` `rgb(12,157,140)``\n• #9d0c66\n``#9d0c66` `rgb(157,12,102)``\nSplit Complementary Color\n• #9d8c0c\n``#9d8c0c` `rgb(157,140,12)``\n• #0c9d8c\n``#0c9d8c` `rgb(12,157,140)``\n• #8c0c9d\n``#8c0c9d` `rgb(140,12,157)``\nTriadic Color\n• #1d9d0c\n``#1d9d0c` `rgb(29,157,12)``\n• #0c9d8c\n``#0c9d8c` `rgb(12,157,140)``\n• #8c0c9d\n``#8c0c9d` `rgb(140,12,157)``\n• #9d0c1d\n``#9d0c1d` `rgb(157,12,29)``\nTetradic Color\n• #07564d\n``#07564d` `rgb(7,86,77)``\n• #086e62\n``#086e62` `rgb(8,110,98)``\n• #0a8577\n``#0a8577` `rgb(10,133,119)``\n• #0c9d8c\n``#0c9d8c` `rgb(12,157,140)``\n• #0eb5a1\n``#0eb5a1` `rgb(14,181,161)``\n• #10ccb6\n``#10ccb6` `rgb(16,204,182)``\n• #11e4cb\n``#11e4cb` `rgb(17,228,203)``\nMonochromatic Color\n\n# Alternatives to #0c9d8c\n\nBelow, you can see some colors close to #0c9d8c. Having a set of related colors can be useful if you need an inspirational alternative to your original color choice.\n\n• #0c9d68\n``#0c9d68` `rgb(12,157,104)``\n• #0c9d74\n``#0c9d74` `rgb(12,157,116)``\n• #0c9d80\n``#0c9d80` `rgb(12,157,128)``\n• #0c9d8c\n``#0c9d8c` `rgb(12,157,140)``\n• #0c9d98\n``#0c9d98` `rgb(12,157,152)``\n• #0c969d\n``#0c969d` `rgb(12,150,157)``\n• #0c8a9d\n``#0c8a9d` `rgb(12,138,157)``\nSimilar Colors\n\n# #0c9d8c Preview\n\nText with hexadecimal color #0c9d8c\n\nThis text has a font color of #0c9d8c.\n\n``<span style=\"color:#0c9d8c;\">Text here</span>``\n#0c9d8c background color\n\nThis paragraph has a background color of #0c9d8c.\n\n``<p style=\"background-color:#0c9d8c;\">Content here</p>``\n#0c9d8c border color\n\nThis element has a border color of #0c9d8c.\n\n``<div style=\"border:1px solid #0c9d8c;\">Content here</div>``\nCSS codes\n``.text {color:#0c9d8c;}``\n``.background {background-color:#0c9d8c;}``\n``.border {border:1px solid #0c9d8c;}``\n\n# Shades and Tints of #0c9d8c\n\nA shade is achieved by adding black to any pure hue, while a tint is created by mixing white to any pure color. In this example, #010b0a is the darkest color, while #f8fefe is the lightest one.\n\n• #010b0a\n``#010b0a` `rgb(1,11,10)``\n• #021d1a\n``#021d1a` `rgb(2,29,26)``\n• #04302b\n``#04302b` `rgb(4,48,43)``\n• #05423b\n``#05423b` `rgb(5,66,59)``\n• #06544b\n``#06544b` `rgb(6,84,75)``\n• #08665b\n``#08665b` `rgb(8,102,91)``\n• #09796c\n``#09796c` `rgb(9,121,108)``\n• #0b8b7c\n``#0b8b7c` `rgb(11,139,124)``\n• #0c9d8c\n``#0c9d8c` `rgb(12,157,140)``\n• #0daf9c\n``#0daf9c` `rgb(13,175,156)``\n• #0fc1ac\n``#0fc1ac` `rgb(15,193,172)``\n• #10d4bd\n``#10d4bd` `rgb(16,212,189)``\n• #12e6cd\n``#12e6cd` `rgb(18,230,205)``\nShade Color Variation\n• #1deed5\n``#1deed5` `rgb(29,238,213)``\n• #30efd9\n``#30efd9` `rgb(48,239,217)``\n• #42f1dc\n``#42f1dc` `rgb(66,241,220)``\n• #54f2df\n``#54f2df` `rgb(84,242,223)``\n• #66f3e3\n``#66f3e3` `rgb(102,243,227)``\n• #78f5e6\n``#78f5e6` `rgb(120,245,230)``\n• #8bf6ea\n``#8bf6ea` `rgb(139,246,234)``\n• #9df8ed\n``#9df8ed` `rgb(157,248,237)``\n• #aff9f0\n``#aff9f0` `rgb(175,249,240)``\n• #c1faf4\n``#c1faf4` `rgb(193,250,244)``\n• #d4fcf7\n``#d4fcf7` `rgb(212,252,247)``\n• #e6fdfa\n``#e6fdfa` `rgb(230,253,250)``\n• #f8fefe\n``#f8fefe` `rgb(248,254,254)``\nTint Color Variation\n\n# Tones of #0c9d8c\n\nA tone is produced by adding gray to any pure hue. In this case, #545655 is the less saturated color, while #06a491 is the most saturated one.\n\n• #545655\n``#545655` `rgb(84,86,85)``\n• #4d5c5a\n``#4d5c5a` `rgb(77,92,90)``\n• #47635f\n``#47635f` `rgb(71,99,95)``\n• #406964\n``#406964` `rgb(64,105,100)``\n• #3a7069\n``#3a7069` `rgb(58,112,105)``\n• #33766e\n``#33766e` `rgb(51,118,110)``\n• #2d7d73\n``#2d7d73` `rgb(45,125,115)``\n• #268378\n``#268378` `rgb(38,131,120)``\n• #208a7d\n``#208a7d` `rgb(32,138,125)``\n• #199082\n``#199082` `rgb(25,144,130)``\n• #139787\n``#139787` `rgb(19,151,135)``\n• #0c9d8c\n``#0c9d8c` `rgb(12,157,140)``\n• #06a491\n``#06a491` `rgb(6,164,145)``\nTone Color Variation\n\n# Color Blindness Simulator\n\nBelow, you can see how #0c9d8c is perceived by people affected by a color vision deficiency. This can be useful if you need to ensure your color combinations are accessible to color-blind users.\n\nMonochromacy\n• Achromatopsia 0.005% of the population\n• Atypical Achromatopsia 0.001% of the population\nDichromacy\n• Protanopia 1% of men\n• Deuteranopia 1% of men\n• Tritanopia 0.001% of the population\nTrichromacy\n• Protanomaly 1% of men, 0.01% of women\n• Deuteranomaly 6% of men, 0.4% of women\n• Tritanomaly 0.01% of the population"
]
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https://answers.everydaycalculation.com/divide-fractions/10-18-divided-by-10-7 | [
"Solutions by everydaycalculation.com\n\n## Divide 10/18 with 10/7\n\n1st number: 10/18, 2nd number: 1 3/7\n\n10/18 ÷ 10/7 is 7/18.\n\n#### Steps for dividing fractions\n\n1. Find the reciprocal of the divisor\nReciprocal of 10/7: 7/10\n2. Now, multiply it with the dividend\nSo, 10/18 ÷ 10/7 = 10/18 × 7/10\n3. = 10 × 7/18 × 10 = 70/180\n4. After reducing the fraction, the answer is 7/18\n\nMathStep (Works offline)",
null,
"Download our mobile app and learn to work with fractions in your own time:"
]
| [
null,
"https://answers.everydaycalculation.com/mathstep-app-icon.png",
null
]
| {"ft_lang_label":"__label__en","ft_lang_prob":0.5893156,"math_prob":0.93249094,"size":424,"snap":"2022-27-2022-33","text_gpt3_token_len":214,"char_repetition_ratio":0.27380952,"word_repetition_ratio":0.0,"special_character_ratio":0.5754717,"punctuation_ratio":0.071428575,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.96108484,"pos_list":[0,1,2],"im_url_duplicate_count":[null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-06-26T11:39:27Z\",\"WARC-Record-ID\":\"<urn:uuid:53f762c3-6238-41d0-a82f-053f8592e024>\",\"Content-Length\":\"8393\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:45a7c473-03fd-4d55-8b4c-b06313c61eb5>\",\"WARC-Concurrent-To\":\"<urn:uuid:3c6fd4c0-f161-4493-ac45-00a725217b26>\",\"WARC-IP-Address\":\"96.126.107.130\",\"WARC-Target-URI\":\"https://answers.everydaycalculation.com/divide-fractions/10-18-divided-by-10-7\",\"WARC-Payload-Digest\":\"sha1:VO2LDHZGKMGD5IIIMDCEREDLKGWSXXKP\",\"WARC-Block-Digest\":\"sha1:NSLBWMRZVWDZ2J5PLYM27GMFTEFEYIO6\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-27/CC-MAIN-2022-27_segments_1656103205617.12_warc_CC-MAIN-20220626101442-20220626131442-00464.warc.gz\"}"} |
https://www.colorhexa.com/18d787 | [
"# #18d787 Color Information\n\nIn a RGB color space, hex #18d787 is composed of 9.4% red, 84.3% green and 52.9% blue. Whereas in a CMYK color space, it is composed of 88.8% cyan, 0% magenta, 37.2% yellow and 15.7% black. It has a hue angle of 154.9 degrees, a saturation of 79.9% and a lightness of 46.9%. #18d787 color hex could be obtained by blending #30ffff with #00af0f. Closest websafe color is: #00cc99.\n\n• R 9\n• G 84\n• B 53\nRGB color chart\n• C 89\n• M 0\n• Y 37\n• K 16\nCMYK color chart\n\n#18d787 color description : Strong cyan - lime green.\n\n# #18d787 Color Conversion\n\nThe hexadecimal color #18d787 has RGB values of R:24, G:215, B:135 and CMYK values of C:0.89, M:0, Y:0.37, K:0.16. Its decimal value is 1628039.\n\nHex triplet RGB Decimal 18d787 `#18d787` 24, 215, 135 `rgb(24,215,135)` 9.4, 84.3, 52.9 `rgb(9.4%,84.3%,52.9%)` 89, 0, 37, 16 154.9°, 79.9, 46.9 `hsl(154.9,79.9%,46.9%)` 154.9°, 88.8, 84.3 00cc99 `#00cc99`\nCIE-LAB 76.4, -61.484, 27.534 29.048, 50.541, 31.145 0.262, 0.456, 50.541 76.4, 67.368, 155.876 76.4, -65.445, 47.882 71.092, -51.477, 23.79 00011000, 11010111, 10000111\n\n# Color Schemes with #18d787\n\n• #18d787\n``#18d787` `rgb(24,215,135)``\n• #d71868\n``#d71868` `rgb(215,24,104)``\nComplementary Color\n• #18d728\n``#18d728` `rgb(24,215,40)``\n• #18d787\n``#18d787` `rgb(24,215,135)``\n• #18c8d7\n``#18c8d7` `rgb(24,200,215)``\nAnalogous Color\n• #d72818\n``#d72818` `rgb(215,40,24)``\n• #18d787\n``#18d787` `rgb(24,215,135)``\n• #d718c8\n``#d718c8` `rgb(215,24,200)``\nSplit Complementary Color\n• #d78718\n``#d78718` `rgb(215,135,24)``\n• #18d787\n``#18d787` `rgb(24,215,135)``\n• #8718d7\n``#8718d7` `rgb(135,24,215)``\n• #68d718\n``#68d718` `rgb(104,215,24)``\n• #18d787\n``#18d787` `rgb(24,215,135)``\n• #8718d7\n``#8718d7` `rgb(135,24,215)``\n• #d71868\n``#d71868` `rgb(215,24,104)``\n• #10925c\n``#10925c` `rgb(16,146,92)``\n• #13a96a\n``#13a96a` `rgb(19,169,106)``\n• #15c079\n``#15c079` `rgb(21,192,121)``\n• #18d787\n``#18d787` `rgb(24,215,135)``\n• #22e694\n``#22e694` `rgb(34,230,148)``\n• #39e99f\n``#39e99f` `rgb(57,233,159)``\n• #50ebaa\n``#50ebaa` `rgb(80,235,170)``\nMonochromatic Color\n\n# Alternatives to #18d787\n\nBelow, you can see some colors close to #18d787. Having a set of related colors can be useful if you need an inspirational alternative to your original color choice.\n\n• #18d757\n``#18d757` `rgb(24,215,87)``\n• #18d767\n``#18d767` `rgb(24,215,103)``\n• #18d777\n``#18d777` `rgb(24,215,119)``\n• #18d787\n``#18d787` `rgb(24,215,135)``\n• #18d797\n``#18d797` `rgb(24,215,151)``\n• #18d7a7\n``#18d7a7` `rgb(24,215,167)``\n• #18d7b7\n``#18d7b7` `rgb(24,215,183)``\nSimilar Colors\n\n# #18d787 Preview\n\nThis text has a font color of #18d787.\n\n``<span style=\"color:#18d787;\">Text here</span>``\n#18d787 background color\n\nThis paragraph has a background color of #18d787.\n\n``<p style=\"background-color:#18d787;\">Content here</p>``\n#18d787 border color\n\nThis element has a border color of #18d787.\n\n``<div style=\"border:1px solid #18d787;\">Content here</div>``\nCSS codes\n``.text {color:#18d787;}``\n``.background {background-color:#18d787;}``\n``.border {border:1px solid #18d787;}``\n\n# Shades and Tints of #18d787\n\nA shade is achieved by adding black to any pure hue, while a tint is created by mixing white to any pure color. In this example, #000302 is the darkest color, while #f1fdf8 is the lightest one.\n\n• #000302\n``#000302` `rgb(0,3,2)``\n• #02150d\n``#02150d` `rgb(2,21,13)``\n• #042718\n``#042718` `rgb(4,39,24)``\n• #063823\n``#063823` `rgb(6,56,35)``\n• #084a2e\n``#084a2e` `rgb(8,74,46)``\n• #0a5b39\n``#0a5b39` `rgb(10,91,57)``\n• #0c6d45\n``#0c6d45` `rgb(12,109,69)``\n• #0e7f50\n``#0e7f50` `rgb(14,127,80)``\n• #10905b\n``#10905b` `rgb(16,144,91)``\n• #12a266\n``#12a266` `rgb(18,162,102)``\n• #14b471\n``#14b471` `rgb(20,180,113)``\n• #16c57c\n``#16c57c` `rgb(22,197,124)``\n• #18d787\n``#18d787` `rgb(24,215,135)``\n• #1de692\n``#1de692` `rgb(29,230,146)``\n• #2fe89a\n``#2fe89a` `rgb(47,232,154)``\n• #40eaa3\n``#40eaa3` `rgb(64,234,163)``\n• #52ecab\n``#52ecab` `rgb(82,236,171)``\n• #63eeb4\n``#63eeb4` `rgb(99,238,180)``\n• #75f0bc\n``#75f0bc` `rgb(117,240,188)``\n• #87f2c5\n``#87f2c5` `rgb(135,242,197)``\n• #98f4cd\n``#98f4cd` `rgb(152,244,205)``\n• #aaf6d6\n``#aaf6d6` `rgb(170,246,214)``\n• #bcf7de\n``#bcf7de` `rgb(188,247,222)``\n• #cdf9e7\n``#cdf9e7` `rgb(205,249,231)``\n• #dffbf0\n``#dffbf0` `rgb(223,251,240)``\n• #f1fdf8\n``#f1fdf8` `rgb(241,253,248)``\nTint Color Variation\n\n# Tones of #18d787\n\nA tone is produced by adding gray to any pure hue. In this case, #747b78 is the less saturated color, while #06e98a is the most saturated one.\n\n• #747b78\n``#747b78` `rgb(116,123,120)``\n• #6b847a\n``#6b847a` `rgb(107,132,122)``\n• #628d7b\n``#628d7b` `rgb(98,141,123)``\n• #58977d\n``#58977d` `rgb(88,151,125)``\n• #4fa07e\n``#4fa07e` `rgb(79,160,126)``\n• #46a980\n``#46a980` `rgb(70,169,128)``\n• #3db281\n``#3db281` `rgb(61,178,129)``\n• #34bb83\n``#34bb83` `rgb(52,187,131)``\n• #2ac584\n``#2ac584` `rgb(42,197,132)``\n• #21ce86\n``#21ce86` `rgb(33,206,134)``\n• #18d787\n``#18d787` `rgb(24,215,135)``\n• #0fe088\n``#0fe088` `rgb(15,224,136)``\n• #06e98a\n``#06e98a` `rgb(6,233,138)``\nTone Color Variation\n\n# Color Blindness Simulator\n\nBelow, you can see how #18d787 is perceived by people affected by a color vision deficiency. This can be useful if you need to ensure your color combinations are accessible to color-blind users.\n\nMonochromacy\n• Achromatopsia 0.005% of the population\n• Atypical Achromatopsia 0.001% of the population\nDichromacy\n• Protanopia 1% of men\n• Deuteranopia 1% of men\n• Tritanopia 0.001% of the population\nTrichromacy\n• Protanomaly 1% of men, 0.01% of women\n• Deuteranomaly 6% of men, 0.4% of women\n• Tritanomaly 0.01% of the population"
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https://www.omnicalculator.com/physics/curie-constant | [
"Number of atoms\nLattice constant\nnm\nMagnetic moment\nμB\nCurie constant\nK*A/(T*m)\n\n# Curie Constant Calculator\n\nBy Miłosz Panfil, PhD\n\nThis Curie constant calculator helps you calculate the Curie constant. The Curie constant characterizes a response of a paramagnetic material to the magnetic field. Keep on reading to learn more about the Curie constant equation and the Curie's law of magnetism.\n\n## Curie's law of magnetism\n\nThe Curie law of magnetism states that the magnetization `M` of a paramagnetic substance is proportional to the Curie constant `C` and the magnetic field `B`. It is inversely proportional to the temperature `T`. In the form of the equation, we can write\n\n`M = C/T * B`.\n\nTo learn more about the Curie's law check the Curie's law calculator. Here we focus on the Curie constant `C` and its dependence on the properties of the material considered.\n\n## Curie constant equation\n\nThe Curie constant characterizes susceptibility of the paramagnetic material to the magnetic field. It depends on the strength of the magnetic moments in the atoms forming the substance and on the density of these moments. The equation is\n\n`C = μ0/(3kB) * N / a³ * μ²`\n\n• `μ0 = 4 * π * 10^(-7) T * m/A` is the permeability of free space,\n• `kB = 1,381 * 10^(-7) J/K` is the Boltzmann constant,\n• `N` is the number of atoms carrying the magnetic moment in a unit cell,\n• `a [m]` is the lattice constant,\n• `μ [J/T]` is the magnetic moment of a single atom.\n\nThe unit of the Curie constant is `[K * A/(T * m)]`. The magnetic moment `μ` is a characteristic number describing a magnetic property of a single atom (or a particle, molecule, etc.). You can learn more about the magnetic moment in quantum mechanics checking the Magnetic moment calculator.\n\n## Curie constant calculator\n\nYou can quickly compute the Curie constant with our calculator. To simplify the computations we set the units of the lattice constant `a` to nanometers. Nanometers are an appropriate unit to describe the atomic world. For the same reason the magnetic moment `μ` is in the Bohr magneton `μB = 9.274 * 10^(−24) J/T` units.\n\nFor example, let us consider we take a crystal formed by atoms on a simple cubic lattice with a lattice constant `a = 0.2 nm`. In a simple cubic lattice there is one atom per unit cell. We assume that each atom carries magnetic moment `μ = 2 μB`. With the Curie constant calculator, we get that the Curie constant `C = 1.3047 K * A/(T * m)`.\n\nMiłosz Panfil, PhD\n\n## Get the widget!\n\nCurie Constant Calculator can be embedded on your website to enrich the content you wrote and make it easier for your visitors to understand your message.\n\nIt is free, awesome and will keep people coming back!",
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https://www.heldermann.de/JCA/JCA22/JCA221/jca22008.htm | [
"",
null,
"Journal Home Page Cumulative Index List of all Volumes Complete Contentsof this Volume Previous Article Journal of Convex Analysis 22 (2015), No. 1, 145--159Copyright Heldermann Verlag 2015 Convex Hypersurfaces with Hyperplanar Intersections of Their Homothetic Copies Valeriu Soltan Dept. of Mathematical Sciences, George Mason University, 4400 University Drive, Fairfax, VA 22030, U.S.A. [email protected] [Abstract-pdf] Extending a well-known characteristic property of ellipsoids, we describe all convex solids $K \\subset \\mathbb{R}^n$, possibly unboun\\-ded, with the following property: for any vector $z \\in \\mathbb{R}^n$ and any scalar $\\lambda \\ne 0$ such that $K \\ne z + \\lambda K$, the intersection of the boundaries of $K$ and $z + \\lambda K$ lies in a hyperplane. This property is related to hyperplanarity of shadow-boundaries of $K$ and central symmetricity of small 2-dimensional sections of $K$. Keywords: Besicovitch, body, convex, ellipse, ellipsoid, convex, quadric, section, shadow-boundary, solid. MSC: 52A20 [ Fulltext-pdf (151 KB)] for subscribers only."
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https://bmcbioinformatics.biomedcentral.com/articles/10.1186/s12859-016-1244-x | [
"# rapidGSEA: Speeding up gene set enrichment analysis on multi-core CPUs and CUDA-enabled GPUs\n\n## Abstract\n\n### Background\n\nGene Set Enrichment Analysis (GSEA) is a popular method to reveal significant dependencies between predefined sets of gene symbols and observed phenotypes by evaluating the deviation of gene expression values between cases and controls. An established measure of inter-class deviation, the enrichment score, is usually computed using a weighted running sum statistic over the whole set of gene symbols. Due to the lack of analytic expressions the significance of enrichment scores is determined using a non-parametric estimation of their null distribution by permuting the phenotype labels of the probed patients. Accordingly, GSEA is a time-consuming task due to the large number of required permutations to accurately estimate the nominal p-value – a circumstance that is even more pronounced during multiple hypothesis testing since its estimate is lower-bounded by the inverse number of samples in permutation space.\n\n### Results\n\nWe present rapidGSEA – a software suite consisting of two tools for facilitating permutation-based GSEA: cudaGSEA and ompGSEA. cudaGSEA is a CUDA-accelerated tool using fine-grained parallelization schemes on massively parallel architectures while ompGSEA is a coarse-grained multi-threaded tool for multi-core CPUs. Nominal p-value estimation of 4,725 gene sets on a data set consisting of 20,639 unique gene symbols and 200 patients (183 cases + 17 controls) each probing one million permutations takes 19 hours on a Xeon CPU and less than one hour on a GeForce Titan X GPU while the established GSEA tool from the Broad Institute (broadGSEA) takes roughly 13 days.\n\n### Conclusion\n\ncudaGSEA outperforms broadGSEA by around two orders-of-magnitude on a single Tesla K40c or GeForce Titan X GPU. ompGSEA provides around one order-of-magnitude speedup to broadGSEA on a standard Xeon CPU. The rapidGSEA suite is open-source software and can be downloaded at https://github.com/gravitino/cudaGSEAas standalone application or package for the R framework.\n\n## Background\n\nHigh-throughput technologies such as microarray or next-generation sequencing enable researchers to routinely measure the expressions of tens of thousands of genes in many patients. Typically, long lists of interesting candidate genes are generated by subsequent computational analyses. However, interpreting these gene lists is challenging. Recognizing that genes act in concert to drive various biological processes, Gene Set Enrichment Analysis (GSEA) was introduced to summarize genomics data using a predefined gene set. Nowadays, GSEA is a heavily used tool in bioinformatics and has been successfully applied to gain insights into the biological function of diseases such as cancer and diabetes.\n\nHowever, the GSEA procedure can be highly time-consuming since significance of a calculated enrichment score is typically tested using a resampling strategy drawing large numbers of permutations. When a whole database of gene sets is used, the amount of required permutations is even higher in order to account for multiple hypothesis testing. Furthermore, size and availability of input data sets continue to increase driven by advances in high-throughput technologies . Thus, developing fast software solutions is of high importance to research. Previous work on accelerating gene set analysis has been limited to cloud computing . We present the rapidGSEA suite – an efficient parallelization of the GSEA method for commonly available multi-core CPUs and CUDA-enabled GPUs. By using a combination of parallelization techniques we can achieve speedups of one order-of-magnitude on Xeon CPUs and around two orders-of-magnitude on a single GPU compared to broadGSEA.\n\n## Implementation\n\nThis section is divided into three parts. First, we give a brief explanation of the sequential GSEA algorithm and its four major processing steps for estimating the nominal p-value of a determined enrichment score using a single gene set. Second, we introduce novel parallelization schemes for single and multiple gene set probing and their explicit implementation optimized for multi-core CPUs and CUDA-enabled GPUs. Finally, we describe the usage of our standalone application and the bundled package for the R framework.\n\n### The sequential algorithm\n\nThe traditional GSEA algorithm operates on a real-valued gene expression matrix D(g i ,p j ) of shape |G|×|P| where g i G denotes |G| unique gene identifiers and p j P enumerates |P| patient identifiers each labelled by a binary phenotype L(p j ){0,1} encoding cases and controls. The computation of the enrichment score statistics can be split into four major stages:\n\n#### Computation of local deviation measures\n\nFor each gene symbol g i (each row of D) a local deviation score Δ(g i ) is computed that encodes the inter-class deviation between cases and controls. As an example, the difference of means between both classes can be employed to express their variability per gene:\n\n$$\\begin{array}{*{20}l} \\Delta(g_{i}) &= \\mu_{i}^{(1)} - \\mu_{i}^{(0)} \\\\ \\mu_{i}^{(1)} &= \\sum\\limits_{j=0}^{|P|-1} \\frac{L(p_{j})}{m^{(1)}} D(g_{i}, p_{j})\\\\ \\mu_{i}^{(0)} &= \\sum\\limits_{j=0}^{|P|-1} \\frac{1-L(p_{j})}{m^{(0)}} D(g_{i}, p_{j}) \\end{array}$$\n\nwhere $$m^{(1)} = \\sum _{j=0}^{|P|-1} L(p_{j})$$ and m (0)=|P|−m (1) denote the number of patients in each class from the set {0,1}. Variations that combine intra-class means and standard deviations e.g.\n\n\\begin{aligned} \\begin{array}{lll} \\text{fold change:} &\\Delta(g_{i}) =\\frac{\\mu_{i}^{(1)}-\\mu_{i}^{(0)}}{\\sigma_{i}^{(1)}+\\sigma_{i}^{(0)}} &\\text{,} \\\\ \\text{t-test:} &\\Delta(g_{i}) = \\frac{\\mu_{i}^{(1)}-\\mu_{i}^{(0)}}{\\sqrt{\\left(\\sigma_{i}^{(1)}\\right)^{2}+\\left(\\sigma_{i}^{(0)}\\right)^{2}}}& \\end{array} \\end{aligned}\n(1)\n\nare common choices for Δ in GSEA implementations. Please note that extensions from binary to real-valued phenotype profiles $$L(p_{j}) \\in \\mathbb {R}$$ using Euclidean distance, Pearson’s product-moment or Spearman’s rank-order correlation coefficient are straightforward and thus will not be discussed further in this paper.\n\n#### Gene ranking\n\nAfter computation of the local deviations, the indices i{0,…,|G|−1} enumerating the gene symbols g i are reordered such that\n\n$$\\begin{array}{*{20}l} \\left(\\Delta\\left(g_{\\sigma(0)}\\right), \\dots, \\Delta\\left(g_{\\sigma(i)}\\right), \\dots\\Delta\\left(g_{\\sigma(|G|-1)}\\right)\\right) \\end{array}$$\n\nis a sorted (usually descending) sequence of local deviation scores. The sequence of reordered gene symbols g σ(i) is called gene ranking according to Δ and will later be used to determine the enrichment score statistic. Figure 1 illustrates the first and second stage of the GSEA algorithm.\n\n#### Enrichment score computation\n\nTo elucidate significant differences in gene regulation across different phenotypes, it is generally insufficient to consider transcription differences Δ(g σ(i)) individually. Each gene can be significantly up- or down regulated by chance alone, or through correlation with processes such as the cell cycle. In principle, information can be gained from clustering genes according to their regulation . Interpretation of the resulting clusters, however, is often unclear. Instead, prior information about gene classes that are assumed to behave correlatedly (e.g. genes on a regulatory pathway), is used in the analysis. Today, this is typically achieved through the framework of GSEA, which considers the significance of the transcription profile of a set of gene symbols SG as a whole as opposed to individual enrichment values.\n\nLet S be a gene set supposedly correlated to the observed phenotypes and σ(i) the aforementioned reordering of gene symbols. The enrichment score E S(S) is then determined as the maximal amplitude of a weighted running sum statistic ρ(k)[−1,1]:\n\n$$\\begin{array}{*{20}l} ES(S) &= \\rho\\left(\\mathop{\\text{argmax}}_{k} |\\rho(k)| \\right) \\ \\ \\quad\\text{where} \\\\ \\rho(k) &= \\sum\\limits_{i=0}^{k} \\left\\{ \\begin{array}{lll} \\frac{1}{\\alpha} \\cdot |\\Delta(g_{\\sigma(i)})|^{q} & \\text{if} & g_{\\sigma(i)} \\in S \\\\ - \\frac{1}{\\beta} & \\text{if} & g_{\\sigma(i)} \\notin S \\end{array} \\right. \\end{array}$$\n\nwith precomputed constants $$\\alpha = \\sum _{g \\in S} |\\Delta (g)|^{q}$$ and β=|G|−|S|. The exponent q≥0 is usually chosen from the set $$\\{0, 1, \\tfrac {3}{2}, 2\\}$$ and controls the leverage of the weights |Δ(g σ(i))|. Please note that the special case q=0 is the well-known Kolmogorov-Smirnov statistic . Figure 2 illustrates an example for the linear-weighted (q=1) computation of E S(S) using a toy data set.\n\n#### Significance estimation\n\nSimilar to Pearson’s correlation coefficient the enrichment score takes values in the interval [−1,1] with |E S(S)|=1 indicating perfect (anti-)correlation and |E S(S)|≈0 implying no dependency between S and the observed phenotypes in terms of the used deviation measure. When E S(S)=±1 all gene symbols gS are situated at the top/bottom of the ranked gene list. In contrast, small values are observed if the gene symbols gS are scattered over the index domain and thus are unlikely to explain the phenotype distribution.\n\nES values have no intrinsic significance, though. A value of E S(S)=0.857, as computed in our toy model in Fig. 2, might correspond to a high or low significance, depending on the probability to arrive at such a value by chance alone. Unfortunately, closed forms for the statistical distribution of enrichment score are inaccessible. Therefore, p-values are typically estimated by sampling the null distribution using a permutation of phenotype labels. Please note that while some GSEA implementations allow to permute gene identifiers instead of phenotype labels [1, 6] to estimate the null distribution, phenotype permutation is often considered the more appropriate choice – genes are expected to feature statistical dependencies within a single patient, while probes gained from distinct patients are less likely to do so. Hence, in the following we only consider phenotype permutation.\n\nFigure 3 depicts the enrichment score computation for a permutation π=(1 4) of the original list of six patients where the columns 1 and 4 of D have been swapped.1 The resulting score E S(S,π)=0.457<0.857=E S(S) suggests that the original value is considerably higher than a randomly sampled one. An exact computation of the p-value – due to absent closed forms for their distribution – would require us to calculate E S(S,π) for all |P|! permutations and finally determine the portion of values which are more extreme than E S(S). GSEA implementations hence usually estimate p-values by sampling in the space of permutations since |P|! is too large even for a moderate number of patients.\n\nWhen probing more than one gene set at once, p-value estimates have to be adjusted for multiple hypothesis testing. As an example, Bonferroni-corrected acceptance levels and family-wise error rates (FWER) are frequently used criteria to evaluate the significance of enrichment scores. The need for a large number of samples in the space of permutation is even more pronounced during multiple hypothesis testing: let eΠ be the identity permutation in the set of n tested permutations Π. Then the p-value estimate for a fixed gene set S is strictly positive and lower-bounded by inverse sample size:\n\n$$\\begin{array}{*{20}l} \\hat p_{S} = \\frac{1}{n}\\sum\\limits_{\\pi\\in\\Pi} \\left(|ES(S, \\pi)| \\geq |ES(S, e)| \\right) \\geq \\frac{1}{n} \\end{array}$$\n\nThe Molecular Signature Database v5.1 contains more than 13,000 gene sets divided into eight major collections. Thus, when testing all gene sets at a Bonferroni-adjusted significance level of $$\\alpha = \\frac {0.01}{13,000}$$ we have to probe more than 1,300,000 permutations in order to allow the result $$\\hat p_{S} < \\alpha$$. For the rest of the paper, we focus on the efficient computation of the enrichment score table E S(S,π) since p-value estimates and other statistics such as FWER can be determined using its entries in a post-processing phase.\n\n### The parallel algorithm\n\nGSEA can be parallelized using coarse-grained computation schemes such as assigning threads to each permutation π or gene set S since all entries in E S(S,π) can be processed independently. This approach will be used in our multi-threaded shared memory implementation of GSEA (ompGSEA): The set of n probed permutations is split into m partitions each of approximate size $$\\frac {n}{m}$$ and afterwards m threads independently operate on the individual chunks. This can easily be achieved in shared memory architectures using OpenMP pragmas. Moreover, extensions to distributed memory architectures using the Message Passing Interface (MPI) are conceivable.\n\nHowever, CUDA-enabled accelerators can maintain up to several thousands of threads (e.g. Titan X/Tesla K40c: 3,072/2,880 cores) but only exhibit a limited amount of RAM (both GPUs provide 12 GB). As a result, fine-grained computation schemes that parallelize the aforementioned building blocks of the GSEA algorithm have to be employed to exploit the full compute capabilities of CUDA-enabled accelerators. In the following, we will present the fine-grained parallelization scheme for each processing stage separately.\n\n#### Computation of local deviation measures\n\nMany local deviation measures used in traditional GSEA e.g. difference of means or fold change can be expressed in terms of intra-class means and standard deviations. Therefore, we have to separately accumulate sums of expression values and their squares for each of the two phenotypes. Although efficient implementations for parallel reduction on CUDA-enabled accelerators are known we instead parallelize the loop over the gene symbols since each row of the data matrix D can be processed independently without the need for expensive synchronization as used in reduction algorithms. Moreover, the number of gene symbols will most likely exceed the number of probed patients and thus the loop over g i is better suited for massively parallel computation. During the calculation of statistical moments we encounter two challenges:\n\nFirst, the numerically stable computation of standard deviations is known to be a stubborn task. On the one hand, when accumulating a large number of entries (here patients) one has to account for numeric stability using cancellation-compensation or two-pass algorithms for the standard deviations. On the other hand, when dealing with only a few patients one-pass or cancellation-compensated online algorithms for the standard deviation might be a proper trade-off between accuracy and speed . rapidGSEA exploits the C++ template engine to provide specialized and user-customizable accumulator functors adaptable to the task’s requirements.\n\nSecond, the gene-wise computation of transcription differences Δ(g i ) accumulates statistical moments along the rows of the matrix D. Using a CUDA thread block of up to 1,024 CUDA threads for a fixed permutation of the phenotype array L(p π(j)) it is advisable to transpose D to guarantee coalesced access to global memory. More specifically, since a warp of 32 threads is executed simultaneously on the GPU their concurrent reads from the same column of D would result in excessive cache misses. In contrast, when transposing D the same access pattern causes consecutive threads to simultaneously access consecutive memory. This change from column-major-order to row-major-order traversal decreases the runtime of this processing step by one order-of-magnitude in our experiments. Since D usually tends to be smaller than 100 MB, we can use a standard bank conflict-free out-of-place algorithm for matrix transposition . Figure 4 depicts the described computation scheme for two CUDA thread blocks each consisting of ten CUDA threads. Please note that the genes are distributed using a block-cyclic distribution if the number of genes exceed the number of threads.\n\nThe sampling of permutations can be accomplished using the pseudo random number generators (PRNG) from the cuRAND library bundled with the CUDA SDK. Unfortunately, cuRAND does not provide host-sided calls for the random number generators defined in the device API. Thus, we implemented the keep it simple stupid (KISS) PRNG for the CPU and GPU in order to provide consistent results across architectures. Both cuRAND’s xorwow PRNG and our KISS implementation pass all tests of the dieharder suite . The permutation of the phenotype labels L(p π(j)) is generated by reordering the original label list L(p j ) in shared memory using a Fisher-Yates shuffle.\n\n#### Gene ranking\n\nUp to this point, the transcription differences Δ(g i ) have been computed for a batch of permutations that fit into the RAM of the GPU. Unfortunately, we cannot directly apply a key-value sort to Δ(g i ) within the same kernel due to the 48 KB limitation of shared memory. Thus, after termination of the previous kernel, we call a device-wide key-value radix sort primitive cub::DeviceSegmentedRadixSort from the CUB library specifically optimized for the efficient sorting of segmented arrays. This approach is up to one order-of-magnitude faster than stacking single device-wide cub::DeviceRadixSort calls for each permutation or aliasing global memory to the block-wide cub::BlockRadixSort primitives. The number of concurrently sorted arrays has been set to 128 as a proper trade-off between runtime and memory consumption. At the end of this stage, we have stored the sorted deviation scores Δ(g σ(i)) and corresponding indices σ(i) for each of the probed permutations in global memory. Figure 4 illustrates the described workflow.\n\n#### Enrichment score computation\n\nThe computation scheme for the running sum statistic is similar to the processing of local deviation scores. For each permutation a CUDA thread block operates on a pair (g σ(i),Δ(g σ(i))) of reordered gene symbols and gene transcription differences. The test whether a gene identifier is part of a gene set g σ(i)S is usually implemented with hash sets on CPUs. Efficient hashing algorithms on CUDA-enabled devices are stated in the literature which typically involve linked lists or binary search in sorted arrays in order to resolve collisions. However, we decided to encode the affiliation of a gene g with a binary bit mask b(g,S). The computation of the bit mask can be delegated to the CPU using STL hashes. Further, the corresponding execution time can be overlapped with the deviation score and gene ranking kernels. As a result, we can determine a gene’s affiliation on the GPU in constant time by reading the corresponding entry of the bit mask from global memory.\n\nEach thread k within a thread block processes one gene set S k . Shared memory can be utilized to avoid slow accesses to global memory since all threads in a warp have to access the same entry from the bit mask b(g σ(i),S k ) in random order. To achieve this, batches of 64 entries of reordered gene transcription differences Δ(g σ(i)) and bit mask entries b(g σ(i),S k ) are consecutively loaded into shared memory (scratchpad) and afterwards processed in order. Due to the large number of genes we again use numerically stable Kahan summation in order to suppress cancellation in floating point arithmetic. Finally, the maximum amplitude of the weighted Kolmogorov-Smirnov statistic is written to the enrichment score table E S(S,π) and consecutively transferred to the host. Figure 5 illustrates the described procedure.\n\n#### Significance estimation\n\nWhen only computing p-value estimates the counting of values in the tails of the null distribution could be accomplished on the GPU using the device-wide reduction primitive cub::DeviceSegmentedReduce from the CUB library. A similar approach for the computation of the FWER is conceivable. However, we decided to copy E S(S,π) to the host in order to provide the full information for consecutive analysis and visualization of sampled distributions.\n\n### Bindings for the R language\n\nThe core algorithm written in CUDA and C++11 is provided as standalone application and additionally as Rcpp-based package for R. The latter includes functions for the reading of gene expression tables (*.gct), class assignment labels (*.cls) and gene sets files (*.gmt) as well as methods for the querying and selection of the used GPU (see user manual).\n\n## Results and discussion\n\nThe performance of rapidGSEA is compared to the broadGSEA Java application in version 2.2.2 on the following platform:\n\n• (CPU) Intel Xeon E5-2660 v3 @ 2.60 GHz GHz (10+10 HT) with 128 GB DDR4 RAM\n\n• (GPU) NVIDIA GeForce GTX Titan X with 12 GB GDDR5 RAM, NVIDIA Tesla K40c with 12 GB GDDR5 RAM disabled ECC, NVCC ver. 7.5\n\n• (Software) Ubuntu 14.04 LTS, GCC ver. 4.8.4, IcedTea ver. 2.6.3 OpenJDK 64-Bit Server VM\n\nIn our experiments, we use gene expression data (GEO: Series GSE19429) consisting of 183 MDS patients and 17 healthy controls where the array spots have been collapsed to |G|=20,639 unique gene symbols by max pooling ambiguous mappings in the Affymetrix Human Genome U133 Plus 2.0 Array (GEO: Platform GPL570) . We further choose the smallest (H: hallmark, 50 gene sets) and the biggest (C: curated, 4726 gene sets) collection from the Molecular Signatures Database 5.0 . The number of tested permutations ranges from 1,024 up to 1,0242 = 1,048,576 samples. Single-precision runs are executed on the GeForce GTX Titan X and double-precision experiments on the Tesla K40c GPU. If not stated otherwise, rapidGSEA and broadGSEA have been configured to read the input data from disk and afterwards to write the full enrichment score table E S(S,π) to the file system in order to ensure fair competition.\n\n### Accuracy and compliance of enrichment scores\n\nWe have evaluated the compliance of computed enrichment scores between broadGSEA and rapidGSEA using the identity permutation on the 50 gene sets of the Hallmark collection under the difference of classes measure. The deviation of computed enrichment scores between rapidGSEA and broadGSEA comply within six digits for both single and double-precision arithmetic (see Fig. 6). Using identical floating point data types the computed scores of both rapidGSEA components, cudaGSEA and ompGSEA, are indistinguishable.\n\nHowever, a comparison of computed histograms E S(S,π) is more complex due to different implementations of random number generators. Thus, we have approximated the probability density functions (PDFs) of the enrichment score distribution using n=1,0242 permutations and $$\\sqrt {n} = 1,024$$ bins uniformly sampling the interval [−1,1]. Afterwards, the approximate cumulative distribution functions (CDFs) are computed by prefix summation. The maximum absolute difference of approximated CDFs, also know as Kolmogorov distance,\n\n$$\\begin{array}{*{20}l} dist = \\max\\limits_{k} | CDF^{\\mathrm{(rapidGSEA)}}_{k} - CDF^{\\mathrm{(broadGSEA)}}_{k} | \\end{array}$$\n\nis then determined for each of the 50 gene sets. Note, the Kolmogorov distance is a reasonable choice since it determines the measurement error of the area under the PDF of the enrichment score distribution and thus relates to the error of the estimated p-value. Figure 7 visualizes the described procedure for one gene set. The minimum/median/maximum absolute deviation between the approximated CDFs produced by rapidGSEA and broadGSEA over the 50 gene sets is given by 0.0005/0.0011/0.0018. When comparing two histograms both computed by broadGSEA with different seeds the same metrics yield 0.0006/0.0011/0.0018. Moreover, in 26 out of 50 cases rapidGSEA produces histograms with a smaller Kolmogorov distances to broadGSEA in contrast to 24 cases where both histograms produced by broadGSEA are more similar. Concluding, the deviations in estimated areas are reasonably small and mainly caused by different samples in permutation space.2\n\n### Scaling over multiple cores\n\nWe perform a strong scalability test of our ompGSEA implementation over multiple cores of the Xeon CPU. Note, ompGSEA is part of the cudaGSEA binary and can be selected using the -cpu flag. The time needed to process the 50 gene sets defined in the H(allmark) collection is measured for a fixed input size of n=16,384 permutations and a variable number of threads. The experiments cover performance measurements for up to ten physical cores each executing a single thread and a hyper-threaded scenario where up to twenty threads are assigned to ten physical cores. When taking measurements on less than ten physical cores we enforce a thread’s affinity using the taskset command in order to avoid rescheduling by the operating system. The obtained runtimes are listed in Table 1 and illustrated in Fig. 8. The first experiment utilizing only physical cores reveals almost linear speedup for ompGSEA with an efficiency of roughly 77 % for ten cores. However, the hyper-threaded variant exhibits slightly super-linear behaviour for up to nine physical cores and an efficiency of 98 % for all cores. Throughout the rest of this paper all reported runtimes of ompGSEA refer to the hyper-threaded ten core scenario running approximately ten times faster than the corresponding single-core application. Please note that the time for writing the enrichment score table E S(S,π) to disk has been neglected during this benchmark.\n\n### Comparison between rapidGSEA and broadGSEA\n\nThe execution time of rapidGSEA and broadGSEA is measured on the aforementioned data set over a wide range of permutations (1,024 up to 1,0242) using the Hallmark (H: 50 gene sets) and Curated (C2: 4,725 gene sets) collections. The experiments include parsing of input files, memory transfers over PCIe when using CUDA and writing the enrichment score table E S(S,π) to spinning disk. The obtained runtimes and speedups are listed in Table 2 and illustrated in Figs. 9 and 10. Numbers in square brackets or dashed lines indicate linearly extrapolated runtimes for broadGSEA in log-log space for large amounts of permutations.\n\nOur multi-threaded implementation ompGSEA outperforms broadGSEA on both gene set collections (H and C2) by at least one order-of-magnitude. Note, although broadGSEA spawns more than twenty threads the majority remains idle during processing. Therefore, broadGSEA cannot benefit from the additional physical cores of the Xeon processor. The same behaviour can be observed on an Intel i7 i3970X CPU with six physical cores.\n\nMoreover, cudaGSEA outperforms broadGSEA by around two orders-of-magnitude with growing speedups for an increasing number of permutations. This can be explained by the thread occupancy of the used GPUs. Both, the GeForce Titan X and the Tesla K40c can store at once tens of thousands of permutations (roughly 70k/35k in single/double-precision) within their 12 GB of RAM. Thus, when probing a small number of permutations the majority of streaming multi-processors remain idle. Furthermore, the parsing of input files and dumping of results takes several seconds and cannot be parallelized on the GPU.\n\n## Conclusions\n\nIn this paper, we have introduced rapidGSEA – a software suite consisting of two tools for facilitating permutation-based GSEA: cudaGSEA and ompGSEA. cudaGSEA is a CUDA-accelerated tool using fine-grained parallelization schemes on massively parallel architectures while ompGSEA is a coarse-grained multi-threaded tool for multi-core CPUs. ompGSEA outperforms the state-of-the-art implementation of GSEA (broadGSEA) by at least one order-of-magnitude in terms of execution times while providing compliant results. Furthermore, cudaGSEA outperforms broadGSEA by around two orders-of-magnitude. The time for probing 1,048,576 permutations on a gene expression data set consisting of 20,639 unique gene symbols and 200 patients can drastically be reduced from roughly 13 days for broadGSEA to less than two hours using rapidGSEA on a commonly available Tesla K40c GPU in double-precision or less than one hour on a GeForce Titan X in single-precision.\n\nA possible direction of future research in order to further reduce runtimes is the parallelization of GSEA on a compute cluster with multiple GPUs attached to each node. Furthermore, extensions of GSEA to consider graph-based (Gene Graph Enrichment Analysis ) or network-based (Network-based GSEA ) correlations between gene symbols and observed phenotypes have gained increasing attention in recent years. It will be interesting to investigate how the parallelization techniques discussed in this paper can be applied to accelerate these extended enrichment methods.\n\n## Availability and requirements\n\nProject name: cudaGSEA Project home page: https://github.com/gravitino/cudaGSEAOperating system(s): Linux Programming language: C++, CUDA, R Other requirements: CUDA-capable GPULicense: GNU LGPL Any restrictions to use by non-academics: None\n\n## Endnotes\n\n1 Please note that throughout this manuscript, we use zero-based indexing.\n\n2 Individual results for each gene set can be found at the github repository of rapidGSEA.\n\n## Abbreviations\n\nAPI:\n\nApplication programming interface\n\nCUDA:\n\nCompute unified device architecture\n\nFWER:\n\nFamily-wise error rate\n\nGSEA:\n\nGene set enrichment analysis\n\nMPI:\n\nMessage passing interface\n\nPCIe:\n\nPeripheral component interconnect express\n\nPRNG:\n\nPseudo random number generator\n\n## References\n\n1. 1\n\nSubramanian, et al.Gene Set Enrichment Analysis: A Knowledge-Based Approach for Interpreting Genome-Wide Expression Profiles. Proc Natl Acad Sci. 2005; 102(43):15545–15550. doi:http://dx.doi.org/10.1073/pnas.0506580102.\n\n2. 2\n\nHung JH, Yang TH, Hu Z, Weng Z, DeLisi C. Gene Set Enrichment Analysis: Performance Evaluation and Usage Guidelines. Brief. Bioinform. 2012; 13(3):281–91.\n\n3. 3\n\nWang X, Cairns MJ. SeqGSEA: a Bioconductor Package for Gene Set Enrichment Analysis of RNA-Seq Data Integrating Differential Expression and Splicing. Bioinformatics. 2014; 30(12):1777–1779. doi:http://dx.doi.org/10.1093/bioinformatics/btu090.\n\n4. 4\n\nZhang L, Gu S, Liu Y, Wang B, Azuaje F. Gene set analysis in the cloud. Bioinformatics. 2012; 28(2):294–5.\n\n5. 5\n\nEisen MB, Spellman PT, Brown PO, Botstein D. Cluster analysis and display of genome-wide expression patterns. Proc Natl Acad Sci. 1998; 95(25):14863–14868. arxiv http://www.pnas.org/content/95/25/14863.full.pdf. Accessed 1 Apr 2016.\n\n6. 6\n\nBackes C, Keller A, Kuentzer J, Kneissl B, Comtesse N, Elnakady YA, Müller R, Meese E, Lenhof HP. GeneTrail-advanced gene set enrichment analysis. Nucleic Acids Research. 2007; 35(suppl 2):186–92.\n\n7. 7\n\nPhipson B, Smyth GK. Permutation P-values Should Never Be Zero: Calculating Exact P-values When Permutations Are Randomly Drawn. Stat Appl Genet Mol Biol. 2010;9(1), Article 39. http://www.degruyter.com/view/j/sagmb.2010.9.1/sagmb.2010.9.1.1585/sagmb.2010.9.1.1585.xml.\n\n8. 8\n\nMolecular Signatures Database. Accessed 1 Apr 2016. http://software.broadinstitute.org/gsea/msigdb.\n\n9. 9\n\nCUB: CUDA Unbound Library. Accessed 1 Apr 2016. https://nvlabs.github.io/cub/.\n\n10. 10\n\nKahan W. Pracniques: Further Remarks on Reducing Truncation Errors. Commun. ACM. 1965; 8(1):40–8. doi:http://dx.doi.org/10.1145/363707.363723.\n\n11. 11\n\nChan TF, Golub GH, LeVeque RJ. Updating Formulae and a Pairwise Algorithm for Computing Sample Variances, Technical report. Stanford: Stanford University; 1979. http://i.stanford.edu/pub/cstr/reports/cs/tr/79/773/CS-TR-79-773.pdf.\n\n12. 12\n\nRuetsch G, Micikevicius P. Optimize Matrix Transpose Technical report. Santa Clara: NVIDIA coporation; 2010. http://docs.nvidia.com/cuda/samples/6_Advanced/transpose/doc/MatrixTranspose.pdf. Accessed 1 Apr 2016.\n\n13. 13\n\ncuRAND: NVIDIA CUDA Random Number Generation Library. [Accessed 1 Apr 2016. https://developer.nvidia.com/curand].\n\n14. 14\n\nMarsaglia G, Tsang WW, et al.Some difficult-to-pass tests of randomness. J Stat Softw. 2002; 7(3):1–9.\n\n15. 15\n\ndieharder: Random Number Generator Testing Suite. Accessed 1 Apr 2016. https://www.phy.duke.edu/~rgb/General/dieharder.php.\n\n16. 16\n\nAlcantara DAF. Efficient hash tables on the gpu, PhD thesis. Davis: University of California at Davis; 2011. AAI3482095.\n\n17. 17\n\nEddelbuettel D, François R. Rcpp: Seamless R and C++ Integration. J Stat Softw. 2011; 40(8):1–18.\n\n18. 18\n\n19. 19\n\nPellagatti, et al.Deregulated Gene Expression Pathways in Myelodysplastic Syndrome Hematopoietic Stem Cells. Leukemia. 2010; 24:756–64.\n\n20. 20\n\nGeistlinger L, Csaba G, Küffner R, Mulder N, Zimmer R. From sets to graphs: towards a realistic enrichment analysis of transcriptomic systems. Bioinformatics. 2011; 27(13):366–73.\n\n21. 21\n\nGlaab E, Baudot A, Krasnogor N, Schneider R, Valencia A. Enrichnet: network-based gene set enrichment analysis. Bioinformatics. 2012; 28(18):451.\n\n## Acknowledgements\n\nPartial funding was gratefully provided by the Center for Computational Science in Mainz.\n\n### Authors’ contributions\n\nBS and AH conceived the study, and participated in its design and coordination. All authors contributed to the writing of the manuscript. CH wrote and evaluated the CPU and GPU implementations. All authors read and approved the final manuscript.\n\n### Competing interests\n\nThe authors declare that they have no competing interests.\n\nNot applicable.\n\n### Ethics approval and consent to participate\n\nThroughout this paper the used gene expression data set is anonymized and has been obtained from NCBI Gene Expression Omnibus (GEO: Series GSE19429). The data has exclusively been used for runtime measurements and compliance evaluation of computed enrichment score values between broadGSEA and rapidGSEA. The original source explicitly states approval granted by appropriate ethics committees: ’The study was approved by the ethics committees (Oxford C00.196, Bournemouth 9991/03/E, Duisburg 2283/03, Stockholm 410/03, Pavia 26264/2002) and informed consent was obtained.’\n\n## Author information\n\nAuthors\n\n### Corresponding author\n\nCorrespondence to Christian Hundt.\n\n## Rights and permissions",
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https://mathoverflow.net/questions/59319/computing-squaring-operations-in-the-adams-spectral-sequence | [
"# Computing squaring operations in the Adams spectral sequence\n\nThis question is about the classical Adams spectral sequence. Squaring operations are defined on its $E_2$ term. I'd like to know how to compute some of the non-trivial operations, such as $Sq^2 ( c_0 ) = h_0 e_0$. I feel like this ought to be doable in the May spectral sequence, but I don't know the details.\n\nI'm aware of some work of Milgram on the subject, but there are some problems with his approach because of indeterminacies of Massey products.\n\nThanks!\n\n• Congratulations on asking the 1000th algebraic topology question! – Dan Ramras Mar 23 '11 at 16:54\n• I think that Chris Douglas and Mike Hill may have thought about something along these lines. I recall a talk that Chris gave about it. I don't think it has been written up. The title of the talk was something like \"Juggling Toda Brackets.\" I suspect what they were doing was at least related to Milgram's work (they were finding ways to deal with indeterminacies in Toda brackets/Massey products). I probably have notes from the talk at home, so if Mike and Chris have forgotten about it, send me an e-mail and I can dig up my notes. – Dan Ramras Mar 23 '11 at 16:58\n• Bob Bruner knows an enormous amount about this, and has computer programs that can do all sorts of useful things. I think that most of this is unpublished. You should ask him. – Neil Strickland Mar 23 '11 at 18:29\n• Because of Bob Bruner's computer calculations, I know what the answers ought to be. But I'm looking for a conceptual way of obtaining the answers. – Dan Isaksen Mar 25 '11 at 13:51\n• I think what Dan would like is a precise statement about the relation between the Sq^i in Ext_A and the Sq^i in Ext_{E^0(A)}, the E_2 term of the May ss. The problem with brackets is indeterminacy. Of course, May ss, or any ss, calculations will suffer the same problem. – Robert Bruner Mar 27 '11 at 3:03\n\nI think for $k>0$ currently no-one knows an efficient algorithmic way to compute the $Sq^k$, e.g., from a minimal resolution. (The $Sq^0$ is easy since it is induced by the \"Frobenius\" map on $A_\\ast$ and one just needs to compute a chain map).\n\nFor the May spectral sequence you might want to check out Nakamura, Osamu. On the squaring operations in the May spectral sequence. Mem. Fac. Sci. Kyushu Univ. Ser. A 26 (1972), no. 2, 293–308 (Google can find this online).\n\n• There is one efficient 'algorithm' for the Sq^i other than Sq^0, and it is due to Nassau! It does require an actual resolution, not just a spectral sequence, and Dan was hoping to avoid this. I will post a brief account as a separate answer because I don't have much room here and can't make it display math here except as raw TeX code. – Robert Bruner Mar 26 '11 at 20:24\n• I have been trying to make my computer brute force these for months, but it has been unbearably slow. This is bad news for me, but explains why I have been thus far unable to do these computations quickly. – Joseph Victor Nov 2 '12 at 18:40\n\nSuppose $x \\in Ext^{s,t}_A(k,k)$ and let $\\mathcal{C} = \\cdots \\to C_s \\to \\cdots \\to C_0 \\to \\to k \\to 0$ be a resolution so that $x$ is represented by a cocycle $C_s \\to \\Sigma^t k$.\n\nTo compute $Sq^i(x)$, find a 'small' extension $0 \\to \\Sigma^t k \\to M_{s-1} \\to \\cdots \\to M_0 \\to k \\to 0$ realizing x and call it $\\mathcal{X}$. Then $x$ is represented by a chain map $\\mathcal{C} \\to \\mathcal{X}$ with the cocycle $C_s \\to \\Sigma^t k$ at one end and the identity of $k$ at the other. The cocommutative Hopf algebra structure of $A$ makes $\\mathcal{X} \\otimes_k \\mathcal{X}$ a complex of $A$-modules, and there is a chain map $\\chi : \\mathcal{C} \\to \\mathcal{X} \\otimes_k \\mathcal{X}$ covering $1_k$ whose other end $C_{2s} \\to \\Sigma^{2t} k$ is a cocycle representing $x^2$. The composite $\\tau \\chi$ of this chain map with the twist map $\\tau : \\mathcal{X} \\otimes \\mathcal{X} \\to \\mathcal{X} \\otimes \\mathcal{X}$ is another lift of $1_k$, so there is a chain homotopy $\\chi_1 : \\mathcal{C} \\to \\mathcal{X} \\otimes \\mathcal{X}$ between them. This gives a cocycle $C_{2s-1} \\to \\Sigma^{2t} k$ which represents $x \\cup_1 x = Sq^{s-1}(x)$. Repeat to get $\\chi_2$, giving a cocycle $C_{2s-2} \\to \\Sigma^{2t} k$ representing $x \\cup_2 x = Sq^{s-2}(x)$, etc. If the extension $\\mathcal{X}$ is small, this is a remarkably effective way to compute these operations, ${\\it{\\text{ if you have a resolution to work with}}}$.\n\nTwo minute exercise: show that $Sq^0(h_0) = h_1$ in $Ext_{A(1)}(F_2,F_2)$ using this method and the resolution\n\n$C_1 = \\Sigma^1 A(1) \\oplus \\Sigma^2 A(1) \\to C_0 = A(1)$ by $\\iota_1 \\mapsto Sq^1$, $\\iota_2 \\mapsto Sq^2$, and\n\n$C_2 = \\Sigma^2 A(1) \\oplus \\Sigma^4 A(1) \\to C_1$ by $\\iota_2 \\mapsto Sq^1 \\iota_1$ and $\\iota_4 \\mapsto Sq^1 Sq^2 \\iota_1 + Sq^2 \\iota_2$.\n\nWhat have we done? The universal case is a $C_2$-equivariant map $\\mathcal{W} \\otimes \\mathcal{C} \\to \\mathcal{C} \\otimes \\mathcal{C}$ extending the diagonal $\\mathcal{C} \\to \\mathcal{C} \\otimes \\mathcal{C}$ coming from the cocommutative coproduct of $A$. This is what May's Springer LNM V. 168 article needs to construct Steenrod operations. Unfortunately, if $\\mathcal{C}$ is a production strength resolution, not just some little toy, then $\\mathcal{C} \\otimes \\mathcal{C}$ is a monster, and we do not want to be trying to lift maps over its differential. (Maybe you, dear reader, see a way out of this. If so, this is great news!)\n\nChristian's clever observation is that we can do this one cocycle at a time, and compute the map $\\mathcal{W} \\otimes \\mathcal{C} \\to \\mathcal{X} \\otimes \\mathcal{X}$ if we can find small extensions $\\mathcal{X}$ representing the cocycles of interest. This step is not algorithmic, because the easy answers are too large, essentially as large as $\\mathcal{C}$ itself. On the other hand it is quite feasible by hand in low degrees. Sean Tilson and I have carried this out for $c_0$, $d_0$, $e_0$, $f_0$ and $n$ (in the BMT notation for $Ext_A(F_2,F_2)$), perhaps another one or two. I suppose we should convert the results from computer code to TeX one day. I was interested in this in order to unambiguously determine which of the two possible elements in $Ext^{4,22}$ found by my computer program was $f_0 = Sq^1(c_0)$ and which was $f_0 + h_1^3 h_4$. Similarly for $f_1 \\in Ext^{4,44}$; this follows by $Sq^0$ from identifying $f_0$, and as Christian notes, $Sq^0$ is easy since that is induced by the Frobenius in the dual of $A$, hence by restriction along the dual of the Frobenius, $A \\to D(A)$, where $D(A)$ is the 'double' of $A$, $D(A)_{2n} = A_n$ and $D(A)_{2n+1} = 0$. On Milnor basis elements this is $Sq(2r_1, \\ldots, 2r_k) \\mapsto Sq(r_1, \\ldots, r_k)$ while basis elements with any odd entries go to 0, so it is easy to automate.\n\nFinally, since what I thought was going to be a paragraph or two has already grown absurdly long, I might as well go whole hog and mention a favorite problem. The complex $\\mathcal{X} \\otimes \\mathcal{X}$ is one way to construct a complex representing $x^2$. The Yoneda composite of $\\mathcal{X}$ with itself is a much 'slimmer' one. Unfortunately, we cannot recognize $\\tau$ in it. If we could, we could then automate the calculation of the $Sq^i$. This is the problem of a description of the squaring operations in terms of Yoneda product rather than tensor product and a solution would be great to have.\n\nYou can avoid (some) indeterminacies by using Baues's secondary Steenrod algebra and it's relation to the $E_2$ and $E_3$ terms of the Adams spectral sequence:\n\nMR2220189 (2008a:55015) Baues, Hans-Joachim The algebra of secondary cohomology operations. Progress in Mathematics, 247. Birkhäuser Verlag, Basel, 2006. xxxii+483 pp. ISBN: 3-7643-7448-9; 978-3-7643-7448-8 (Reviewer: David Chataur), 55S20 (18G10 55T15)\n\nMR2193337 (2006k:55031) Baues, Hans-Joachim; Jibladze, Mamuka Secondary derived functors and the Adams spectral sequence. Topology 45 (2006), no. 2, 295–324. (Reviewer: J. P. C. Greenlees), 55T15 (18D05 18G10 55S20 55U35)\n\nSo far, this has been successfully applied to the computation of tons of Massey products (see the first reference), but one can also deal with squaring operations along the lines of this paper and\n\nMR2482810 (2009i:55015) Baues, Hans-Joachim; Muro, Fernando Toda brackets and cup-one squares for ring spectra. Comm. Algebra 37 (2009), no. 1, 56–82. (Reviewer: Tyler D. Lawson), 55Q35 (55P43)\n\nSorry for this shameless self-propaganda, and also for not being more explicit here, but I find no way to answer this question in a short enough but explicit way.\n\nP.D. I know this should have been a comment, but it is too long for that.\n\n• Fernando, What would be needed to get cup-2 and higher by these methods? – Robert Bruner Mar 27 '11 at 10:13\n\nThis is an answer Bob's question after my previous answer. It doesn't fit in a comment.\n\nAs a spectrum, $\\mathrm{End}(H\\mathbb{Z}/2)$ is known to be a product of Eilenberg-MacLane spectra, so it is the classifying spectrum $\\mathbb{H} C_{*}$ of a chain complex $C_{*}$, i.e. it is in the image of the Eilenbeg-MacLane functor $\\mathbb{H}$ from chain complexes to spectra. The work of Shipley shows that this functor is well behaved with respect to multipliciative structures, since the symmetric monoidal model category of chain complexes is Quillen equivalent to that of $H\\mathbb{Z}$-algebra spectra. In particular $\\mathbb{H}$ takes chain algebras to ring spectra. However, we know that the ring spectrum $\\mathrm{End}(H\\mathbb{Z}/2)$ is not an $H\\mathbb{Z}$-algebra, so it is the classifying spectrum of a chain complex, but not the classifying ring spectrum of a chain algebra.\n\nYour question would be answered in full generality if we understood what a ring spectrum structure over the classifying spectrum of a chain complex is, i.e. what structure on a chain complex $C_{*}$ is equivalent to an $S$-algebra structure on $\\mathbb{H} C_{*}$.\n\nThis is an achievable task, since there are nice models for the functor $\\mathbb{H}$ taking values on symmetric spectra, and if $X$ is a symmetric spectrum, it is easy to describe a ring spectrum structure in terms of maps of simplicial sets $X_p\\wedge X_q\\rightarrow X_{p+q}$. The outcome should be some sort of non-additive multiplication on the chain complex $C_*$.\n\nNow, let me remind you that Leif Kristensen constructed a long time ago a chain complex $C_{*}$ whose cohomology is the Steenrod algebra (reference below). This chain complex is endowed with a multiplication, which induces the product on the Steenrod algebra in cohomology. However, Kristensen's multiplication is not a chain algebra structure on $C_{*}$ since it is left distributive but not right distributive. This could be an approximation to a solution to the previous problem.\n\nSo far, I've been very lazy to pursue this project. There'd be a lot of work to do, but the outcome can be very valuable and surprising, and it looks like a feasible project.\n\nMR0159333 (28 #2550) Kristensen, Leif On secondary cohomology operations. Math. Scand. 12 1963 57–82. (Reviewer: J. F. Adams)\n\n• Fernando, This is great. I'm sorry I only now saw it. I guess I didn't get notified because it was a new answer rather than a comment. Anyway, Thanks! There are other interesting operations that are distributive on one side only, with some kind of additional structure on the other side: unstable composition and H_infinity homotopy operations are the ones I know. Bob – Robert Bruner Feb 28 '12 at 0:59"
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.84671426,"math_prob":0.99460214,"size":4384,"snap":"2019-43-2019-47","text_gpt3_token_len":1491,"char_repetition_ratio":0.16118722,"word_repetition_ratio":0.011004127,"special_character_ratio":0.32527372,"punctuation_ratio":0.08846585,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9987773,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-11-13T11:44:44Z\",\"WARC-Record-ID\":\"<urn:uuid:1afc7662-b884-4e0e-a8d8-293dee17a55b>\",\"Content-Length\":\"147598\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:96de2cea-996e-4251-bc2b-6b0e6a61b796>\",\"WARC-Concurrent-To\":\"<urn:uuid:db203cd0-a527-4079-b956-bd38213b0719>\",\"WARC-IP-Address\":\"151.101.129.69\",\"WARC-Target-URI\":\"https://mathoverflow.net/questions/59319/computing-squaring-operations-in-the-adams-spectral-sequence\",\"WARC-Payload-Digest\":\"sha1:OIKXFEL3Z5ZBZYZI5EX7QKM2HFH2W4ZR\",\"WARC-Block-Digest\":\"sha1:S4Y3S3GTZF2SHDXH43ZUCZGJKODGP6QG\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-47/CC-MAIN-2019-47_segments_1573496667260.46_warc_CC-MAIN-20191113113242-20191113141242-00196.warc.gz\"}"} |
https://docs.informatica.com/data-integration/powercenter/10-5/transformation-guide/external-procedure-transformation/developing-com-procedures/developing-com-procedures-with-visual-basic/step-3--add-a-method-to-the-class.html | [
"Search\n\n1. Preface\n2. Working with Transformations\n3. Aggregator Transformation\n4. Custom Transformation\n5. Custom Transformation Functions\n8. Expression Transformation\n9. External Procedure Transformation\n10. Filter Transformation\n11. HTTP Transformation\n12. Identity Resolution Transformation\n13. Java Transformation\n14. Java Transformation API Reference\n15. Java Expressions\n16. Java Transformation Example\n17. Joiner Transformation\n18. Lookup Transformation\n19. Lookup Caches\n20. Dynamic Lookup Cache\n21. Normalizer Transformation\n22. Rank Transformation\n23. Router Transformation\n24. Sequence Generator Transformation\n25. Sorter Transformation\n26. Source Qualifier Transformation\n27. SQL Transformation\n28. Using the SQL Transformation in a Mapping\n29. Stored Procedure Transformation\n30. Transaction Control Transformation\n31. Union Transformation\n32. Unstructured Data Transformation\n33. Update Strategy Transformation\n34. XML Transformations",
null,
"# Step 3. Add a Method to the Class\n\nPlace the pointer inside the Code window and enter the following text:\n```Public Function FV( _\n\nRate As Double, _\n\nnPeriods As Long, _\n\nPayment As Double, _\n\nPresentValue As Double, _\n\nPaymentType As Long _\n\n) As Double\nDim v As Double\n\nv = (1 + Rate) ^ nPeriods\n\nFV = -( _\n\n(PresentValue * v) + _\n\n(Payment * (1 + (Rate * PaymentType))) * ((v - 1) / Rate) _\n)\nEnd Function```\nThis Visual Basic FV function performs the same operation as the C++ FV function in Developing COM Procedures with Visual Basic.",
null,
"Updated July 27, 2022\n\nResources"
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"https://docs.informatica.com/etc/designs/informaticadita-com/images/h2l_bkgrnd_img.png",
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"https://docs.informatica.com/etc/designs/informaticadita-com/images/date_picker_icon_u119.png",
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.66754866,"math_prob":0.9525017,"size":358,"snap":"2022-40-2023-06","text_gpt3_token_len":115,"char_repetition_ratio":0.1779661,"word_repetition_ratio":0.0,"special_character_ratio":0.37430167,"punctuation_ratio":0.092307694,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.96414256,"pos_list":[0,1,2,3,4],"im_url_duplicate_count":[null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-02-02T00:53:56Z\",\"WARC-Record-ID\":\"<urn:uuid:3a03a3c1-dcfc-4c55-acc0-52c47e379930>\",\"Content-Length\":\"448486\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:3a427cfd-6129-4cc6-b882-aed56fc7f936>\",\"WARC-Concurrent-To\":\"<urn:uuid:8599544d-8839-4b2e-beec-55d81a3802e6>\",\"WARC-IP-Address\":\"104.64.208.130\",\"WARC-Target-URI\":\"https://docs.informatica.com/data-integration/powercenter/10-5/transformation-guide/external-procedure-transformation/developing-com-procedures/developing-com-procedures-with-visual-basic/step-3--add-a-method-to-the-class.html\",\"WARC-Payload-Digest\":\"sha1:TF4KMUODWPPQI7AU2PMW5KHP6J3PDUXG\",\"WARC-Block-Digest\":\"sha1:NTLQCIAAHNM2JM4MRMESDAS43C2EZ33X\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-06/CC-MAIN-2023-06_segments_1674764499954.21_warc_CC-MAIN-20230202003408-20230202033408-00320.warc.gz\"}"} |
https://rustgym.com/leetcode/1136 | [
"1136. Parallel Courses\n\nYou are given an integer `n` which indicates that we have `n` courses, labeled from `1` to `n`. You are also given an array `relations` where `relations[i] = [a, b]`, representing a prerequisite relationship between course `a` and course `b`: course `a` has to be studied before course `b`.\n\nIn one semester, you can study any number of courses as long as you have studied all the prerequisites for the course you are studying.\n\nReturn the minimum number of semesters needed to study all courses. If there is no way to study all the courses, return `-1`.\n\nExample 1:",
null,
"```Input: n = 3, relations = [[1,3],[2,3]]\nOutput: 2\nExplanation: In the first semester, courses 1 and 2 are studied. In the second semester, course 3 is studied.\n```\n\nExample 2:",
null,
"```Input: n = 3, relations = [[1,2],[2,3],[3,1]]\nOutput: -1\nExplanation: No course can be studied because they depend on each other.\n```\n\nConstraints:\n\n• `1 <= n <= 5000`\n• `1 <= relations.length <= 5000`\n• `1 <= a, b <= n`\n• `a != b`\n• All the pairs `[a, b]` are unique.\n\n1136. Parallel Courses\n``````struct Solution;\nuse std::collections::VecDeque;\n\nimpl Solution {\nfn minimum_semesters(n: i32, relations: Vec<Vec<i32>>) -> i32 {\nlet n = n as usize;\nlet mut adj = vec![vec![]; n];\nlet mut degree = vec![0; n];\nfor edge in relations {\nlet x = edge as usize - 1;\nlet y = edge as usize - 1;\ndegree[y] += 1;\n}\nlet mut visited = vec![false; n];\nlet mut queue = VecDeque::new();\nfor i in 0..n {\nif degree[i] == 0 {\nvisited[i] = true;\nqueue.push_back(i);\n}\n}\nlet mut res = 0;\nwhile !queue.is_empty() {\nlet n = queue.len();\nres += 1;\nfor _ in 0..n {\nlet u = queue.pop_front().unwrap();\ndegree[v] -= 1;\nif !visited[v] && degree[v] == 0 {\nvisited[v] = true;\nqueue.push_back(v);\n}\n}\n}\n}\nif visited.into_iter().all(|x| x) {\nres\n} else {\n-1\n}\n}\n}\n\n#[test]\nfn test() {\nlet n = 3;\nlet relations = vec_vec_i32![[1, 3], [2, 3]];\nlet res = 2;\nassert_eq!(Solution::minimum_semesters(n, relations), res);\nlet n = 3;\nlet relations = vec_vec_i32![[1, 2], [2, 3], [3, 1]];\nlet res = -1;\nassert_eq!(Solution::minimum_semesters(n, relations), res);\n}\n``````"
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"https://assets.leetcode.com/uploads/2021/02/24/course1graph.jpg",
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"https://assets.leetcode.com/uploads/2021/02/24/course2graph.jpg",
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.74522805,"math_prob":0.9963582,"size":2013,"snap":"2022-27-2022-33","text_gpt3_token_len":631,"char_repetition_ratio":0.13837731,"word_repetition_ratio":0.033519555,"special_character_ratio":0.37456533,"punctuation_ratio":0.25369978,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9974447,"pos_list":[0,1,2,3,4],"im_url_duplicate_count":[null,4,null,4,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-07-07T07:49:09Z\",\"WARC-Record-ID\":\"<urn:uuid:48fdbc12-696e-4242-b18e-49929bad5eec>\",\"Content-Length\":\"13792\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:d45a759d-1951-4ab1-a5c3-f8aa14dcac03>\",\"WARC-Concurrent-To\":\"<urn:uuid:085374dd-916b-4c1c-9881-a7a251541ad8>\",\"WARC-IP-Address\":\"35.188.52.69\",\"WARC-Target-URI\":\"https://rustgym.com/leetcode/1136\",\"WARC-Payload-Digest\":\"sha1:ZRAQKSLF5ETRTWN2J2MLIK5B2ZZ4RINU\",\"WARC-Block-Digest\":\"sha1:2ISRUYYSZGTNFOE3A32X5SMNXDIGNZKI\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-27/CC-MAIN-2022-27_segments_1656104683708.93_warc_CC-MAIN-20220707063442-20220707093442-00744.warc.gz\"}"} |
https://zio.dev/reference/test/dynamic-test-generation/ | [
"Version: ZIO 2.x\n\n# Dynamic Test Generation\n\nTests in ZIO are dynamic. Meaning that they are not required to be statically defined at compile time. They can be generated at runtime effectfully.\n\nAssume we have implemented the `add` operator which adds two numbers:\n\n``def add(a: Int, b: Int): Int = ???``\n\nWe want to test this function using the following test data inside the `resources` directory:\n\nsrc/test/resources/test-data.csv\n``0, 0, 01, 0, 10, 1, 10, -1, -1-1, 0, -11, 1, 21, -1, 0-1, 1, 0``\n\nLet's load it and create a bunch of tests using this test data:\n\n``import zio._import zio.test._import zio.test.testdef loadTestData: Task[List[((Int, Int), Int)]] = ZIO.attemptBlocking( scala.io.Source .fromResource(\"test-data.csv\") .getLines() .toList .map(_.split(',').map(_.trim)) .map(i => ((i(0).toInt, i(1).toInt), i(2).toInt)) ) def makeTest(a: Int, b: Int)(expected: Int): Spec[Any, Nothing] = test(s\"test add(\\$a, \\$b) == \\$expected\") { assertTrue(add(a, b) == expected) }def makeTests: ZIO[Any, Throwable, List[Spec[Any, Nothing]]] = loadTestData.map { testData => testData.map { case ((a, b), expected) => makeTest(a, b)(expected) } }``\n\nNow we are ready to run all generated tests:\n\n``import zio._import zio.test._object AdditionSpec extends ZIOSpecDefault { override def spec = suite(\"add\")(makeTests)}``\n\nHere is the test runner's output:\n\n``+ add + test add(0, 0) == 0 + test add(1, 0) == 1 + test add(0, -1) == -1 + test add(0, 1) == 1 + test add(-1, 1) == 0 + test add(1, -1) == 0 + test add(1, 1) == 2 + test add(-1, 0) == -18 tests passed. 0 tests failed. 0 tests ignored.``"
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.5209009,"math_prob":0.96797544,"size":1553,"snap":"2023-14-2023-23","text_gpt3_token_len":515,"char_repetition_ratio":0.13428017,"word_repetition_ratio":0.011952192,"special_character_ratio":0.36252415,"punctuation_ratio":0.24184783,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99724853,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-03-20T12:54:08Z\",\"WARC-Record-ID\":\"<urn:uuid:7d1bda64-2307-4bc9-8dee-dc82c883605d>\",\"Content-Length\":\"59437\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:0dfcb2e4-d648-4393-8b72-162ead931182>\",\"WARC-Concurrent-To\":\"<urn:uuid:fbe5561d-42ae-45e2-9c2b-c14539522ed8>\",\"WARC-IP-Address\":\"35.229.48.116\",\"WARC-Target-URI\":\"https://zio.dev/reference/test/dynamic-test-generation/\",\"WARC-Payload-Digest\":\"sha1:FC26QCRJV4IH33W5LVABGJBBDLDP4Z5M\",\"WARC-Block-Digest\":\"sha1:ORMD37JWGGI3KCHTQGBCSTJEHL7CQO6F\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-14/CC-MAIN-2023-14_segments_1679296943483.86_warc_CC-MAIN-20230320114206-20230320144206-00427.warc.gz\"}"} |
https://www.real-world-physics-problems.com/physics-formulas.html | [
"# Useful Physics Formulas\n\nCertain physics formulas are more commonly used than others, and these are the formulas that are convenient to have at your fingertips.\n\nI put together a set of pages below which give the formulas used in classical mechanics that I have found to be most useful over the years. These are the formulas that are most likely to be applied when you solve non-textbook (i.e. real world) physics problems. The formulas cut across all the relevant areas of classic mechanics, and as such, are pretty much all you need to solve a wide variety of problems for one-dimensional, two-dimensional, and three-dimensional dynamics.\n\nWhere necessary, brief descriptions and graphical illustrations are given to convey the correct meaning of the formulas and equations, and how they apply.\n\nClick on the links below to see the formulas and equations for those particular subject areas.\n\nMotion Along A Straight Line\nMotion Along A Circular Path\nCircular Motion\nProjectile Motion\nNewton's Laws Of Motion\nTorque\nWork And Energy\nRotational Inertia\nCenter Of Mass\nFriction\nRolling\nMomentum And Collisions\nA Closer Look At Velocity And Acceleration\nRigid Body Dynamics\nParallel Axis And Parallel Plane Theorem"
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https://socratic.org/questions/how-do-you-solve-f-x-x-2-x-3-using-the-quadratic-formula | [
"# How do you solve f(x)=x^2-x+3 using the quadratic formula?\n\nJul 19, 2017\n\nSee a solution process below:\n\n#### Explanation:\n\nFor $f \\left(x\\right) = a {x}^{2} + b x + c$, the values of $x$ which are the solutions to the equation for $0$ are given by:\n\n$x = \\frac{- b \\pm \\sqrt{{b}^{2} - 4 a c}}{2 a}$\n\nSubstituting $1$ for $a$; $- 1$ for $b$ and $3$ for $c$ gives:\n\n$x = \\frac{- \\left(- 1\\right) \\pm \\sqrt{{\\left(- 1\\right)}^{2} - \\left(4 \\cdot 1 \\cdot 3\\right)}}{2 \\cdot 1}$\n\n$x = \\frac{1 \\pm \\sqrt{1 - 12}}{2}$\n\n$x = \\frac{1 \\pm \\sqrt{- 11}}{2}$"
]
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https://www.gurufocus.com/term/ROC/PG/ROC-Percentage/Procter%20&%20Gamble%20Co | [
"Switch to:\n\n# Procter & Gamble Co ROC %\n\n: 10.78% (As of Jun. 2020)\nView and export this data going back to 1950. Start your Free Trial\n\nROC % measures how well a company generates cash flow relative to the capital it has invested in its business. It is also called ROIC %. Procter & Gamble Co's annualized return on capital (ROC %) for the quarter that ended in Jun. 2020 was 10.78%.\n\nAs of today (2020-08-06), Procter & Gamble Co's WACC % is 3.08%. Procter & Gamble Co's ROC % is 12.54% (calculated using TTM income statement data). Procter & Gamble Co generates higher returns on investment than it costs the company to raise the capital needed for that investment. It is earning excess returns. A firm that expects to continue generating positive excess returns on new investments in the future will see its value increase as growth increases.\n\n## Procter & Gamble Co ROC % Historical Data\n\n* All numbers are in millions except for per share data and ratio. All numbers are in their local exchange's currency.\n\n Procter & Gamble Co Annual Data Jun11 Jun12 Jun13 Jun14 Jun15 Jun16 Jun17 Jun18 Jun19 Jun20 ROC %",
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"8.80 9.52 8.97 8.36 12.34\n\n## Procter & Gamble Co ROC % Calculation\n\nProcter & Gamble Co's annualized Return on Capital (ROC %) for the fiscal year that ended in Jun. 2020 is calculated as:\n\n ROC % (A: Jun. 2020 ) = NOPAT / Average Invested Capital = Operating Income * ( 1 - Tax Rate % ) / ( (Invested Capital (A: Jun. 2019 ) + Invested Capital (A: Jun. 2020 )) / count ) = 15706 * ( 1 - 17.25% ) / ( (106733 + 103896) / 2 ) = 12996.715 / 105314.5 = 12.34 %\n\nwhere\n\n Invested Capital (A: Jun. 2019 ) = Total Assets - Accounts Payable & Accrued Expense - Excess Cash = Total Assets - Accounts Payable & Accrued Expense - ( Cash, Cash Equivalents, Marketable Securities - max(0, Total Current Liabilities - Total Current Assets + Cash, Cash Equivalents, Marketable Securities )) = 115095 - 15900 - ( 10287 - max(0, 30011 - 22473 + 10287 )) = 106733\n\n Invested Capital (A: Jun. 2020 ) = Total Assets - Accounts Payable & Accrued Expense - Excess Cash = Total Assets - Accounts Payable & Accrued Expense - ( Cash, Cash Equivalents, Marketable Securities - max(0, Total Current Liabilities - Total Current Assets + Cash, Cash Equivalents, Marketable Securities )) = 120700 - 21793 - ( 16181 - max(0, 32976 - 27987 + 16181 )) = 103896\n\nProcter & Gamble Co's annualized Return on Capital (ROC %) for the quarter that ended in Jun. 2020 is calculated as:\n\n ROC % (Q: Jun. 2020 ) = NOPAT / Average Invested Capital = Operating Income * ( 1 - Tax Rate % ) / ( (Invested Capital (Q: Mar. 2020 ) + Invested Capital (Q: Jun. 2020 )) / count ) = 13924 * ( 1 - 19.5% ) / ( (104121 + 103896) / 2 ) = 11208.82 / 104008.5 = 10.78 %\n\nwhere\n\n Invested Capital (Q: Mar. 2020 ) = Total Assets - Accounts Payable & Accrued Expense - Excess Cash = Total Assets - Accounts Payable & Accrued Expense - ( Cash, Cash Equivalents, Marketable Securities - max(0, Total Current Liabilities - Total Current Assets + Cash, Cash Equivalents, Marketable Securities )) = 118560 - 20195 - ( 15393 - max(0, 32896 - 27140 + 15393 )) = 104121\n\n Invested Capital (Q: Jun. 2020 ) = Total Assets - Accounts Payable & Accrued Expense - Excess Cash = Total Assets - Accounts Payable & Accrued Expense - ( Cash, Cash Equivalents, Marketable Securities - max(0, Total Current Liabilities - Total Current Assets + Cash, Cash Equivalents, Marketable Securities )) = 120700 - 21793 - ( 16181 - max(0, 32976 - 27987 + 16181 )) = 103896\n\nNote: The Operating Income data used here is four times the quarterly (Jun. 2020) data.\n\n* All numbers are in millions except for per share data and ratio. All numbers are in their local exchange's currency.\n\nProcter & Gamble Co (NYSE:PG) ROC % Explanation\n\nROC % measures how well a company generates cash flow relative to the capital it has invested in its business. It is also called ROIC %. The reason book values of debt and equity are used is because the book values are the capital the company received when issuing the debt or receiving the equity investments.\n\nThere are four key components to this definition. The first is the use of operating income or EBIT rather than net income in the numerator. The second is the tax adjustment to this operating income or EBIT, computed as a hypothetical tax based on an effective or marginal tax rate. The third is the use of book values for invested capital, rather than market values. The final is the timing difference; the capital invested is from the end of the prior year whereas the operating income or EBIT is the current year's number.\n\nWhy is ROC % important?\n\nBecause it costs money to raise capital. A firm that generates higher returns on investment than it costs the company to raise the capital needed for that investment is earning excess returns. A firm that expects to continue generating positive excess returns on new investments in the future will see its value increase as growth increases, whereas a firm that earns returns that do not match up to its cost of capital will destroy value as it grows.\n\nAs of today, Procter & Gamble Co's WACC % is 3.08%. Procter & Gamble Co's ROC % is 12.54% (calculated using TTM income statement data). Procter & Gamble Co generates higher returns on investment than it costs the company to raise the capital needed for that investment. It is earning excess returns. A firm that expects to continue generating positive excess returns on new investments in the future will see its value increase as growth increases.\n\nBe Aware\n\nLike ROE % and ROA %, ROC % is calculated with only 12 months of data. Fluctuations in the company's earnings or business cycles can affect the ratio drastically. It is important to look at the ratio from a long term perspective."
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http://old.robowiki.net/robowiki?Skilgannon/PreciseIntersect | [
"#",
null,
"Skilgannon/PreciseIntersect\n\nRobo Home | Skilgannon | Changes | Preferences | AllPages\n\nLike all my code, this is designed to run as fast as possible. At first I was working on a distance-based method (similar to the check at the beginning), but I realized there would be a bucketload of special cases, so I ditched that and went with standard square/circle intersection. All my own work, although after writing I did go over Krabb's and Rednaxela's implementations to check for bugs. It seems great minds think alike =) although I have a strange love of arrays ;-)\n\n```\nimport java.awt.geom.*;\nimport java.util.*;\npublic class PreciseUtils{\n\n//high speed test to determine if the full method should be run this tick\npublic static boolean intersects(Point2D.Double botLocation, Wave wave){\ndouble[] distSq = new double[]{\nwave.fireLocation.distanceSq(botLocation.x - 18, botLocation.y + 18),\nwave.fireLocation.distanceSq(botLocation.x + 18, botLocation.y + 18),\nwave.fireLocation.distanceSq(botLocation.x + 18, botLocation.y - 18),\nwave.fireLocation.distanceSq(botLocation.x - 18, botLocation.y - 18)};\n\n//faster? I'm not sure\n// Arrays.sort(distSq);\n// return (sqr(wave.distanceTraveled + bulletVelocity) > distSq\n// && sqr(wave.distanceTraveled) < distSq);\n\nreturn (sqr(wave.distanceTraveled) >\nMath.min(Math.min(distSq, distSq), Math.min(distSq,distSq))\n&& sqr(wave.distanceTraveled - wave.bulletVelocity) < Math.max(Math.max(distSq, distSq), Math.max(distSq,distSq)));\n\n}\npublic static double[] getIntersectionRange(Point2D.Double botLocation, Wave wave){\ndouble[] yBounds = new double[]{botLocation.y - 18, botLocation.y + 18};\ndouble[] xBounds = new double[]{botLocation.x - 18, botLocation.x + 18};\n\ndouble[] radii = new double[]{wave.distanceTraveled, wave.distanceTraveled - wave.bulletVelocity};\n\nArrayList<Point2D.Double> intersects = new ArrayList<Point2D.Double>();\nfor(int i = 0; i < 2; i++)\nfor(int j = 0; j < 2; j++){\nPoint2D.Double[] testPoints = vertIntersect(wave.fireLocation.x, wave.fireLocation.y, radii[i], xBounds[j]);\nfor(int k = 0; k < testPoints.length; k++)\nif(inBounds(testPoints[k].y, yBounds))\n}\n\nfor(int i = 0; i < 2; i++)\nfor(int j = 0; j < 2; j++){\nPoint2D.Double[] testPoints = horizIntersect(wave.fireLocation.x, wave.fireLocation.y, radii[i], yBounds[j]);\nfor(int k = 0; k < testPoints.length; k++)\nif(inBounds(testPoints[k].x, xBounds))\n}\nfor(int i = 0; i < 2; i++)\nfor(int j = 0; j < 2; j++){\nPoint2D.Double testCorner = new Point2D.Double(xBounds[i], yBounds[j]);\ndouble distSq = testCorner.distanceSq(wave.fireLocation);\n}\ndouble antiClockAngle = Double.POSITIVE_INFINITY;\ndouble clockAngle = Double.NEGATIVE_INFINITY;\ndouble absBearing = angle(wave.fireLocation, botLocation);\nfor(Point2D.Double p: intersects){\ndouble angDiff = fastRelativeAngle(angle(wave.fireLocation,p) - absBearing);\nif(angDiff > clockAngle)\nclockAngle = angDiff;\nif(angDiff < antiClockAngle)\nantiClockAngle = angDiff;\n}\nreturn new double[]{fastAbsoluteAngle(antiClockAngle + absBearing), fastAbsoluteAngle(clockAngle + absBearing)};\n}\nstatic boolean inBounds(double q,double[] xBounds){\nreturn q >= bounds && q <= bounds ;\n}\n\n//assumes between -PI*2 and PI*2\npublic static double fastRelativeAngle(double angle) {\nreturn angle<-Math.PI?angle+Math.PI*2:angle>Math.PI?angle-Math.PI*2:angle;\n}\n\n//assumes between -PI*2 and PI*4\npublic static double fastAbsoluteAngle(double angle){\nreturn angle > Math.PI*2 ? angle - Math.PI*2 : angle < 0? angle + Math.PI*2 : angle;\n}\n\nstatic Point2D.Double[] vertIntersect(double centerX, double centerY, double r, double intersectX){\ndouble deltaX = centerX - intersectX;\ndouble sqrtVal = r*r - deltaX*deltaX;\nif(sqrtVal < 0)\nreturn new Point2D.Double[]{};\n\n// commented out because having this decision would\n// take more execution time than the rare occurence\n// that a perfect 0 discriminant would save, and\n// the code downstream can deal with it anyways\n// if(sqrtVal == 0)\n// return new Point2D.Double[]{\n// new Point2D.Double(intersectX, centerY)};\n\nsqrtVal = Math.sqrt(sqrtVal);\nreturn new Point2D.Double[]{\nnew Point2D.Double(intersectX, centerY + sqrtVal),\nnew Point2D.Double(intersectX, centerY - sqrtVal)};\n}\nstatic Point2D.Double[] horizIntersect(double centerX, double centerY, double r, double intersectY){\ndouble deltaY = centerY - intersectY;\ndouble sqrtVal = r*r - deltaY*deltaY;\nif(sqrtVal < 0)\nreturn new Point2D.Double[]{};\n\nif(sqrtVal == 0)\nreturn new Point2D.Double[]{\nnew Point2D.Double(centerX, intersectY)};\n\nsqrtVal = Math.sqrt(sqrtVal);\nreturn new Point2D.Double[]{\nnew Point2D.Double(centerX + sqrtVal, intersectY),\nnew Point2D.Double(centerX - sqrtVal, intersectY),};\n}\n\npublic static double sqr(double d){\nreturn d*d;}\n\npublic static double angle(Point2D.Double source, Point2D.Double target) {\nreturn Math.atan2(target.x - source.x, target.y - source.y);\n}\n\npublic class Wave{\ndouble distanceTraveled;\ndouble bulletVelocity;\nPoint2D.Double fireLocation;\n}\n}\n```\n\nHmm, nice. This looks like a pretty solid implementation to me, and at a glance looks like it would probably notably quicker than mine. I'm quiet curious to see what affect this will have on Druss's targeting ability :-) -- Rednaxela\n\nI've actually made a couple small changes, which would mostly affect the execution speed. I'm curious how you layered all the ranges on top of each other to find a firing angle. Is the methodology anything similar to this?\n\n``` GFRange[] ranges = new GFRange[cluster.size()];\ndouble[] indices = new double[ranges.length*2];\nfor(int i = 0; i < ranges.length; i++){\nranges[i] = iterator.next();\nindices[i*2] = ranges[i].min;\nindices[i*2 + 1] = ranges[i].max;\n}\n\nArrays.sort(indices);\nint[] indicesScores = new int[indices.length - 1];\nfor(int i = 0; i < ranges.length; i++){\nint minIndex = Arrays.binarySearch(indices,ranges[i].min);\nint maxIndex = Arrays.binarySearch(indices,ranges[i].max);\nwhile(minIndex<maxIndex)\nindicesScores[minIndex++]++;\n}\nint maxIndex = indicesScores.length/2;\nfor(int i = 0; i < indicesScores.length; i++){\nif(indicesScores[i] > indicesScores[maxIndex])\nmaxIndex = i;\n}\ndouble fireGF = (indices[maxIndex] + indices[maxIndex + 1])/2;\n\nreturn fireGF*MEA*lateralDirection;\n```\n\nWell, the way I did it should be in effect similar to that. I use a class I call \"MovementProfile\" that stores overlayed ranges and has a variety of functions for adding/retrieving data from it. See the MovementProfile file in RougeDC's code or [here]. It's a bit messy and not all of the utility functions are actually used currently/anymore but I have them around in case I still need them. The utility functions besides getting the best GF that I am using are mostly there for the purpose of getting data about the profile for my CrowdTargeting weighting system to use (because I don't like VirtualBullets). Basically, it stores a sorted array where each element contains 1) an amount of increase/decrease and 2) The precise GF that increase/decrease occurs at. The reason I store amounts of increases/decreases instead of storing the overall value of an index, is that it makes incrementally adding to the profile far simpler and it's not as if searching for the highest point is much slower. -- Rednaxela\n\nRobo Home | Skilgannon | Changes | Preferences | AllPages"
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https://www.mathworks.com/matlabcentral/answers/428564-why-array-only-saves-last-run-of-results-from-nested-for-loop?s_tid=prof_contriblnk | [
"# Why array only saves last run of results from nested FOR Loop?\n\n31 views (last 30 days)\nHello guys, I have been seeking for a solution for this, but I can't find it. I created a for loops and store values for each loops in array. However, the only results from last run are save to array. Could you please explain me why and how to solve it? Here is my code, it is about create a file name with a format of 'HT-yyyy-mm-dd-hh-mi.dat'. yyyy refers as year, mm is month, dd is day, hh is hour, and mi is minute. 'My guess is that the array I preallocated is redefined again in each loops, so only the last run is saved. But how to fix it I have no idea.\nDate_matrix=strings(No_of_file_1D,1,numdays);\nfor ii=Num_date_i:Num_date_j\nfolder_ii=datestr(ii,'yyyy-mm-dd');\n[yy,mm,dd] = ymd(datetime(folder_ii, 'ConvertFrom', 'datenum'));\nnumdays_i=(ii - Num_date_i)+1;\nfor hh=0:1:24\nhh_i=hh;\nfor mi=10:10:60\nmi_i=mi;\nindex_save=(numdays_i)*No_of_file_1D+hh*(60/Time_interval)+(mi)/10;\nif mi==60\nmi_i=0;\nhh_i=hh+1;\nelse\nend\nif hh==24\nhh_i=00;\ndd=dd+1;\nelse\nend\nfile_name_ii=sprintf('%s%04d%c%02d%c%02d%c%02d%c%02d%s','HT-',yy,'-',mm,'-',...\ndd,'-',hh_i,'-',mi_i,'.dat');\nDate_matrix(index_save,1,numdays_i)=file_name_ii;\nend\nend\nend\n\nStephen23 on 7 Nov 2018\nEdited: Stephen23 on 8 Nov 2018\n\"I created a for loops...\"\nYes, you did.\n\"...and store values for each loops in array.\"\nNo, you did not.\n\"Could you please explain me why...\"\nBecause you did not use any indexing to store the output data in any array.\n\"...and how to solve it?\"\nPreallocate array/s before the loop, and within the loop use indexing to store your calculated value/s in the array/s. That is all. Here is a simple example of how to use indexing to store values in an array:\nA = nan(1,4);\nfor k = 1:4\nA(k) = sqrt(k);\nend\nDepending on the sizes of the data you want to store you might want to use a cell array or some dimensions of an array (e.g. rows of a matrix). This depends on the data sizes and classes, which you have not told us anything about.\nYour variable names, e.g. folder_ii and file_name_ii with the loop variable in them, hint that you were trying to magically change the variable name in the loop. Magically accessing variable names is one way that beginners force themselves into writing slow, complex, buggy code that is hard to debug. Read this to know why:\nThe indexing that I showed you is much simpler, is less liable to bugs, and is much much more efficient. Note that the introductory tutorials show how to use loops and indexing, which is why these tutorials are recommended for all beginners (who need to learn basic MATLAB concepts):\nDanupon Subanapong on 8 Nov 2018\nThank you so much!!\n\nR2018a\n\n### Community Treasure Hunt\n\nFind the treasures in MATLAB Central and discover how the community can help you!\n\nStart Hunting!"
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https://www.jiskha.com/questions/14760/can-someone-please-find-me-a-couple-of-websites-to-practice-for-elementary-standardized | [
"# referance\n\ncan someone please find me a couple of websites, to practice for elementary standardized test\n\nIt all depends on which standardized test you want to practice for.\n\nTAKS (if you're in Texas): (Broken Link Removed)\n\nStandardized test-prep in general: http://school.familyeducation.com/educational-testing/study-skills/34555.html\n\n=)\n\n1. 👍\n2. 👎\n3. 👁\n\n## Similar Questions\n\n1. ### Math\n\nFind the mode of the set of data 59,61,54,51,38 46,56,52,63,40,44 1.B 2.D 3.C 4.A 5.D 6.C 7.B 8.C 9.C 10.C 11.B 12.D 13.B 14.A 15.A 16.A 17.D 18.A 19.C 20.D Math lesson:10 Unit:2 Practice test 100% answers\n\n2. ### math\n\nA standardized test is designed so that scores have a mean of 50 and a standard deviation of 4. What is the probability that a test score is above 54?\n\n3. ### statistics\n\nThe distribution of scores on a standardized aptitude test is approximately normal with a mean of 490 and a standard deviation of 95. What is the minimum score needed to be in the top 25% on this test? Carry your intermediate\n\n4. ### Math\n\nOn a standardized test, Phyllis scored 84, exactly one standard deviation about the mean. If the standard deviation for the test is 6, what is the mean score for the test?\n\nthis is for a practice test. i just need the explanation on how to solve them and help solving them. or for a couple of them to check my answers. write each expression using a single exponent. 1) x^9/x^2 A)x^11** B)x^7 C)7 D)x^5\n\n2. ### math\n\nA couple intends to have four children. Assume that having a boy and girl is equally likely event. a) List sample space b) Find the probability that couple has two boys and two girls; c) Find the probability that couple has at\n\n3. ### math Standard deviation\n\nSix students took a national standardized test for which the average score was 700 and the standard deviation was 150. If a student’s z-value is 1.75 on the test, what was the student’s score?\n\n4. ### statistics\n\nthe admissions policy at a certain university requires that incoming students score in the upper 20% on a standardized test. if the mean score on the test is 510 and the standard deviation of the scores is 80, what is the minimum\n\n1. ### algebra 2\n\nThe average score on a standardized test is 750 points with a standard deviation of 50 points. What is the probability that a student scores more than 700 on the standardized test?\n\n2. ### Math\n\nThe scores on a standardized test are normally distributed with the mean of 750 and standard deviation 70. Find the probability that a score is more than 890.\n\n3. ### statistics\n\nA researcher hypothesizes that individuals who listen to classical music will score differently from the general population on a test of spatial ability. On a standardized test of spatial ability, (u = 58). A random sample of 14\n\n4. ### statistics\n\n10th grade New York public school students taking a standardized English test produced test scores that were normally distributed with a mean of 85 and a standard deviation of 4. Let x be a random variable that represents the test"
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.91013676,"math_prob":0.91021967,"size":2702,"snap":"2021-21-2021-25","text_gpt3_token_len":681,"char_repetition_ratio":0.18272795,"word_repetition_ratio":0.026694044,"special_character_ratio":0.26535898,"punctuation_ratio":0.10862069,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9957395,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-05-07T12:25:39Z\",\"WARC-Record-ID\":\"<urn:uuid:b3cceb8b-67fc-418e-9569-8d7f830ad477>\",\"Content-Length\":\"17364\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:22e85f30-8537-4805-8359-b4ac85b135d5>\",\"WARC-Concurrent-To\":\"<urn:uuid:02da636b-432d-4c56-a0a5-fa01750853b5>\",\"WARC-IP-Address\":\"66.228.55.50\",\"WARC-Target-URI\":\"https://www.jiskha.com/questions/14760/can-someone-please-find-me-a-couple-of-websites-to-practice-for-elementary-standardized\",\"WARC-Payload-Digest\":\"sha1:EDAGOHK2NSXAKX5QIZOX2UFIAQ4GCHA5\",\"WARC-Block-Digest\":\"sha1:KJFGOO3ALIFOIDE3CC77R6L2K2LX5IO6\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-21/CC-MAIN-2021-21_segments_1620243988793.99_warc_CC-MAIN-20210507120655-20210507150655-00174.warc.gz\"}"} |
https://gsdhelp.info/jurlewicz-skoczylas-algebra-liniowa-1-99/ | [
"# JURLEWICZ SKOCZYLAS ALGEBRA LINIOWA 1 PDF\n\nZad”ii. Przestrzenie Pierwszy tydzieri. liniowe. Podstawowe definicje ()#. Podprzestrzcnie pncstrzliniowej. (L2).. I. I. ad Uza.s;,. dnir z. Study of the basic concepts of algebra with the purpose of solving systems of linear equations. C3. Learning . T. Jurlewicz, Z. Skoczylas, Algebra liniowa 1 . Story time just got better with Prime Book Box, a subscription that delivers hand- picked children’s books every 1, 2, or 3 months — at 40% off List Price.",
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"Give example of the canonical form of an antisymmetric matrix. Ability to solve equations and inequalities. Basis of linear space. Give examples of inner products algebra liniowa skoczylas orthonormal basis.\n\nCalculus and linear algebra.\n\n## Some basic information about the module\n\nThe evaluation of the lecture ilniowa the evaluation jurlewcz a multiple-choice test to algebra liniowa skoczylas the learning outcomes in skoczy,as of: The evaluation of the lecture is the evaluation of a multiple-choice test to algebra liniowa skoczylas the learning outcomes in terms of: Additional information registration calendar, class conductors, localization skoczulas schedules of classesmight be available in the USOSweb system:. Faculty of Mathematics and Computer Science.\n\nFRANK ADELSTEIN PERVASIVE COMPUTING PDF\n\nDiscussion practice class, 28 hours more information Lecture, 28 hours more information. Complex numbers Representation of a complex number: The positive evaluation of the two colloquia is algebra liniowa skoczylas prerequisite for admission to the test.",
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"Given the matrix of an operator find eigenvalues and eigenvectors. Howard, Calculus with analytic geometry, 3rd ed. Give example of the canonical form of an antisymmetric matrix. Algebra liniowa skoczylas pdf. Rank of a matrix, determinant of a square matrix. Positive final grade is awarded only when positive results of liniwoa were obtained.",
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"Matrices and systems of linear equations. Whether you think based on download Algebra Liniowa 5: Basic knowledge of trigonometry. Lecture, discussion, working in groups, heuristic talk, directed reasoning, self-study. Explain algebra liniowa skoczylas relation between symmetric billinear forms linilwa quadratic forms. Explain the geometrical meaning of transformations that shift a conic into canonical form. Faculty of Mathematics and Computer Science.\n\nPdf Pelamis wave energy converter. The main aim of study: Differential equations Differential equations: Com host jurlewicz skoczylas — algebra liniowa z geometria analityczna przykady. Examples of geometric applications of definite integral. The download Algebra liniowa 5: Representation of a complex number: Sequences Bounded sequence, monotone, arithmetic, and geometric sequence, limit of a sequence and its properties arithmetic operations on limits, squeeze theorem, etc.\n\nLGB 51750 PDF\n\nRaw algebra liniowa skoczylas clone embed report skoczykas text 5. Rectangular and trygonometric form of a complex number. Copyright by University of Lodz. Lecture, discussion, working in groups, heuristic talk, directed reasoning, self-study.\n\n### Rok I – Ebooki z informatyki za darmo\n\nBasic mathematical knowledge of secondary school. Libiowa danych jest w trybie tylko do odczytu.\n\nFunctins and their properties.",
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.7143959,"math_prob":0.8666742,"size":3476,"snap":"2021-04-2021-17","text_gpt3_token_len":820,"char_repetition_ratio":0.12932028,"word_repetition_ratio":0.09406953,"special_character_ratio":0.2013809,"punctuation_ratio":0.17326732,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.95101696,"pos_list":[0,1,2,3,4,5,6,7,8,9,10],"im_url_duplicate_count":[null,2,null,6,null,3,null,2,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-04-11T00:43:29Z\",\"WARC-Record-ID\":\"<urn:uuid:82247899-3943-4b06-8e19-ff2a76f94264>\",\"Content-Length\":\"41112\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:41fac444-6960-43da-a4ee-9e06942666db>\",\"WARC-Concurrent-To\":\"<urn:uuid:41dd52d4-95bb-4232-9dc8-5ba143e83cd7>\",\"WARC-IP-Address\":\"172.67.199.83\",\"WARC-Target-URI\":\"https://gsdhelp.info/jurlewicz-skoczylas-algebra-liniowa-1-99/\",\"WARC-Payload-Digest\":\"sha1:RGXXOTGYQW4OXR3DBR67QKASEDYCZWDY\",\"WARC-Block-Digest\":\"sha1:UGOFJTXBWYHX5I4OJ7KNQRQLDEA4KD3T\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-17/CC-MAIN-2021-17_segments_1618038060603.10_warc_CC-MAIN-20210411000036-20210411030036-00008.warc.gz\"}"} |
https://answers.everydaycalculation.com/divide-fractions/1-98-divided-by-3-6 | [
"Solutions by everydaycalculation.com\n\nDivide 1/98 with 3/6\n\n1/98 ÷ 3/6 is 1/49.\n\nSteps for dividing fractions\n\n1. Find the reciprocal of the divisor\nReciprocal of 3/6: 6/3\n2. Now, multiply it with the dividend\nSo, 1/98 ÷ 3/6 = 1/98 × 6/3\n3. = 1 × 6/98 × 3 = 6/294\n4. After reducing the fraction, the answer is 1/49\n\nDivide Fractions Calculator\n\n÷\n\nUse fraction calculator with our all-in-one calculator app: Download for Android, Download for iOS"
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.5916622,"math_prob":0.9941993,"size":292,"snap":"2019-43-2019-47","text_gpt3_token_len":143,"char_repetition_ratio":0.25694445,"word_repetition_ratio":0.0,"special_character_ratio":0.50342464,"punctuation_ratio":0.077922076,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9953277,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-10-15T16:34:04Z\",\"WARC-Record-ID\":\"<urn:uuid:f170fd38-d482-44f8-bf09-53e30748e353>\",\"Content-Length\":\"7811\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:9c4e2812-ee96-4921-841c-a66f46dcd873>\",\"WARC-Concurrent-To\":\"<urn:uuid:942afed3-fa20-47f4-a0c9-5ee1227f1d2b>\",\"WARC-IP-Address\":\"96.126.107.130\",\"WARC-Target-URI\":\"https://answers.everydaycalculation.com/divide-fractions/1-98-divided-by-3-6\",\"WARC-Payload-Digest\":\"sha1:BHOGJ25G7GW4U6XLIXTKPONFXO4ZMWTC\",\"WARC-Block-Digest\":\"sha1:I4HNN5H7MMIPP6Y3OJ73QAQZPXJ2H7YT\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-43/CC-MAIN-2019-43_segments_1570986660067.26_warc_CC-MAIN-20191015155056-20191015182556-00469.warc.gz\"}"} |
https://brainmass.com/engineering/electrical-engineering/bjt-ac-dc-loadline-analysis-15792 | [
"Explore BrainMass\n\n### Explore BrainMass\n\nNot what you're looking for? Search our solutions OR ask your own Custom question.\n\nThis content was COPIED from BrainMass.com - View the original, and get the already-completed solution here!\n\nThe problem says to Find Icq, Vceq; plot dc and ac loadlines, calculate the small signal voltage gain, then determine the input and output resistances (Rib and Ro).\n\nBeta=120. Va=infinity.\n\n#### Solution Preview\n\nIn the lastscan.jpg, this is what I have of the problem so far. I did a Thevenin equivalent circuit to find Icq and Vceq. However, I don't understand how the load lines are ...\n\n#### Solution Summary\n\nBJT AC/DC load-line analysis is analyzed. The expert calculates the small signal voltage gain and the input and output resistances.\n\n\\$2.49"
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.8688672,"math_prob":0.89971197,"size":916,"snap":"2021-43-2021-49","text_gpt3_token_len":231,"char_repetition_ratio":0.09210526,"word_repetition_ratio":0.0,"special_character_ratio":0.22816594,"punctuation_ratio":0.16483517,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9739316,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-11-29T02:54:42Z\",\"WARC-Record-ID\":\"<urn:uuid:64b1ab1d-7cc6-45a1-b60a-73a452064c92>\",\"Content-Length\":\"331450\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:1bd608bd-8b1d-4696-a4f8-c8f3a2f11898>\",\"WARC-Concurrent-To\":\"<urn:uuid:cae0cd75-5472-4a0f-96e2-6d58fc7322f4>\",\"WARC-IP-Address\":\"172.67.75.38\",\"WARC-Target-URI\":\"https://brainmass.com/engineering/electrical-engineering/bjt-ac-dc-loadline-analysis-15792\",\"WARC-Payload-Digest\":\"sha1:S5NQNNHEGRAZLPFJANJSDN63BXP4TQ3L\",\"WARC-Block-Digest\":\"sha1:QFSBH5447FPL4JI5RT45W6COFB2B4CUW\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-49/CC-MAIN-2021-49_segments_1637964358685.55_warc_CC-MAIN-20211129014336-20211129044336-00519.warc.gz\"}"} |
https://ddcode.net/2019/06/21/text-classification-principle-of-naive-bayesian-algorithm/ | [
"## Order\n\nThis paper mainly studies how Naive Bayesian algorithm classifies text.\n\n## Bayesian algorithm\n\nBayesian method converts the probability of “belonging to a certain type under certain characteristics” into the probability of “belonging to a certain type under certain characteristics”, which belongs to supervised learning.\n\n``后验概率 ＝ 先验概率 ｘ 调整因子``\n\nThis is the meaning of Bayesian inference. We first estimate a “prior probability” and then add the experimental results to see whether the experiment strengthens or weakens the “prior probability”, thus obtaining a “posterior probability” closer to the truth.\n\n### Formula\n\n``p(yi|x) = p(yi) * p(x|yi) / p(x)``\n\nP (belonging to category yi| belonging to category yi) = p (belonging to category yi) * p (belonging to category yi) /p (belonging to category x)\n\nAccording to the formula, the probability of “belonging to a certain category under certain characteristics” can be converted into the probability of “belonging to a certain category under certain characteristics”.\n\n• prior probability\n\nWhere p(yi) is called prior probability, that is, the probability of yi event occurring before x event occurs\n\n• posterior probability\n\nP(yi|x) is called posterior probability, that is, the probability of yi event occurring after the occurrence of x event, which is an observable value\n\nP(x|yi)/p(x) is the adjustment factor and also becomes the possibility function (`Likelyhood`), making the estimated probability closer to the real probability\n\n## Naive Bayesian algorithm\n\nNaive Bayesian theory originates from the independence of random variables: in terms of text classification, from the perspective of Naive Bayesian, the relationship between two words in a sentence is independent of each other, that is, each dimension in the feature vector of an object is independent of each other. This is the ideological basis of naive Bayesian theory. The process is as follows\n\n``````- 第一阶段，训练数据生成训练样本集：TF-IDF。\n- 第二阶段，对每个类别计算P(yi)。\n- 第三阶段，对每个特征属性计算所有类别下的条件概率p(ai|yi)。\n- 第四阶段，对每个类别计算p(x|yi)p(yi)。\n- 第五阶段，以p(x|yi)p(yi)的最大项作为x的所属类别。\n``````\n\n### Problem\n\nSuppose x is a text to be classified, which is characterized by {a1, a2, …, am}; Known category set {y1, y2, … yn}; Find the category x belongs to\n\n### Basic idea\n\n``如果p(yi|x) = max{p(y1|x),p(y2|x),p(y3|x),...,p(yn|x)}，则x属于yi类别``\n\n### How to calculate p(yi|x)\n\nUsing Bayesian formula\n\n``p(yi|x) = p(x|yi)*p(yi) / p(x)``\n\nThe problem is converted to calculating p(x|yi) for each categoryP(yi) in p(x|yi)The maximum term of p(yi) is taken as the category of x\n\n### Calculate p(x|yi)\n\nSince Naive Bayes assumes that each feature is independent of each other, therefore\n\n``p(x|yi) = p(a1|yi)*p(a2|yi)...*p(am|yi)``\n\n### Calculate p(ai|yi)\n\nWhile p(ai|yi) can be determined by training set (`There are already good categories.`), the conditional probability p(ai|yi) of various features under each category is calculated.\n\nSince then, the category p(yi|x) to which x belongs is solved step by step, which can be obtained by calculating the conditional probability p(ai|yi) of various features under each category in the training set.\n\nIn the training process, the influencing factor p(x|yi)=p(a1|yi) of the adjustment factor is calculated according to the training set.p(a2|yi) …P(am|yi), so the quality of the training set directly affects the accuracy of the prediction results.\n\n## TF-IDF\n\nTf-idf (termfrequency-inverse document frequency) is a common weighting method for information retrieval and data mining.\n\n``TF-IDF = TF * IDF``\n\nThe main idea of TF-IDF is that if a word or phrase appears frequently in one article TF and rarely in other articles (IDF value is large), then it is considered that the word or phrase has good classification ability and is suitable for classification.\n\n### TF\n\nIt means Term Frequency, that is, the frequency of a word appearing in a document.\n\n``TFi = Ndti / Ndt1..n``\n\nThat is, the sum of the number of occurrences of the word in the document/the number of occurrences of all words in the document\n\n### IDF\n\nIt means Inverse Document Frequency, which is an adjustment coefficient of the importance of a word and measures whether a word is a common word or not. If a word is rare, but it appears many times in this article, then it probably reflects the characteristics of this article, which is exactly the keyword we need.\n\nIDF for a particular word can be obtained by dividing the total number of documents by the number of documents containing the word, and then taking the logarithm of the quotient obtained.\n\n``IDF = log(Nd/Ndt)``\n\nNd is the total number of files and Ndt is the number of files containing the word\n\nIf a word is very commonly used, the Ndt will become larger, then the IDF value will become smaller, thus the TF-IDF value will become smaller, thus weakening the characteristics of commonly used words.\n\n## Summary\n\nNaive Bayesian algorithm solves the problem step by step and finally solves it through training set, which is worth studying and considering."
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.8739744,"math_prob":0.97934747,"size":5451,"snap":"2020-45-2020-50","text_gpt3_token_len":1362,"char_repetition_ratio":0.13456948,"word_repetition_ratio":0.071428575,"special_character_ratio":0.2135388,"punctuation_ratio":0.08091908,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99931645,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-11-27T09:00:35Z\",\"WARC-Record-ID\":\"<urn:uuid:42b16120-a54b-4fe0-8cda-08a685557a39>\",\"Content-Length\":\"34561\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:7c660f99-aa7f-4cbf-83b7-47a0f3213e49>\",\"WARC-Concurrent-To\":\"<urn:uuid:fef26a40-d061-40de-b0d4-d9bdfbf1a83b>\",\"WARC-IP-Address\":\"47.240.29.120\",\"WARC-Target-URI\":\"https://ddcode.net/2019/06/21/text-classification-principle-of-naive-bayesian-algorithm/\",\"WARC-Payload-Digest\":\"sha1:CN7T2ULE54AOUVJBJJCWDFVUCGDBV5AG\",\"WARC-Block-Digest\":\"sha1:HKI2MGMFN4NQWETB57ZZSYORBWHH7GED\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-50/CC-MAIN-2020-50_segments_1606141191511.46_warc_CC-MAIN-20201127073750-20201127103750-00134.warc.gz\"}"} |
https://www.trendnewzd.com/5608/colorful-solid-granny-square-dishcloth-crochet-pattern/ | [
"Wednesday , 22 March 2023\n\n# COLORFUL SOLID GRANNY SQUARE DISHCLOTH CROCHET PATTERN\n\nHello!! Today I want to show about COLORFUL SOLID GRANNY SQUARE DISHCLOTH CROCHET PATTERN.\n\n\n\nPATTERN\nMake a slip knot and ch 4, insert hook into the first chain and join with a slip stitch to make a circle.\n\nROUND 1: COLOR A – Ch 5, working into the center of the ring make 3 dc, ch 2, 3 dc into the ring, ch 2, 3 dc into the ring, ch 2, 2 dc into the ring, join using a sl into the third ch from the ch 5 we started the round with. (will have 3 dc per side after the round is complete)\n\nROUND 2: COLOR A – Ch 3, [2 dc into the corner space, ch 2, 2 dc into the corner space, 5 dc into each of the dc below], repeat pattern for next 2 sides. After the 2 dc in the last corner, make 2 dc into the dc below, join using a sl into the third ch from the ch 3 we started the round with. (will have 7 dc per side after the round is complete)\n\n\n\nROUND 3: COLOR B – Ch 3, 2 dc into each of the dc below, [2 dc into the corner space, ch 2, 2 dc into the corner space, 7 dc into each of the dc below], repeat pattern for next 2 sides. After the 2 dc in the last corner, make 4 dc into the dc below, join using a sl into the third ch from the ch 3 we started the round with. (will have 11 dc per side after the round is complete)\n\nROUND 4: COLOR B – Ch 3, 4 dc into each of the dc below, [2 dc into the corner space, ch 2, 2 dc into the corner space, 11 dc into each of the dc below], repeat pattern for next 2 sides. After the 2 dc in the last corner, make 6 dc into the dc below, join using a sl into the third ch from the ch 3 we started the round with. (will have 15 dc per side after the round is complete)\n\n\n\nROUND 3: COLOR B – Ch 3, 2 dc into each of the dc below, [2 dc into the corner space, ch 2, 2 dc into the corner space, 7 dc into each of the dc below], repeat pattern for next 2 sides. After the 2 dc in the last corner, make 4 dc into the dc below, join using an sl into the third ch from the ch 3 we started the round with. (will have 11 dc per side after the round is complete)\n\nROUND 4: COLOR B – Ch 3, 4 dc into each of the dc below, [2 dc into the corner space, ch 2, 2 dc into the corner space, 11 dc into each of the dc below], repeat pattern for next 2 sides. After the 2 dc in the last corner, make 6 dc into the dc below, join using a sl into the third ch from the ch 3 we started the round with. (will have 15 dc per side after the round is complete)\n\n\n\nROUND 7: COLOR D – Ch 3, 10 dc into each of the dc below, [2 dc into the corner space, ch 2, 2 dc into the corner space, 23 dc into each of the dc below], repeat pattern for next 2 sides. After the 2 dc in the last corner, make 12dc into the dc below, join using a sl into the third ch from the ch 3 we started the round with. (will have 27 dc per side after the round is complete)\n\nROUND 8: COLOR D – Ch 3, 12 dc into each of the dc below, [2 dc into the corner space, ch 2, 2 dc into the corner space, 27 dc into each of the dc below], repeat pattern for next 2 sides. After the 2 dc in the last corner, make 14dc into the dc below, join using a sl into the third ch from the ch 3 we started the round with. (will have 31 dc per side after the round is complete)\n\n\n\nVideo Tutorial :"
]
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.8849765,"math_prob":0.98708415,"size":3134,"snap":"2023-14-2023-23","text_gpt3_token_len":915,"char_repetition_ratio":0.23354633,"word_repetition_ratio":0.7741007,"special_character_ratio":0.29897895,"punctuation_ratio":0.11764706,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.96696657,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-03-23T05:14:44Z\",\"WARC-Record-ID\":\"<urn:uuid:39446b01-eb4e-4dad-af09-53572ff0d222>\",\"Content-Length\":\"54882\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:c5b8061a-bbc2-41af-99c9-b9d60ec759e3>\",\"WARC-Concurrent-To\":\"<urn:uuid:c7805ad8-6246-443d-b52e-691454361700>\",\"WARC-IP-Address\":\"198.187.31.103\",\"WARC-Target-URI\":\"https://www.trendnewzd.com/5608/colorful-solid-granny-square-dishcloth-crochet-pattern/\",\"WARC-Payload-Digest\":\"sha1:RWJFMRZEFM6M5O5GOXJV6AC3EFIF4Q3F\",\"WARC-Block-Digest\":\"sha1:XBQNYONIPVKBR7CWLZFHZHMDBB7G6N6Y\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-14/CC-MAIN-2023-14_segments_1679296944996.49_warc_CC-MAIN-20230323034459-20230323064459-00322.warc.gz\"}"} |
https://www.teachexcel.com/excel-tutorial/find-the-next-blank-row-with-vba-macros-in-excel_1261.html?nav=sim_side_col_forum | [
"# Find the Next Blank Row with VBA Macros in Excel\n\nAuthor:\n\nLearn how to find the next empty cell in a range in Excel using VBA and Macros. This method will skip any blanks that are between your data but not at the very end of it.\n\nLook at this example:",
null,
"We want the VBA to locate cell C7 and NOT cell C5.\n\nThe VBA to do this is quite simple:\n\n``````Sub Find_Next_Empty_Row()\n\n'selects the next empty row\n'ignores blanks inbetween the data set\n\nRange(\"A\" & Rows.Count).End(xlUp).Offset(1).Select\n\nEnd Sub\n``````\n\nThere are two comments in there and only one actual line of code:\n\n``````Range(\"A\" & Rows.Count).End(xlUp).Offset(1).Select\n``````\n\nThis is where the magic happens and now Ill explain it.\n\nThe A is the column where we want to get the next usable blank cell. So, if we wanted to check for the next cell in column D then you change the A to a D.\n\nThe Rows.Count part just counts how many rows are in Excel and adds them there so that it will not miss any rows when checking for blank ones. This is important since the row count in Excel changed from version 2003 to 2007, and so this makes the code a bit more versatile.\n\nThe End(xlUp) part tells Excel to return the first cell with data that it finds when lookup up the worksheet starting at the last row in the worksheet.\n\nSo far, this will find the last cell that has data in it in the column, but we need to get the very next, or empty, cell and this is what Offset(1) does for us. The 1 tells Offset how many rows it should offset and since we want the very next row, 1 works for us.\n\nAfter that, we can do whatever we want with this cell. At this point, Excel knows where we are talking about. In this example, I used a simple Select to select the empty cell so you can verify that the code works but, like I said, you can do anything here; you dont have to actually select the cell first.\n\nThis is a very versatile piece of code and it should help save you a lot of time.\n\nMake sure to download the Excel file for this tutorial so you can see the macro in action and play around with it.\n\nI hope this was helpful! :)\n\n## Question? Ask it in our Excel Forum\n\nSimilar Content on TeachExcel\nDo Something Every so Many Rows with a Macro in Excel\nTutorial: How to have a macro do something on a set interval of rows; for instance, input a value ev...\nGuide to Creating Charts with a Macro in Excel\nTutorial: How to add, edit, and position charts in Excel using VBA. This tutorial covers what to do ...\nInput Form to Get Data and Store it in Another Tab in Excel\nTutorial: How to make a user input form in Excel and have the data stored on another worksheet at th...\nBreak out of or Exit Different Types of Loops in VBA Macros in Excel\nTutorial: How to Exit, End, or Break out of loops in Excel. This tutorial includes breaking out of W...\nWhat is a Macro in Excel?\nTutorial: This is the first step to learning about Macros for Excel and how to use them. What is a...\nRun a Macro from Another Macro in Excel\nTutorial: I will show you how to run a macro from another macro in Excel. This means that you can...\nTutorial Details"
]
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null,
"https://www.teachexcel.com/images/uploads/362197ee9dda8175f891755dfb62e34c.png",
null
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https://www.nagwa.com/en/videos/508185672514/ | [
"# Video: Finding the Maximum Value of a Function Involving a Rational Function Using the Quotient Rule\n\nGiven that the volume of a hot air balloon grows according to the relation 𝑓(𝑡) = (7000𝑡/(𝑡² + 49)) + 4000, where the time is measured in hours, determine its maximum volume.\n\n05:55\n\n### Video Transcript\n\nGiven that the volume of a hot-air balloon grows according to the relation 𝑓 of 𝑡 equals 7000𝑡 over 𝑡 squared plus 49 plus 4000, where the time is measured in hours, determine its maximum volume.\n\nWe’ve been given an equation then that governs the growth of this hot-air balloon. And we’re asked to determine its maximum volume. Well, this will take place when the volume of the hot-air balloon is no longer growing, which means that the rate of change of this volume with respect to time will be equal to zero. We know that the rate of change of a quantity is given by its first derivative. So in order to determine the maximum volume, we first need to determine the time at which 𝑓 prime of 𝑡, the first derivative of this function, is equal to zero.\n\nWe can use differentiation in order to find an expression for 𝑓 prime of 𝑡. Looking at our function 𝑓 of 𝑡, we see that it is a quotient of two differentiable functions plus a constant. So we’re going to need to use the quotient rule to help us differentiate. The quotient rule tells us that, for two differentiable functions 𝑢 and 𝑣, the derivative of their quotient, 𝑢 over 𝑣, is equal to 𝑣𝑢 prime minus 𝑢𝑣 prime all over 𝑣 squared.\n\nWe’re therefore going to let 𝑢 equal the function in the numerator of the quotient, that’s 7000𝑡, and 𝑣 equal the function in the denominator. That’s 𝑡 squared plus 49. We need to find each of their individual derivatives with respect to 𝑡, which we can do using the power rule of differentiation. The derivative of 𝑢 with respect to 𝑡, 𝑢 prime, is equal to 7000. And the derivative of to 𝑣 with respect to 𝑡, 𝑣 prime, is equal to two 𝑡.\n\nNow, we can substitute into the formula for the quotient rule. 𝑓 prime of 𝑡 is equal to 𝑣𝑢 prime, that’s 𝑡 squared plus 49 multiplied by 7000, minus 𝑢𝑣 prime. That’s 7000𝑡 multiplied by two 𝑡. And this is all divided by 𝑣 squared. That’s 𝑡 squared plus 49 all squared. Now, remember, this just gives the derivative of the first part of our function 𝑓 of 𝑡. We also need to remember to differentiate that constant of 4000. But as the derivative of a constant is just zero, it actually makes no contribution to our derivative. We can then distribute the parentheses in the numerator and collect like terms, to give that 𝑓 prime of 𝑡 is equal to 343000 minus 7000𝑡 squared over 𝑡 squared plus 49 squared.\n\nNow, remember, we said that the maximum volume of this hot-air balloon would be achieved when its rate of change with respect to time is equal to zero. So next, we’re going to set our expression for the first derivative equal to zero and solve for 𝑡. We have the equation then, 343000 minus 7000𝑡 squared over 𝑡 squared plus 49 squared is equal to zero. Now, for a quotient to be equal to zero, it must be true that the numerator of the quotient is equal to zero. So actually, we can simplify our equation. We’re just left with 343000 minus 7000𝑡 squared is equal to zero. Adding 7000𝑡 squared to each side and then dividing by 7000 gives 𝑡 squared is equal to 49.\n\nWe can solve this equation by square rooting. And as 𝑡 represents time, we’re only going to take the positive value. The positive square root of 49 is seven. So we have that the maximum volume of the hot-air balloon will be achieved when 𝑡 equals seven. That’s after seven hours. Next, we need to determine what the maximum volume actually is. And we can do this by substituting 𝑡 equals seven into our equation 𝑓 of 𝑡. We obtain 7000 multiplied by seven over seven squared plus 49 plus 4000. This is equal to 49000 over 98 plus 4000, which is all equal to 4500. We found then that the maximum volume of the hot-air balloon is 4500 cubic units.\n\nNow, we only found one value for 𝑡 when solving 𝑓 prime of 𝑡 is equal to zero. So this is the only critical point of our function 𝑓 of 𝑡. But we should confirm that it is indeed a maximum. And we can do this using the first derivative test. What we’re going to do is consider the shape of the curve around this critical point by evaluating the slope of the curve. Now, a critical point occurred when 𝑡 was equal to seven. So we’ll choose values of 𝑡 either side of this. Let’s choose six and eight. We know that when 𝑡 is equal to seven, the first derivative in the slope is equal to zero. When 𝑡 is equal to six, the first derivative is equal to 343000 minus 7000 multiplied by six squared over six squared plus 49 squared, which is equal to 12.595. So we find that the slope of the curve is positive when 𝑡 equals six.\n\nOn the other side of the critical point, when 𝑡 is equal to eight, the first derivative is equal to 343000 minus 7000 multiplied by eight squared over eight squared plus 49 squared, which is equal to negative 8.223. So the slope of the curve is negative on this side of the critical point. So the slope changes from positive to zero to negative. And so we can see from a sketch of this that the critical point at 𝑡 equals seven is indeed a local maximum.\n\nWe’ve completed the problem then. We found the maximum volume of the hot-air balloon to be 4500 cubic units. And using the first derivative test, we’ve confirmed that it is indeed a local maximum."
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https://www.coolstuffshub.com/weight/convert/pennyweights-to-cubic-yards/ | [
"# Convert Pennyweights to Cubic yards (dwt to yd³ Conversion)\n\n1 pennyweight is equal to 8.45272 × 10-7 cubic yards.\n\n1 dwt = 8.45272 × 10-7 yd³\n\n## How to convert pennyweights to cubic yards?\n\nTo convert pennyweights to cubic yards, multiply the value in pennyweights by 8.45272 × 10-7.\n\nYou can use the conversion formula :\ncubic yards = pennyweights × 8.45272 × 10-7\n\nTo calculate, you can also use our pennyweights to cubic yards converter, which is a much faster and easier option as compared to calculating manually.\n\n## How many cubic yards are in a pennyweight?\n\nThere are 8.45272 × 10-7 cubic yards in a pennyweight.\n\n• 1 pennyweight = 8.45272 × 10-7 cubic yards\n• 2 pennyweights = 1.690544 × 10-6 cubic yards\n• 3 pennyweights = 2.535816 × 10-6 cubic yards\n• 4 pennyweights = 3.381088 × 10-6 cubic yards\n• 5 pennyweights = 4.22636 × 10-6 cubic yards\n• 10 pennyweights = 8.45272 × 10-6 cubic yards\n• 100 pennyweights = 8.45272 × 10-5 cubic yards\n\n## Examples to convert dwt to yd³\n\nExample 1:\nConvert 50 dwt to yd³.\n\nSolution:\nConverting from pennyweights to cubic yards is very easy.\nWe know that 1 dwt = 8.45272 × 10-7 yd³.\n\nSo, to convert 50 dwt to yd³, multiply 50 dwt by 8.45272 × 10-7 yd³.\n\n50 dwt = 50 × 8.45272 × 10-7 yd³\n50 dwt = 4.22636 × 10-5 yd³\n\nTherefore, 50 pennyweights converted to cubic yards is equal to 4.22636 × 10-5 yd³.\n\nExample 2:\nConvert 125 dwt to yd³.\n\nSolution:\n1 dwt = 8.45272 × 10-7 yd³\n\nSo, 125 dwt = 125 × 8.45272 × 10-7 yd³\n125 dwt = 0.000105659 yd³\n\nTherefore, 125 dwt converted to yd³ is equal to 0.000105659 yd³.\n\nFor faster calculations, you can simply use our dwt to yd³ converter.\n\n## Pennyweights to cubic yards conversion table\n\nPennyweights Cubic yards\n0.001 dwt 8.45272 × 10-10 yd³\n0.01 dwt 8.45272 × 10-9 yd³\n0.1 dwt 8.45272 × 10-8 yd³\n1 dwt 8.45272 × 10-7 yd³\n2 dwt 1.690544 × 10-6 yd³\n3 dwt 2.535816 × 10-6 yd³\n4 dwt 3.381088 × 10-6 yd³\n5 dwt 4.22636 × 10-6 yd³\n6 dwt 5.071632 × 10-6 yd³\n7 dwt 5.916904 × 10-6 yd³\n8 dwt 6.762176 × 10-6 yd³\n9 dwt 7.607448 × 10-6 yd³\n10 dwt 8.45272 × 10-6 yd³\n20 dwt 1.690544 × 10-5 yd³\n30 dwt 2.535816 × 10-5 yd³\n40 dwt 3.381088 × 10-5 yd³\n50 dwt 4.22636 × 10-5 yd³\n60 dwt 5.071632 × 10-5 yd³\n70 dwt 5.916904 × 10-5 yd³\n80 dwt 6.762176 × 10-5 yd³\n90 dwt 7.607448 × 10-5 yd³\n100 dwt 8.45272 × 10-5 yd³"
]
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https://www.mathworks.com/matlabcentral/profile/authors/18604279 | [
"Community Profile",
null,
"# Atharva Shirke\n\nLast seen: 1 year ago Active since 2020\n\n#### Statistics\n\n•",
null,
"•",
null,
"#### Content Feed\n\nView by\n\nSolved\n\nTimes 2 - START HERE\nTry out this test problem first. Given the variable x as your input, multiply it by two and put the result in y. Examples:...\n\n2 years ago\n\nSolved\n\nDoubling elements in a vector\nGiven the vector A, return B in which all numbers in A are doubling. So for: A = [ 1 5 8 ] then B = [ 1 1 5 ...\n\n2 years ago\n\nSolved\n\nVector creation\nCreate a vector using square brackets going from 1 to the given value x in steps on 1. Hint: use increment.\n\n2 years ago\n\nSolved\n\nCreate a vector\nCreate a vector from 0 to n by intervals of 2.\n\n2 years ago\n\nSolved\n\nFind max\nFind the maximum value of a given vector or matrix.\n\n2 years ago\n\nSolved\n\nGet the length of a given vector\nGiven a vector x, the output y should equal the length of x.\n\n2 years ago\n\nSolved\n\nSelect every other element of a vector\nWrite a function which returns every other element of the vector passed in. That is, it returns the all odd-numbered elements, s...\n\n2 years ago\n\nSolved\n\nFind the sum of all the numbers of the input vector\nFind the sum of all the numbers of the input vector x. Examples: Input x = [1 2 3 5] Output y is 11 Input x ...\n\n2 years ago"
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.6549121,"math_prob":0.98202676,"size":1602,"snap":"2022-40-2023-06","text_gpt3_token_len":487,"char_repetition_ratio":0.1583229,"word_repetition_ratio":0.052805282,"special_character_ratio":0.26716605,"punctuation_ratio":0.10982659,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99668753,"pos_list":[0,1,2,3,4,5,6],"im_url_duplicate_count":[null,1,null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-02-08T11:35:43Z\",\"WARC-Record-ID\":\"<urn:uuid:9eba0a5f-bec3-4c67-903c-0f82da77ef9d>\",\"Content-Length\":\"86270\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:8b2d3303-ed59-41f6-8433-a3d21fca0a09>\",\"WARC-Concurrent-To\":\"<urn:uuid:26e635e6-8077-499c-b2bd-dfb045aec7b1>\",\"WARC-IP-Address\":\"104.86.80.92\",\"WARC-Target-URI\":\"https://www.mathworks.com/matlabcentral/profile/authors/18604279\",\"WARC-Payload-Digest\":\"sha1:3E6KSPYM2UV2GEMWOZI57JMEUS6TPISU\",\"WARC-Block-Digest\":\"sha1:RM3JLNNOIFM5LD5W6TH5XKQDPC27AYKP\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-06/CC-MAIN-2023-06_segments_1674764500758.20_warc_CC-MAIN-20230208092053-20230208122053-00830.warc.gz\"}"} |
https://online-unit-converter.com/speed/convert-kilometers-per-hour-to-miles-per-hour/176-5-kmh-to-mph/ | [
"# 176.5 km/h to mph converter. How many miles per hour are in 176.5 kilometers per hour?\n\n## 176.5 km/h to mph\n\nThe question “What is 176.5 km/h to mph?” is the same as “How many miles per hour are in 176.5 km/h?” or “Convert 176.5 kilometers per hour to miles per hour” or “What is 176.5 kilometers per hour to miles per hour?” or “176.5 kilometers per hour to mph”. Read on to find free km/h to miles per hour converter and learn how to convert 176.5 km/h to mph. You’ll also learn how to convert 176.5 km/h to mph.\n\nAnswer: There are 109.6720154 mph in 176.5 km/h.\n\nAlternatively, you can say “176.5 km/h equals to 109.6720154 mph” or “176.5 km/h = 109.6720154 mph” or “176.5 kilometers per hour is 109.6720154 miles per hour”.\n\n## kilometers per hour to miles per hour conversion formula\n\nA mile per hour is equal to 1.609344 kilometers per hour. A kilometer per hour equals 0.6213711922 miles per hour.\nTo convert 176.5 kilometers per hour to miles per hour you can use one of the formulas:\n\nFormula 1\nMultiply 176.5 km/h by 0.6213711922.\n176.5 * 0.6213711922 = 109.6720154 mph\n\nFormula 2\nDivide 176.5 km/h by 1.609344.\n176.5 / 1.609344 = 109.6720154 mph\n\nHint: no need to use a formula. Use our free km/h to mph converter.\n\n## Alternative spelling of 176.5 km/h to mph\n\nMany of our visitor spell kilometers per hour and miles per hour differently. Below we provide all possible spelling options.\n\n• Spelling options with “kilometers per hour”: 176.5 kilometers per hour to mph, 176.5 kilometers per hour to miles per hour, 176.5 kilometers per hour in mph, 176.5 kilometers per hour in miles per hour.\n• Spelling options with “km/h”: 176.5 km/h to mph, 176.5 km/h to miles per hour, 176.5 km/h in mph, 176.5 km/h in miles per hour.\n• Spelling options with “in”: 176.5 km/h in mph, 176.5 km/h in miles per hour, 176.5 kilometers per hour in mph, 176.5 kilometers per hour in miles per hour.\n\n## FAQ on 176.5 km/h to mph conversion\n\nHow many miles per hour are in 176.5 kilometers per hour?\n\nThere are 109.6720154 miles per hour in 176.5 kilometers per hour.\n\n176.5 km/h to mph?\n\n176.5 km/h is equal to 109.6720154 mph. There are 109.6720154 miles per hour in 176.5 kilometers per hour.\n\nWhat is 176.5 km/h to mph?\n\n176.5 km/h is 109.6720154 mph. You can use a rounded number of 109.6720154 for convenience. In this case you can say that 176.5 km/h is 109.67 mph.\n\nHow to convert 176.5 km/h to mph?\n\nUse our free kilometers per hour to miles per hour converter or multiply176.5 km/h by 0.6213711922.\n176.5 * 0.6213711922 = 109.6720154 mph"
]
| [
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https://www.piping-designer.com/index.php/properties/thermodynamics/264-compressibility | [
"# Compressibility\n\non . Posted in Thermodynamics",
null,
"Compressibility, abbreviated as $$\\beta$$ (Greek symbol beta), also called coefficient of compressibility, a dimensionless number, measures the change in volume under external forces for any liquid. Compressibility of a fluid increases with pressure and temperature and results in loss of volume output of pumps. In control systems, compression of fluid provides a mass spring condition that limits system response.\n\n## Compressibility formula\n\n$$\\large{ \\beta = \\frac{1}{K} }$$\nSymbol English Metric\n$$\\large{ \\beta }$$ (Greek symbol beta) = compressibility $$\\large{ dimensionless }$$\n$$\\large{ K }$$ = bulk modulus $$\\large{\\frac{lbm}{in^2}}$$ $$\\large{ Pa }$$\n\n## Compressibility formula\n\n$$\\large{ \\beta = -\\; \\frac{1}{V_i} \\; \\frac{\\Delta V}{\\Delta p} }$$\nSymbol English Metric\n$$\\large{ \\beta }$$ (Greek symbol beta) = compressibility $$\\large{ dimensionless }$$\n$$\\large{ V_i }$$ = initial volume $$\\large{ in^3 }$$ $$\\large{ mm^3 }$$\n$$\\large{ \\Delta V }$$ = volume change $$\\large{ in^3 }$$ $$\\large{ mm^3 }$$\n$$\\large{ \\Delta p }$$ = pressure change $$\\large{\\frac{lbm}{in^2}}$$ $$\\large{ Pa }$$",
null,
""
]
| [
null,
"https://www.piping-designer.com/images/properties/fluid_mechanics/flow/compressibility_3.jpg",
null,
"https://www.piping-designer.com/images/Piping%20Designer%20Gallery/Piping-Designer_Logo_1.jpg",
null
]
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https://arifindustri.wordpress.com/tag/random-variate-generator/ | [
"Tag Archives: random variate generator\n\nRandomization Of Random Sampling Using Inverse-Transform Random Variate Generator Based On Hypergeometric Distribution\n\nRANDOMIZATION OF RANDOM SAMPLING USING INVERSE-TRANSFORM RANDOM VARIATE GENERATOR BASED ON HYPERGEOMETRIC DISTRIBUTION Author : ARIF RAHMAN Abstract : Implementing random sampling, it should use a good randomization to take a number of samples from a given population, in order … Continue reading\n\nInverse-Transform Random Variate Generator Based On Geometric Distribution Generator For Randomization Of Random Sampling\n\nINVERSE-TRANSFORM RANDOM VARIATE GENERATOR BASED ON GEOMETRIC DISTRIBUTION GENERATOR FOR RANDOMIZATION OF RANDOM SAMPLING Author : ARIF RAHMAN Abstract : Taking a number of samples in a given population using random sampling requires a good randomization to avoid a biased … Continue reading"
]
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https://wiki.seg.org/wiki/Dictionary:Hanning_function/zh | [
"# 汉宁函数\n\n${\\frac {1}{2}}+{\\frac {\\cos {\\phi }}{2}},\\quad -\\pi <\\phi <\\pi$",
null,
",\n$0,\\quad \\phi <-\\pi {\\text{ or }}\\phi >\\pi$",
null,
".",
null,
""
]
| [
null,
"https://en.wikipedia.org/api/rest_v1/media/math/render/svg/dd3f82bda8bc8f68a3ef0198a7d8120093df42f2",
null,
"https://en.wikipedia.org/api/rest_v1/media/math/render/svg/92cda0bda8bdaa28945a9a300f0bc83334e0cfd7",
null,
"https://wiki.seg.org/images/7/76/Segw12.jpg",
null
]
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https://meniny.cn/posts/priority_of_operations_cpp/ | [
"",
null,
"C++运算符优先级\nPrecedence Operator Description Associativity\n1 `::` Scope resolution Left-to-right\n2 `++` `--` Suffix/postfix increment and decrement\n`()` Function call\n`[]` Array subscripting\n`.` Element selection by reference\n`−>` Element selection through pointer\n3 `++` `--` Prefix increment and decrement Right-to-left\n`+` `−` Unary plus and minus\n`!` `~` Logical NOT and bitwise NOT\n`(type)` Type cast\n`*` Indirection (dereference)\n`&` Address-of\n`sizeof` Size-of\n`new`, `new[]` Dynamic memory allocation\n`delete`, `delete[]` Dynamic memory deallocation\n4 `.*` `->*` Pointer to member Left-to-right\n5 `*` `/` `%` Multiplication, division, and remainder\n6 `+` `−` Addition and subtraction\n7 `<<` `>>` Bitwise left shift and right shift\n8 `<` `<=` For relational operators < and ≤ respectively\n`>` `>=` For relational operators > and ≥ respectively\n9 `==` `!=` For relational = and ≠ respectively\n10 `&` Bitwise AND\n11 `^` Bitwise XOR (exclusive or)\n12 `|` Bitwise OR (inclusive or)\n13 `&&` Logical AND\n14 `||` Logical OR\n15 `?:` Ternary conditional Right-to-Left\n16 `=` Direct assignment (provided by default for C++ classes)\n`+=` `−=` Assignment by sum and difference\n`*=` `/=` `%=` Assignment by product, quotient, and remainder\n`<<=` `>>=` Assignment by bitwise left shift and right shift\n`&=` `^=` `|=` Assignment by bitwise AND, XOR, and OR\n17 `throw` Throw operator (for exceptions)\n18 `,` Comma Left-to-right"
]
| [
null,
"https://meniny.cn/assets/images/posts/cpp.jpg",
null
]
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https://de.webqc.org/molecularweightcalculated-190417-4.html | [
"",
null,
"#### Chemical Equations Balanced on 04/17/19\n\n Molecular weights calculated on 04/16/19 Molecular weights calculated on 04/18/19\nCalculate molecular weight\n 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137\nMolar mass of FeSO4*7H2O is 278.01456\nMolar mass of (C11H9N5O)2CuCl2 is 588,89612\nMolar mass of (C11H9N5O)2CuCl2(H2O)2 is 624,92668\nMolar mass of SO4 is 96,0626\nMolar mass of SO4 is 96,0626\nMolar mass of (C11H9N5O)CuCl2 is 361,67406\nMolar mass of (C11H9N5O)2CuCl2(H2O)4 is 660,95724\nMolar mass of C5H11O4Na is 158,12820928\nMolar mass of hydrogen is 2.01588\nMolar mass of C5H9O3Na is 140,11292928\nMolar mass of C5H11O4Na is 158,12820928\nMolar mass of SO3 is 80,0632\nMolar mass of CuCl2 2(CH3)2SO is 921.64544\nMolar mass of Fe3(CO)12 is 503,6562\nMolar mass of NH3 is 17,03052\nMolar mass of CuCl2 2(CH3)2SO is 921.64544\nMolar mass of Co(acac)2 is 259.164835\nMolar mass of H3PO4 is 97,995182\nMolar mass of CuCl2 is 134.452\nMolar mass of CaMg(CO3)2 is 184.4008\nMolar mass of Al(acac)3 is 327.3289986\nMolar mass of Cu(C6H5N5)2(C2N3)2 is 489,9038\nMolar mass of Na2S2O3*5*H2O is 176,12301856\nMolar mass of Cu(C6H5N5)2(C2N3)2(H2O)2 is 525,93436\nMolar mass of c10h8 is 128.17052\nMolar mass of Cu(C6H5N5)2(C2N3)2(H2O) is 507,91908\nMolar mass of Ca(acac)2 is 240.30964\nMolar mass of Cl is 35,453\nMolar mass of H2SO4 is 98,07848\nMolar mass of Ce is 140.116\nMolar mass of Zn is 65.38\nMolar mass of p205 is 6349.62121\nMolar mass of CuCl2 is 134.452\nMolar mass of p2o5 is 141.944524\nMolar mass of CH4 is 16.04246\nMolar mass of CH3 is 15.03452\nMolar mass of Cu(C4N2H6S3)2(C2N3)2 is 552,22668\nMolar mass of Mn is 54.938045\nMolar mass of (CH3)2SO is 78.13344\nMolar mass of Cu(C4N2H6S3)2(C2N3)2(H2O)2 is 588,25724\nMolar mass of CaSO4 is 136,1406\nMolar mass of Mg is 24.305\nMolar mass of Cu(C4N2H6S3)4(C2N3)2 is 908,82436\nMolar mass of K2CO3 is 138,2055\nMolar mass of CuO is 79,5454\nMolar mass of Cu(C4N2H6S3)4(C2N3)2H2O is 926,83964\nMolar mass of Cu(C4N2H6S3)4(C2N3)2 is 908,82436\nMolar mass of FeCl2 is 126,751\nMolar mass of Mg(NO3)2 is 148,3148\nMolar mass of Ne is 20.1797\n 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137\nCalculate molecular weight\n Molecular weights calculated on 04/16/19 Molecular weights calculated on 04/18/19\nMolecular masses on 04/10/19\nMolecular masses on 03/18/19\nMolecular masses on 04/17/18\nMit der Benutzung dieser Webseite akzeptieren Sie unsere Allgemeinen Geschäftsbedingungen und Datenschutzrichtlinien.\n© 2020 webqc.org Alle Rechte vorbehalten.\n Periodensystem der Elemente Einheiten-Umrechner Chemie Werkzeuge Chemie-Forum Chemie FAQ Konstanten Symmetrie Chemie Links Link zu uns Kontaktieren Sie uns Eine bessere Übersetzung vorschlagen? Wählen Sie eine SpracheDeutschEnglishEspañolFrançaisItalianoNederlandsPolskiPortuguêsРусский中文日本語한국어 Wie zitieren? WebQC.Org Online-Schulung kostenlose Hausaufgabenhilfe Chemie Probleme Fragen und Antworten"
]
| [
null,
"https://de.webqc.org/images/logo.png",
null
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http://www.stepbystep.com/difference-between-vectors-and-scalars-100072/ | [
"# Difference Between Vectors and Scalars\n\nThe basic difference between a scalar and vector quantity can be identified by the parameters which define these quantities.\n\nA scalar quantity is a number, which indicates the magnitude of anything under consideration – a medium. The vector, on the other hand, not only needs a magnitude but also direction. There are many applications of these quantities in science, especially physics.\n\nIn this article we will explain these two quantities with the help of appropriate examples.\n\n### Instructions\n\n• 1\n\nScalar\n\nA scalar is a one dimensional description of any quantity, representing only its magnitude. It does not give any indication about the direction of motion or flow. The basic example of a scalar quantity is speed.\n\nWhen we say that a car is running at a speed of 80 kilometers per hours, we don’t give any indication about the direction of motion.\n\nSimilarly, when an object is at 120 degree Fahrenheit, we only get a measure of the magnitude of hotness; it does not indicate the direction in which the heat of the object will flow.\n\nTime is also a major example of scalar quantities. When we express the time in months, days, hours, minutes and seconds, it only indicates a quantity.\n\nSimilarly a volume of any object only gives the indication of the space occupied by it, and no description about the direction of motion.\n\nImage courtesy: faculty.plattsburgh.edu",
null,
"• 2\n\nVector\n\nVector quantities need both magnitude and direction to be defined properly.\n\nConsidering the above example of speed, when a person moves at a certain speed, there is no direction involved, but when he/she moves with a velocity – rate of change of speed – then there is a direction attached, indicating whether the speed is low or high.\n\nIn order to be defined a vector quantity though; there are a certain pre requisites.\n\nThere should be an indication of direction, for example, forward, backward, towards left or right, or at an angle etc.\n\nThere should be an initial reference point, which can be used to determine the direction easily. This reference point should have its centre in either north, south, east or west quadrant so that the direction can be easily referenced.\n\nThus when we said that a car was moving at 80 kilometers per hours to define a scalar quantity, if we add 80 kilometers per hours towards north, it becomes a vector quantity.\n\nOther examples of vector quantities are moments, forces and accelerations.\n\nImage courtesy: edinformatics.com",
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"http://www.stepbystep.com/wp-content/uploads/2013/05/Difference-Between-Vectors-and-Scalars1.gif",
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"http://www.stepbystep.com/wp-content/uploads/2013/05/Difference-Between-Vectors-and-Scalars.gif",
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"http://assets.pinterest.com/images/pidgets/pinit_fg_en_rect_gray_20.png",
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https://www.msdotnet.co.in/2014/06/sql-server-interview-questions-and.html | [
"# Sql Server Interview Questions and Answers Part 4\n\n1.) Where we generally use Identity column Constraints on a table ?\nIf any user wants,one column value should incremented automatically without any intervention of users then we generally use this constraints on the column.\n\n2.) How to create a table and add an Identity column on that table?\nSyntax:-\nIdentity (<seed> < Incremented >)\nExample:-\n`create table student(eid ``int` `identity(0,1),sname varchar(30), sage ``int``)`\n``` ```\n3.) How to add Identity column in Existing table ?\n`alter table student add sid ``int` `identity(0,1)`\n` OR`\n `ALTER TABLE student ADD ToBeIdentity INT IDENTITY(0,1)`\n``` ```\n4.) How to ON/OFF Identity column constraints in sql server using command ?\n`set` `identity_insert student on`\n`set` `identity_insert student off `\n``` ```\n`5.)Can we copy data from one table to another table using sql command ? `\nYes.If Number of column in both table is same.\n\n6.) How to create a new table at run time and copy the data from a existing table using sql command ?\n\n`select *into student2 from student1`\nNote:-Here student2 is a new table and student1 is a existing table.\n\n7.) How to copy data in two existing table in sql server using command ?\nSyntax:-\ninsert [first table_name] select *from[second table_name]\nExample:-\n`insert students select *from studentdetails`\n\n8.) How to copy the data in two existing table if sequence of column name of both table are same ?\nSyntax:-\n `insert into [first table_name](column2,column3,column1)select(column2,column3,column1)from [Second table_name]` ``` ``` `Example:-` `insert into student2(sid,sname,sage)select sid, sname,sage from student1`\n\n9.) What are the two sql statement used to remove the records from database ?\nSyntax:\ndelete from Table_name where Clause.\nExample:\ndelete from student where sid=101;\nSyntax:\ntruncate table Table_name;\nExample:\ntruncate table student;\n9.) What is difference between delete and Truncate statement in sql server ?\n\n• In case of delete each row is deleted one by one and deleted entry is made within log file but In case of truncate table structure is drop and recreated rows are not deleted physically.\n• Truncate is faster than delete.\n• Within delete statement we can specify 'Where' clause but in truncate statement we can not specify the 'Where' clauses.\n• In case of Identity column the difference is,if we run delete statement then insert a new record then identity column value will be the next value of the last inserted values.But if we run truncate statement and insert a new record the next value seed value.\n• In case of Foreign key constraint the difference is, if child table does not have any record then we can run delete statement against parents table but we can not run truncate statement against parent table.\n• Delete is DML Statement and truncate is DDL Statement.\n• If record is deleted that can be recover but if record is truncated then can not recover.\n\n1.",
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"1.",
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"https://www.blogger.com/img/blogger_logo_round_35.png",
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https://answers.everydaycalculation.com/subtract-fractions/4-2-minus-2-63 | [
"Solutions by everydaycalculation.com\n\n## Subtract 2/63 from 4/2\n\n1st number: 2 0/2, 2nd number: 2/63\n\n4/2 - 2/63 is 124/63.\n\n#### Steps for subtracting fractions\n\n1. Find the least common denominator or LCM of the two denominators:\nLCM of 2 and 63 is 126\n\nNext, find the equivalent fraction of both fractional numbers with denominator 126\n2. For the 1st fraction, since 2 × 63 = 126,\n4/2 = 4 × 63/2 × 63 = 252/126\n3. Likewise, for the 2nd fraction, since 63 × 2 = 126,\n2/63 = 2 × 2/63 × 2 = 4/126\n4. Subtract the two like fractions:\n252/126 - 4/126 = 252 - 4/126 = 248/126\n5. After reducing the fraction, the answer is 124/63\n6. In mixed form: 161/63\n\nMathStep (Works offline)",
null,
"Download our mobile app and learn to work with fractions in your own time:"
]
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"https://answers.everydaycalculation.com/mathstep-app-icon.png",
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https://www.helpingwithmath.com/printables/worksheets/basic_facts/wor0201basic_facts02b.htm | [
"# Basic Facts: Doubles & Number +1\n\n------ Note: The Information above this point will not be sent to your printer --------\n\nPractice Doubles and Number +1\n\n 2 + 2 = 4 6 + 1 = 7 5 + 1 = 6 9 + 1 = 10 1 + 8 = 9 5 + 5 = 10 3 + 1 = 4 1 + 7 = 8 1 + 9 = 10 4 + 1 = 5 1 + 2 = 3 5 + 5 = 10 7 + 1 = 8 2 + 2 = 4 8 + 1 = 9 1 + 5 = 6 5 + 5 = 10 1 + 6 = 7 2 + 1 = 3 2 + 2 = 4 1 + 1 = 2 5 + 5 = 10 1 + 3 = 4 1 + 4 = 5\n\nPractice Doubles and Number +1\n5\n+ 5\n10\n1\n+ 4\n5\n8\n+ 1\n9\n2\n+ 2\n4\n6\n+ 1\n7\n1\n+ 9\n10\n1\n+ 5\n6\n2\n+ 2\n4\n3\n+ 1\n4\n1\n+ 7\n8\n1\n+ 4\n5\n5\n+ 5\n10\n1\n+ 1\n2\n2\n+ 1\n3\n\n------ Note: The Information below this point will not be sent to your printer --------\n\nOther Worksheets in this Series: Number +1 | Number Plus One | Doubles & Number +1 |Number +1 & Doubles | Backward 1 | Addition and Subtraction | Addition and Subtraction | Number +1 , Doubles, & backwards 1 | Number +1 , Doubles, & backwards 1 | Adding Zero | Zeros & Doubles | Basic Facts Worksheet | Adding Doubles | Review Worksheet | Doubles Subtraction | Addition & Subtraction | Review | Right Next To Each Other | Subtraction | Addition and Subtraction Facts | Doubles & Right Next To Each Other | Adding & Subtracting Facts | Adding 2: 2 Ladder | 2 Ladder & Doubles | Practice All Tricks\n\n## Related Resources\n\nThe various resources listed below are aligned to the same standard, (1OA06) taken from the CCSM (Common Core Standards For Mathematics) as the Addition and subtraction Worksheet shown above.\n\nAdd and subtract within 20, demonstrating fluency for addition and subtraction within 10. Use strategies such as counting on; making ten (e.g., 8 + 6 = 8 + 2 + 4 = 10 + 4 = 14); decomposing a number leading to a ten (e.g., 13 - 4 = 13 - 3 - 1 = 10 - 1 = 9); using the relationship between addition and subtraction (e.g., knowing that 8 + 4 = 12, one knows 12 - 8 = 4); and creating equivalent but easier or known sums (e.g., adding 6 + 7 by creating the known equivalent 6 + 6 + 1 = 12 + 1 = 13)."
]
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https://mathmaine.com/2016/11/05/one-step-linear-equation-problems/ | [
"# Practice Problems: One Step Linear Equations\n\nThe solution to each of the following twenty problems is 12. Focus on finding the most helpful algebraic step to take a reader from the problem as stated to the solution, and be sure you can explain why that step leads to a solution.\n\n1.",
null,
"$x+7~=~19$\n2.",
null,
"$3+x~=~15$\n3.",
null,
"$11~=~-1+x$\n4.",
null,
"$3x~=~12+2x$\n5.",
null,
"$x-2~=~10$\n6.",
null,
"$-18+x~=~-6$\n7.",
null,
"$9~=~x-3$\n8.",
null,
"$-5x~=~-6x+12$\n9.",
null,
"$2x~=~24$\n10.",
null,
"$84~=~7x$\n11.",
null,
"$x\\cdot 5~=~60$\n12.",
null,
"$36~=~x\\cdot 3$\n13.",
null,
"$\\dfrac{x}{3}~=~4$\n14.",
null,
"$6~=~\\dfrac{x}{2}$\n15.",
null,
"$\\dfrac{1}{4}x~=~3$\n16.",
null,
"$\\dfrac{4}{3}~=~x \\cdot \\dfrac{1}{9}$\n\n– Read each of the following, and decide on a variable you will use to answer the question.\n– Then write an algebraic equation that describes the situation using the variable you chose.\n– Then solve for the variable to answer the question:\n\n17. Alicia is 15 years old. She is 3 years older than her younger brother Tony. How old is Tony?\n18. Mahmoud has 24 apples. He has twice as many as Janice. How many apples does Janice have?\n19. Gabriella’s immediate family has 8 family members. She has 4 fewer than Darius. How many immediate family members does Darius have?\n20. Lawrence has 6 books. He has half as many as Aisha. How many books does Aisha have?\n\nThis site uses Akismet to reduce spam. Learn how your comment data is processed."
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"https://s0.wp.com/latex.php",
null,
"https://s0.wp.com/latex.php",
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https://forum.allaboutcircuits.com/threads/plotting-signal-sequences-on-matlab.58718/ | [
"# Plotting signal sequences on Matlab\n\n#### tquiva\n\nJoined Oct 19, 2010\n176\nI'm trying to plot these signals onto Matlab but I'm not exactly sure how to.\n\nThe problem says that a, b, and c are chosen so that the sequences have unit energy. I'm not sure how to go about doing this, but here's my Matlab code so far. I get an error that my variable n is undefined. How would I define it and all the other variables?\n\nRich (BB code):\n% Signal bbox\nn=4; a=1;\nstar=-15:0.01:15;\nfor x=1:length(star);\nif abs(star(x))<=n\nbbox(x)=a;\nelse\nbbox(x)=0;\nend;\nend;\nplot(star,bbox),xlabel('*'),ylabel('bbox[*]'),title('Signal bbox[*]')\n;\n\n% Signal cox\nn=4; b=1;\nstar=-15:0.01:15;\nfor x=1:length(star);\nif abs(star(x))<=n\ncox(x)=b*cos(pi*star/n);\nelse\ncox(x)=0;\nend;\nend;\nplot(star,cox),xlabel('*'),ylabel('cox[*]'),title('Signal cox[*]')\n\n% Signal c2ox\nn=4; c=1;\nstar=-15:0.01:15;\nfor x=1:length(star);\nif abs(star(x))<=n\nc2ox(x)=c*(cos(pi*star/n))^2;\nelse\nc2ox(x)=0;\nend;\nend;\nplot(star,c2ox),xlabel('*'),ylabel('c2ox[*]'),title('Signal c2ox[*]')\n\nLast edited:\n\n#### t_n_k\n\nJoined Mar 6, 2009\n5,447\n\nThe variable 'n' is to be a positive integer. As you have defined the array 'star' with 0.01 increments from -15 onwards to +15 the elements of 'star' will not all be positive integers - some being negative and / or non-integer. I'm assuming these are meant to be discrete signal sequences - although I may have misinterpreted the notation.\n\nYou've also got a couple of errors such as in the line\n\ncox(x)=b*cos(pi*star/n);\n\nwhich should be\n\ncox(x)=b*cos(pi*star(x)/n);\n\nand also the line\n\nc2ox(x)=c*(cos(pi*star/n))^2;\n\nwhich should be\n\nc2ox(x)=c*(cos(pi*star(x)/n))^2;\n\nTo set the variables a, b & c to achieve unit energy you would have to do something like equating the function\n\n$$\\int_0^{n'} (f(x))^2dx=1$$\n\nfor each case\n\nMy guess is that (in one or more of the three cases) increasing values of n produce increasingly better agreement between the discretely sampled energy case and the continuous energy case - as exemplified in the generic integral function above. There is an exception to this statement regarding one of the three cases which I think you might readily deduce yourself.\n\nLast edited:\n\n#### tquiva\n\nJoined Oct 19, 2010\n176\nTo set the variables a, b & c to achieve unit energy you would have to do something like equating the function\n\n$$\\int_0^{n'} (f(x))^2dx=1$$\n\nfor each case\n\nMy guess is that (in one or more of the three cases) increasing values of n produce increasingly better agreement between the discretely sampled energy case and the continuous energy case - as exemplified in the generic integral function above. There is an exception to this statement regarding one of the three cases which I think you might readily deduce yourself.\nThank you for the tips on my Matlab script. However, I'm still confused on how to choose the a, b, c values to obtain unit energy.\n\nSo for the function cox_n\n[*] , I would square this and integrate it? But what does the limits of 0 to n' mean? I'm still a little lost. Could you please help me out? I really appreciate your help.\n\nI integrated the cox_n function with the given integral, and received the following answer:",
null,
"I didn't use any limits, but this is a rather length answer. What am I doing wrong?\n\nBut for the function, bbox, a=1? Is that correct?\n\nLast edited:\n\n#### t_n_k\n\nJoined Mar 6, 2009\n5,447\nTake the case of ....\n\n$$cox_{n} [*]=\\{^{bcos(\\pi * /n), \\ \\ |*| \\leq n}_{0, \\ \\ \\ |*|\\gt n^{'}}$$\n\nchange the variable n' to x (to avoid confusion with n & n' terms) such that\n\n$$f(x)=bcos(\\frac{\\pi x}{n})$$\n\nNow evaluate the energy of f(x) ...\n\n$$E\\_{f(x)}=\\int^{x=n}_{0}{(f(x))^2 dx}$$\n\n$$E\\_{f(x)}=\\int^{x=n}_{0}{(bcos(\\frac{\\pi x}{n})^2 dx}$$\n\n$$E\\_{f(x)}=\\int^{x=n}_{0}{b^2cos^2(\\frac{\\pi x}{n}) dx}$$\n\n$$E\\_{f(x)}=\\int^{x=n}_{0}{b^2(\\frac{1+cos(\\frac{2 \\pi x}{n})}{2}) dx}$$\n\nwhich after integration on the limits reduces to\n\n$$E\\_{f(x)}=\\frac{nb^2}{2}$$\n\nequating this to unit energy (or 1) gives\n\n$$b=sqrt{\\frac{2}{n}}$$\n\nIn a similar manner it would be true that for the bbox function\n\n$$a=\\frac{1}{sqrt{n}}$$\n\nFinding c is a bit more complex - you should try to show that for unit energy\n\n$$c=sqrt{\\frac{8}{3n}}$$\n\nIn your Matlab code you could add computations for the three sequences using the a, b & c values above to evaluate the energy content for each sequence. The values should converge to '1' as n increases - you assign the n values as you wish. If you include n=0 in the energy evaluation, then you would possibly need to replace the 'n' value in the energy computation functions with (n+1) in each case. See how it goes anyway.\n\n#### tquiva\n\nJoined Oct 19, 2010\n176\nthank you so much. I worked through all the steps, and I finally understand how to obtain the a,b,c values. I inputted these into Matlab code, and this is now my most up-to-date script:\n\nRich (BB code):\n% Signal bbox\nn=4; a=1/sqrt(n);\nstar=0:0.01:15;\nfor x=1:length(star);\nif abs(star(x))<=n\nbbox(x)=a;\nelse\nbbox(x)=0;\nend;\nend;\nplot(star,bbox),xlabel('*'),ylabel('bbox\n[*]'),title('Signal bbox\n[*]')\n\n% Signal cox\nn=4; b=sqrt(2/n);\nfor x=1:length(star);\nif abs(star(x))<=n\ncox(x)=b*cos(pi*star(x)/n);\nelse\ncox(x)=0;\nend;\nend;\nplot(star,cox),xlabel('*'),ylabel('cox\n[*]'),title('Signal cox\n[*]')\n\n% Signal c2ox\nn=4; c=sqrt(8/(3*n));\nfor x=1:length(star);\nif abs(star(x))<=n\nc2ox(x)=c*(cos(pi*star(x)/n))^2;\nelse\nc2ox(x)=0;\nend;\nend;\nplot(star,c2ox),xlabel('*'),ylabel('c2ox\n[*]'),title('Signal c2ox\n[*]')\nDid I do it correctly?\n\nFor part b of the problem, I am asked measure the dissimilarity between a pair of signals. If I'm not mistaken, this means that I would find the difference in energies of the signals? Such as using the integration that you suggested for each signal to obtain the energies, then subtracting the from another... Is my conclusion correct?\n\n#### t_n_k\n\nJoined Mar 6, 2009\n5,447\nConsider the code snippet ....\n\nn=4; a=1/sqrt(n);\nstar=0:0.01:15;\nfor x=1:length(star);\nif abs(star(x))<=n\nbbox(x)=a;\nelse\nbbox(x)=0;\nend;\nend;\n\nI would re-cast it as\n\nn=1000; a=1/sqrt(n+1);\nstar=0:1:n+500;\nfor x=1:length(star);\nif abs(star(x))<=n\nbbox(x)=a;\nelse\nbbox(x)=0;\nend;\nend;\n\nThe energy would be found from something like ....\n\nsum(bbox.*bbox)\n\nwhere \".*\" means element by element multiplication - i.e. basically squaring the individual element values. I'm not sure if sum() is a Matlab function - I don't use Matlab. By sum() I just mean adding up all the elements of the squaring process.\n\nIn reality you don't need to concern yourself with the sequence values outside of values greater than 'n', but it is informative to note the sequence forms over a range greater than 'n'.\n\nAs I said in previous post you set 'n' to whatever you wish in your code.\n\n#### t_n_k\n\nJoined Mar 6, 2009\n5,447\nThe energy values can probably be computed by a simpler relationship such as\n\nE_bbox=sum(bbox^2)\n\nif Matlab allows that overall function ...\n\nand for the cox sequence ...\n\nE_cox=sum(cox^2)\n\nand so on."
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https://homework.cpm.org/category/CON_FOUND/textbook/a2c/chapter/12/lesson/12.1.1/problem/12-7 | [
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"### Home > A2C > Chapter 12 > Lesson 12.1.1 > Problem12-7\n\n12-7.\n\nWrite a formula for the $n$th term of each of the following series.\n\n1. $3+10+17+$ …\n\nThe terms are increasing by $7$ each time.\n\n$t(1)=3$\n\n$t(n)=3+7(n-1)$\n\nSimplified, this becomes $t(n)=7n-4$.\n\n1. $20+11+2+$ …\n\nWhat is added to each term to get the next?\n\nWhat is the first term?"
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null,
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",
null
]
| {"ft_lang_label":"__label__en","ft_lang_prob":0.9274553,"math_prob":0.99947315,"size":367,"snap":"2022-27-2022-33","text_gpt3_token_len":105,"char_repetition_ratio":0.1322314,"word_repetition_ratio":0.37142858,"special_character_ratio":0.28610355,"punctuation_ratio":0.13414635,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99995327,"pos_list":[0,1,2,3,4],"im_url_duplicate_count":[null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-08-08T00:42:43Z\",\"WARC-Record-ID\":\"<urn:uuid:9635b998-5a35-47d9-ba64-ebe3c7339464>\",\"Content-Length\":\"41254\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:aaf36056-8cdb-42f1-852d-246e3f6e02b8>\",\"WARC-Concurrent-To\":\"<urn:uuid:91199733-6855-4e3c-ac8a-30d30e387c92>\",\"WARC-IP-Address\":\"172.67.70.60\",\"WARC-Target-URI\":\"https://homework.cpm.org/category/CON_FOUND/textbook/a2c/chapter/12/lesson/12.1.1/problem/12-7\",\"WARC-Payload-Digest\":\"sha1:H3MFRCYRGWHE3RKUVDFLL5WB56I6FUUI\",\"WARC-Block-Digest\":\"sha1:2LDY7PQLFAZWESULIDBT4SHLMDP33Y6Z\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-33/CC-MAIN-2022-33_segments_1659882570741.21_warc_CC-MAIN-20220808001418-20220808031418-00182.warc.gz\"}"} |
https://ccrma.stanford.edu/~jos/pasp/Mass_Moment_Inertia_Tensor.html | [
"Next | Prev | Up | Top | Index | JOS Index | JOS Pubs | JOS Home | Search\n\n### MassMoment of Inertia Tensor\n\nAs derived in the previous section, the moment of inertia tensor, in 3D Cartesian coordinates, is a three-by-three matrix",
null,
"that can be multiplied by any angular-velocity vector to produce the corresponding angular momentum vector for either a point mass or a rigid mass distribution. Note that the origin of the angular-velocity vector",
null,
"is always fixed at",
null,
"in the space (typically located at the center of mass). Therefore, the moment of inertia tensor",
null,
"is defined relative to that origin.\n\nThe moment of inertia tensor can similarly be used to compute the mass moment of inertia for any normalized angular velocity vector",
null,
"as",
null,
"(B.22)\n\nSince rotational energy is defined as",
null,
"(see Eq.(B.7)), multiplying Eq.(B.22) by",
null,
"gives the following expression for the rotational kinetic energy in terms of the moment of inertia tensor:",
null,
"(B.23)\n\nWe can show Eq.(B.22) starting from Eq.(B.14). For a point-mass",
null,
"located at",
null,
", we have",
null,
"where again",
null,
"denotes the three-by-three identity matrix, and",
null,
"(B.24)\n\nwhich agrees with Eq.(B.20). Thus we have derived the moment of inertia",
null,
"in terms of the moment of inertia tensor",
null,
"and the normalized angular velocity",
null,
"for a point-mass",
null,
"at",
null,
".\n\nFor a collection of",
null,
"masses",
null,
"located at",
null,
", we simply sum over their masses to add up the moments of inertia:",
null,
"Finally, for a continuous mass distribution, we integrate as usual:",
null,
"where",
null,
"is the total mass.\n\nSubsections\nNext | Prev | Up | Top | Index | JOS Index | JOS Pubs | JOS Home | Search\n\n[How to cite this work] [Order a printed hardcopy] [Comment on this page via email]"
]
| [
null,
"https://ccrma.stanford.edu/~jos/pasp/img557.png",
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"https://ccrma.stanford.edu/~jos/pasp/img110.png",
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"https://ccrma.stanford.edu/~jos/pasp/img557.png",
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"https://ccrma.stanford.edu/~jos/pasp/img2883.png",
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"https://ccrma.stanford.edu/~jos/pasp/img2885.png",
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"https://ccrma.stanford.edu/~jos/pasp/img2886.png",
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"https://ccrma.stanford.edu/~jos/pasp/img3.png",
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"https://ccrma.stanford.edu/~jos/pasp/img108.png",
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"https://ccrma.stanford.edu/~jos/pasp/img2887.png",
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"https://ccrma.stanford.edu/~jos/pasp/img2888.png",
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"https://ccrma.stanford.edu/~jos/pasp/img2889.png",
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"https://ccrma.stanford.edu/~jos/pasp/img236.png",
null,
"https://ccrma.stanford.edu/~jos/pasp/img557.png",
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"https://ccrma.stanford.edu/~jos/pasp/img111.png",
null,
"https://ccrma.stanford.edu/~jos/pasp/img3.png",
null,
"https://ccrma.stanford.edu/~jos/pasp/img108.png",
null,
"https://ccrma.stanford.edu/~jos/pasp/img17.png",
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"https://ccrma.stanford.edu/~jos/pasp/img652.png",
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"https://ccrma.stanford.edu/~jos/pasp/img2288.png",
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"https://ccrma.stanford.edu/~jos/pasp/img2890.png",
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"https://ccrma.stanford.edu/~jos/pasp/img2891.png",
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"https://ccrma.stanford.edu/~jos/pasp/img2892.png",
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.9068239,"math_prob":0.94015014,"size":879,"snap":"2021-43-2021-49","text_gpt3_token_len":171,"char_repetition_ratio":0.14285715,"word_repetition_ratio":0.0,"special_character_ratio":0.18998863,"punctuation_ratio":0.085365854,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99841183,"pos_list":[0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28,29,30,31,32,33,34,35,36,37,38,39,40,41,42,43,44,45,46,47,48,49,50],"im_url_duplicate_count":[null,null,null,null,null,1,null,null,null,2,null,4,null,1,null,1,null,1,null,null,null,null,null,1,null,6,null,2,null,null,null,null,null,null,null,null,null,null,null,null,null,6,null,2,null,4,null,2,null,1,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-11-30T21:31:16Z\",\"WARC-Record-ID\":\"<urn:uuid:93586bb3-14f2-41ec-b1c2-d0ea491a64d0>\",\"Content-Length\":\"17730\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:54be99f9-9df8-41b2-b13b-89d7f3295733>\",\"WARC-Concurrent-To\":\"<urn:uuid:9e2f4730-009c-40df-a6f2-7003193e497c>\",\"WARC-IP-Address\":\"171.64.197.141\",\"WARC-Target-URI\":\"https://ccrma.stanford.edu/~jos/pasp/Mass_Moment_Inertia_Tensor.html\",\"WARC-Payload-Digest\":\"sha1:VVKXRBZWOQDZZTY4N5C4VN2BCLGQKGRU\",\"WARC-Block-Digest\":\"sha1:RCY53O3HKWL7C6MKITQ2SRR5ZYOMX6SH\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-49/CC-MAIN-2021-49_segments_1637964359073.63_warc_CC-MAIN-20211130201935-20211130231935-00575.warc.gz\"}"} |
https://docs.vmware.com/en/Site-Recovery-Manager/5.5/com.vmware.srm.install_config.doc/GUID-4A34D0C9-8CC1-46C4-96FF-3BF7583D3C4F.html | [
"To determine the bandwidth that vSphere Replication requires to replicate virtual machines efficiently, you calculate the average data change rate within an RPO period divided by the link speed.\n\nIf you have groups of virtual machines that have different RPO periods, you can determine the replication time for each group of virtual machines. For example, you might have four groups with RPO of 15 minutes, one hour, four hours, and 24 hours. Factor in all the different RPOs in the environment, the subset of virtual machines in your environment that is replicated, the change rate of the data within that subset, the amount of data changes within each configured RPO, and the link speeds in your network.\n\n## Prerequisites\n\nExamine how data change rate, traffic rates, and the link speed meet the RPO. Then look at the aggregate of each group.\n\n## Procedure\n\n1. Identify the average data change rate within the RPO by calculating the average change rate over a longer period then dividing it by the RPO.\n2. Calculate how much traffic this data change rate generates in each RPO period."
]
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https://www.hsslive.co.in/2022/07/haryana-board-class-7th-mathematics-syllabus-pdf.html | [
"# HSSlive: Plus One & Plus Two Notes & Solutions for Kerala State Board\n\n## Sunday, July 17, 2022\n\nHaryana Board Class 7th Mathematics Syllabus 2022 PDF Download - BSEH has released the syllabus of Haryana Board Class 7th Mathematics Syllabus 2022 along with the official notification on the official website - bseh.org.in. The Haryana Board Class 7th Mathematics 2022 Syllabus pdf comprises the subject-wise topics which will be asked in the class 7 Mathematics exam. Students of Haryana State Board Class 7th Mathematics can download the syllabus PDF from this page.\n\n## BSEH Class 7 Mathematics Syllabus 2022-23 PDF\n\nUsing the Haryana Board Class 7th Mathematics Syllabus 2022 PDF, students can prepare their study schedule and exam preparation strategy. As the Haryana Board exam date has been released, candidates can plan their schedule according to it, therefore, they can prepare their syllabus of Haryana Board Class 7th Mathematics Exam 2022 accordingly. Along with the Haryana Board Class 7th Mathematics 2022 syllabus, candidates can also check the official BSEH exam pattern and the previous year's BSEH Class 7th Mathematics question papers.\n\n## Haryana State Board Class 7th Mathematics Syllabus 2022 PDF Online\n\n Name of the Board Haryana Board Name of the Grade 7 Subjects Mathematics Year 2022-23 Format PDF/DOC Provider hsslive.co.in Official Website bseh.org.in\n\n## How To Download Haryana State Board Class 7th Mathematics Syllabus 2022 PDF Online?\n\n1. Visit the official website - bseh.org.in.\n2. Look for BSEH Class 7th Mathematics Syllabus 2022.\n3. Now check for Haryana Board Class 7 Mathematics Syllabus 2022 PDF.\n\n## Haryana Board Class 7th Mathematics Syllabus 2022-23 PDF\n\nStudents of download the Haryana Board Class 7th Mathematics Syllabus 2022-23 PDF online using the links provided below:\n\n Chapter Name of the Chapter Chapter 1 IntegersIntroduction of IntegersProperties of Addition and Subtraction of IntegersMultiplication of IntegersMultiplication of a Positive and Negative IntegerMultiplication of two Negative IntegerProperties of Multiplication of IntegersDivision of IntegersProperties of Division of Integers Chapter 2 Fractions and DecimalsAddition and Subtraction of FractionsMultiplication of FractionMultiplication of a Fraction by a Whole NumberMultiplication of a Fraction by a FractionDivision of FractionDivision of Whole Number by a FractionReciprocal of FractionDivision of a fraction by a Whole NumberDivision of Fraction by Another FractionMultiplication of Decimal NumbersMultiplication of Decimal Numbers by 10, 100 and 1000Division of Decimal NumbersDivision of Decimals by 10, 100 and 1000Division of a Decimal Number by a Whole NumberDivision of a Decimal Number by Another Decimal Number Chapter 3 Data HandlingCollection of DataOrganisation of DataArithmetic MeanModeMode of Large DataMedianUse of Bar Graphs with a Different PurposeChance and Probability Chapter 4 Simple EquationsWhat is EquationSolving an EquationMore EquationsFrom Solution to EquationApplications of Simple Equations To Practical Situations Chapter 5 Lines and AnglesIntroduction to Lines and AnglesRelated anglesComplementary AnglesSupplementary AnglesAdjacent AnglesLinear PairVertically Opposite AnglesPairs of LinesIntersecting LinesTransversalAngles made by a TransversalTransversal of Parallel LinesChecking for Parallel Lines Chapter 6 The Triangle and its PropertiesMedians of a TriangleAltitudes of a TriangleExterior Angle of a Triangle and Its PropertyAngle Sum Property of a TriangleTwo Special Triangles: Equilateral and IsoscelesSum of the Lengths of Two Sides of A TriangleRight-Angled Triangles and Pythagoras Property Chapter 7 Congruence of TrianglesCongruence of Plane FiguresCongruence Among Line SegmentsCongruence of AnglesCongruence of TrianglesCriteria For Congruence of TrianglesCongruence Among Right-Angled Triangles Chapter 8 Comparing QuantitiesEquivalent RatiosPercentage-Another way of comparing quantitiesConverting fraction numbers to percentageConverting Decimals to PercentageConverting percentage to Fraction or DecimalsUse of PercentageInterpreting PercentageConverting Percentage to “How Many”Ratios to PercentIncrease or Decrease as PercentPrice Related to an Item or Buying and SellingProfit or Loss as a PercentageCharge Given on Borrowed Money or Simple InterestInterest for Multiple Years Chapter 9 Rational NumbersNeed For Rational NumbersWhat Are Rational NumbersPositive And Negative Rational NumbersRationals Numbers On a Number LineRational Numbers in Standard FormComparison of Rational NumbersRational Numbers Between Two Rational NumbersOperations On Rational NumbersAddition of Rational NumbersSubtraction of Rational NumbersMultiplication of Rational NumbersDivision of Rational Numbers Chapter 10 Practical GeometryConstruction of a Line Parallel To a Given Line, Through a Point Not on The LineConstruction of TriangleConstructing a Triangle When the Lengths of its Three Sides are Known (SSS Criterion)Constructing a Triangle When the Lengths of Two Sides And The Measure of The Angle Between Them Are Known (SAS Criterion)Constructing a Triangle When The Measures of Two of Its Angles and The Length of The Side Inclined Between Them Is Given (ASA Criterion)Constructing a Right-Angled Triangle When The Length of One Leg and Its Hypotenuse are Given (RHS Criterion) Chapter 11 Perimeter and AreaSquares and RectanglesTriangles as Parts of RectanglesGeneralising for Other Congruent Parts of RectanglesArea of a ParallelogramArea of TriangleCircumference of a CircleArea of CircleConversion of UnitsApplications Chapter 12 Algebraic ExpressionsHow are Expressions FormedTerms of An ExpressionLike and Unlike TermsMonomials, Binomials, Trinomials and PolynomialsAddition and Subtraction of Algebraic ExpressionsFinding The Value of An ExpressionUsing Algebraic Expressions – Formulas and Rules Chapter 13 Exponents and PowersExponentsLaws of ExponentsMultiplying Powers With The Same BaseDividing Powers With The Same BaseTaking Power of a PowerMultiplying Powers With The Same BaseDividing Powers With The Same ExponentsMiscellaneous Examples Using The Laws of ExponentsDecimal Number SystemExpressing Large Numbers in The Standard Form Chapter 14 Symmetry Chapter 15 Visualising Solid ShapesIntroduction to Plane Figures and Solid FiguresFaces, Edges and VerticesNets for Building 3-D ShapesDrawing Solids on a Flat SurfaceOblique SketchesIsometric SketchesVisualising Solid ObjectsViewing Different Sections of a SolidOne way to View an Object By Cutting or SlicingAnother Way is By Shadow Play\n\n## Key Benefits Of Solving Haryana Board Class 7th Mathematics Syllabus 2022-23 PDF\n\nThere are several benefits of Haryana Board Class 7th Mathematics Syllabus 2022 PDF\n\n• It enhances the speed of solving questions and time management skills.\n• It helps candidates to gain insight into the BSEH Class 7th Mathematics Syllabus 2022 PDF.\n• Understand the type of questions and marking scheme.\n• Helps in analyzing the preparation level.\n• Gives an idea about the real exam scenario.\n• Enhances exam temperament and boosts confidence.\n• Gives an idea about the topics important for examination point of view.\n• Haryana State Board Class 7th Mathematics Syllabus 2022 PDF helps in understanding the Exam pattern and its level of difficulty.\n\n## FAQ about BSEH Class 7th Mathematics Syllabus 2022 PDF\n\n#### What is the Haryana Board Class 7th Mathematics Syllabus 2022-23??\n\nThe Haryana Board Class 7th Mathematics Syllabus 2022 PDF comprises the subject-wise topics which will be asked in the exam.\n\n#### Is it necessary to go through the BSEH Class 7th Mathematics Syllabus 2022?\n\nCandidates, if they want to score higher marks and stay ahead in the competition, should not ignore the syllabus. They should read the syllabus thoroughly. This will help in developing a strong preparation strategy and candidates will also gain valuable insights into the exam pattern, important chapters and topics, weightage of marks, objective of the entire course, etc.\nShare:"
]
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.75724214,"math_prob":0.5058401,"size":8055,"snap":"2022-27-2022-33","text_gpt3_token_len":1939,"char_repetition_ratio":0.1854428,"word_repetition_ratio":0.13279352,"special_character_ratio":0.19292365,"punctuation_ratio":0.031720858,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.95817465,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-08-11T18:39:31Z\",\"WARC-Record-ID\":\"<urn:uuid:9354bc83-2f17-4371-98b9-85ec561e84d3>\",\"Content-Length\":\"255444\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:881e86f7-5010-487e-a8ea-db44bfb64174>\",\"WARC-Concurrent-To\":\"<urn:uuid:70d63ae2-b691-4730-a54c-9f72e4bcec56>\",\"WARC-IP-Address\":\"104.21.49.201\",\"WARC-Target-URI\":\"https://www.hsslive.co.in/2022/07/haryana-board-class-7th-mathematics-syllabus-pdf.html\",\"WARC-Payload-Digest\":\"sha1:LR4KBAZZIKKG6HE55DYGVJEVWFSOXMRT\",\"WARC-Block-Digest\":\"sha1:HTOPOHB7UZJBXPUORJWJB7Y5Y4QOI52Z\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-33/CC-MAIN-2022-33_segments_1659882571483.70_warc_CC-MAIN-20220811164257-20220811194257-00001.warc.gz\"}"} |
https://web.cs.elte.hu/archive/Gyori/ | [
"## Generating intervals\n\n#### András Benczúr, Jörg Förster, Zoltán Király\n\nGyőri's theorem says that for a system of subintervals of a path the minimum size of a generating system (a set of intervals such that all given subinterval is the union of some of them) is equal to the maximum size of an independent subsystem (ie. subintervals I(1), I(2), ..., I(t) such that they cannot be generated by less then t generators, because for all 1<i<=t there is a point in I(i) such that not any of I(1), ..., I(i-1) contains that point).\n\nRecently András Frank developed a new combinatorial algorithm for finding the minimum generator system and the maximum independent set even for intervals of a cycle. We implemented his algorithm.\n\nFor reading about the theoretical background, history and the implementation itself read the Thesis of Jörg (unzipped version) or our paper in progress (unzipped version).\n\nThere are three programs here, the first is for path, the second is for cycle and the third is for rectangles in a vertically convex region. All three was written by Jörg Förster.\n\nThe program Gyori in directory Path (the compiled code is for Solaris 2.5.1) computes a minimum generating system and a maximum independent subsystem of a family of a mainpath's subpaths (or analogously for a system of a maininterval`s subintervals). We consider the mainpath and all the subpaths drawn horizontally and having integer endpoints. A makefile is included, however, the paths for the libraries (especially for xlib) and include files may have to be reset to work on your system.\n\nYou can either copy the program and/or sourcecode to your directory. When you execute the file 'Gyori' you are prompted for an input file.\n\nAs an input the program needs a file in the following format:\n\n```X Y -- The 1st line: the left and right coordinates of the mainpath.\nsize -- The 2nd line: where size is the no. of subpaths.\nx y -- All following lines: left and right coord. of the subpaths.\nx y\nx y\n.\n.\n.```\n\nSuch an input can also be generated with the program 'testgen' where you are prompted for some parameters of the mainpath and subfamily and a seed for generating an arbitrary system. The program 'testgen' writes the input in the file 'test.in'.\n\nThe next input you are prompted for by 'Gyori' is if you want to enter a freebie generator (Type 'n' if you don't know what to do!) If you chose yes, you are asked to type in the name of the file containing the freebie generator.\n\nThis file has to have the following format:\n\n```size -- The 1st line: where size is the no. of freebie generator.\nx y -- All following lines: left and right coord. of the freebie generator.\nx y\nx y\n.\n.\n.\n```\n\nThe last question you are asked is if the interval system should be showed graphically in a separate window. Answer yes ONLY IF you have an xlib connection to the server the program is running on. In this window you can obtain online help by pushing the 'H'-key of the keyboard. By clicking on any orange 'edge' the directly following members of the poset are being highlighted.\n\nThe Output of the program is written to the files Generator and Independent.\n\nIn directory Circ the \"almost-finished\" program Gyori (the compiled code is for Solaris 2.5.1) does the same thing but for intervals of a cycle.\n\nSee also the graphical program in the directory Rectangle, which demonstrates Győri`s theorem for a rectangle cover of a vertically convex polygon.\n\nGood luck!"
]
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https://issues.apache.org/jira/browse/SPARK-21190 | [
"",
null,
"SPIP: Vectorized UDFs in Python\n\nDetails\n\n• Type:",
null,
"Sub-task\n• Status: Resolved\n• Priority:",
null,
"Major\n• Resolution: Fixed\n• Affects Version/s: 2.2.0\n• Fix Version/s:\n• Component/s:\n• Labels:\n• Target Version/s:\n\nDescription\n\nBackground and Motivation\nPython is one of the most popular programming languages among Spark users. Spark currently exposes a row-at-a-time interface for defining and executing user-defined functions (UDFs). This introduces high overhead in serialization and deserialization, and also makes it difficult to leverage Python libraries (e.g. numpy, Pandas) that are written in native code.\n\nThis proposal advocates introducing new APIs to support vectorized UDFs in Python, in which a block of data is transferred over to Python in some columnar format for execution.\n\nTarget Personas\nData scientists, data engineers, library developers.\n\nGoals\n\n• Support vectorized UDFs that apply on chunks of the data frame\n• Low system overhead: Substantially reduce serialization and deserialization overhead when compared with row-at-a-time interface\n• UDF performance: Enable users to leverage native libraries in Python (e.g. numpy, Pandas) for data manipulation in these UDFs\n\nNon-Goals\nThe following are explicitly out of scope for the current SPIP, and should be done in future SPIPs. Nonetheless, it would be good to consider these future use cases during API design, so we can achieve some consistency when rolling out new APIs.\n\n• Define block oriented UDFs in other languages (that are not Python).\n• Define aggregate UDFs\n• Tight integration with machine learning frameworks\n\nProposed API Changes\nThe following sketches some possibilities. I haven’t spent a lot of time thinking about the API (wrote it down in 5 mins) and I am not attached to this design at all. The main purpose of the SPIP is to get feedback on use cases and see how they can impact API design.\n\nA few things to consider are:\n\n1. Python is dynamically typed, whereas DataFrames/SQL requires static, analysis time typing. This means users would need to specify the return type of their UDFs.\n\n2. Ratio of input rows to output rows. We propose initially we require number of output rows to be the same as the number of input rows. In the future, we can consider relaxing this constraint with support for vectorized aggregate UDFs.\n\n3. How do we handle null values, since Pandas doesn't have the concept of nulls?\n\nProposed API sketch (using examples):\n\nUse case 1. A function that defines all the columns of a DataFrame (similar to a “map” function):\n\n```@spark_udf(some way to describe the return schema)\ndef my_func_on_entire_df(input):\n\"\"\" Some user-defined function.\n\n:param input: A Pandas DataFrame with two columns, a and b.\n:return: :class: A Pandas data frame.\n\"\"\"\ninput[c] = input[a] + input[b]\nInput[d] = input[a] - input[b]\nreturn input\n\nspark.range(1000).selectExpr(\"id a\", \"id / 2 b\")\n.mapBatches(my_func_on_entire_df)\n```\n\nUse case 2. A function that defines only one column (similar to existing UDFs):\n\n```@spark_udf(some way to describe the return schema)\ndef my_func_that_returns_one_column(input):\n\"\"\" Some user-defined function.\n\n:param input: A Pandas DataFrame with two columns, a and b.\n:return: :class: A numpy array\n\"\"\"\nreturn input[a] + input[b]\n\nmy_func = udf(my_func_that_returns_one_column)\n\ndf = spark.range(1000).selectExpr(\"id a\", \"id / 2 b\")\ndf.withColumn(\"c\", my_func(df.a, df.b))\n```\n\nOptional Design Sketch\nI’m more concerned about getting proper feedback for API design. The implementation should be pretty straightforward and is not a huge concern at this point. We can leverage the same implementation for faster toPandas (using Arrow).\n\nOptional Rejected Designs\nSee above.\n\nAttachments\n\n1. SPIPVectorizedUDFsforPython (1).pdf\n109 kB\nReynold Xin\n\nPeople\n\n• Assignee:",
null,
"Bryan Cutler\nReporter:",
null,
"Reynold Xin"
]
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null,
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null,
"https://issues.apache.org/jira/images/icons/priorities/major.svg",
null,
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http://vogelvriendenbocholtz.nl/iron_ore/1450835143/trapezium-dome-calculator.html | [
"## trapezium dome calculator\n\n• ### An online calculator to calculate the area of a Trapezoid.",
null,
"How to use the calculator. Enter the sides a and b and the height of the trapezoid as positive real numbers and press \"enter\".\n\n• ### Pins from geo-dome.co.uk on Pinterest",
null,
"geo-dome.co.uk. Dome calculation tools. Daniel Franks. Small Homes. Pin it. Like.Swimming Pools, Design Drawing, Portfolio Categories, Trapezium Design, Helmets Dome, Design Portfolio, Spaces...\n\n• ### Tarpezoid | MathCaptain.com | Trapezoid Problems",
null,
"A Parallelogram. Areas of Parallelograms. Related Calculators. Trapezoid Calculator.\n\n• ### Dome Calculator",
null,
"Explore the reverse dome trapezium oct calculator jan . Portal to geodesic end . Project, we have a angles in cubic feetcob dome correct length .\n\n• ### Pro Calculators - Trapezoid Area Calculator",
null,
"Recent Posts. Window Shade Calculator. Window Quote Calculator. Wheelchair Ramp Calculator.\n\n• ### Calculate area of a trapezoid, its median, perimeter and sides",
null,
"The trapezoid area calculator shows the area, median and perimeter formulas. Calculate with different measurement units.\n\n• ### Trapezoid calculator",
null,
"Trapezium wiki artikel. Related calculators: Parallelogram calculator Rectangle calculator Rhombus calculator Square calculator.\n\n• ### Area of a trapezoid. Definition and formula - Math Open Reference",
null,
"Area of a trapezoid (\"trapezium\" in British usage).Calculator. Recall that the bases are the two parallel sides of the trapezoid.\n\n• ### Trapezium dome design",
null,
"The trapezium dome has very obvious advantages for joining with non-dome architecture, uniformity of panels, mating better with doors and windows, etc.\n\n• ### Trapezoid Calculator",
null,
"Trapezoid Calculator. Scroll Down for instructions and definitions. To use the calculator, you need to know the lengths of all 4 trapezoid sides.\n\n• ### Desert Domes - The Dome Calculator",
null,
"Go to the Reverse Dome Calculator if you know one of the strut lengths, and you want to calculate the radius and the lengths of the other struts.\n\n• ### This calculator is designed to give the area of any trapezoid.",
null,
"Trapezoid Area Calculator. This calculator requires the use of Javascript enabled and capable browsers.\n\n• ### Trapezium Calculator - Calculate Area and Perimeter...",
null,
"Trapezium Calculator. Area of Trapezium/Trapezoid: [ ½×(sum of parallel sides)×(distance between them) ].\n\n• ### Trapezoid Lesson | Free Math Help",
null,
"Definition and examples of a trapezoid. This geometry lesson explains that a trapezoid has four sides, with two of themCALCULATORS. Equation SolverFactoring CalculatorDerivative Calculatormore...\n\n• ### trapezium definition, meaning - what is trapezium in the British English...",
null,
"trapezium - definition, meaning, audio pronunciation, synonyms and more. What is trapezium? a flat shape with four sides, where two of the sides are parallel: See more incylinder. cylindrical. dome.\n\n• ### trapezium sub enclosure calculator?",
null,
"trapezium sub enclosure calculator? (too old to reply).anyone knows if there is a good tutorial + software calculator on designing a trapezium box? math is killing me...\n\n• ### Trapezium Calculator - Calculate Area and Perimeter...",
null,
"Trapezium Calculator. A quadrilateral with one set of parallel sides.Perimeter of Trapezium/Trapezoid:[(sum of all four sides)].\n\n• ### Dome calculation tools",
null,
"Trapezium calculation tool.Punch 5900mm radius into the 3v dome calculator to get an idea of what you need.\n\n• ### How to calculate the area of a trapezium using the formula and our calculator.",
null,
"Calculate Area Trapezium. Trapezium Surface Calculator. OTHER CALCULATORSWhat is the area of a trapezium or trapezoid?\n\n• ### Trapezoidal rule - Wikipedia, the free encyclopedia",
null,
"In numerical analysis, the trapezoidal rule (also known as the trapezoid rule or trapezium rule) is a technique for approximating the definite integral. The trapezoidal rule works by approximating the region under the graph of the function as a trapezoid and calculating its area. It follows that.\n\n• ### Area of a Trapezoid Calculator, Trapezoid Calculator...",
null,
"The total space inside the boundary of the trapezoid is known as the area of the trapezoid. Area is measured in terms of square unit. Area of a Trapezoid Calculator (or simply Trapezoid Calculator)...\n\n• ### How to Calculate the area of a trapezoid « Math",
null,
"How to Calculate faster than a calculator. How to Solve percentages without a calculator.How to Quickly find the area of a trapezoid. How to Calculate the area of irregular shapes.\n\n• ### Area Calculators - Triangle, Cone, Parallelogram, Rhombus, Trapezium, Circle...",
null,
"Area Calculators. Enter value, select unit and click on calculate. Result will be displayed.Calculate Area Of a Trapezium. Enter your values\n\n• ### Trapezoid Area Calculator | Calculate Trapezoid Area",
null,
"The Trapezoid Area Calculator will calculate the area of a trapezoid if you enter in the height, the length of the top, and the length of the top ofIn British English the trapezoid is called the trapezium.\n\n• ### Online calculator. Trapezoid perimeter.",
null,
"Trapezoid (Trapezium) - is quadrilateral with at least one pair of parallel sides.This free online calculator will help you to find the perimeter of a trapezoid.\n\n• ### dome shape? | Forum",
null,
"I tried out the trapezium dome calculator, and came to pretty much my only objection to the concept - the calculator gives the sizes required for a 16 segment dome, and the...\n\n• ### Geodesic dome calculator - Acidome.ru",
null,
"Geodesic dome calculator. How to use it →. Facebook group →.\n\n• ### This is also a reverse Trapezium calculator",
null,
"This is also a reverse Trapezium calculator. It will calculate any size of dome and display the lengths for each strut, total amount of material required...\n\n• ### Area of a Trapezoid Calculator",
null,
"The online Area of a Trapezoid Calculator is used to help you find the area of a trapezoid.The trapezoid is equivalent to the British definition of trapezium. Area of a Trapezoid Formula.\n\n• ### Trapezoid Calculator -- EndMemo",
null,
"Trapezoid Calculator. Parallel Side a» G-Force RPM Calculator. » Chemical Molecular Weight Calculator. » Mole, Moles to Grams Calculator.\n\n• ### CalcTool: Trapezium properties calculator",
null,
"Geometry (2d) online calculation: Trapezium properties - Area, angles, and other lengths.Trapezium properties. Main Index; Mathematics; Geometry (2d)."
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https://math.libretexts.org/Courses/Monroe_Community_College/MTH_104_Intermediate_Algebra/8%3A_Roots_and_Radicals/8.2%3A_Simplify_Expressions_with_Roots/8.2E%3A_Exercises | [
"$$\\newcommand{\\id}{\\mathrm{id}}$$ $$\\newcommand{\\Span}{\\mathrm{span}}$$ $$\\newcommand{\\kernel}{\\mathrm{null}\\,}$$ $$\\newcommand{\\range}{\\mathrm{range}\\,}$$ $$\\newcommand{\\RealPart}{\\mathrm{Re}}$$ $$\\newcommand{\\ImaginaryPart}{\\mathrm{Im}}$$ $$\\newcommand{\\Argument}{\\mathrm{Arg}}$$ $$\\newcommand{\\norm}{\\| #1 \\|}$$ $$\\newcommand{\\inner}{\\langle #1, #2 \\rangle}$$ $$\\newcommand{\\Span}{\\mathrm{span}}$$\n\n# 8.1E: Exercises\n\n$$\\newcommand{\\vecs}{\\overset { \\rightharpoonup} {\\mathbf{#1}} }$$ $$\\newcommand{\\vecd}{\\overset{-\\!-\\!\\rightharpoonup}{\\vphantom{a}\\smash {#1}}}$$$$\\newcommand{\\id}{\\mathrm{id}}$$ $$\\newcommand{\\Span}{\\mathrm{span}}$$ $$\\newcommand{\\kernel}{\\mathrm{null}\\,}$$ $$\\newcommand{\\range}{\\mathrm{range}\\,}$$ $$\\newcommand{\\RealPart}{\\mathrm{Re}}$$ $$\\newcommand{\\ImaginaryPart}{\\mathrm{Im}}$$ $$\\newcommand{\\Argument}{\\mathrm{Arg}}$$ $$\\newcommand{\\norm}{\\| #1 \\|}$$ $$\\newcommand{\\inner}{\\langle #1, #2 \\rangle}$$ $$\\newcommand{\\Span}{\\mathrm{span}}$$ $$\\newcommand{\\id}{\\mathrm{id}}$$ $$\\newcommand{\\Span}{\\mathrm{span}}$$ $$\\newcommand{\\kernel}{\\mathrm{null}\\,}$$ $$\\newcommand{\\range}{\\mathrm{range}\\,}$$ $$\\newcommand{\\RealPart}{\\mathrm{Re}}$$ $$\\newcommand{\\ImaginaryPart}{\\mathrm{Im}}$$ $$\\newcommand{\\Argument}{\\mathrm{Arg}}$$ $$\\newcommand{\\norm}{\\| #1 \\|}$$ $$\\newcommand{\\inner}{\\langle #1, #2 \\rangle}$$ $$\\newcommand{\\Span}{\\mathrm{span}}$$\n\n## Simplifying Expressions with Roots\n\nIn the following exercises, simplify.\n\n1. a. $$\\sqrt{64}$$ b. $$-\\sqrt{81}$$\n\na. $$8$$ b. $$-9$$\n\n2. a. $$\\sqrt{169}$$ b. $$-\\sqrt{100}$$\n\n3. a. $$\\sqrt{196}$$ b. $$-\\sqrt{1}$$\n\na. $$14$$ b. $$-1$$\n\n4. a. $$\\sqrt{144}$$ b. $$-\\sqrt{121}$$\n\n5. a. $$\\sqrt{\\frac{4}{9}}$$ b. $$-\\sqrt{0.01}$$\n\na. $$\\frac{2}{3}$$ b. $$-0.1$$\n\n6. a. $$\\sqrt{\\frac{64}{121}}$$ b. $$-\\sqrt{0.16}$$\n\n7. a. $$\\sqrt{-121}$$ b. $$-\\sqrt{289}$$\n\na. not a real number b. $$-17$$\n\n8. a. $$-\\sqrt{400}$$ b. $$\\sqrt{-36}$$\n\n9. a. $$-\\sqrt{225}$$ b. $$\\sqrt{-9}$$\n\na. $$-15$$ b. not a real number\n\n10. a. $$\\sqrt{-49}$$ b. $$-\\sqrt{256}$$\n\n11. a. $$\\sqrt{216}$$ b. $$\\sqrt{256}$$\n\na. $$6$$ b. $$4$$\n\n12. a. $$\\sqrt{27}$$ b. $$\\sqrt{16}$$ c. $$\\sqrt{243}$$\n\n13. a. $$\\sqrt{512}$$ b. $$\\sqrt{81}$$ c. $$\\sqrt{1}$$\n\na. $$8$$ b. $$3$$ b. $$1$$\n\n14. a. $$\\sqrt{125}$$ b. $$\\sqrt{1296}$$ c. $$\\sqrt{1024}$$\n\n15. a. $$\\sqrt{-8}$$ b. $$\\sqrt{-81}$$ c. $$\\sqrt{-32}$$\n\na. $$-2$$ b. not a real number c. $$-2$$\n\n16. a. $$\\sqrt{-64}$$ b. $$\\sqrt{-16}$$ c. $$\\sqrt{-243}$$\n\n17. a. $$\\sqrt{-125}$$ b. $$\\sqrt{-1296}$$ c. $$\\sqrt{-1024}$$\n\na. $$-5$$ b. not a real number c. $$-4$$\n\n18. a. $$\\sqrt{-512}$$ b. $$\\sqrt{-81}$$ c. $$\\sqrt{-1}$$\n\nIn the following exercises, estimate each root by giving the interval of two consecutive whole numbers in which the root lies.\n\n19. a. $$\\sqrt{70}$$ b. $$\\sqrt{71}$$\n\na. $$8<\\sqrt{70}<9$$ b. $$4<\\sqrt{71}<5$$\n\n20. a. $$\\sqrt{55}$$ b. $$\\sqrt{119}$$\n\n21. a. $$\\sqrt{200}$$ b. $$\\sqrt{137}$$\n\na. $$14<\\sqrt{200}<15$$ b. $$5<\\sqrt{137}<6$$\n\n22. a. $$\\sqrt{172}$$ b. $$\\sqrt{200}$$\n\nIn the following exercises, approximate each root and round to two decimal places.\n\n23. a. $$\\sqrt{19}$$ b. $$\\sqrt{89}$$ c. $$\\sqrt{97}$$\n\na. $$\\approx 4.36$$ b. $$\\approx 4.46$$ c. $$\\approx 3.14$$\n\n24. a. $$\\sqrt{21}$$ b. $$\\sqrt{93}$$ c. $$\\sqrt{101}$$\n\n25. a. $$\\sqrt{53}$$ b. $$\\sqrt{147}$$ c. $$\\sqrt{452}$$\n\na. $$\\approx 7.28$$ b. $$\\approx 5.28$$ c. $$\\approx 4.61$$\n\n26. a. $$\\sqrt{47}$$ b. $$\\sqrt{163}$$ c. $$\\sqrt{527}$$\n\n### Simplify Variable Expressions with Roots\n\nIn the following exercises, simplify using absolute values as necessary.\n\n27. a. $$\\sqrt{u^{5}}$$ b. $$\\sqrt{v^{8}}$$\n\na. $$u$$ b. $$|v|$$\n\n28. a. $$\\sqrt{a^{3}}$$ b. $$\\sqrt{b^{9}}$$\n\n29. a. $$\\sqrt{y^{4}}$$ b. $$\\sqrt{m^{7}}$$\n\na. $$|y|$$ b. $$m$$\n\n30. a. $$\\sqrt{k^{8}}$$ b. $$\\sqrt{p^{6}}$$\n\n31. a. $$\\sqrt{x^{6}}$$ b. $$\\sqrt{y^{16}}$$\n\na. $$|x^{3}|$$ b. $$y^{8}$$\n\n32. a. $$\\sqrt{a^{14}}$$ b. $$\\sqrt{w^{24}}$$\n\n33. a. $$\\sqrt{x^{24}}$$ b. $$\\sqrt{y^{22}}$$\n\na. $$x^{12}$$ b. $$|y^{11}|$$\n\n34. a. $$\\sqrt{a^{12}}$$ b. $$\\sqrt{b^{26}}$$\n\n35. a. $$\\sqrt{x^{9}}$$ b. $$\\sqrt{y^{12}}$$\n\na. $$x^{3}$$ b. $$|y^{3}|$$\n\n36. a. $$\\sqrt{a^{10}}$$ b. $$\\sqrt{b^{27}}$$\n\n37. a. $$\\sqrt{m^{8}}$$ b. $$\\sqrt{n^{20}}$$\n\na. $$m^{2}$$ b. $$n^{4}$$\n\n38. a. $$\\sqrt{r^{12}}$$ b. $$\\sqrt{s^{30}}$$\n\n39. a. $$\\sqrt{49 x^{2}}$$ b. $$-\\sqrt{81 x^{18}}$$\n\na. $$7|x|$$ b. $$-9|x^{9}|$$\n\n40. a. $$\\sqrt{100 y^{2}}$$ b. $$-\\sqrt{100 m^{32}}$$\n\n41. a. $$\\sqrt{121 m^{20}}$$ b. $$-\\sqrt{64 a^{2}}$$\n\na. $$11m^{10}$$ b. $$-8|a|$$\n\n42. a. $$\\sqrt{81 x^{36}}$$ b. $$-\\sqrt{25 x^{2}}$$\n\n43. a. $$\\sqrt{16 x^{8}}$$ b. $$\\sqrt{64 y^{12}}$$\n\na. $$2x^{2}$$ b. $$2y^{2}$$\n\n44. a. $$\\sqrt{-8 c^{9}}$$ b. $$\\sqrt{125 d^{15}}$$\n\n45. a. $$\\sqrt{216 a^{6}}$$ b. $$\\sqrt{32 b^{20}}$$\n\na. $$6a^{2}$$ b. $$2b^{4}$$\n\n46. a. $$\\sqrt{128 r^{14}}$$ b. $$\\sqrt{81 s^{24}}$$\n\n47. a. $$\\sqrt{144 x^{2} y^{2}}$$ b. $$\\sqrt{169 w^{8} y^{10}}$$ c. $$\\sqrt{8 a^{51} b^{6}}$$\n\na. $$12|x y|$$ b. $$13 w^{4}\\left|y^{5}\\right|$$ c. $$2 a^{17} b^{2}$$\n\n48. a. $$\\sqrt{196 a^{2} b^{2}}$$ b. $$\\sqrt{81 p^{24} q^{6}}$$ c. $$\\sqrt{27 p^{45} q^{9}}$$\n\n49. a. $$\\sqrt{121 a^{2} b^{2}}$$ b. $$\\sqrt{9 c^{8} d^{12}}$$ c. $$\\sqrt{64 x^{15} y^{66}}$$\n\na. $$11|ab|$$ b. $$3c^{4}d^{6}$$ c. $$4x^{5}y^{22}$$\n\n50. a. $$\\sqrt{225 x^{2} y^{2} z^{2}}$$ b. $$\\sqrt{36 r^{6} s^{20}}$$ c. $$\\sqrt{125 y^{18} z^{27}}$$\n\n### Writing Exercises\n\n51. Why is there no real number equal to $$\\sqrt{-64}$$?\n\nSince the square of any real number is positive, it's not possible for a real number to square to $$-64$$.\n\n52. What is the difference between $$9^{2}$$ and $$\\sqrt{9}$$?\n\n53. Explain what is meant by the $$n^{th}$$ root of a number.\n\nIf you raise this root to the $$n^{th}$$ power, it will give you back the original number (under the radical).\n\n54. Explain the difference of finding the $$n^{th}$$ root of a number when the index is even compared to when the index is odd.\n\n### Self Check\n\na. After completing the exercises, use this checklist to evaluate your mastery of the objectives of this section."
]
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http://cxh99.com/2021/07/17/78203.shtml | [
" 飞狐十年不败预警选股公式公式、代码源码-其他软件公式 －程序化交易（CXH99.COM）\n\n# 飞狐十年不败预警选股公式公式、代码源码[其他软件公式]\n\nV1赋值:10日内最低价的最低值\nV2赋值:20日内最高价的最高值\nLYF1赋值:((收盘价-V1)/(V2-V1))*((-100))的3日指数移动平均\nVARG5赋值:((收盘价)/(5日前的收盘价)>=1.15)\nVARG6赋值:若VARG5则将最近5周期置为1\n\nV1:=LLV(LOW,10);\nV2:=HHV(HIGH,20);\nlyf1:=EMA(((CLOSE-V1)/(V2-V1))*((-100)),3);\nVARg5:=((CLOSE)/(REF(CLOSE,5))>=1.15);\nVARg6:=BACKSET(VARg5,5);\n\nAND ((CLOSE)/(REF(CLOSE,1))>1.098) AND (REF(VOL,LOW)>REF(VOL,34))\nAND (REF(VOL,LOW)>REF(VOL,1)) AND (LYF1>(-93))\nAND (REF(VOL,LOW)>REF(VOL,8))),60,0);\ncross(入市日 ,(58));",
null,
"(注：由于人数限制，QQ或微信请选择方便的一个联系我们就行，加好友时请简单备注下您的需求，否则无法通过。谢谢您！)\n\n【字体： 】【打印文章】【查看评论\n\n没有相关内容"
]
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"http://cxh99.com/UploadFiles/Article/2012/12/201212012011408343.png",
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| {"ft_lang_label":"__label__zh","ft_lang_prob":0.76601213,"math_prob":0.9997433,"size":1083,"snap":"2021-43-2021-49","text_gpt3_token_len":804,"char_repetition_ratio":0.11399444,"word_repetition_ratio":0.0,"special_character_ratio":0.38504156,"punctuation_ratio":0.2081448,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9856722,"pos_list":[0,1,2],"im_url_duplicate_count":[null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-10-23T14:20:40Z\",\"WARC-Record-ID\":\"<urn:uuid:80efb9af-086e-453c-b118-d8c7ff067fb9>\",\"Content-Length\":\"42966\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:aa7a3256-d85f-4a04-bc01-7fb692e27ca7>\",\"WARC-Concurrent-To\":\"<urn:uuid:7274c0b7-ed95-41f9-8af6-1d41e92e1e51>\",\"WARC-IP-Address\":\"122.114.122.51\",\"WARC-Target-URI\":\"http://cxh99.com/2021/07/17/78203.shtml\",\"WARC-Payload-Digest\":\"sha1:KT6IOQGXQZ5LWFGTVBLFAJE4YU6C6VOX\",\"WARC-Block-Digest\":\"sha1:HHYTQV3NFNGVFEEWCH4CYPLRHLTWQZCD\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-43/CC-MAIN-2021-43_segments_1634323585696.21_warc_CC-MAIN-20211023130922-20211023160922-00107.warc.gz\"}"} |
https://www.gcflcm.com/lcm-of-69-and-76 | [
"# What is the Least Common Multiple of 69 and 76?\n\nLeast common multiple or lowest common denominator (lcd) can be calculated in two way; with the LCM formula calculation of greatest common factor (GCF), or multiplying the prime factors with the highest exponent factor.\n\nLeast common multiple (LCM) of 69 and 76 is 5244.\n\nLCM(69,76) = 5244\n\nLCM Calculator and\nand\n\n## Least Common Multiple of 69 and 76 with GCF Formula\n\nThe formula of LCM is LCM(a,b) = ( a × b) / GCF(a,b).\nWe need to calculate greatest common factor 69 and 76, than apply into the LCM equation.\n\nGCF(69,76) = 1\nLCM(69,76) = ( 69 × 76) / 1\nLCM(69,76) = 5244 / 1\nLCM(69,76) = 5244\n\n## Least Common Multiple (LCM) of 69 and 76 with Primes\n\nLeast common multiple can be found by multiplying the highest exponent prime factors of 69 and 76. First we will calculate the prime factors of 69 and 76.\n\n### Prime Factorization of 69\n\nPrime factors of 69 are 3, 23. Prime factorization of 69 in exponential form is:\n\n69 = 31 × 231\n\n### Prime Factorization of 76\n\nPrime factors of 76 are 2, 19. Prime factorization of 76 in exponential form is:\n\n76 = 22 × 191\n\nNow multiplying the highest exponent prime factors to calculate the LCM of 69 and 76.\n\nLCM(69,76) = 31 × 231 × 22 × 191\nLCM(69,76) = 5244"
]
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.8593294,"math_prob":0.99726254,"size":1200,"snap":"2021-31-2021-39","text_gpt3_token_len":377,"char_repetition_ratio":0.17809364,"word_repetition_ratio":0.09821428,"special_character_ratio":0.36,"punctuation_ratio":0.104,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9999976,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-08-06T01:01:42Z\",\"WARC-Record-ID\":\"<urn:uuid:a1a84b59-7db0-439a-8067-7dad92aed926>\",\"Content-Length\":\"21741\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:e2aded8f-af6a-4494-88c0-a0ff39cf3e9d>\",\"WARC-Concurrent-To\":\"<urn:uuid:934bdf99-c26a-4428-aec4-57f873960fb3>\",\"WARC-IP-Address\":\"104.154.21.107\",\"WARC-Target-URI\":\"https://www.gcflcm.com/lcm-of-69-and-76\",\"WARC-Payload-Digest\":\"sha1:NRAFB6W7GLZ6VSIJ5M5ACBLZSFQEHGDR\",\"WARC-Block-Digest\":\"sha1:KZ2OMRGS776C7GVZ5VJXJJVSAEWADFEC\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-31/CC-MAIN-2021-31_segments_1627046152085.13_warc_CC-MAIN-20210805224801-20210806014801-00553.warc.gz\"}"} |
https://codereview.stackexchange.com/questions/26333/trying-to-improve-a-working-regex/26334 | [
"Trying to improve a working regex\n\nI am extracting data from a text file. Some of the lines from which I want to extract the data consist of a text description with single spaces, followed by a multiple-space gap preceding four fields containing the data, each separated by multiple spaces. A field might either contain the indicator \"N/A\" or else it will contain an integer < 10,000 (possibly comma-ed) such as 15 or 7,151 followed by a valid percentage in parentheses. The percentage will always have a single decimal point; for example, (0.0%) or (19.8%) or (100.0%) . If the field contains \"N/A\", then I want to write out \"NA,NA\" and if the field contains a number and percentage, then I want to write those two values out separated by a comma.\n\nAt the moment, I use the following regex to describe a single field:\n\n$naNumberGroup = qr/(N\\/A|(([0-9]{1,3}(,[0-9]{3})*) $$([0-9]{1,2}\\.[0-9])\\%$$))/ and then the following code to get the various pieces from the current line, which is$line :\n\nif( $line =~ m/$naNumberGroup +$naNumberGroup +$naNumberGroup +$naNumberGroup/ ) { if(\"N/A\" eq$1) {\nprint \"NA,NA\";\n} else {\nprint \",$3,$5\";\n}\n\nif(\"N/A\" eq $6) { print \",NA,NA\"; } else { print \",$8,$10\"; } if(\"N/A\" eq$11) {\nprint \",NA,NA\";\n} else {\nprint \",$13,$15\";\n}\n\nif(\"N/A\" eq $16) { print \",NA,NA\"; } else { print \",$18,$20\"; } print \"\\n\"; } It seems horribly clumsy; for example, it's easy to make a mistake in counting the parentheses and getting the pairs of fields correctly referenced ... but I am unsure of even what sorts of things I should be looking at to improve it (assuming that's possible). I would appreciate some guidance or comment. Even, \"it seems fine\" would at least indicate that I shouldn't waste time on improving it! An example line of text is: Adults who actively pursue work opportunities 197 (82.8%) 30 (12.6%) N/A N/A The description at the beginning of the line changes depending on the data. The output that I want for this line is: 197,82.8,30,12,6,NA,NA,NA,NA Similarly, if the line were: Adults who actively pursue work opportunities 197 (82.8%) N/A 30 (12.6%) N/A then I want the output: 197,82.8,NA,NA,30,12.6,NA,NA migrated from stackoverflow.comMay 19 '13 at 8:34 This question came from our site for professional and enthusiast programmers. • Can you provide sample input and expected output ? Also a single improvement to make your regex shorter would be to use \\d instead of [0-9]. – HamZa DzCyberDeV May 18 '13 at 11:14 • Can't you just split the string by /\\s{2,}+(?!$$)/ pattern, then work with fields as normal arrays? As for groups mismatch, well, you can use named capture groups (with ?<name> notation). – raina77ow May 18 '13 at 11:18 2 Answers Using \"split\" is an option as other people have mentioned. However, using a regex has the added benefit of validating the input data while parsing, so is still worth considering depending on your use-case. A regex is better used in loop here since we're matching the same pattern repeatedly. And you should use non-capturing parentheses for the bits you're not interested in capturing. E.g. changing nothing else your code would look like this: naNumberGroup = qr/(N\\/A|(?:([0-9]{1,3}(?:,[0-9]{3})*) \\(([0-9]{1,2}\\.[0-9])\\%$$))/; my @outFields; while ($line =~ m/\\s\\s+$naNumberGroup/g) { if(\"N/A\" eq$1) {\npush @outFields, 'NA', 'NA';\n} else {\npush @outFields, $2,$3;\n}\n}\nprint join(',', @outfields),\"\\n\";\n\nIt's worth noting that your code as-is would preserve any commas in the input, therefore breaking your output. And \"100.0%\" isn't handled.\n\nIf you're wanting to improve readability and maintainability of your regexes, here are some additional things worth changing:\n\n1. Use the /x modifier to improve readability/maintainability.\n2. Use more intermediate variables to build your regexes.\n3. Avoid having to escape slashes by using qr{...} instead of qr/.../\n\nE.g.\n\nmy $numberGroup = qr{ (?<number> [0-9,]+ ) # Number with optional commas [ ] # Single space $$(?<percent> [0-9]+\\.[0-9] ) %$$ # Percentage in parens }x; my$naNumberGroup = qr{\n[ ]{2,} # Two or more spaces\n(?: $numberGroup | N/A ) # No need to capture \"N/A\" }x; my @outFields; while ($line =~ m/$naNumberGroup/xg) { my$number = $+{number} // 'NA'; my$percent = $+{percent} // 'NA';$number =~ tr/,//d; # Strip commas\npush @outFields, $number,$percent;\n}\nif (scalar @outFields == 8) {\nprint join(',', @outFields),\"\\n\";\n} else {\n# Description line, or invalid line. You may be able to use\n# another regex to determine which.\n}\n• That's what longer regex should always look like! btw, % doesn't need backslashing. – mpapec May 18 '13 at 20:49\n• Thanks - good point re % - I've updated the answer. – Simon Poole May 18 '13 at 20:55\n• Many thanks for all the suggestions. I knew about the 'split' command but have never used it, so the suggestions about how to use it in this context are very educational as well as directly helpful. The suggestions about how to improve the formulation and use of the regex itself are more along the lines of my original expectation, and also very helpful. Thanks again. – user02814 May 19 '13 at 5:51\n\nSplitting is much better solution than regex, as someone already mentioned.\n\nmy $line = \"197 (82.8%) N/A 30 (12.6%) N/A\"; my$result =\njoin \",\",\nmap {\ntr|()%||d;\n$_ eq \"N/A\" ? qw(NA NA) :$_;\n}\nsplit /\\s+/, $line; print \"$result\\n\";\n\ngives\n\n197,82.8,NA,NA,30,12.6,NA,NA\n• Nice solution - simplifying the problem to: delete excessive characters and replace N/A to NA NA. Really nice. – jm666 May 18 '13 at 18:16"
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.8795792,"math_prob":0.9217845,"size":2383,"snap":"2019-43-2019-47","text_gpt3_token_len":704,"char_repetition_ratio":0.1063472,"word_repetition_ratio":0.048469387,"special_character_ratio":0.34158623,"punctuation_ratio":0.18148148,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.96748847,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-10-17T09:43:57Z\",\"WARC-Record-ID\":\"<urn:uuid:290998f8-fafb-4d05-863c-533aa2cee058>\",\"Content-Length\":\"150839\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:ca08f31b-388a-4512-bfef-aa7b331deac3>\",\"WARC-Concurrent-To\":\"<urn:uuid:17c6a2fd-4e2b-46b0-95b0-4e91c85f8c63>\",\"WARC-IP-Address\":\"151.101.129.69\",\"WARC-Target-URI\":\"https://codereview.stackexchange.com/questions/26333/trying-to-improve-a-working-regex/26334\",\"WARC-Payload-Digest\":\"sha1:ML3HJXGJXZOILLERM34XMSBRJE4SRMFK\",\"WARC-Block-Digest\":\"sha1:GZEIJ3NYANWWQ2X7XFKR3NTLXVJXZDCB\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-43/CC-MAIN-2019-43_segments_1570986673250.23_warc_CC-MAIN-20191017073050-20191017100550-00307.warc.gz\"}"} |
http://seaperch.org/page-resources/how-things-work-how-things-float/ | [
"# How Things Float\n\nHow Things Work series\n\nLast Updated: September 2020",
null,
"Overview\n\nWhy do things float or sink? Objects float when they are positively buoyant, or less dense than the fluid in which they are immersed. Archimedes Principle can explain this phenomenon – “When an object is immersed fully or partially in a fluid, it experiences an upward force that is equal to the weight of the fluid displaced by it.” Objects need to have a greater ratio of empty space to mass than the fluid. Basic understanding of buoyancy, density, fluids, and the Archimedes Principle are necessary in order to understand how things float.",
null,
"Buoyancy\n\nBuoyancy is a force that moves an object upward. This upward force occurs when the object is immersed (either fully or partially) in a fluid that has measurable density. The fluid can be a liquid or a gas (not a solid). Buoyant force is measured in Newtons (N) in the International System of Units (SI).",
null,
"Types of Buoyancy",
null,
"",
null,
"1. NEGATIVE BUOYANCY – An object is negatively buoyant when it is denser than the fluid it displaces. The object will sink because its weight is greater than the buoyant force.\n2. NEUTRAL BUOYANCY - An object is neutrally buoyant when its density is equal to the density of the fluid in which it is immersed, resulting in the buoyant force balancing the force of gravity that would otherwise cause the object to sink or rise. An object that has neutral buoyancy will neither sink nor rise.\n3. POSITIVE BUOYANCY – An object is positively buoyant when it is lighter than the fluid it displaces. The object will float because the buoyant force is greater than the object’s weight.",
null,
"Density\n\nDensity is a measure of how much matter occupies a given amount of space. High density substances have tightly packed particles and low density substances are made up of loosely packed particles.",
null,
"The density, of a substance is its mass per unit volume. The symbol most often used for density is ρ, the Greek symbol Rho. Mathematically, density is defined as mass divided by volume: where ρ is the density, m is the mass, and v is the volume.\n\nρ = m / v\n\nCompare a sponge and a rock of the same size. The rock is more dense because the sponge has hundreds of holes in the same volume of space. In the case of the sponge, the mass of the sponge as well as the air that fills the holes of the sponge is measured. Using this same principal, a boat hull full of air will float, but a boat hull full of water will sink. The density of the air-filled boat is much less than a boat hull filled with water.",
null,
"Fluids\n\nFluids are a state of matter, such as liquid or gas, in which the component particles (generally molecules) can move past one another. Fluids flow easily and conform to the shape of their containers. It is common to describe liquids (water, juice, coolant) as fluids, and gases (such air, nitrogen, propane) as a gas. However, liquids and gases share common properties, such as compressibility, density, pressure, buoyancy and viscosity, and they are scientifically classified together as fluids.",
null,
"Archimedes Principle\n\nArchimedes’ principle states that the upward buoyant force that is exerted on a body immersed in a fluid, whether fully or partially submerged, is equal to the weight of the fluid that the body displaces.\n\nMass out of fluid – Mass in fluid = Density of fluid * Volume of solid\n\nBuoyancy (N) = ρ * v\n\n1st part: When an object is immersed fully or partially in a fluid, it experiences an upward force that can be seen and felt.",
null,
"",
null,
"2nd part: The upward force is equal to the weight of the fluid displaced by the object.",
null,
"",
null,
"Archimedes, a famous Greek scientist, was given the task of determining whether the king’s golden wreath, a gift to the gods, was made of real gold or some alloy. Archimedes wanted to melt the gold material into a known volume and weigh it, thus determining the density. But the king would not allow this. Archimedes thought about this for some time, until one day while bathing, he noticed the water level increased as he entered the tub, proportionally to his volume. This meant that he could put the difficult-to-measure crown in water and instantly know the volume of the crown by measuring the displacement of water. At this point, Archimedes ran through the streets shouting “Eureka! Eureka!”\n\nArchimedes determined that by measuring the weight of the water increase when the wreath was submerged, he was actually measuring the buoyant force acting on the wreath. And from this, he could divide by the density of water, and find the volume of the crown. Fun Fact: Buoy = float (in Latin).",
null,
"Vocabulary\n\nArchimedes Principle: The relationship between buoyancy and displaced fluid: An immersed object is buoyed up by a force equal to the weight of the fluid it displaces.\n\nBuoyancy: An upward force acting on an immersed or floating object by the supporting fluid.\n\nEquilibrium: A state of balance between opposing forces.\n\nForce: A push or a pull. An influence on a body or system, causing or tending to cause a change in movement or shape.\n\nMass: The amount of matter that is contained by an object.\n\nNegative Buoyancy: Exists when the weight of the body is greater than the weight of an equal volume of the displaced fluid. The body sinks.\n\nNeutral Buoyancy: Exists when the weight of the body is equal to the weight of an equal volume of the displaced fluid. The body remains suspended – neither rising nor sinking – unless acted upon by an outside force.\n\nNewton (N): The derived unit used to measure force in the International System of Units (SI). One newton is equal to the force needed to accelerate a mass of one kilogram one meter per second per second.\n\nPositive Buoyancy: Exists when the weight of the body is less than the weight of an equal volume of the displaced fluid.\n\nPressure: The force per unit of surface area; exerted perpendicular to the surface; measured in Pascals.\n\nVolume: The amount of space enclosed by a shape or object; how much 3-dimensional space (length, width, and height) it occupies.\n\nWeight: The downward force caused by gravity on an object.",
null,
"1. A dog jumps into a pool. The water holds him up because he only weighs 10N in water. On land, he weighs 45N. What is the buoyant force?\n2. A person weighs 250N while swimming in the Dead Sea. When outside of the water, they weigh 600N. What is the buoyant force acting on them? Will they sink or float?\n3. A small fish weighing 27 g displaced a volume of water that weighed 25N. the weight of the fish was 30N. What is the buoyant force on the fish and will it sink or float?\n4. A rock has a mass of 76 g and displaced 80 mL of water when it was submerged. What is the volume of the rock?",
null,
"",
null,
"Resources\n\nVideos:\n\nHow Submarines Work (How Stuff Works): Provides a background on submarines from how they dive and surface to how they maintain hospitable conditions for the people who live and work on the vessel.\n\nMythbusters: “Let’s Talk Buoyancy” (Discovery Go): A video where the Mythbusters explain buoyancy and relate it to a pirate myth of turning a rowboat into a submarine.\n\nBuoyancy Playground (University of Colorado – Boulder): A buoyancy simulator supported by the University of Colorado Boulder that allows you to manipulate different experimental parameters including mass, volume, and density of the object and the fluid. The simulator provides quantitative measurements and allows you to overlay force vectors onto the simulation.\n\nUnderwater Exploration (National Geographic): A National Geographic website that provides information on the many expeditions and discoveries made by National Geographic explorers. Includes articles, videos, and other multimedia sources.\n\nFloat or Sink (Learning Junction – YouTube): A video on the concept of floating or sinking (density) for young children.\n\nBuoyancy: What Makes Something Float or Sink (Kids Want to Know – YouTube): A short overview video about buoyancy.\n\nLesson Plans:\n\nExploring Buoyancy – Design Challenge Learning (The Tech: Museum of Innovation)\n\nLesson: Estimating Buoyancy (University of Colorado – Teach Engineering STEM Curriculum for K-12)\n\nLesson: Archimedes’ Principle, Pascal’s Law and Bernoulli’s Principle (University of Colorado – Teach Engineering STEM Curriculum for K-12)",
null,
"",
null,
"## Related Resources",
null,
"#### Electricity\n\nThis resource is a part of the How Things Work series.",
null,
"#### Electric Motors\n\nThis resource is a part of the How Things Work series.",
null,
"#### Microcontrollers\n\nThis resource is a part of the How Things Work series.",
null,
"#### PVC\n\nThis resource is a part of the How Things Work series.",
null,
"#### Relays\n\nThis resource is a part of the How Things Work series.",
null,
"#### Switches\n\nThis resource is a part of the How Things Work series."
]
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null,
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https://link.springer.com/article/10.1140/epjc/s10052-016-4582-y | [
"First search for dark matter annihilations in the Earth with the IceCube detector\n\nIceCube Collaboration\n• M. G. Aartsen\n• K. Abraham\n• M. Ackermann\n• J. A. Aguilar\n• M. Ahlers\n• M. Ahrens\n• D. Altmann\n• K. Andeen\n• T. Anderson\n• I. Ansseau\n• G. Anton\n• M. Archinger\n• C. Argüelles\n• J. Auffenberg\n• S. Axani\n• X. Bai\n• S. W. Barwick\n• V. Baum\n• R. Bay\n• J. J. Beatty\n• J. Becker Tjus\n• K.-H. Becker\n• S. BenZvi\n• D. Berley\n• E. Bernardini\n• A. Bernhard\n• D. Z. Besson\n• G. Binder\n• D. Bindig\n• M. Bissok\n• E. Blaufuss\n• S. Blot\n• C. Bohm\n• M. Börner\n• F. Bos\n• D. Bose\n• S. Böser\n• O. Botner\n• J. Braun\n• L. Brayeur\n• H.-P. Bretz\n• S. Bron\n• A. Burgman\n• T. Carver\n• M. Casier\n• E. Cheung\n• D. Chirkin\n• A. Christov\n• K. Clark\n• L. Classen\n• S. Coenders\n• G. H. Collin\n• D. F. Cowen\n• R. Cross\n• M. Day\n• J. P. A. M. de André\n• C. De Clercq\n• E. del Pino Rosendo\n• H. Dembinski\n• S. De Ridder\n• P. Desiati\n• K. D. de Vries\n• G. de Wasseige\n• M. de With\n• T. DeYoung\n• J. C. Díaz-Vélez\n• V. di Lorenzo\n• H. Dujmovic\n• J. P. Dumm\n• M. Dunkman\n• B. Eberhardt\n• T. Ehrhardt\n• B. Eichmann\n• P. Eller\n• S. Euler\n• P. A. Evenson\n• S. Fahey\n• A. R. Fazely\n• J. Feintzeig\n• J. Felde\n• K. Filimonov\n• C. Finley\n• S. Flis\n• C.-C. Fösig\n• A. Franckowiak\n• E. Friedman\n• T. Fuchs\n• T. K. Gaisser\n• J. Gallagher\n• L. Gerhardt\n• K. Ghorbani\n• W. Giang\n• M. Glagla\n• T. Glauch\n• T. Glüsenkamp\n• A. Goldschmidt\n• G. Golup\n• J. G. Gonzalez\n• D. Grant\n• Z. Griffith\n• C. Haack\n• A. Haj Ismail\n• A. Hallgren\n• F. Halzen\n• E. Hansen\n• B. Hansmann\n• T. Hansmann\n• K. Hanson\n• D. Hebecker\n• D. Heereman\n• K. Helbing\n• R. Hellauer\n• S. Hickford\n• J. Hignight\n• G. C. Hill\n• K. D. Hoffman\n• R. Hoffmann\n• K. Holzapfel\n• K. Hoshina\n• F. Huang\n• M. Huber\n• K. Hultqvist\n• S. In\n• A. Ishihara\n• E. Jacobi\n• G. S. Japaridze\n• M. Jeong\n• K. Jero\n• B. J. P. Jones\n• M. Jurkovic\n• A. Kappes\n• T. Karg\n• A. Karle\n• U. Katz\n• M. Kauer\n• A. Keivani\n• J. L. Kelley\n• J. Kemp\n• A. Kheirandish\n• M. Kim\n• T. Kintscher\n• J. Kiryluk\n• T. Kittler\n• S. R. Klein\n• G. Kohnen\n• R. Koirala\n• H. Kolanoski\n• R. Konietz\n• L. Köpke\n• C. Kopper\n• S. Kopper\n• D. J. Koskinen\n• M. Kowalski\n• K. Krings\n• M. Kroll\n• G. Krückl\n• C. Krüger\n• J. Kunnen",
null,
"• S. Kunwar\n• N. Kurahashi\n• T. Kuwabara\n• M. Labare\n• J. L. Lanfranchi\n• M. J. Larson\n• F. Lauber\n• D. Lennarz\n• M. Lesiak-Bzdak\n• M. Leuermann\n• J. Leuner\n• L. Lu\n• J. Lünemann\n• G. Maggi\n• K. B. M. Mahn\n• S. Mancina\n• M. Mandelartz\n• R. Maruyama\n• K. Mase\n• R. Maunu\n• F. McNally\n• K. Meagher\n• M. Medici\n• M. Meier\n• A. Meli\n• T. Menne\n• G. Merino\n• T. Meures\n• S. Miarecki\n• L. Mohrmann\n• T. Montaruli\n• M. Moulai\n• R. Nahnhauer\n• U. Naumann\n• G. Neer\n• H. Niederhausen\n• S. C. Nowicki\n• D. R. Nygren\n• A. Obertacke Pollmann\n• A. Olivas\n• T. Palczewski\n• H. Pandya\n• D. V. Pankova\n• P. Peiffer\n• Ö. Penek\n• J. A. Pepper\n• C. Pérez de los Heros\n• D. Pieloth\n• E. Pinat\n• P. B. Price\n• G. T. Przybylski\n• M. Quinnan\n• C. Raab\n• L. Rädel\n• M. Rameez\n• K. Rawlins\n• R. Reimann\n• B. Relethford\n• M. Relich\n• E. Resconi\n• W. Rhode\n• M. Richman\n• B. Riedel\n• S. Robertson\n• M. Rongen\n• C. Rott\n• T. Ruhe\n• D. Ryckbosch\n• D. Rysewyk\n• L. Sabbatini\n• S. E. Sanchez Herrera\n• A. Sandrock\n• J. Sandroos\n• S. Sarkar\n• K. Satalecka\n• M. Schimp\n• P. Schlunder\n• T. Schmidt\n• S. Schoenen\n• S. Schöneberg\n• L. Schumacher\n• D. Seckel\n• S. Seunarine\n• D. Soldin\n• M. Song\n• G. M. Spiczak\n• C. Spiering\n• M. Stahlberg\n• T. Stanev\n• A. Stasik\n• J. Stettner\n• A. Steuer\n• T. Stezelberger\n• A. Stößl\n• R. Ström\n• N. L. Strotjohann\n• G. W. Sullivan\n• M. Sutherland\n• H. Taavola\n• J. Tatar\n• F. Tenholt\n• S. Ter-Antonyan\n• A. Terliuk\n• G. Tešić\n• S. Tilav\n• P. A. Toale\n• M. N. Tobin\n• S. Toscano\n• D. Tosi\n• M. Tselengidou\n• A. Turcati\n• E. Unger\n• M. Usner\n• J. Vandenbroucke\n• N. van Eijndhoven\n• S. Vanheule\n• M. van Rossem\n• J. van Santen\n• J. Veenkamp\n• M. Vehring\n• M. Voge\n• E. Vogel\n• M. Vraeghe\n• C. Walck\n• A. Wallace\n• M. Wallraff\n• N. Wandkowsky\n• Ch. Weaver\n• M. J. Weiss\n• C. Wendt\n• S. Westerhoff\n• B. J. Whelan\n• S. Wickmann\n• K. Wiebe\n• C. H. Wiebusch\n• L. Wille\n• D. R. Williams\n• L. Wills\n• M. Wolf\n• T. R. Wood\n• E. Woolsey\n• K. Woschnagg\n• D. L. Xu\n• X. W. Xu\n• Y. Xu\n• J. P. Yanez\n• G. Yodh\n• S. Yoshida\n• M. Zoll\nOpen Access\nRegular Article - Experimental Physics\n\nAbstract\n\nWe present the results of the first IceCube search for dark matter annihilation in the center of the Earth. Weakly interacting massive particles (WIMPs), candidates for dark matter, can scatter off nuclei inside the Earth and fall below its escape velocity. Over time the captured WIMPs will be accumulated and may eventually self-annihilate. Among the annihilation products only neutrinos can escape from the center of the Earth. Large-scale neutrino telescopes, such as the cubic kilometer IceCube Neutrino Observatory located at the South Pole, can be used to search for such neutrino fluxes. Data from 327 days of detector livetime during 2011/2012 were analyzed. No excess beyond the expected background from atmospheric neutrinos was detected. The derived upper limits on the annihilation rate of WIMPs in the Earth and the resulting muon flux are an order of magnitude stronger than the limits of the last analysis performed with data from IceCube’s predecessor AMANDA. The limits can be translated in terms of a spin-independent WIMP–nucleon cross section. For a WIMP mass of 50 GeV this analysis results in the most restrictive limits achieved with IceCube data.\n\nKeywords\n\nDark Matter Atmospheric Neutrino Annihilation Rate Dark Matter Annihilation Muon Flux\nThese keywords were added by machine and not by the authors. This process is experimental and the keywords may be updated as the learning algorithm improves.\n\n1 Introduction\n\nA large number of observations, like rotation curves of galaxies and the cosmic microwave background temperature anisotropies, suggests the existence of an unknown component of matter , commonly referred to as dark matter. However, despite extensive experimental efforts, no constituents of dark matter have been discovered yet. A frequently considered dark matter candidate is a Weakly Interacting Massive Particle . Different strategies are pursued to search for these particles: at colliders, dark matter particles could be produced , in direct detection experiments, nuclear recoils from a massive target could be observed [4, 5, 6, 7], and indirect detection experiments search for a signal of secondary particles produced by self-annihilating dark matter [8, 9, 10, 11, 12].\n\nGamma-ray telescopes provide very strong constraints on the thermally averaged annihilation cross section from observations of satellite dwarf spheroidal galaxies . However, neutrinos are the only messenger particles that can be used to probe for dark matter in close-by massive baryonic bodies like the Sun or the Earth. In these objects dark matter particles from the Galactic halo can be accumulated after becoming bound in the gravitational potential of the Solar system as it passes through the Galaxy . The WIMPs may then scatter weakly on nuclei in the celestial bodies and lose energy. Over time, this leads to an accumulation of dark matter in the center of the bodies. The accumulated dark matter may then self-annihilate at a rate that is proportional to the square of its density, generating a flux of neutrinos with a spectrum that depends on the annihilation channel and WIMP mass. The annihilation would also contribute to the energy deposition in the Earth. A comparison of the expected energy deposition with the measured heat flow allows one to exclude strongly interacting dark matter .\n\nThe expected neutrino event rates and energies depend on the specific nature of dark matter, its local density and velocity distribution, and the chemical composition of the Earth. Different scenarios yield neutrino-induced muon fluxes between $$10^{-8}$$ and $$10^5$$ per km$$^2$$ per year for WIMPs with masses in the GeV–TeV range . The large uncertainty on the neutrino flux leaves a discovery potential for the searches for dark matter. The AMANDA [17, 18] and Super-K collaborations have already ruled out muon fluxes above $${\\sim }10^3$$ per km$$^2$$ per year for masses larger than some 100 GeV. The ANTARES collaboration has recently presented the results of a similar search using 5 years of data . The possibility of looking for even smaller fluxes with the much bigger IceCube neutrino observatory motivates the continued search for neutrinos coming from WIMP annihilations in the center of the Earth. This search is sensitive to the spin-independent WIMP–nucleon cross section and complements IceCube searches for dark matter in the Sun , the Galactic center and halo and in dwarf spheroidal galaxies .\n\n2 The IceCube Neutrino Telescope\n\nThe IceCube telescope, situated at the geographic South Pole, is designed to detect the Cherenkov radiation produced by high energy neutrino-induced charged leptons traveling through the detector volume. By recording the number of Cherenkov photons and their arrival times, the direction and energy of the charged lepton, and consequently that of the parent neutrino, can be reconstructed.\n\nIceCube consists of approximately 1 km$$^3$$ volume of ice instrumented with 5160 digital optical modules (DOMs) in 86 strings, deployed between 1450 and 2450 m depth . Each DOM contains a 25.3 cm diameter Hamamatsu R7081-02 photomultiplier tube connected to a waveform recording data acquisition circuit. The inner strings at the center of IceCube comprise DeepCore , a more densely instrumented sub-array equipped with higher quantum efficiency DOMs.\n\nWhile the large ice overburden above the detector provides a shield against downward going, cosmic ray induced muons with energies $$\\lesssim$$ 500 GeV at the surface, most analyses focus on upward going neutrinos employing the entire Earth as a filter. Additionally, low energy analyses use DeepCore as the fiducial volume and the surrounding IceCube strings as an active veto to reduce penetrating muon backgrounds. The search for WIMP annihilation signatures at the center of the Earth takes advantage of these two background rejection techniques as the expected signal will be vertically up-going and of low energy.\n\n3 Neutrinos from dark matter annihilations in the center of the Earth\n\nWIMPs annihilating in the center of the Earth will produce a unique signature in IceCube as vertically up-going muons. The number of detected neutrino-induced muons depends on the WIMP annihilation rate $$\\Gamma _A$$. If the capture rate C is constant in time t, $$\\Gamma _A$$ is given by \n\\begin{aligned} \\Gamma _A = \\frac{C}{2} \\tanh ^2\\left( \\frac{t}{\\tau }\\right) , \\quad \\tau = \\left( CC_A\\right) ^{-1/2}. \\end{aligned}\n(1)\nThe equilibrium time $$\\tau$$ is defined as the time when the annihilation rate and the capture rate are equal. $$C_A$$ is a constant depending on the WIMP number density. For the Earth, the equilibrium time is of the order of $$10^{11}$$ years if the spin-independent WIMP–nucleon cross section is $$\\sigma _{\\chi -N}^\\text {SI} \\sim 10^{-43}\\, \\mathrm{cm}^2$$ . The age of the Solar system is $$t_\\circ \\approx 4.5\\times 10^{9}$$ years and so $$t_\\circ /\\tau \\ll 1$$. We thus expect that $$\\Gamma _A \\propto C^2$$, i.e. the higher the capture rate, the higher the annihilation rate and thus the neutrino-induced muon flux.",
null,
"Fig. 1 Rate at which dark matter particles are captured to the interior of the Earth for a scattering cross section of $$\\sigma _\\mathrm{SI} = 10^{-44}$$ cm$$^2$$. The peaks correspond to resonant capture on the most abundant elements in the Earth : $$^{56}$$Fe, $$^{16}$$O, $$^{28}$$Si and $$^{24}$$Mg and their isotopes\n\nThe rate at which WIMPs are captured in the Earth depends on their mass (which is unknown), their velocity in the halo (which cannot be measured observationally, and therefore needs to be estimated through simulations) and their local density (which can be estimated from observations). The exact value of the local dark matter density is still under debate , with estimations ranging from $${\\sim } 0.2$$ to $${\\sim }0.5~\\mathrm{GeV}/\\mathrm {cm}^3$$. We take a value of 0.3 GeV/cm$$^3$$ as suggested in for the results presented in this paper in order to compare to the results of other experiments. If the WIMP mass is nearly identical to that of one of the nuclear species in the Earth, the capture rate will increase considerably, as shown in Fig. 1.\n\nThe capture rate could be higher if the velocity distribution of WIMPs with respect to the Earth is lower, as only dark matter with lower velocities can be captured by the Earth. The velocity distribution of dark matter in the halo is uncertain, as it is very sensitive to theoretical assumptions. The simplest halo model is the Standard Halo Model (SHM), a smooth, spherically symmetric density component with a non-rotating Gaussian velocity distribution . Galaxy formation simulations indicate, however, that additional macrostructural components, like a dark disc [35, 36, 37], are required. This would affect the velocity distribution, especially at low velocities, and, consequently, the capture rate in the Earth.\n\nThe signal simulations that are used in the analysis are performed using WimpSim , which describes the capture and annihilation of WIMPs inside the Earth, collects all neutrinos that emerge and lets these propagate through the Earth to the detector. The code includes neutrino interactions and neutrino oscillations in a complete three-flavor treatment. Eleven benchmark masses between 10 GeV and 10 TeV were simulated for different annihilation channels: the annihilation into $$b\\bar{b}$$ leads to a soft neutrino energy spectrum, while a hard channel is defined by the annihilation into $$W^+W^-$$ for WIMP masses larger than the rest mass of the W bosons and annihilation into $$\\tau ^+\\tau ^-$$ for lower WIMP masses.\n\n4 Background\n\nAs signal neutrinos originate near the center of the Earth, they induce a vertically up-going signal in the detector. This is, however, a special direction in the geometry of IceCube, as the strings are also vertical. While in other point source searches, a signal-free control region of the same detector acceptance can be defined by changing the azimuth, this is not possible for an Earth WIMP analysis. Consequently, a reliable background estimate can only be derived from simulation.\n\nTwo types of background have to be taken into account: the first type consists of atmospheric muons produced by cosmic rays in the atmosphere above the detector. Although these particles enter the detector from above, a small fraction will be reconstructed incorrectly as up-going. The cosmic ray interactions in the atmosphere that produce these particles are simulated by CORSIKA .\n\nThe second type of background consists of atmospheric neutrinos. This irreducible background is coming from all directions and is simulated with GENIE for neutrinos with energies below 190 GeV and with NuGeN for higher energies.\n\n5 Event selection\n\nThis analysis used the data taken in the first year of the fully deployed detector (from May 2011 to May 2012) with a livetime of 327 days. During the optimization of the event selection, only 10% of the complete dataset was used to check the agreement with the simulations. The size of this dataset is small enough to not reveal any potential signal, and hence allows us to maintain statistical blindness.\n\nTo be sensitive to a wide range of WIMP masses, the analysis is split into two parts that are optimized separately. The high energy event selection aims for an optimal sensitivity for WIMP masses of 1 TeV and the $$\\chi \\chi \\rightarrow W^+W^-$$ channel. The event selection for the low energy part is optimized for 50 GeV WIMPs annihilating into tau leptons. Because the capture rate for WIMPs of this mass shows a maximum (see Fig. 1), the annihilation and thus the expected neutrino rate are also maximal. As the expected neutrino energy for 50 GeV WIMPs is lower than 50 GeV, the DeepCore detector is crucial in this part of the analysis. Both samples are analyzed for the hard and the soft channel.\n\nThe data are dominated by atmospheric muons (kHz rate), which can be reduced via selection cuts, as explained below. These cuts lower the data rate by six orders of magnitude, to reach the level where the data are mainly consisting of atmospheric neutrino events (mHz rate). Since atmospheric neutrino events are indistinguishable from signal if they have the same direction and energy as signal neutrino events, a statistical analysis is performed on the final neutrino sample, to look for an excess coming from the center of the Earth ($$\\mathrm{zenith} =180^\\circ$$).\n\nThe first set of selection criteria, based on initial track reconstructions , is applied on the whole dataset, i.e. before splitting it into a low and a high energy sample. This reduces the data rate to a few Hz, so that more precise (and more time-consuming) reconstructions can be used to calculate the energy on which the splitting will be based. These initial cuts consist of a selection of online filters that tag up-going events, followed by cuts on the location of the interaction vertex and the direction of the charged lepton. These variables are not correlated with the energy of the neutrino and have thus similar efficiencies for different WIMP masses.\n\nThe variables that are used for cuts at this level are the reconstructed zenith angle, the reconstructed interaction vertex and the average temporal development of hits in the vertical (z) direction. The zenith angle cut is relatively loose to retain a sufficiently large control region in which the agreement between data and background simulation can be tested. An event is removed if the reconstructed direction points more than 60$$^\\circ$$ from the center of the Earth (i.e. the zenith is required to be larger than 120$$^\\circ$$). In this way the agreement between data and background simulation can be tested in a signal-free zenith region between 120$$^\\circ$$ and 150$$^\\circ$$ (see zenith distribution in Fig. 5). The other cut values are chosen by looping over all possible combinations and checking which combination brings down the background to the Hz level, while removing as little signal as possible.",
null,
"Fig. 2 Reconstructed energy distributions for neutrinos induced by 50 GeV and 1 TeV WIMPs trapped in the Earth. The vertical dashed line shows where the dataset is split. The error bars show statistical uncertainties\n\nAfter this first cut level, the data rate is reduced to $$\\sim$$3 Hz, while 30–60% of the signal (depending on WIMP mass and channel) is kept. The data is still dominated by atmospheric muons at this level. Now that the rate is sufficiently low, additional reconstructions can be applied to the data .\n\nThe distribution of the reconstructed energies for 50 GeV and 1 TeV WIMP signal events are shown in Fig. 2. The peak at $$\\sim$$750 GeV is an artifact of the energy reconstruction algorithm used in this analysis: if the track is not contained in the detector, the track length cannot be reconstructed and is set to a default value of 2 km. The track length is used to estimate the energy of the produced muon, while the energy of the hadronic cascade is reconstructed separately and can exceed the muon energy. Events showing this artifact are generally bright events, so their classification into the high energy sample is desired. The reconstructed energy is not used for other purposes than for splitting the data. A division at 100 GeV, shown as a vertical line in this figure, is used to split the dataset into low and high energy samples which are statistically independent and are optimized and analyzed separately.",
null,
"Fig. 3 BDT score distributions at pre-BDT level for the low energy analysis (left) and for the high energy analysis using the Pull-Validation method (right). Signal distributions are upscaled to be visible in the plot. Signal and backgrounds are compared to experimental data from 10% of the first year of IC86 data. For the atmospheric neutrinos, all flavors are taken into account. In gray, the sum of all simulated background is shown. The vertical lines indicate the final cut value used in each analysis, where high scores to the right of the line are retained\n\nBoth analyses use Boosted Decision Trees (BDTs) to classify background and signal events. This machine learning technique is designed to optimally separate signal from background after an analysis-specific training by assigning a score between −1 (background-like) and $$+$$1 (signal-like) to each event. In order to train a reliable BDT, the simulation must reproduce the experimental data accurately. Therefore a set of pre-BDT cuts are performed. Demanding a minimum of hits in a time window between −15 and 125 ns of the expected photon arrival time at each DOM, and a cut on the zenith of a more accurate reconstruction on causally connected hits improves the agreement between data and simulation. By comparing the times and distances of the first hits, the number of events with noise hits can be reduced. The last cut variable at this step is calculated by summing the signs of the differences between the z-coordinates of two temporally succeeding hits, which reduces further the amount of misreconstructed events. After these cuts, the experimental data rates are of the order of 100 mHz, and the data are still dominated by atmospheric muons. The BDTs are then trained on variables that show good agreement between data and simulation and have low correlation between themselves.\n\nIn the low energy optimization, the BDT training samples consist of simulated 50 GeV WIMP events and experimental data for the signal and background, respectively. Because the opening angle between the neutrino and its daughter lepton is inversely correlated to the energy of the neutrino, WIMP neutrino-induced muons in the high energy analysis are narrowly concentrated into vertical zenith angles, whereas in the low energy analysis they are spread over a wider range of zenith angles. Consequently, if the BDT for the high energy optimization was trained on simulated 1 TeV WIMP events, straight vertical events would be selected. This would make a comparison between data and simulation in a signal-free region more difficult. Instead, in the high energy analysis an isotropic muon neutrino simulation weighted to the energy spectrum of 1 TeV signal neutrinos is used to train a BDT.\n\nCoincident events of neutrinos and atmospheric muons can affect the data rate. Their influence is larger at low energies, as the atmospheric neutrino flux decreases steeply with increasing energy. In the low energy analysis, this effect cannot be neglected. As the amount of available simulated coincident events was limited, individual correction factors for the components of atmospheric background simulation are applied to take this effect into account. These correction factors are calculated by scaling the BDT score distributions of the simulated background to the experimental data. Only events with a reconstructed zenith of less than 132$$^\\circ$$ are used to determine the correction factors. With this choice, the background cannot be incorrectly adjusted to a signal that could be contained in the experimental data, as 95% of WIMP induced events have a larger zenith.\n\nThe distributions of the BDT scores for the low energy and high energy analyses are shown in Fig. 3. Cuts on the BDT score are chosen such that the sensitivities of the analyses are optimal. The sensitivities are calculated with a likelihood ratio hypothesis test based on the values of the reconstructed zenith, using the Feldman–Cousins unified approach . The required probability densities for signal and background are both calculated from simulations, as this analysis cannot make use of an off-source region. The background sample that is left after the cut on the BDT score mainly consists of atmospheric neutrinos and only has a small number of atmospheric muon events.\n\nTwo different smoothing methods are used to deal with the problem of small simulation statistics. The high energy analysis uses Pull-Validation , a method to improve the usage of limited statistics: a large number of BDTs (200 in the case of the present analysis) are trained on small subsets that are randomly resampled from the complete dataset. The variation of the BDT output between the trainings can be interpreted as a probability density function (PDF) for each event. This PDF can be used to calculate a weight that is applied to each event instead of making a binary cut decision. With this method, not only the BDT score distribution is smoothed (Fig. 3-right), but also the distributions that are made after a cut on the BDT score. In particular, the reconstructed zenith distribution used in the likelihood calculation is smooth, as events that would be removed when using a single BDT could now be kept, albeit with a smaller weight.\n\nThe low energy analysis tackles the problem of poor statistics of the atmospheric muon background simulation in a different way. In this part of the analysis, only a single BDT is trained (Fig. 3-left), and after the cut on the BDT score, the reconstructed zenith distribution is smoothed using a Kernel Density Estimator (KDE) [47, 48] with gaussian kernel and choosing an optimal bandwidth .\n\nThe event rates at different cut levels are summarized in Table 1.\nTable 1\n\nRates for experimental data, simulated atmospheric muons and atmospheric neutrinos of all favors, and signal efficiencies for WIMP masses of 50 GeV and 1 TeV, respectively, at different cut levels. Level 2 refers to the predefined common starting level, level 3 shows the event rates after the first set of cuts and the split into a high (HE) and a low energy (LE) sample and level 4 indicates the final analysis level after additional cuts and the BDT selection. Note that due to the Pull-Validation procedure, all events in the high energy sample at final level contain a weight. The effective data rates are shown\n\nCut level\n\nData rate (Hz)\n\nAtm. $$\\mu$$ rate (Hz)\n\nAtm. $$\\nu$$ rate (Hz)\n\nSignal eff.\n\nLE\n\nHE\n\nLE\n\nHE\n\nLE\n\nHE\n\nLE (%)\n\nHE (%)\n\n2\n\n670\n\n650\n\n0.027\n\n100\n\n3\n\n1.39\n\n1.35\n\n1.03\n\n0.97\n\n$$2.5\\times 10^{-3}$$\n\n$$2.0\\times 10^{-3}$$\n\n40.8\n\n45.1\n\n4\n\n$$2.8\\times 10^{-4}$$\n\n$$5.6\\times 10^{-4}$$\n\n$$8.0\\times 10^{-5}$$\n\n$$6.3\\times 10^{-5}$$\n\n$$2.0\\times 10^{-4}$$\n\n$$4.6\\times 10^{-4}$$\n\n15.6\n\n17.0\n\n6 Shape analysis\n\nAfter the event selection, the data rate is reduced to 0.28 mHz for the low energy selection and 0.56 mHz for the high energy selection. Misreconstructed atmospheric muons are almost completely filtered out and the remaining data sample consists mainly of atmospheric neutrinos. To analyze the dataset for an additional neutrino signal coming from the center of the Earth, we define a likelihood test that has been used in several IceCube analyses before (e.g. [21, 22]). Based on the background ($$f_\\mathrm{bg}$$) and signal distribution($$f_\\mathrm{s}$$) of space angles $$\\Psi$$ between the reconstructed muon track and the Earth center (i.e. the reconstructed zenith angle), the probability to observe a value $$\\Psi$$ for a single event is\n\\begin{aligned} f(\\Psi |\\mu )=\\frac{\\mu }{n_\\mathrm{obs}}f_s(\\Psi )+\\left( 1-\\frac{\\mu }{n_\\mathrm{obs}}\\right) f_\\mathrm{bg}(\\Psi ). \\end{aligned}\n(2)\nHere, $$\\mu$$ specifies the number of signal events in a set of $$n_\\mathrm{obs}$$ observed events. The likelihood to observe a certain number of events at specific space angles $$\\Psi _i$$ is defined as\n\\begin{aligned} \\mathcal {L}=\\prod _{i}^{n_\\mathrm{obs}}f(\\Psi _i|\\mu ). \\end{aligned}\n(3)\nFollowing the procedure in , the ranking parameter\n\\begin{aligned} \\mathcal {R}(\\mu )=\\frac{\\mathcal {L}(\\mu )}{\\mathcal {L}(\\hat{\\mu })} \\end{aligned}\n(4)\nis used as test statistic for the hypothesis testing, where $$\\hat{\\mu }$$ is the best fit of $$\\mu$$ to the observation. A critical ranking $$\\mathcal {R}^{90}$$ is defined for each signal strength, so that 90% of all experiments have a ranking larger than $$\\mathcal {R}^{90}$$. This is determined by $$10^4$$ pseudo experiments for each injected signal strength. The sensitivity is defined as the expectation value for the upper limit in the case that no signal is present. This is determined by generating $$10^4$$ pseudo experiments with no signal injected.",
null,
"Fig. 4 Effect of the assumed uncertainty on the sensitivity of the volumetric flux. The example shows 50 GeV WIMPs annihilating into $$\\tau ^+\\tau ^-$$. The points show the estimated sensitivity and include a correction for coincident muons, while the band indicates one standard deviation\n\n7 Systematic uncertainties\n\nDue to the lack of a control region, the background estimation has to be derived from simulation. Therefore, systematic uncertainties of the simulated datasets were carefully studied. The effects of the uncertainties were quantified by varying the respective input parameters in the simulations.",
null,
"Fig. 5 Reconstructed zenith distributions of 1 year of IC86 data compared to the simulated background distributions. For the atmospheric neutrinos, all flavors are taken into account. In the low energy analysis (left) the distributions were smoothed by a KDE and in the high energy analysis (right) the Pull-Validation method was used. Signal distributions are upscaled to be visible in the plot. The gray areas indicate the total predicted background distributions with 1 sigma uncertainties, including statistical and systematic uncertainties\nTable 2\n\nUpper limits at 90% confidence level on the number of signal events $$\\mu _\\mathrm{s}$$, the WIMP annihilation rate inside the Earth $$\\Gamma _A$$, the muon flux $$\\Phi _\\mu$$ and the spin-independent cross section $$\\sigma _\\mathrm{SI}$$, assuming an annihilation cross section of $$\\langle \\sigma _A v \\rangle = 3\\times 10^{-26}\\,\\mathrm{cm}^{3}\\,\\mathrm{s}^{-1}$$. Soft channel refers to annihilation into $$b\\bar{b}$$, while hard channel is defined by annihilation into $$W^+W^-$$ for WIMP masses larger than the rest mass of the W bosons and annihilation into $$\\tau ^+\\tau ^-$$ for lower WIMP masses. Systematic errors are included\n\nWIMP mass (GeV/c$$^2$$)\n\n$$\\mu _\\mathrm{s}$$ (year$$^{-1}$$)\n\n$$\\Gamma _A$$ (s$$^{-1}$$)\n\n$$\\Phi _\\mu$$ (km$$^{-2}$$ year$$^{-1}$$)\n\n$$\\sigma _\\mathrm{SI}$$ (cm$$^{2}$$)\n\nHard channel\n\nSoft channel\n\nHard channel\n\nSoft channel\n\nHard channel\n\nSoft channel\n\nHard channel\n\n10\n\n586\n\n$$3.01\\times 10^{16}$$\n\n$$1.54\\times 10^{4}$$\n\n$$2.5\\times 10^{-38}$$\n\n20\n\n209\n\n$$0.90\\times 10^{15}$$\n\n$$3.57\\times 10^{3}$$\n\n$$6.0\\times 10^{-41}$$\n\n35\n\n202\n\n405\n\n$$2.35\\times 10^{14}$$\n\n$$4.05\\times 10^{16}$$\n\n$$2.52\\times 10^{3}$$\n\n$$8.70\\times 10^{3}$$\n\n$$1.1\\times 10^{-41}$$\n\n50\n\n189\n\n253\n\n$$1.12\\times 10^{14}$$\n\n$$7.88\\times 10^{15}$$\n\n$$1.62\\times 10^{2}$$\n\n$$3.85\\times 10^{3}$$\n\n$$2.8\\times 10^{-43}$$\n\n100\n\n148\n\n172\n\n$$3.25\\times 10^{13}$$\n\n$$5.24\\times 10^{14}$$\n\n$$8.12\\times 10^{2}$$\n\n$$1.36\\times 10^{3}$$\n\n$$1.0\\times 10^{-41}$$\n\n250\n\n14.9\n\n128\n\n$$9.06\\times 10^{11}$$\n\n$$4.22\\times 10^{13}$$\n\n$$1.51\\times 10^{2}$$\n\n$$7.30\\times 10^{2}$$\n\n$$1.3\\times 10^{-41}$$\n\n500\n\n11.9\n\n11.8\n\n$$1.40\\times 10^{11}$$\n\n$$3.49\\times 10^{12}$$\n\n87.6\n\n$$2.14\\times 10^{2}$$\n\n$$1.7\\times 10^{-41}$$\n\n1000\n\n9.3\n\n10.6\n\n$$3.25\\times 10^{10}$$\n\n$$5.38\\times 10^{11}$$\n\n71.6\n\n$$1.05\\times 10^{2}$$\n\n$$2.0\\times 10^{-41}$$\n\n3000\n\n7.1\n\n8.1\n\n$$4.68\\times 10^{9}$$\n\n$$6.88\\times 10^{10}$$\n\n65.0\n\n66.6\n\n$$3.0\\times 10^{-41}$$\n\n5000\n\n6.6\n\n7.5\n\n$$2.12\\times 10^{9}$$\n\n$$3.28\\times 10^{10}$$\n\n64.1\n\n60.3\n\n$$3.8\\times 10^{-41}$$\n\n10000\n\n5.8\n\n6.8\n\n$$8.06\\times 10^{8}$$\n\n$$1.47\\times 10^{10}$$\n\n64.7\n\n57.6\n\n$$5.1\\times 10^{-41}$$\n\nDifferent types of detector related uncertainties have to be considered. The efficiency of the DOM to detect Cherenkov photons is not exactly known. To estimate the effect of this uncertainty, three simulated datasets with 90, 100 and 110% of the nominal efficiency were investigated. With these datasets, the sensitivity varies by ±10% for both event selections of the analysis. Taking anisotropic scattering in the South Pole ice into account , has an effect of −10% in the high and the low energy selection. The reduced scattering length of photons in the refrozen ice of the holes leads to an uncertainty of −10% in both selections. Furthermore, the uncertainty on the scattering and absorption lengths influences the result by ±10% for the low energy and ±5% for the high energy selection.\n\nBesides the detector related uncertainties, the uncertainties on the models of the background physics are taken into account. The uncertainty of the atmospheric flux can change the rates by ±30%, as determined e.g. in . For low energies, uncertainties on neutrino oscillation parameters are significant. This effect has been studied in a previous analysis and influences the event rates by ±6%. The effect of the uncertainty of the neutrino–nucleon cross section has been studied in the same analysis. It depends on the neutrino energy and is conservatively estimated as ±6% for the low and ±3% for the high energy sample. Finally, the rate of coincidences of atmospheric neutrinos and atmospheric muons has a large impact on the low energy analysis. While in the baseline data sets, coincident events were not simulated, a comparison with a test simulation that includes coincident events shows an effect of −30% on the final event rates.\n\nAdding these uncertainties in quadrature results in a total of $$+$$34%/−48% in the low energy analysis and $$+$$32%/−35% for high energies. For the limit calculation, they are taken into account by using a semi-bayesian extension to the Feldman–Cousins approach . Technically, it is realized by randomly varying the expectation value of each pseudo-experiment by a gaussian of the corresponding uncertainty. As an illustration, the effect of this procedure is shown in Fig. 4 for different uncertainties.\n\n8 Results\n\nAs mentioned in Sect. 5, only 10% of the data were used for quality checks during the optimization of the analysis chain. Half of this subsample was used to train the BDTs and therefore these events could not be used for the later analysis. After the selection criteria were completely finalized, the zenith distributions of the remaining 95% of the dataset were examined (Fig. 5). No statistically significant excess above the expected atmospheric background was found from the direction of the center of the Earth.\n\nUsing the method described in Sect. 6, upper limits at the 90% confidence level on the volumetric flux\n\\begin{aligned} \\Gamma _{\\mu \\rightarrow \\nu }=\\frac{\\mu _\\mathrm{s}}{t_\\mathrm{live}\\cdot V_\\mathrm{eff}} \\end{aligned}\n(5)\nwere calculated from the high and the low energy sample for WIMP masses between 10 GeV and 10 TeV in the hard and in the soft channel. Here $$\\mu _\\mathrm{s}$$ denotes the upper limit on the number of signal neutrinos, $$t_\\mathrm{live}$$ the livetime and $$V_\\mathrm{eff}$$ the effective volume of the detector. Using the package WimpSim , the volumetric flux was converted into the WIMP annihilation rate inside the Earth $$\\Gamma _A$$ and the resulting muon flux $$\\Phi _\\mu$$. The obtained 90% C.L. limits are shown in Fig. 6 and listed in Table 2. For each mass and channel, the result with the most restricting limit is shown.",
null,
"Fig. 6 Top individual upper limits at 90% confidence level (solid lines) on the muon flux $$\\Phi _\\mu$$ for the low and high energy analysis. Systematic uncertainties are included. For the soft channel, $$\\chi \\chi \\rightarrow b\\bar{b}$$ is assumed with 100% braching ratio, while for the hard channel the annihilation $$\\chi \\chi \\rightarrow \\tau ^{+}\\tau ^{-}$$ for masses $$\\le$$ 50 GeV and $$\\chi \\chi \\rightarrow W^{+}W^{-}$$ for higher masses is assumed. A flux with mixed branching ratios will be between these extremes. The dashed lines and the bands indicate the corresponding sensitivities with one sigma uncertainty. Bottom the combined best upper limits (solid line) and sensitivities (dashed line) with 1 sigma uncertainty (green band) on the annihilation rate in the Earth $$\\Gamma _A$$ for 1 year of IC86 data as a function of the WIMP mass. For each WIMP mass, the sample (high energy or low energy) which yields the best sensitivity is used. Systematic uncertainties are included. The dotted line shows the latest upper limit on the annihilation rate, which was calculated with AMANDA data [17, 18]",
null,
"Fig. 7 Upper limits at 90% confidence level on $$\\sigma _{\\chi -N}^\\mathrm{SI}$$ as a function of the annihilation cross section for 50 GeV WIMPs annihilating into $$\\tau ^+\\tau ^-$$ and for 1 TeV WIMPs annihilating into $$W^+W^-$$. Systematic uncertainties are included. As a comparison, the limits of LUX are shown as dashed lines. The red vertical line indicates the thermal annihilation cross section. Also indicated are IceCube limits on the annihilation cross section for the respective models \ns",
null,
"Fig. 8 Upper limits at 90% confidence level on $$\\sigma _{\\chi -N}^\\mathrm{SI}$$ as a function of the WIMP mass assuming a WIMP annihilation cross section of $$\\langle \\sigma _A v \\rangle = 3\\times 10^{-26}\\,\\mathrm{cm}^{3}\\,\\mathrm{s}^{-1}$$. For WIMP masses above the rest mass of the W bosons, annihilation into $$W^+W^-$$ is assumed and annihilation into $$\\tau ^+\\tau ^-$$ for lower masses. Systematic uncertainties are included. The result is compared to the limits set by SuperCDMSlite , LUX , Super-K and by a Solar WIMP analysis of IceCube in the 79-string configuration . The displayed limits are assuming a local dark matter density of $$\\rho _\\chi =0.3$$ GeV cm$$^{-3}$$. A larger density, as suggested e.g. by , would scale all limits linearly\n\nFurthermore, limits on the spin-independent WIMP–nucleon cross section $$\\sigma _{\\chi -N}^\\mathrm{SI}$$ can be derived. In contrast to dark matter accumulated in the Sun, the annihilation rate in the Earth and $$\\sigma _{\\chi -N}^\\mathrm{SI}$$ are not directly linked. As no equilibrium between WIMP capture and annihilation can be assumed, the annihilation rate depends on $$\\sigma _{\\chi -N}^\\mathrm{SI}$$ and on the annihilation cross section $$\\langle \\sigma _A v \\rangle$$. Fig. 7 shows the limits in the $$\\sigma _{\\chi -N}^\\text {SI}$$ - $$\\langle \\sigma _A v \\rangle$$ plane for two WIMP masses. If a typical value for the natural scale $$\\langle \\sigma _A v \\rangle = 3\\times 10^{-26}\\,\\mathrm{cm}^{3}\\,\\mathrm{s}^{-1}$$, for which the WIMP is a thermal relic , is assumed as annihilation cross section, from the limits in Fig. 7 one can derive upper limits on the spin-independent WIMP–nucleon scattering cross section as function of the WIMP mass. While the limits in Table 2 correspond to the investigated benchmark masses, in Fig. 8, interpolated results were taken into account, showing the effect of the resonant capture on the most abundant elements in the Earth.\n\nWe note that Solar WIMP, Earth WIMP, and direct searches have very different dependences on astrophysical uncertainties. A change in the WIMP velocity distribution has minor effects on Solar WIMP bounds [54, 55], while Earth WIMPs and direct searches are far more susceptible to it. In particular the existence of a dark disk could enhance Earth WIMP rates by several orders of magnitude while leaving direct bounds largely unchanged. The limits presented here assume a standard halo and are conservative with respect to the existence of a dark disk.\n\n9 Summary\n\nUsing 1 year of data taken by the fully completed detector, we performed the first IceCube search for neutrinos produced by WIMP dark matter annihilations in the center of the Earth. 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Phys. 1307, 016 (2013)\n\nFunded by SCOAP3\n\nAuthors and Affiliations\n\n• M. G. Aartsen\n• 2\n• K. Abraham\n• 34\n• M. Ackermann\n• 52\n• 16\n• J. A. Aguilar\n• 12\n• M. Ahlers\n• 30\n• M. Ahrens\n• 42\n• D. Altmann\n• 24\n• K. Andeen\n• 32\n• T. Anderson\n• 48\n• I. Ansseau\n• 12\n• G. Anton\n• 24\n• M. Archinger\n• 31\n• C. Argüelles\n• 14\n• J. Auffenberg\n• 1\n• S. Axani\n• 14\n• X. Bai\n• 40\n• S. W. Barwick\n• 27\n• V. Baum\n• 31\n• R. Bay\n• 7\n• J. J. Beatty\n• 18\n• 19\n• J. Becker Tjus\n• 10\n• K.-H. Becker\n• 51\n• S. BenZvi\n• 49\n• D. Berley\n• 17\n• E. Bernardini\n• 52\n• A. Bernhard\n• 34\n• D. Z. Besson\n• 28\n• G. Binder\n• 7\n• 8\n• D. Bindig\n• 51\n• M. Bissok\n• 1\n• E. Blaufuss\n• 17\n• S. Blot\n• 52\n• C. Bohm\n• 42\n• M. Börner\n• 21\n• F. Bos\n• 10\n• D. Bose\n• 44\n• S. Böser\n• 31\n• O. Botner\n• 50\n• J. Braun\n• 30\n• L. Brayeur\n• 13\n• H.-P. Bretz\n• 52\n• S. Bron\n• 25\n• A. Burgman\n• 50\n• T. Carver\n• 25\n• M. Casier\n• 13\n• E. Cheung\n• 17\n• D. Chirkin\n• 30\n• A. Christov\n• 25\n• K. Clark\n• 45\n• L. Classen\n• 35\n• S. Coenders\n• 34\n• G. H. Collin\n• 14\n• 14\n• D. F. Cowen\n• 47\n• 48\n• R. Cross\n• 49\n• M. Day\n• 30\n• J. P. A. M. de André\n• 22\n• C. De Clercq\n• 13\n• E. del Pino Rosendo\n• 31\n• H. Dembinski\n• 36\n• S. De Ridder\n• 26\n• P. Desiati\n• 30\n• K. D. de Vries\n• 13\n• G. de Wasseige\n• 13\n• M. de With\n• 9\n• T. DeYoung\n• 22\n• J. C. Díaz-Vélez\n• 30\n• V. di Lorenzo\n• 31\n• H. Dujmovic\n• 44\n• J. P. Dumm\n• 42\n• M. Dunkman\n• 48\n• B. Eberhardt\n• 31\n• T. Ehrhardt\n• 31\n• B. Eichmann\n• 10\n• P. Eller\n• 48\n• S. Euler\n• 50\n• P. A. Evenson\n• 36\n• S. Fahey\n• 30\n• A. R. Fazely\n• 6\n• J. Feintzeig\n• 30\n• J. Felde\n• 17\n• K. Filimonov\n• 7\n• C. Finley\n• 42\n• S. Flis\n• 42\n• C.-C. Fösig\n• 31\n• A. Franckowiak\n• 52\n• E. Friedman\n• 17\n• T. Fuchs\n• 21\n• T. K. Gaisser\n• 36\n• J. Gallagher\n• 29\n• L. Gerhardt\n• 7\n• 8\n• K. Ghorbani\n• 30\n• W. Giang\n• 23\n• 30\n• M. Glagla\n• 1\n• T. Glauch\n• 1\n• T. Glüsenkamp\n• 52\n• A. Goldschmidt\n• 8\n• G. Golup\n• 13\n• J. G. Gonzalez\n• 36\n• D. Grant\n• 23\n• Z. Griffith\n• 30\n• C. Haack\n• 1\n• A. Haj Ismail\n• 26\n• A. Hallgren\n• 50\n• F. Halzen\n• 30\n• E. Hansen\n• 20\n• B. Hansmann\n• 1\n• T. Hansmann\n• 1\n• K. Hanson\n• 30\n• D. Hebecker\n• 9\n• D. Heereman\n• 12\n• K. Helbing\n• 51\n• R. Hellauer\n• 17\n• S. Hickford\n• 51\n• J. Hignight\n• 22\n• G. C. Hill\n• 2\n• K. D. Hoffman\n• 17\n• R. Hoffmann\n• 51\n• K. Holzapfel\n• 34\n• K. Hoshina\n• 30\n• 53\n• F. Huang\n• 48\n• M. Huber\n• 34\n• K. Hultqvist\n• 42\n• S. In\n• 44\n• A. Ishihara\n• 15\n• E. Jacobi\n• 52\n• G. S. Japaridze\n• 4\n• M. Jeong\n• 44\n• K. Jero\n• 30\n• B. J. P. Jones\n• 14\n• M. Jurkovic\n• 34\n• A. Kappes\n• 35\n• T. Karg\n• 52\n• A. Karle\n• 30\n• U. Katz\n• 24\n• M. Kauer\n• 30\n• A. Keivani\n• 48\n• J. L. Kelley\n• 30\n• J. Kemp\n• 1\n• A. Kheirandish\n• 30\n• M. Kim\n• 44\n• T. Kintscher\n• 52\n• J. Kiryluk\n• 43\n• T. Kittler\n• 24\n• S. R. Klein\n• 7\n• 8\n• G. Kohnen\n• 33\n• R. Koirala\n• 36\n• H. Kolanoski\n• 9\n• R. Konietz\n• 1\n• L. Köpke\n• 31\n• C. Kopper\n• 23\n• S. Kopper\n• 51\n• D. J. Koskinen\n• 20\n• M. Kowalski\n• 9\n• 52\n• K. Krings\n• 34\n• M. Kroll\n• 10\n• G. Krückl\n• 31\n• C. Krüger\n• 30\n• J. Kunnen\n• 13\nEmail author\n• S. Kunwar\n• 52\n• N. Kurahashi\n• 39\n• T. Kuwabara\n• 15\n• M. Labare\n• 26\n• J. L. Lanfranchi\n• 48\n• M. J. Larson\n• 20\n• F. Lauber\n• 51\n• D. Lennarz\n• 22\n• M. Lesiak-Bzdak\n• 43\n• M. Leuermann\n• 1\n• J. Leuner\n• 1\n• L. Lu\n• 15\n• J. Lünemann\n• 13\n• 41\n• G. Maggi\n• 13\n• K. B. M. Mahn\n• 22\n• S. Mancina\n• 30\n• M. Mandelartz\n• 10\n• R. Maruyama\n• 37\n• K. Mase\n• 15\n• R. Maunu\n• 17\n• F. McNally\n• 30\n• K. Meagher\n• 12\n• M. Medici\n• 20\n• M. Meier\n• 21\n• A. Meli\n• 26\n• T. Menne\n• 21\n• G. Merino\n• 30\n• T. Meures\n• 12\n• S. Miarecki\n• 7\n• 8\n• L. Mohrmann\n• 52\n• T. Montaruli\n• 25\n• M. Moulai\n• 14\n• R. Nahnhauer\n• 52\n• U. Naumann\n• 51\n• G. Neer\n• 22\n• H. Niederhausen\n• 43\n• S. C. Nowicki\n• 23\n• D. R. Nygren\n• 8\n• A. Obertacke Pollmann\n• 51\n• A. Olivas\n• 17\n• 12\n• T. Palczewski\n• 46\n• H. Pandya\n• 36\n• D. V. Pankova\n• 48\n• P. Peiffer\n• 31\n• Ö. Penek\n• 1\n• J. A. Pepper\n• 46\n• C. Pérez de los Heros\n• 50\n• D. Pieloth\n• 21\n• E. Pinat\n• 12\n• P. B. Price\n• 7\n• G. T. Przybylski\n• 8\n• M. Quinnan\n• 48\n• C. Raab\n• 12\n• L. Rädel\n• 1\n• M. Rameez\n• 20\n• K. Rawlins\n• 3\n• R. Reimann\n• 1\n• B. Relethford\n• 39\n• M. Relich\n• 15\n• E. Resconi\n• 34\n• W. Rhode\n• 21\n• M. Richman\n• 39\n• B. Riedel\n• 23\n• S. Robertson\n• 2\n• M. Rongen\n• 1\n• C. Rott\n• 44\n• T. Ruhe\n• 21\n• D. Ryckbosch\n• 26\n• D. Rysewyk\n• 22\n• L. Sabbatini\n• 30\n• S. E. Sanchez Herrera\n• 23\n• A. Sandrock\n• 21\n• J. Sandroos\n• 31\n• S. Sarkar\n• 20\n• 38\n• K. Satalecka\n• 52\n• M. Schimp\n• 1\n• P. Schlunder\n• 21\n• T. Schmidt\n• 17\n• S. Schoenen\n• 1\n• S. Schöneberg\n• 10\n• L. Schumacher\n• 1\n• D. Seckel\n• 36\n• S. Seunarine\n• 41\n• D. Soldin\n• 51\n• M. Song\n• 17\n• G. M. Spiczak\n• 41\n• C. Spiering\n• 52\n• M. Stahlberg\n• 1\n• T. Stanev\n• 36\n• A. Stasik\n• 52\n• J. Stettner\n• 1\n• A. Steuer\n• 31\n• T. Stezelberger\n• 8\n• 8\n• A. Stößl\n• 52\n• R. Ström\n• 50\n• N. L. Strotjohann\n• 52\n• G. W. Sullivan\n• 17\n• M. Sutherland\n• 18\n• H. Taavola\n• 50\n• 5\n• J. Tatar\n• 7\n• 8\n• F. Tenholt\n• 10\n• S. Ter-Antonyan\n• 6\n• A. Terliuk\n• 52\n• G. Tešić\n• 48\n• S. Tilav\n• 36\n• P. A. Toale\n• 46\n• M. N. Tobin\n• 30\n• S. Toscano\n• 13\n• D. Tosi\n• 30\n• M. Tselengidou\n• 24\n• A. Turcati\n• 34\n• E. Unger\n• 50\n• M. Usner\n• 52\n• J. Vandenbroucke\n• 30\n• N. van Eijndhoven\n• 13\n• S. Vanheule\n• 26\n• M. van Rossem\n• 30\n• J. van Santen\n• 52\n• J. Veenkamp\n• 34\n• M. Vehring\n• 1\n• M. Voge\n• 11\n• E. Vogel\n• 1\n• M. Vraeghe\n• 26\n• C. Walck\n• 42\n• A. Wallace\n• 2\n• M. Wallraff\n• 1\n• N. Wandkowsky\n• 30\n• Ch. Weaver\n• 23\n• M. J. Weiss\n• 48\n• C. Wendt\n• 30\n• S. Westerhoff\n• 30\n• B. J. Whelan\n• 2\n• S. Wickmann\n• 1\n• K. Wiebe\n• 31\n• C. H. Wiebusch\n• 1\n• L. Wille\n• 30\n• D. R. Williams\n• 46\n• L. Wills\n• 39\n• M. Wolf\n• 42\n• T. R. Wood\n• 23\n• E. Woolsey\n• 23\n• K. Woschnagg\n• 7\n• D. L. Xu\n• 30\n• X. W. Xu\n• 6\n• Y. Xu\n• 43\n• J. P. Yanez\n• 52\n• G. Yodh\n• 27\n• S. Yoshida\n• 15\n• M. Zoll\n• 42\n1. 1.III. Physikalisches InstitutRWTH Aachen UniversityAachenGermany\n3. 3.Department of Physics and AstronomyUniversity of Alaska AnchorageAnchorageUSA\n4. 4.CTSPSClark-Atlanta UniversityAtlantaUSA\n5. 5.School of Physics and Center for Relativistic AstrophysicsGeorgia Institute of TechnologyAtlantaUSA\n6. 6.Department of PhysicsSouthern UniversityBaton RougeUSA\n7. 7.Department of PhysicsUniversity of CaliforniaBerkeleyUSA\n8. 8.Lawrence Berkeley National LaboratoryBerkeleyUSA\n9. 9.Institut für PhysikHumboldt-Universität zu BerlinBerlinGermany\n10. 10.Fakultät für Physik & AstronomieRuhr-Universität BochumBochumGermany\n11. 11.Physikalisches InstitutUniversität BonnBonnGermany\n12. 12.Science Faculty CP230Université Libre de BruxellesBrusselsBelgium\n13. 13.Dienst ELEMVrije Universiteit BrusselBrusselsBelgium\n14. 14.Department of PhysicsMassachusetts Institute of TechnologyCambridgeUSA\n15. 15.Department of Physics and Institute for Global Prominent ResearchChiba UniversityChibaJapan\n16. 16.Department of Physics and AstronomyUniversity of CanterburyChristchurchNew Zealand\n17. 17.Department of PhysicsUniversity of MarylandCollege ParkUSA\n18. 18.Department of Physics and Center for Cosmology and Astro-Particle PhysicsOhio State UniversityColumbusUSA\n19. 19.Department of AstronomyOhio State UniversityColumbusUSA\n20. 20.Niels Bohr InstituteUniversity of CopenhagenCopenhagenDenmark\n21. 21.Department of PhysicsTU Dortmund UniversityDortmundGermany\n22. 22.Department of Physics and AstronomyMichigan State UniversityEast LansingUSA\n23. 23.Department of PhysicsUniversity of AlbertaEdmontonCanada\n24. 24.Erlangen Centre for Astroparticle PhysicsFriedrich-Alexander-Universität Erlangen-NürnbergErlangenGermany\n25. 25.Département de physique nucléaire et corpusculaireUniversité de GenèveGenevaSwitzerland\n26. 26.Department of Physics and AstronomyUniversity of GentGentBelgium\n27. 27.Department of Physics and AstronomyUniversity of CaliforniaIrvineUSA\n28. 28.Department of Physics and AstronomyUniversity of KansasLawrenceUSA\n29. 29.Department of AstronomyUniversity of WisconsinMadisonUSA\n30. 30.Department of Physics and Wisconsin IceCube Particle Astrophysics CenterUniversity of WisconsinMadisonUSA\n31. 31.Institute of PhysicsUniversity of MainzMainzGermany\n32. 32.Department of PhysicsMarquette UniversityMilwaukeeUSA\n33. 33.Université de MonsMonsBelgium\n34. 34.Physik-departmentTechnische Universität MünchenGarchingGermany\n35. 35.Institut für KernphysikWestfälische Wilhelms-Universität MünsterMünsterGermany\n36. 36.Department of Physics and Astronomy, Bartol Research InstituteUniversity of DelawareNewarkUSA\n37. 37.Department of PhysicsYale UniversityNew HavenUSA\n38. 38.Department of PhysicsUniversity of OxfordOxfordUK\n40. 40.Physics DepartmentSouth Dakota School of Mines and TechnologyRapid CityUSA\n41. 41.Department of PhysicsUniversity of WisconsinRiver FallsUSA\n42. 42.Department of Physics, Oskar Klein CentreStockholm UniversityStockholmSweden\n43. 43.Department of Physics and AstronomyStony Brook UniversityStony BrookUSA\n44. 44.Department of PhysicsSungkyunkwan UniversitySuwonKorea\n45. 45.Department of PhysicsUniversity of TorontoTorontoCanada\n46. 46.Department of Physics and AstronomyUniversity of AlabamaTuscaloosaUSA\n47. 47.Department of Astronomy and AstrophysicsPennsylvania State UniversityUniversity ParkUSA\n48. 48.Department of PhysicsPennsylvania State UniversityUniversity ParkUSA\n49. 49.Department of Physics and AstronomyUniversity of RochesterRochesterUSA\n50. 50.Department of Physics and AstronomyUppsala UniversityUppsalaSweden\n51. 51.Department of PhysicsUniversity of WuppertalWuppertalGermany\n52. 52.DESYZeuthenGermany\n53. 53.Earthquake Research InstituteUniversity of TokyoBunkyoJapan\n\nPersonalised recommendations",
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https://code.orgmode.org/bzg/org-mode/commit/1829aa79b51c462032a270431217e98e63c37ecc | [
"### org-list: use list structure to update checkboxes and cookies\n\n```* lisp/org-list.el (org-toggle-checkbox): use structures to fix\ncheckboxes of a list\n(org-update-checkbox-count): use structures to update cookies```",
null,
"Nicolas Goaziou 8 years ago\nparent\ncommit\n1829aa79b5\n1 changed files with 191 additions and 173 deletions\n1. 191 173\nlisp/org-list.el\n\n#### + 191 - 173 lisp/org-list.el View File\n\n ``@@ -1798,77 +1798,91 @@ If the cursor is in a headline, apply this to all checkbox items`` `` in the text below the heading, taking as reference the first item`` `` in subtree, ignoring drawers.\"`` `` (interactive \"P\")`` ``- ;; Bounds is a list of type (beg end single-p) where single-p is t`` ``- ;; when `org-toggle-checkbox' is applied to a single item. Only`` ``- ;; toggles on single items will return errors.`` ``- (let* ((bounds`` ``- (cond`` ``- ((org-region-active-p)`` ``- (let ((rbeg (region-beginning))`` ``- (rend (region-end)))`` ``- (save-excursion`` ``- (goto-char rbeg)`` ``- (if (org-search-forward-unenclosed org-item-beginning-re rend 'move)`` ``- (list (point-at-bol) rend nil)`` ``- (error \"No item in region\")))))`` ``- ((org-on-heading-p)`` ``- ;; In this case, reference line is the first item in`` ``- ;; subtree outside drawers`` ``- (let ((pos (point))`` ``- (limit (save-excursion (outline-next-heading) (point))))`` ``- (save-excursion`` ``- (goto-char limit)`` ``- (org-search-backward-unenclosed \":END:\" pos 'move)`` ``- (org-search-forward-unenclosed`` ``- org-item-beginning-re limit 'move)`` ``- (list (point) limit nil))))`` ``- ((org-at-item-p)`` ``- (list (point-at-bol) (1+ (point-at-eol)) t))`` ``- (t (error \"Not at an item or heading, and no active region\"))))`` ``- (beg (car bounds))`` ``- ;; marker is needed because deleting or inserting checkboxes`` ``- ;; will change bottom point`` ``- (end (copy-marker (nth 1 bounds)))`` ``- (single-p (nth 2 bounds))`` ``- (ref-presence (save-excursion`` ``- (goto-char beg)`` ``- (org-at-item-checkbox-p)))`` ``- (ref-status (equal (match-string 1) \"[X]\"))`` ``- (act-on-item`` ``- (lambda (ref-pres ref-stat)`` ``- (if (equal toggle-presence '(4))`` ``- (cond`` ``- ((and ref-pres (org-at-item-checkbox-p))`` ``- (replace-match \"\"))`` ``- ((and (not ref-pres)`` ``- (not (org-at-item-checkbox-p))`` ``- (org-at-item-p))`` ``- (goto-char (match-end 0))`` ``- ;; Ignore counter, if any`` ``- (when (looking-at \"\\\\(?:\\\\[@\\\\(?:start:\\\\)?[0-9]+\\\\][ \\t]*\\\\)?\")`` ``- (goto-char (match-end 0)))`` ``- (let ((desc-p (and (org-at-item-description-p)`` ``- (cdr (assq 'checkbox org-list-automatic-rules)))))`` ``- (cond`` ``- ((and single-p desc-p)`` ``- (error \"Cannot add a checkbox in a description list\"))`` ``- ((not desc-p) (insert \"[ ] \"))))))`` ``- (let ((blocked (org-checkbox-blocked-p)))`` ``- (cond`` ``- ((and blocked single-p)`` ``- (error \"Checkbox blocked because of unchecked box in line %d\" blocked))`` ``- (blocked nil)`` ``- ((org-at-item-checkbox-p)`` ``- (replace-match`` ``- (cond ((equal toggle-presence '(16)) \"[-]\")`` ``- (ref-stat \"[ ]\")`` ``- (t \"[X]\"))`` ``- t t nil 1))))))))`` ``- (save-excursion`` ``- (goto-char beg)`` ``- (while (< (point) end)`` ``- (funcall act-on-item ref-presence ref-status)`` ``- (org-search-forward-unenclosed org-item-beginning-re end 'move)))`` ``+ (save-excursion`` ``+ (let* (singlep`` ``+ block-item`` ``+ lim-up`` ``+ lim-down`` ``+ (orderedp (ignore-errors (org-entry-get nil \"ORDERED\")))`` ``+ (bounds`` ``+ ;; In a region, start at first item in region`` ``+ (cond`` ``+ ((org-region-active-p)`` ``+ (let ((limit (region-end)))`` ``+ (goto-char (region-beginning))`` ``+ (if (org-search-forward-unenclosed org-item-beginning-re`` ``+ limit t)`` ``+ (setq lim-up (point-at-bol))`` ``+ (error \"No item in region\"))`` ``+ (setq lim-down (copy-marker limit))))`` ``+ ((org-on-heading-p)`` ``+ ;; On an heading, start at first item after drawers`` ``+ (let ((limit (save-excursion (outline-next-heading) (point))))`` ``+ (forward-line 1)`` ``+ (when (looking-at org-drawer-regexp)`` ``+ (re-search-forward \"^[ \\t]*:END:\" limit nil))`` ``+ (if (org-search-forward-unenclosed org-item-beginning-re`` ``+ limit t)`` ``+ (setq lim-up (point-at-bol))`` ``+ (error \"No item in subtree\"))`` ``+ (setq lim-down (copy-marker limit))))`` ``+ ;; Just one item: set singlep flag`` ``+ ((org-at-item-p)`` ``+ (setq singlep t)`` ``+ (setq lim-up (point-at-bol)`` ``+ lim-down (point-at-eol)))`` ``+ (t (error \"Not at an item or heading, and no active region\"))))`` ``+ ;; determine the checkbox going to be applied to all items`` ``+ ;; within bounds`` ``+ (ref-checkbox`` ``+ (progn`` ``+ (goto-char lim-up)`` ``+ (let ((cbox (and (org-at-item-checkbox-p) (match-string 1))))`` ``+ (cond`` ``+ ((equal toggle-presence '(16)) \"[-]\")`` ``+ ((equal toggle-presence '(4))`` ``+ (unless cbox \"[ ]\"))`` ``+ ((equal \"[ ]\" cbox) \"[X]\")`` ``+ (t \"[ ]\"))))))`` ``+ ;; When an item is found within bounds, grab the full list at`` ``+ ;; point structure, then: 1. set checkbox of all its items`` ``+ ;; within bounds to ref-checkbox; 2. fix checkboxes of the whole`` ``+ ;; list; 3. move point after the list.`` ``+ (goto-char lim-up)`` ``+ (while (and (< (point) lim-down)`` ``+ (org-search-forward-unenclosed`` ``+ org-item-beginning-re lim-down 'move))`` ``+ (let* ((struct (org-list-struct))`` ``+ (struct-copy (mapcar (lambda (e) (copy-alist e)) struct))`` ``+ (parents (org-list-struct-parent-alist struct))`` ``+ (bottom (copy-marker (org-list-get-bottom-point struct)))`` ``+ (items-to-toggle (org-remove-if`` ``+ (lambda (e) (or (< e lim-up) (> e lim-down)))`` ``+ (mapcar 'car (cdr struct)))))`` ``+ (mapc (lambda (e) (org-list-set-checkbox`` ``+ e struct`` ``+ ;; if there is no box at item, leave as-is`` ``+ ;; unless function was called with C-u prefix`` ``+ (let ((cur-box (org-list-get-checkbox e struct)))`` ``+ (if (or cur-box (equal toggle-presence '(4)))`` ``+ ref-checkbox`` ``+ cur-box))))`` ``+ items-to-toggle)`` ``+ (setq block-item (org-list-struct-fix-box struct parents orderedp))`` ``+ ;; Report some problems due to ORDERED status of subtree. If`` ``+ ;; only one box was being checked, throw an error, else,`` ``+ ;; only signal problems.`` ``+ (cond`` ``+ ((and singlep block-item (> lim-up block-item))`` ``+ (error`` ``+ \"Checkbox blocked because of unchecked box at line %d\"`` ``+ (org-current-line block-item)))`` ``+ (block-item`` ``+ (message`` ``+ \"Checkboxes were removed due to unchecked box at line %d\"`` ``+ (org-current-line block-item))))`` ``+ (goto-char bottom)`` ``+ (org-list-struct-apply-struct struct struct-copy))))`` `` (org-update-checkbox-count-maybe)))`` `` `` `` (defun org-reset-checkbox-state-subtree ()`` ``@@ -1901,110 +1915,114 @@ information.\")`` `` `` `` (defun org-update-checkbox-count (&optional all)`` `` \"Update the checkbox statistics in the current section.`` ``-This will find all statistic cookies like [57%] and [6/12] and update them`` ``-with the current numbers. With optional prefix argument ALL, do this for`` ``-the whole buffer.\"`` ``+This will find all statistic cookies like [57%] and [6/12] and`` ``+update them with the current numbers.`` ``+`` ``+With optional prefix argument ALL, do this for the whole buffer.\"`` `` (interactive \"P\")`` `` (save-excursion`` ``- (let ((cstat 0))`` ``- (catch 'exit`` ``- (while t`` ``- (let* ((buffer-invisibility-spec (org-inhibit-invisibility)) ; Emacs 21`` ``- (beg (condition-case nil`` ``- (progn (org-back-to-heading) (point))`` ``- (error (point-min))))`` ``- (end (copy-marker (save-excursion`` ``- (outline-next-heading) (point))))`` ``- (re-cookie \"\\\\(\\\\(\\\\[[0-9]*%\\\\]\\\\)\\\\|\\\\(\\\\[[0-9]*/[0-9]*\\\\]\\\\)\\\\)\")`` ``- (re-box \"^[ \\t]*\\\\([-+*]\\\\|[0-9]+[.)]\\\\)[ \\t]+\\\\(?:\\\\[@\\\\(?:start:\\\\)?[0-9]+\\\\][ \\t]*\\\\)?\\\\(\\\\[[- X]\\\\]\\\\)\")`` ``- beg-cookie end-cookie is-percent c-on c-off lim new`` ``- curr-ind next-ind continue-from startsearch list-beg list-end`` ``- (recursive`` ``- (or (not org-hierarchical-checkbox-statistics)`` ``- (string-match \"\\\\\"`` ``- (or (ignore-errors`` ``- (org-entry-get nil \"COOKIE_DATA\"))`` ``- \"\")))))`` ``- (goto-char end)`` ``- ;; find each statistics cookie`` ``- (while (and (org-search-backward-unenclosed re-cookie beg 'move)`` ``- (not (save-match-data`` ``- (and (org-on-heading-p)`` ``- (string-match \"\\\\\"`` ``- (downcase`` ``- (or (org-entry-get`` ``- nil \"COOKIE_DATA\")`` ``- \"\")))))))`` ``- (setq beg-cookie (match-beginning 1)`` ``- end-cookie (match-end 1)`` ``- cstat (+ cstat (if end-cookie 1 0))`` ``- startsearch (point-at-eol)`` ``- continue-from (match-beginning 0)`` ``- is-percent (match-beginning 2)`` ``- lim (cond`` ``- ((org-on-heading-p) (outline-next-heading) (point))`` ``- ;; Ensure many cookies in the same list won't imply`` ``- ;; computing list boundaries as many times.`` ``- ((org-at-item-p)`` ``- (unless (and list-beg (>= (point) list-beg))`` ``- (setq list-beg (org-list-top-point)`` ``- list-end (copy-marker`` ``- (org-list-bottom-point))))`` ``- (org-get-end-of-item list-end))`` ``- (t nil))`` ``- c-on 0`` ``- c-off 0)`` ``- (when lim`` ``- ;; find first checkbox for this cookie and gather`` ``- ;; statistics from all that are at this indentation level`` ``- (goto-char startsearch)`` ``- (if (org-search-forward-unenclosed re-box lim t)`` ``- (progn`` ``- (beginning-of-line)`` ``- (setq curr-ind (org-get-indentation))`` ``- (setq next-ind curr-ind)`` ``- (while (and (bolp) (org-at-item-p)`` ``- (if recursive`` ``- (<= curr-ind next-ind)`` ``- (= curr-ind next-ind)))`` ``- (when (org-at-item-checkbox-p)`` ``- (if (member (match-string 1) '(\"[ ]\" \"[-]\"))`` ``- (setq c-off (1+ c-off))`` ``- (setq c-on (1+ c-on))))`` ``- (if (not recursive)`` ``- ;; org-get-next-item goes through list-enders`` ``- ;; with proper limit.`` ``- (goto-char (or (org-get-next-item (point) lim) lim))`` ``- (end-of-line)`` ``- (when (org-search-forward-unenclosed`` ``- org-item-beginning-re lim t)`` ``- (beginning-of-line)))`` ``- (setq next-ind (org-get-indentation)))))`` ``- (goto-char continue-from)`` ``- ;; update cookie`` ``- (when end-cookie`` ``- (setq new (if is-percent`` ``- (format \"[%d%%]\" (/ (* 100 c-on)`` ``- (max 1 (+ c-on c-off))))`` ``- (format \"[%d/%d]\" c-on (+ c-on c-off))))`` ``- (goto-char beg-cookie)`` ``- (insert new)`` ``- (delete-region (point) (+ (point) (- end-cookie beg-cookie))))`` ``- ;; update items checkbox if it has one`` ``- (when (and (org-at-item-checkbox-p)`` ``- (> (+ c-on c-off) 0))`` ``- (setq beg-cookie (match-beginning 1)`` ``- end-cookie (match-end 1))`` ``- (delete-region beg-cookie end-cookie)`` ``- (goto-char beg-cookie)`` ``- (cond ((= c-off 0) (insert \"[X]\"))`` ``- ((= c-on 0) (insert \"[ ]\"))`` ``- (t (insert \"[-]\")))))`` ``- (goto-char continue-from)))`` ``- (unless (and all (outline-next-heading)) (throw 'exit nil))))`` ``- (when (interactive-p)`` ``- (message \"Checkbox statistics updated %s (%d places)\"`` ``- (if all \"in entire file\" \"in current outline entry\") cstat)))))`` ``+ (let ((cookie-re \"\\\\(\\\\(\\\\[[0-9]*%\\\\]\\\\)\\\\|\\\\(\\\\[[0-9]*/[0-9]*\\\\]\\\\)\\\\)\")`` ``+ (box-re \"^[ \\t]*\\\\([-+*]\\\\|[0-9]+[.)]\\\\)[ \\t]+\\\\(?:\\\\[@\\\\(?:start:\\\\)?[0-9]+\\\\][ \\t]*\\\\)?\\\\(\\\\[[- X]\\\\]\\\\)\")`` ``+ (recursivep`` ``+ (or (not org-hierarchical-checkbox-statistics)`` ``+ (string-match \"\\\\\"`` ``+ (or (ignore-errors`` ``+ (org-entry-get nil \"COOKIE_DATA\"))`` ``+ \"\"))))`` ``+ (bounds (if all`` ``+ (cons (point-min) (point-max))`` ``+ (cons (or (ignore-errors (org-back-to-heading) (point))`` ``+ (point-min))`` ``+ (save-excursion (outline-next-heading) (point)))))`` ``+ (count-boxes`` ``+ (function`` ``+ ;; add checked boxes and boxes of all types in all`` ``+ ;; structures in STRUCTS to c-on and c-all, respectively.`` ``+ ;; This looks at RECURSIVEP value. If ITEM is nil, count`` ``+ ;; across the whole structure, else count only across`` ``+ ;; subtree whose ancestor is ITEM.`` ``+ (lambda (item structs)`` ``+ (mapc`` ``+ (lambda (s)`` ``+ (let* ((pre (org-list-struct-prev-alist s))`` ``+ (items`` ``+ (if recursivep`` ``+ (or (and item (org-list-get-subtree item s pre))`` ``+ (mapcar 'car s))`` ``+ (or (and item (org-list-get-all-children item s pre))`` ``+ (org-list-get-all-items`` ``+ (org-list-get-top-point s) s pre))))`` ``+ (cookies (delq nil (mapcar`` ``+ (lambda (e)`` ``+ (org-list-get-checkbox e s))`` ``+ items))))`` ``+ (setq c-all (+ (length cookies) c-all)`` ``+ c-on (+ (org-count \"[X]\" cookies) c-on))))`` ``+ structs))))`` ``+ cookies-list backup-end structs-backup)`` ``+ (goto-char (car bounds))`` ``+ ;; 1. Build an alist for each cookie found within BOUNDS. The`` ``+ ;; key will be position at beginning of cookie and values`` ``+ ;; ending position, format of cookie, number of checked boxes`` ``+ ;; to report, and total number of boxes.`` ``+ (while (re-search-forward cookie-re (cdr bounds) t)`` ``+ (save-excursion`` ``+ (let ((c-on 0) (c-all 0))`` ``+ (save-match-data`` ``+ ;; There are two types of cookies: those at headings and those`` ``+ ;; at list items.`` ``+ (cond`` ``+ ((and (org-on-heading-p)`` ``+ (string-match \"\\\\\"`` ``+ (downcase`` ``+ (or (org-entry-get nil \"COOKIE_DATA\") \"\")))))`` ``+ ;; This cookie is at an heading, but specifically for`` ``+ ;; todo, not for checkboxes: skip it.`` ``+ ((org-on-heading-p)`` ``+ (setq backup-end (save-excursion`` ``+ (outline-next-heading) (point)))`` ``+ ;; This cookie is at an heading. Grab structure of`` ``+ ;; every list containing a checkbox between point and`` ``+ ;; next headline, and save them in STRUCTS-BACKUP`` ``+ (while (org-search-forward-unenclosed box-re backup-end 'move)`` ``+ (let* ((struct (org-list-struct))`` ``+ (bottom (org-list-get-bottom-point struct)))`` ``+ (setq structs-backup (cons struct structs-backup))`` ``+ (goto-char bottom)))`` ``+ (funcall count-boxes nil structs-backup))`` ``+ ((org-at-item-p)`` ``+ ;; This cookie is at an item. Look in STRUCTS-BACKUP`` ``+ ;; to see if we have the structure of list at point in`` ``+ ;; it. Else compute the structure.`` ``+ (let ((item (point-at-bol)))`` ``+ (if (and backup-end (< item backup-end))`` ``+ (funcall count-boxes item structs-backup)`` ``+ (setq end-entry bottom`` ``+ structs-backup (list (org-list-struct)))`` ``+ (funcall count-boxes item structs-backup))))))`` ``+ ;; Build the cookies list, with appropriate information`` ``+ (setq cookies-list (cons (list (match-beginning 1) ; cookie start`` ``+ (match-end 1) ; cookie end`` ``+ (match-beginning 2) ; percent?`` ``+ c-on ; checked boxes`` ``+ c-all) ; total boxes`` ``+ cookies-list)))))`` ``+ ;; 2. Apply alist to buffer, in reverse order so positions stay`` ``+ ;; unchanged after cookie modifications.`` ``+ (mapc (lambda (cookie)`` ``+ (let* ((beg (car cookie))`` ``+ (end (nth 1 cookie))`` ``+ (percentp (nth 2 cookie))`` ``+ (checked (nth 3 cookie))`` ``+ (total (nth 4 cookie))`` ``+ (new (if percentp`` ``+ (format \"[%d%%]\" (/ (* 100 checked)`` ``+ (max 1 total)))`` ``+ (format \"[%d/%d]\" checked total))))`` ``+ (goto-char beg)`` ``+ (insert new)`` ``+ (delete-region (point) (+ (point) (- end beg)))))`` ``+ cookies-list))))`` `` `` `` (defun org-get-checkbox-statistics-face ()`` `` \"Select the face for checkbox statistics.``"
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http://www.finance-lib.com/financial-term-yield-to-maturity.html | [
"",
null,
"Financial Terms Yield to maturity\n\n# Definition of Yield to maturity",
null,
"## Yield to maturity\n\nThe percentage rate of return paid on a bond, note or other fixed income security if you\nbuy and hold it to its maturity date. The calculation for YTM is based on the coupon rate, length of time to\nmaturity and market price. It assumes that coupon interest paid over the life of the bond will be reinvested at\nthe same rate.\n\n## Yield to Maturity\n\nThe measure of the average rate of return that will be earned on a\ndebt security held until it matures\n\n## Yield to maturity\n\nA measure of the average rate of return that will be earned\non a bond if held to maturity.\n\n## yield to maturity\n\nInterest rate for which the present value of the bond’s payments equals the price.\n\n# Related Terms:\n\n## Basis price\n\nPrice expressed in terms of yield to maturity or annual rate of return.\n\n## Benchmark interest rate\n\nAlso called the base interest rate, it is the minimum interest rate investors will\ndemand for investing in a non-Treasury security. It is also tied to the yield to maturity offered on a\ncomparable-maturity Treasury security that was most recently issued (\"on-the-run\").\n\n## Bond-equivalent yield\n\nThe annualized yield to maturity computed by doubling the semiannual yield.\n\n## Internal rate of return\n\na. The average annual yield earned by an investment during the period held.\nb. The effective rate of interest on a loan.\nc. The discount rate in discounted cash flow analysis.\nd. The rate that adjusts the value of future cash receipts earned by an investment so that interest earned equals the original cost.\nSee yield to maturity.",
null,
"## Reoffering yield\n\nIn a purchase and sale, the yield to maturity at which the underwriter offers to sell the bonds\nto investors.\n\n## Yield\n\nThe interest rate that makes the present value of a stream of future payments associated with an asset equal to the current price of that asset. Also called yield to maturity. See also current yield.\n\n## yield curve\n\nGraph of the relationship between time to maturity and yield to maturity.\n\n## Yield to worst\n\nThe bond yield computed by using the lower of either the yield to maturity or the yield to call\non every possible call date.\n\n## Annual percentage yield (APY)\n\nThe effective, or true, annual rate of return. The APY is the rate actually\nearned or paid in one year, taking into account the affect of compounding. The APY is calculated by taking\none plus the periodic rate and raising it to the number of periods in a year. For example, a 1% per month rate\nhas an APY of 12.68% (1.01^12).\n\n## Average maturity\n\nThe average time to maturity of securities held by a mutual fund. Changes in interest rates\nhave greater impact on funds with longer average life.\n\n## Balloon maturity\n\nAny large principal payment due at maturity for a bond or loan with or without a a sinking\nfund requirement.\n\n## Bond equivalent yield\n\nBond yield calculated on an annual percentage rate method. Differs from annual\neffective yield.\n\n## Bond Equivalent Yield\n\nBond yield calculated on an annual percentage rate method\n\n## Capital gains yield\n\nThe price change portion of a stock's return.\n\n## Convenience yield\n\nThe extra advantage that firms derive from holding the commodity rather than the future.\n\n## Coupon equivalent yield\n\nTrue interest cost expressed on the basis of a 365-day year.\n\n## Current maturity\n\nCurrent time to maturity on an outstanding debt instrument.\nCurrent / noncurrent method\nUnder this currency translation method, all of a foreign subsidiary's current\nassets and liabilities are translated into home currency at the current exchange rate while noncurrent assets\nand liabilities are translated at the historical exchange rate, that is, the rate in effect at the time the asset was\nacquired or the liability incurred.\n\n## Current yield\n\nFor bonds or notes, the coupon rate divided by the market price of the bond.\n\n## current yield\n\nAnnual coupon payments divided by bond price.\n\n## Current Yield\n\nThe percentage return on a financial asset based on the current price of the asset, without reference to any expected change in the price of the asset. This contrasts with yield-to-maturity, for which the calculation includes expected price changes. See also yield.\n\n## Dividend yield (Funds)\n\nIndicated yield represents return on a share of a mutual fund held over the past 12\nmonths. Assumes fund was purchased 1 year ago. Reflects effect of sales charges (at current rates), but not\nredemption charges.\n\n## dividend yield ratio\n\nCash dividends paid by a business over the most\nrecent 12 months (called the trailing 12 months) divided by the current\nmarket price per share of the stock. This ratio is reported in the daily\nstock trading tables in the Wall Street Journal and other major newspapers.\n\n## Dividend yield (Stocks)\n\nIndicated yield represents annual dividends divided by current stock price.\n\n## Earnings yield\n\nThe ratio of earnings per share after allowing for tax and interest payments on fixed interest\ndebt, to the current share price. The inverse of the price/earnings ratio. It's the Total Twelve Months earnings\ndivided by number of outstanding shares, divided by the recent price, multiplied by 100. The end result is\nshown in percentage.\n\n## Effective annual yield\n\nAnnualized interest rate on a security computed using compound interest techniques.\n\n## Effective Annual Yield\n\nAnnualized rate of return on a security computed using compound\ninterest techniques\n\n## Equivalent bond yield\n\nAnnual yield on a short-term, non-interest bearing security calculated so as to be\ncomparable to yields quoted on coupon securities.\n\n## Equivalent taxable yield\n\nThe yield that must be offered on a taxable bond issue to give the same after-tax\nyield as a tax-exempt issue.\n\n## Flattening of the yield curve\n\nA change in the yield curve where the spread between the yield on a long-term\nand short-term Treasury has decreased. Compare steepening of the yield curve and butterfly shift.\n\n## Held-to-Maturity Security\n\nA debt security for which the investing entity has both the positive\nintent and the ability to hold until maturity.\n\nSee:junk bond.\n\n## Indicated yield\n\nThe yield, based on the most recent quarterly rate times four. To determine the yield, divide\nthe annual dividend by the price of the stock. The resulting number is represented as a percentage. See:\ndividend yield.\n\n## labor yield variance\n\n(standard mix X actual hours X standard rate) - (standard mix X standard hours X standard rate);\nit shows the monetary impact of using more or fewer total hours than the standard allowed\n\n## Liquid yield option note (LYON)\n\nZero-coupon, callable, putable, convertible bond invented by Merrill\n\n## Liquid yield option note (LYON)\n\nZero-coupon, callable, putable, convertible bond invented by Merrill Lynch & Co.\n\n## material yield variance\n\n(standard mix X actual quantity X standard price) - (standard mix X standard quantity X standard price);\nit computes the difference between the\nactual total quantity of input and the standard total quantity\nallowed based on output and uses standard mix and\nstandard prices to determine variance\n\n## Maturity\n\nFor a bond, the date on which the principal is required to be repaid. In an interest rate swap, the\ndate that the swap stops accruing interest.\n\n## Maturity\n\nThe date or the number of days until a security is due to be paid or\na loan is to be repaid\n\n## Maturity\n\nTime at which a bond can be redeemed for its face value.\n\n## Maturity\n\nThe time when a policy or annuity reaches the end of its span.\n\n## Maturity date\n\nThe date when the issuer returns the final face value of a bond\n\n## Maturity Date\n\nDate on which a debt is due for payment.\n\n## Maturity factoring\n\nFactoring arrangement that provides collection and insurance of accounts receivable.\n\n## Maturity phase\n\nA phase of company development in which earnings continue to grow at the rate of the\ngeneral economy. Related: Three-phase DDM.\n\nExtra average return from investing in longversus short-term Treasury securities.\n\nThe spread between any two maturity sectors of the bond market.\n\n## Maturity value\n\nRelated: par value.\n\n## Non-parallel shift in the yield curve\n\nA shift in the yield curve in which yields do not change by the same\nnumber of basis points for every maturity. Related: Parallel shift in the yield curve.\n\n## Original maturity\n\nmaturity at issue. For example, a five year note has an original maturity of 5 years; one\nyear later it has a maturity of 4 years.\n\n## Par yield curve\n\nThe yield curve of bonds selling at par, or face, value.\n\n## Parallel shift in the yield curve\n\nA shift in the yield curve in which the change in the yield on all maturities is\nthe same number of basis points. In other words, if the 3 month T-bill increases 100 basis points (one\npercent), then the 6 month, 1 year, 5 year, 10 year, 20 year, and 30 year rates increase by 100 basis points as\nwell.\nRelated: Non-parallel shift in the yield curve.\n\n## process quality yield\n\nthe proportion of good units that resulted from the activities expended\n\n## Production yield variance\n\nThe difference between the actual and budgeted proportions\nof product resulting from a production process, multiplied by the standard unit cost.\n\n## Projected maturity date\n\nWith CMOs, final payment at the end of the estimated cash flow window.\n\n## Pure yield pickup swap\n\nMoving to higher yield bonds.\n\n## Realized compound yield\n\nyield assuming that coupon payments are invested at the going market interest\nrate at the time of their receipt and rolled over until the bond matures.\n\nThe ratio of the yield spread to the yield level.\n\n## Remaining maturity\n\nThe length of time remaining until a bond's maturity.\n\n## Required yield\n\nGenerally referring to bonds, the yield required by the marketplace to match available returns\nfor financial instruments with comparable risk.\n\n## Return-to-maturity expectations\n\nA variant of pure expectations theory which suggests that the return that an\ninvestor will realize by rolling over short-term bonds to some investment horizon will be the same as holding\na zero-coupon bond with a maturity that is the same as that investment horizon.\n\n## Riding the yield curve\n\nBuying long-term bonds in anticipation of capital gains as yields fall with the\ndeclining maturity of the bonds.\n\nSee Zero curve.\n\n## Stated maturity\n\nFor the CMO tranche, the date the last payment would occur at zero CPR.\n\n## Steepening of the yield curve\n\nA change in the yield curve where the spread between the yield on a long-term\nand short-term Treasury has increased. Compare flattening of the yield curve and butterfly shift.\n\n## Term to maturity\n\nThe time remaining on a bond's life, or the date on which the debt will cease to exist and\nthe borrower will have completely paid off the amount borrowed. See: maturity.\n\n## Term to Maturity\n\nPeriod of time from the present to the redemption date of a bond.\n\n## Time to maturity\n\nThe time remaining until a financial contract expires. Also called time until expiration.\n\n## Weighted average maturity\n\nThe WAM of a MBS is the weighted average of the remaining terms to maturity\nof the mortgages underlying the collateral pool at the date of issue, using as the weighting factor the balance\nof each of the mortgages as of the issue date.\n\n## Weighted average portfolio yield\n\nThe weighted average of the yield of all the bonds in a portfolio.\n\n## Weighted average remaining maturity\n\nThe average remaining term of the mortgages underlying a MBS.\n\n## Yield\n\nThe percentage rate of return paid on a stock in the form of dividends, or the effective rate of interest\npaid on a bond or note.\n\n## yield\n\nthe quantity of output that results from a specified input\n\n## Yield\n\na. Measure of return on an investment, stated as a percentage of price.\nyield can be computed by dividing return by purchase price, current market\nvalue, or other measure of value.\nb. Income from a bond expressed as an\nannualized percentage rate.\nc. The nominal annual interest rate that gives a\nfuture value of the purchase price equal to the redemption value of the security.\nAny coupon payments determine part of that yield.\n\n## Yield curve\n\nThe graphical depiction of the relationship between the yield on bonds of the same credit quality\nbut different maturities. Related: Term structure of interest rates. Harvey (1991) finds that the inversions of\nthe yield curve (short-term rates greater than long term rates) have preceded the last five U.S. recessions. The\nyield curve can accurately forecast the turning points of the business cycle.\n\n## Yield Curve\n\nA graphical representation of the level of interest rates for\nsecurities of differing maturities at a specific point of time\n\n## Yield curve\n\nGraph of yields (vertical axis) of a particular type of security\nversus the time to maturity (horizontal axis). This curve usually slopes\nsecurities that have a longer time to maturity. The benchmark yield curve is\nfor U.S. Treasury securities with maturities ranging from three months to 30\nyears. See Term structure.\n\n## Yield curve\n\nA graph showing how the yield on bonds varies with time to maturity.\n\n## Yield curve option-pricing models\n\nModels that can incorporate different volatility assumptions along the\nyield curve, such as the Black-Derman-Toy model. Also called arbitrage-free option-pricing models.\n\n## Yield curve strategies\n\nPositioning a portfolio to capitalize on expected changes in the shape of the Treasury yield curve.\n\n## Yield ratio\n\nThe quotient of two bond yields.\n\n## yield ratio\n\nthe expected or actual relationship between input and output\n\nStrategies that involve positioning a portfolio to capitalize on expected changes in\nyield spreads between sectors of the bond market.\n\n## Yield to call\n\nThe percentage rate of a bond or note, if you were to buy and hold the security until the call date.\nThis yield is valid only if the security is called prior to maturity. Generally bonds are callable over several\nyears and normally are called at a slight premium. The calculation of yield to call is based on the coupon rate,\nlength of time to the call and the market price.\n\n## Zero curve, zero-coupon yield curve\n\nA yield curve for zero-coupon bonds;\nzero rates versus maturity dates. Since the maturity and duration (Macaulay\nduration) are identical for zeros, the zero curve is a pure depiction of supply/\ndemand conditions for loanable funds across a continuum of durations and\nmaturities. Also known as spot curve or spot yield curve.\n\n## Back-up\n\n1) When bond yields and prices fall, the market is said to back-up.\n2) When an investor swaps out of one security into another of shorter current maturity he is said to back up.\n\n## Bootstrapping\n\nA process of creating a theoretical spot rate curve , using one yield projection as the basis for\nthe yield of the next maturity.\n\n## Corporate taxable equivalent\n\nRate of return required on a par bond to produce the same after-tax yield to\nmaturity that the premium or discount bond quoted would.\n\n## Current coupon\n\nA bond selling at or close to par, that is, a bond with a coupon close to the yields currently\noffered on new bonds of a similar maturity and credit risk.\n\n## Duration\n\nThe expected life of a fixed-income security considering its coupon\nyield, interest payments, maturity, and call features. As market interest rates\nrise, the duration of a financial instrument decreases. See Macaulay duration.\n\n## Floating-rate contract\n\nA guaranteed investment contract where the credit rating is tied to some variable\n(\"floating\") interest rate benchmark, such as a specific-maturity Treasury yield.\n\n## Forward rate\n\nThe future interest rate of a bond inferred from the term\nstructure, especially from the yield curve of zero-coupon bonds, calculated from\nthe growth factor of an investment in a zero held until maturity.\n\n## Macaulay duration\n\nA widely used measure of price sensitivity to yield\nchanges developed by Frederick Macaulay in 1938. It is measured in years and\nis a weighted average-time-to-maturity of an instrument. The Macaulay\nduration of an income stream, such as a coupon bond, measures how long, on\naverage, the owner waits before receiving a payment. It is the weighted\naverage of the times payments are made, with the weights at time T equal to\nthe present value of the money received at time T.\n\n## Market segmentation theory or preferred habitat theory\n\nA biased expectations theory that asserts that the\nshape of the yield curve is determined by the supply of and demand for securities within each maturity sector.\n\n## Substitution swap\n\nA swap in which a money manager exchanges one bond for another bond that is similar in\nterms of coupon, maturity, and credit quality, but offers a higher yield.\n\nExcess of the yields to maturity on long-term bonds over those of short-term bonds.\n\n## Term structure\n\nThe relationship between the yields on fixed-interest\nsecurities and their maturity dates. Expectation of changes in interest rates\naffects term structure, as do liquidity preferences and hedging pressure. A\nyield curve is one representation in the term structure."
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https://blog.vacs.fr/vacs/wikis/view/1/VACS:Projects:Hardware:LPS | [
"Software\n\nHardware\n\n## LPS Hardware Description\n\nVersion 1 by Stephane Carrez\n\n## Architecture Overview",
null,
"LPS Hardware Architecture\n\n• red lines indicate hot points\n• black lines indicate the ground, signal or control lines\n• blue lines indicate external connection (remote control and programming)\n• strong line will draw more current\n\n## Transformers",
null,
"Toric Transformer\nTwo toric transformers are used: one for the positive voltage and one for the negative voltage. This choice was motivated by:\n\n• the low profile of these transformers\n• the need for at least 4x9V supply to optimize the 4 output supplies while reducing the total heat dissipation (for low voltage outputs)\n• the room available in the final LPS case\n\nEach transformer will deliver 80VA on 2x9V. This represents a maximum current arround 4.4A. The toric transformer used for positive voltage is also used for the power supply of LPS itself. This means that it is possible to use a single toric transformer but LPS will only deliver positive voltage.\n\n## LPS Power Supply\n\nThe power supply controllers and the master power supply board need themselves a power supply. A positive power supply controller uses the following supplies:\n\n+24V, -5V:A power supply controller contains operational amplifiers to drive the output power supply transistor. The +24V and -5V was chosen so that the outputs of the operational amplifiers will have a full range of 0 to more than 18V. The AOP that is used is a TL074 whose maximum supply voltage is 36V.\n+5V:The ATmega8 and other control logic power supply.\n\nThe negative power supply controller will use a different voltage for the operational amplifiers. A +5V and -24V is used so that the outputs can follow the required voltage to drive the power supply transistors.\n\nTo obtain the LPS power supplies we use the toric transformer of the positive power supply. The +24V and -24V are obtained by a voltage multiplier from the 2x9V toric transformers. In theory it can reach 36V so the decoupling capacitor is chosen to support at least 50V.\n\n## Rectifiers\n\nThe rectifier board contains the diodes to convert the alternating current into direct current. Four diodes are used to convert the 2x9V into two positive unregulated outputs. The diodes form a bridge rectifier connected to the 2x9V transformer outputs which are in serial.\n\nTwo 22000uF capacitors are present on each output to produce a steady DC. The DC outputs may vary between 11V and 13V due to the toric transformer. Indeed the toric transformer has an output that could be 10% higher without any load.\n\n18 \\times \\sqrt{2} - 1.4V = 24V\n\n11.7V .. 12.9V 23.4V .. 25.7V\n\nUnder a maximum load of 3A (thus overloading the transformer) the output voltage swing will be arround 2.7V for the 25V output and 1.3V for the 12V output (at 100Hz).\n\ndV = \\frac{I \\times 0.007}{C}\n\n## Positive Power Supply Controller Board\n\nThe Power Supply Controller Board controls two power supplies. One is a fixed voltage and the other is variable.\n\n### Fixed Power Supply\n\nThis power supply has a fixed output voltage that is adjusted by a small trimmer at the rear of LPS. The trimmer will not allow a full range of output voltage. This power supply is either connected to the 11V unregulated supply or to the 22V supply. In the first case, it will allow a fixed power supply in the range 0..9V and in the second case a fixed power supply in the range 9..20V.\n\n### Variable Power Supply\n\nThe variable power supply allows a full voltage range of 0 to 20V. The voltage is controlled by a 8-bit DAC that controls a current between 0 and 2mA. That current is sinked from a resistor that is connected to the output voltage. The higher the current the higher the output voltage. The resistor value is chosen with the following formulae: R = \\frac{Vmax}{2mA}\n\nVmax = 20V => R = 10K\n\nThe current reference is created from the 2.5V reference voltage. The DAC resistor is chosen with the following formulae: R = \\frac{Vref}{Iref} Vref = 2.5V, Iref = 2mA => R = 1250\n\n### Measuring the current\n\nTo measure the output current a resistor is used in the path of the power transistor controlling the output voltage. In fact the resistor is composed of two parallel networks composed of a resistor followed by a high power MOSFET transistor. Each MOSFET transistor can be switched on or off. When both are off, the corresponding output voltage is switched off. This feature is used by the protection system controlled by the power supply controller. To compute the current we just need to measure the voltage of the resistor. We must take into account the Rdson of each MOSFET because they are in the current path. The IRF9540 has an Rdson arround 0.117 ohm and some others have an Rdson that can reach 0.48 ohm. In that later case it is possible to remove the resistor to only have two parallel MOSFETs. The table gives the different values of R1 when at least one MOSFET is on.\n\nR1 Rdson max I max V max (Rdson = 0) V max (Rdson max)\n0.66 (3W) 0.117 2.13A 1.4V 1.65V\n0.22 (3W) 0.117 3.69A 0.81V 1.24V\n0.165 0.117 4.90A/5.28A 0.81V 1.24V\n\nTwo ways can be used to measure the current or voltage arround R1.",
null,
"Current Measurement\nThe current is measured with an operational amplifier. The output current i creates a voltage drop in the resistor R1 that is used to measure the output current. The operational amplifier drives the transistor Q1 to maintain the same voltage drop on resistor R2. The current ic which flows in R2 is therefore proportional to the output current i. The current ic goes in R3 and creates a voltage which depends on i. The voltage on R3 is defined by the formula:\n\n R3 * R1\nVs = --------- * i\nR2\n\n\nThe operational amplifier has its inputs which can range from 0 to the maximum output voltage supported by the power supply. Therefore we must provide it a power supply that is 1 or 2 volts greater than the maximum output voltage.\n\nA second issue more important occurs when the output power supply has a low voltage and high output current. We have a constraint that Vs i.\n\nR1 R2 R3 i Vs\n0.66 2.2K 15K 1.2A 5V\n0.165 2.2K 15K 4.8A 5.4V",
null,
"Current Measurement\nThe second method is not really inovative. We use an operational amplifier in differential mode. If we assume that R2 = R4 and R3 = R5, the output voltage Vs is defined by the formula:\n\n R3 * R1\nVs = -------- * i\nR2\n\n\nThe gain created by R3/R2 is chosen to be near 4 to maximize the output voltage swing for the analog converter.\n\nOne advantage of the differential solution is that we do not have the limitation or constraint on the output voltage. It also offers a low impedance result due to the operational amplifier output."
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"https://blog.vacs.fr/vacs/wikis/images/1/-1/256x/study_current_measure.png",
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"https://blog.vacs.fr/vacs/wikis/images/1/-1/256x/study_current_measure2.png",
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https://docs.w3cub.com/c/numeric/complex/cmplx | [
"/C\n\n# CMPLXF, CMPLX, CMPLXL\n\nDefined in header `<complex.h>`\n`float complex CMPLXF( float real, float imag );`\n(since C11)\n`double complex CMPLX( double real, double imag );`\n(since C11)\n`long double complex CMPLXL( long double real, long double imag );`\n(since C11)\n\nEach of these macros expands to an expression that evaluates to the value of the specified complex type, with the real part having the value of `real` (converted to the specified argument type) and the imaginary part having the value of `imag` (converted to the specified argument type).\n\nThe expressions are suitable for use as initializers for objects with static or thread storage duration, as long as the expressions `real` and `imag` are also suitable.\n\n### Parameters\n\n real - the real part of the complex number to return imag - the imaginary part of the complex number to return\n\n### Return value\n\nA complex number composed of `real` and `imag` as the real and imaginary parts.\n\nThese macros are implemented as if the imaginary types are supported (even if they are otherwise not supported and `_Imaginary_I` is actually undefined) and as if defined as follows:\n\n```#define CMPLX(x, y) ((double complex)((double)(x) + _Imaginary_I * (double)(y)))\n#define CMPLXF(x, y) ((float complex)((float)(x) + _Imaginary_I * (float)(y)))\n#define CMPLXL(x, y) ((long double complex)((long double)(x) + \\\n_Imaginary_I * (long double)(y)))```\n\n### Example\n\n```#include <stdio.h>\n#include <complex.h>\n\nint main(void)\n{\ndouble complex z = CMPLX(0.0, -0.0);\nprintf(\"z = %.1f%+.1fi\\n\", creal(z), cimag(z));\n}```\n\nOutput:\n\n`z = 0.0-0.0i`\n• C11 standard (ISO/IEC 9899:2011):\n• 7.3.9.3 The CMPLX macros (p: 197)"
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http://www.popgen.dk/angsd/index.php?title=SYKmaf&diff=cur&oldid=1845 | [
"ANGSD: Analysis of next generation Sequencing Data\n\nLatest tar.gz version is (0.938/0.939 on github), see Change_log for changes, and download it here.\n\n# Difference between revisions of \"SYKmaf\"\n\n## ML estimator with known minor\n\nFirst infer the Major and Minor allele and then use BFGS (-doMaf 1) optimazation or the EM algorithm (-doMaf 2) to estimate the allele frequencies.",
null,
"$L(D|f) \\propto \\prod_i^N p(D_i|f) = \\prod_i^N \\sum_{g\\in\\{0,1,2\\}}p(D_i|G=g)p(G=g|f)$",
null,
"$\\hat{f}=argmax_{f} L(D|f)$\n\n## ML estimator with unknown minor\n\nFirst infer the Major allele and then use BFGS (-doMaf 4) optimazation or the EM algorithm (-doMaf 8) to estimate the allele frequencies. Here only the Major allele needs to be known and the uncertaincy of infering the minor allele is modelled.\n\nLet",
null,
"$\\{M,m\\}$ denote the major an minor allele assuming adiallelic site, then the maximum likelihood estimate of this pair is found using the likelihood function",
null,
"$P(D|M,f) = \\prod_i P(D_i|M,f) = \\sum_m \\sum_{A_1,A_2} P(D_i|G=A_1A_2)p(G=A_1A_2|m,M)p(m),$"
]
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"http://www.popgen.dk/angsd/images/math/4/5/1/4514da6ddb7b67a884d7d65989ec34dd.png ",
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"http://www.popgen.dk/angsd/images/math/9/b/4/9b4562c91c7636c3d1512ff86a15f126.png ",
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"http://www.popgen.dk/angsd/images/math/e/f/e/efe71df648873c03a1a6ab17e1a5e2a4.png ",
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"http://www.popgen.dk/angsd/images/math/f/6/5/f65ce2b03f6d3d66f6d7e47e7eb8b39e.png ",
null
]
| {"ft_lang_label":"__label__en","ft_lang_prob":0.84730375,"math_prob":0.99098194,"size":1091,"snap":"2022-27-2022-33","text_gpt3_token_len":265,"char_repetition_ratio":0.14535418,"word_repetition_ratio":0.21387284,"special_character_ratio":0.23006417,"punctuation_ratio":0.08163265,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9984757,"pos_list":[0,1,2,3,4,5,6,7,8],"im_url_duplicate_count":[null,null,null,null,null,null,null,6,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-08-13T18:30:04Z\",\"WARC-Record-ID\":\"<urn:uuid:f8da1530-34f8-4b91-a2ec-574c70685698>\",\"Content-Length\":\"25851\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:a790dcc9-5a12-4ac2-af89-50bfa0d83583>\",\"WARC-Concurrent-To\":\"<urn:uuid:aedaec2e-b885-4997-8699-5e95fa7056a8>\",\"WARC-IP-Address\":\"130.225.98.44\",\"WARC-Target-URI\":\"http://www.popgen.dk/angsd/index.php?title=SYKmaf&diff=cur&oldid=1845\",\"WARC-Payload-Digest\":\"sha1:ONVPEOT5U646B5AEDV47UNMDKSVTP24V\",\"WARC-Block-Digest\":\"sha1:X5WV6UHBBJTOWG5X5MNOMNMBWEKP27BT\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-33/CC-MAIN-2022-33_segments_1659882571982.99_warc_CC-MAIN-20220813172349-20220813202349-00533.warc.gz\"}"} |
https://e-baketabam.ir/shop/business-investment/P2751-the-power-law-venture-capital-and-the-making-of-the-new-future-sebastian-mallaby.html | [
"",
null,
"۰\nسبد خرید",
null,
"هر روز با کتابهای بیشتر",
null,
"# The Power Law: Venture Capital and the Making of the New Future | Sebastian Mallaby\n\nکد محصول: eSHB-2678\n۳۸,۳۸۰ تومان۱۹,۱۹۰ تومان",
null,
"",
null,
"",
null,
"افزودن به سبد خرید\n• درباره کتاب\n• مطالعه راحت\n• بخشی از کتاب\n• نظرات\n•",
null,
"• ## تخفیف ویژه | اولین سفارش\n\n<% if (product.thumbnail) { %>",
null,
"<% } %>\n\n### <%- product.title %>\n\n<% if (Array.isArray(product.attributes)) { %>\n<% _.forEach(product.attributes, function(attribute, index) { %>\n• <%- attribute.name %>: <%- attribute.value %>\n• <% }); %>\n<% } %><% if (product.price_label) { %>\n<%- (product.price_label.toString()) %>\n<% } else { %><% if (product.in_stock == 1) { %>\n<% if (product.sale_price) { %><%- (product.price.toString().formatNumber().convertToLocalNumber() + currency_sign) %><%- (product.sale_price.toString().formatNumber().convertToLocalNumber() + currency_sign) %><% } else { %><%- (product.price.toString().formatNumber().convertToLocalNumber() + currency_sign) %><% } %>\n<% } else { %>\nاتمام موجودی\n<% } %><% } %>\n<% if (product.ribbon) { %>\n<%- product.ribbon %>\n<% } %><% if (product.sale_amount) { %><% if (product.in_stock==1) { %>\n<% if (product.sale_type==2) { %> <%- ((product.sale_amount).toString().formatNumber().convertToLocalNumber() + currency_sign) %><% } else { %> <%- (product.sale_amount.toString().formatNumber().convertToLocalNumber()) %> درصد <% } %>\n<% } %><% } %>",
null,
"رمز عبورتان را فراموش کردهاید؟\n\nثبت کلمه عبور خود را فراموش کردهاید؟ لطفا شماره همراه یا آدرس ایمیل خودتان را وارد کنید. شما به زودی یک ایمیل یا اس ام اس برای ایجاد کلمه عبور جدید، دریافت خواهید کرد.\n\nبازگشت به بخش ورود\n\nکد دریافتی را وارد نمایید.\n\nبازگشت به بخش ورود\n\n### مشاهده سفارش\n\n<%- order.customer_name.toString() %>\n<%- order.id.toString().convertToLocalNumber() %>\n<%- order.customer_province.toString() %>-<%- order.customer_city.toString() %>-<%- order.customer_address.toString() %>\n<%- order.customer_mobile.toString().convertToLocalNumber() %>\n<%- order.shipping_name.toString() %>\n<%- (Number(order.total_shipping).toString().formatNumber().convertToLocalNumber() + currency_sign) %>\n<%- order.payment_method_name.toString() %>\n<%- (Number(order.total).toString().formatNumber().convertToLocalNumber() + currency_sign) %>\n<% if(order.tracking_number) { %>\n<%- order.tracking_number.toString() %>\n<% } %>\nنام محصول\nتعداد\nقیمت واحد\nقیمت کل\nتخفیف\nقیمت نهایی\n<% \\$.each(products, function(index,product) { %>",
null,
"<%- product.name %>\n<%- product.quantity %>\n<%- (Number(product.original_price).toString().formatNumber().convertToLocalNumber() + currency_sign) %>\n<%- (Number(product.original_price*product.quantity).toString().formatNumber().convertToLocalNumber() + currency_sign) %>\n<%- (Number(product.discount).toString().formatNumber().convertToLocalNumber() + currency_sign) %>\n<%- (Number(product.total).toString().formatNumber().convertToLocalNumber() + currency_sign) %>\n<% }); %>\n<% if(!orders.length) { %>\n\nشما هنوز هیچ سفارشی ثبت نکردهاید.\n\n<% } else { %>\n• شماره سفارش\nتاریخ سفارش\nپرداخت\nوضعیت\nجمع نهایی\n• <% \\$.each(orders, function(index,order) { %>\n• <%- order.id.toString().convertToLocalNumber() %>"
]
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null,
"https://e-baketabam.ir/uploads/83433fff0fed4ca4ba52b5bd170c3ae6.w_25,h_25,r_k.png",
null,
"https://e-baketabam.ir/uploads/99028db945bf4bf7a2c8b65ad16c749d.w_210,h_105,r_k.png",
null,
"https://e-baketabam.ir/uploads/437304dddcf34502861d4e6526b672fe.w_30,h_30,r_k.png",
null,
"https://e-baketabam.ir/uploads/b796015d438543f18f71c21366dfa1ac.png",
null,
"https://e-baketabam.ir/uploads/d1fd020fc36742e69af0263a5f7ce165.png",
null,
"https://e-baketabam.ir/uploads/00c7c4f7d86249c5b10ba3f2828fb326.png",
null,
"https://e-baketabam.ir/uploads/e530f8d2a9e046aca6e39d7cda2270f4.w_1170,h_2071,r_k.png",
null,
"https://e-baketabam.ir/shop/business-investment/<%- (product.thumbnail.toString().thumb(748, 90)) %>",
null,
"https://e-baketabam.ir/uploads/339129f1b95741988d4adbdc26e692ab.w_90,h_30,r_k.png",
null,
"https://e-baketabam.ir/shop/business-investment/<%- product.thumbnail %>",
null
]
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https://www.hackmath.net/en/word-math-problems/9th-grade-(14y)?page_num=54 | [
"# Examples for 9th grade - page 54\n\n1. 5 pumps",
null,
"Five pump to pump 3 hours 1600 hl of water. How long pumped same amount of water 4 pumps?\n2. Average speed",
null,
"What is the average speed you have to move the way around the world in 80 days? (Path along the equator, round to km/h).\n3. Alcohol 2",
null,
"Two types of alcohol one 63% and second 75% give 20 liters of 69% alcohol. How many liters of each type are in the mixture?\n4. Colour - billboard",
null,
"Shelftalker has the shape of a parallelogram. Its length is 4.9 m and the corresponding height is 3.5 meters. Calculate how much (kg) paint must purchase to redecoration if 1 kg cover 4 m2 of shelftalker surface.\n5. Triangle IRT",
null,
"In isosceles right triangle ABC with right angle at vertex C is coordinates: A (-1, 2); C (-5, -2) Calculate the length of segment AB.\n6. Sugar cubes",
null,
"The glass has 600 ml of tea, which represents 80% of the volume of the glass. If you put twenty regular sugar cubes of 2 cm in the tea, how many ml of tea are poured?\n7. Tank 9",
null,
"The tank with volume V is filled with one pump for three hours and by second pump for 5 hours. When both pumps will run simultaneously calculate: a) how much of the total volume of the tank is filled in one hour b) for how long is the tank full\n8. Cube 9",
null,
"What was the original edge length of the cube if after cutting 39 small cubes with an edge length 2 dm left 200 dm3?\n9. ABS",
null,
"What is the value of ? ?\n10. Cube volume",
null,
"The cube has a surface of 384 cm2. Calculate its volume.\n11. Painting school",
null,
"Redecoration school had scheduled for four crew to 10 days. After two days a single worker ill. For how many days the school was painted when remain workers working with the same tempo?\n12. Slope form",
null,
"Find the equation of a line given the point X(8, 1) and slope -2.8. Arrange your answer in the form y = ax + b, where a, b are the constants.\n13. Square - increased perimeter",
null,
"How many times is increased perimeter of the square, where its sides increases by 150%? If the perimeter of square will increase twice, how much% increases the content area of the square?\n14. Classs mean",
null,
"Class size is 30 students, math was not worse than sign 2. Determine the number of pupils who were sign 1 if the class had a 1.4 sign average.\n15. Holiday SMS",
null,
"Kucera and Malek start from the holiday together. They agreed that if they be 100 km apart, will send a SMS. Kucera traveling at 60 km/h., Malek 90 km/h. They started at 14 hours PM. At what time they send a message?\n16. Rope slack",
null,
"Between two streets 20 m away give the lamp, which is in the middle and hanging 60 cm bellow the taut rope. Can be done with 20.5 meters rope?\n17. Triangle",
null,
"The triangle has known all three sides: a=5.5 m, b=5.3 m, c= 7.8 m. Calculate area of this triangle.\n18. Copy typist",
null,
"Typist for 12 hours rewritten 15% of the manuscript. After how many hours he will done 35% of the manuscript?\n19. Seven tractors",
null,
"Seven tractor plowed 308 ha field in four days. How many hectares of fields will plowe 9 tractors in 3 days at the same power?\n20. Rhombus and diagonals",
null,
"The a rhombus area is 150 cm2 and the ratio of the diagonals is 3:4. Calculate the length of its height.\n\nDo you have an interesting mathematical word problem that you can't solve it? Enter it, and we can try to solve it.\n\nTo this e-mail address, we will reply solution; solved examples are also published here. Please enter the e-mail correctly and check whether you don't have a full mailbox.\n\nPlease do not submit problems from current active competitions such as Mathematical Olympiad, correspondence seminars etc..."
]
| [
null,
"https://www.hackmath.net/thumb/63/t_1663.jpg",
null,
"https://www.hackmath.net/thumb/17/t_3917.jpg",
null,
"https://www.hackmath.net/thumb/46/t_1146.jpg",
null,
"https://www.hackmath.net/thumb/61/t_1961.jpg",
null,
"https://www.hackmath.net/thumb/40/t_2040.jpg",
null,
"https://www.hackmath.net/thumb/6/t_4806.jpg",
null,
"https://www.hackmath.net/thumb/96/t_2096.jpg",
null,
"https://www.hackmath.net/thumb/6/t_2306.jpg",
null,
"https://www.hackmath.net/thumb/0/t_700.jpg",
null,
"https://www.hackmath.net/thumb/99/t_2399.jpg",
null,
"https://www.hackmath.net/thumb/52/t_1552.jpg",
null,
"https://www.hackmath.net/thumb/35/t_1135.jpg",
null,
"https://www.hackmath.net/thumb/19/t_2919.jpg",
null,
"https://www.hackmath.net/thumb/91/t_2091.jpg",
null,
"https://www.hackmath.net/thumb/80/t_1780.jpg",
null,
"https://www.hackmath.net/thumb/16/t_1616.jpg",
null,
"https://www.hackmath.net/thumb/95/t_1195.jpg",
null,
"https://www.hackmath.net/thumb/35/t_1535.jpg",
null,
"https://www.hackmath.net/thumb/64/t_1564.jpg",
null,
"https://www.hackmath.net/thumb/6/t_4406.jpg",
null
]
| {"ft_lang_label":"__label__en","ft_lang_prob":0.8960403,"math_prob":0.96382964,"size":3771,"snap":"2019-43-2019-47","text_gpt3_token_len":1200,"char_repetition_ratio":0.09636316,"word_repetition_ratio":0.0,"special_character_ratio":0.3813312,"punctuation_ratio":0.0931677,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9935907,"pos_list":[0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28,29,30,31,32,33,34,35,36,37,38,39,40],"im_url_duplicate_count":[null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-11-21T00:33:24Z\",\"WARC-Record-ID\":\"<urn:uuid:13ed0825-a269-4957-9422-3b5036147c1a>\",\"Content-Length\":\"33458\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:a1975879-b371-4981-82f0-a09141de49f8>\",\"WARC-Concurrent-To\":\"<urn:uuid:e2adc55b-6e48-4f1a-80dd-2a75d62295a0>\",\"WARC-IP-Address\":\"104.24.104.91\",\"WARC-Target-URI\":\"https://www.hackmath.net/en/word-math-problems/9th-grade-(14y)?page_num=54\",\"WARC-Payload-Digest\":\"sha1:EFVMWGD43TOF5JGYFAVB2ARQCMCJSS7P\",\"WARC-Block-Digest\":\"sha1:DBCWIF5DE3BUC4ETFPRR6IDQNYCSYXFR\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-47/CC-MAIN-2019-47_segments_1573496670643.58_warc_CC-MAIN-20191121000300-20191121024300-00377.warc.gz\"}"} |
https://newszmint.com/heap-sort-program-in-c-language/ | [
"## Heap Sort program in C Language\n\nHeap Sort program in C Language algorithm implementation – Heap sort is two types first is min heap and second is max heap processes the elements by creating the min heap or max heap using the elements of the given array. Min heap or max heap represents the ordering of the array in which root element represents the minimum or maximum element of the array.\n\n## Complexity\n\n• Best Case : Time Complexity Ω(n log (n))\n• Average Case: Time Complexity θ(n log (n))\n• Worst case: Time Complexity O(n log (n))\n• Space Complexity O(1)\n\nIn this program user would be asked to enter the number of elements along with the element values and then the programs would sort them in ascending order by using merge sorting algorithm logic.\n\n## Heap Sort algorithm C Program implementation\n\n``````#include<stdio.h>\nint temp;\n\nvoid heapify(int arr[], int size, int i)\n{\nint largest = i;\nint left = 2*i + 1;\nint right = 2*i + 2;\n\nif (left < size && arr[left] >arr[largest])\nlargest = left;\n\nif (right < size && arr[right] > arr[largest])\nlargest = right;\n\nif (largest != i)\n{\ntemp = arr[i];\narr[i]= arr[largest];\narr[largest] = temp;\nheapify(arr, size, largest);\n}\n}\n\nvoid heapSort(int arr[], int size)\n{\nint i;\nfor (i = size / 2 - 1; i >= 0; i--)\nheapify(arr, size, i);\nfor (i=size-1; i>=0; i--)\n{\ntemp = arr;\narr= arr[i];\narr[i] = temp;\nheapify(arr, i, 0);\n}\n}\n\nvoid main()\n{\nint arr[] = {2, 11, 3, 4, 5, 2, 3, 101,24, 3};\nint i;\nint size = sizeof(arr)/sizeof(arr);\n\nheapSort(arr, size);\n\nprintf(\"printing sorted elements here\\n\");\nfor (i=0; i<size; ++i)\nprintf(\"%d\\n\",arr[i]);\n} ``````\n\n## OUTPUT\n\n``````\n\nprinting sorted elements here\n\n2\n2\n3\n3\n3\n4\n5\n11\n24\n101\n\n``````"
]
| [
null
]
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https://slideplayer.com/slide/6040961/ | [
"",
null,
"# Rules for Multiplying and Dividing Integers\n\n## Presentation on theme: \"Rules for Multiplying and Dividing Integers\"— Presentation transcript:\n\nRules for Multiplying and Dividing Integers\nIntro to Algebra\n\nLet’s Review What are integers? How do you add two positives?\nHow do you add two negatives? How do you add a positive and negative? What is the first thing you should do when you have a subtraction problem, and how do you do it? What do we get if we add opposites?\n\nMultiplying and Dividing Integers\nThe rules are simple! Multiply or Divide the Numbers Normally Count the number of Negatives If the amount of Negatives is Even, Your answer is Positive! If the amount of Negatives is Odd, Your answer is Negative!\n\nHere is a way to remember.\nTo remember whether your answer will be positive or negative when multiplying or dividing two numbers, you can use Mr. Multiplivision!\n\nTo use Mr. Multiplivision\nYou will use Mr. Multiplivision when multiplying or dividing two numbers, and at least one of them is negative. Cover up the two signs you are multiplying or dividing.\n\nDetermine if the answer will be positive or negative.\n(-5) · (-25) 16 · 24 (-48) ÷ (-7) (-2) · 18 72 ÷ (-9)"
]
| [
null,
"https://slideplayer.com/static/blue_design/img/slide-loader4.gif",
null
]
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http://www.onlinemathpro.com/tutorials/graphing-quadratic-functions-1---locating-the-vertex | [
"Tutorials > \n\n### Graphing Quadratic Functions 1 - Locating the Vertex\n\n Recall: 1) The graph of a quadratic function is called a Parabola. 2) The Vertex of a parabola is the point at which the parabola stops falling and begins rising (or vise versa). In other words, it is the point at which the parabola officially “turns the corner”. {It can also be thought of as the only point on the parabola at which the slope is equal to zero.} 3) There are three common forms of the equation of a quadratic function: · Vertex form: y = a(x – h)2 + k · Standard form: y = ax2 + bx + c · Intercept form: y = a(x – p)(x – q) When graphing a parabola, the first step is to locate and plot the vertex. HOW TO LOCATE THE VERTEX – EQUATION IN VERTEX FORM: 1) The x-coordinate is the OPPOSITE of the constant term of the binomial inside of the parentheses. 2) The y-coordinate is the SAME as the constant term being added (generally to the right) to the squared expression. If no such term can be seen, the y-coordinate of the vertex is 0. EXAMPLES WITH EXPLANATIONS: (Click on image to enlarge)",
null,
"HOW TO LOCATE THE VERTEX – EQUATION IN STANDARD FORM: 1) Use the formula: -b / 2a to determine the x-coordinate of the vertex 2) Substitute this value into the function and evaluate in order to determine the y-coordinate of the vertex. EXAMPLES WITH EXPLANATIONS: (Click on images to enlarge)",
null,
"HOW TO LOCATE THE VERTEX - EQUATION IN INTERCEPT FORM: 1) The x-coordinate is halfway between p and q. 2) Substitute this value into the function and evaluate in order to determine the y-coordinate of the vertex. EXAMPLES WITH EXPLANATIONS: (Click on images to enlarge)",
null,
""
]
| [
null,
"http://www.onlinemathpro.com/_/rsrc/1326411968921/tutorials/graphing-quadratic-functions-1---locating-the-vertex/Slide1.JPG",
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"http://www.onlinemathpro.com/_/rsrc/1326412024606/tutorials/graphing-quadratic-functions-1---locating-the-vertex/Slide2.JPG",
null,
"http://www.onlinemathpro.com/_/rsrc/1326413817335/tutorials/graphing-quadratic-functions-1---locating-the-vertex/Slide7.JPG",
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https://carriageandhorses.net/free-math-worksheets-for-4th-grade-word-problems/ | [
"# Fantastic Free Math Worksheets For 4th Grade Word Problems Image Inspirations",
null,
"Fantastic free math worksheets for 4th grade word problems image inspirations fun science reading.",
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"Uncategorized fantastic free math worksheets for 4th grade word problems image inspirationsing pages educational worksheet printableing4free com .",
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]
| [
null,
"https://carriageandhorses.net/wp-content/uploads/2021/07/fantastic-free-math-worksheets-for-4th-grade-word-problems-image-inspirations-fun-science-reading.jpg",
null,
"https://carriageandhorses.net/wp-content/uploads/2021/07/4th-grade-math-word-problems-free-worksheets-with-answers-mashup-fantastic-for-image-inspirations-uncategorized.png",
null,
"https://carriageandhorses.net/wp-content/uploads/2021/07/grade-mixed-word-problem-worksheets-k5-learning-free-math-for-4th-problems-fantastic-image.gif",
null,
"https://carriageandhorses.net/wp-content/uploads/2021/07/readingrksheets-for-4th-grade-printable-free-grammar-math-and.png",
null,
"https://carriageandhorses.net/wp-content/uploads/2021/07/easter-math-problems-freeth-grade-jumpstart-uncategorized-fantastic-worksheets-for-word-image-inspirations.jpg",
null,
"https://carriageandhorses.net/wp-content/uploads/2021/07/free-math-worksheetsor-4th-grade-word-problems-shopping-k5-learning-reading-printable-science.gif",
null,
"https://carriageandhorses.net/wp-content/uploads/2021/07/word-problems-worksheets-dynamically-created-uncategorized-free-math-for-4th-grade-common-core-algebra.png",
null,
"https://carriageandhorses.net/wp-content/uploads/2021/07/4th-grade-math-word-problems-free-worksheets-with-answers-mashup-fantastic-for-image-inspirations-snip20190611_1.png",
null,
"https://carriageandhorses.net/wp-content/uploads/2021/07/free-math-worksheets-for-4th-grade-wordems-uncategorized-fantastic-image-inspirations-multiplication-dozens.png",
null,
"https://carriageandhorses.net/wp-content/uploads/2021/07/grade-word-problems-worksheet-4th-problem-worksheets-printable-k5-learning-free-for-science-reading.gif",
null,
"https://carriageandhorses.net/wp-content/uploads/2021/07/fantastic-free-math-worksheets-for-4thde-word-problems-image-inspirations-multiple-step-problem-science.jpg",
null,
"https://carriageandhorses.net/wp-content/uploads/2021/07/free-worksheets-for-4th-grade-math-word-problems-common-core-algebra.jpg",
null,
"https://carriageandhorses.net/wp-content/uploads/2021/07/5th-uncategorized-grade-math-word-problems-free-worksheets-with-answers-mashup-reading-for-4th-common-core.jpg",
null,
"https://carriageandhorses.net/wp-content/uploads/2021/07/fantastic-free-math-worksheets-for-4th-grade-word-problems-image-inspirations-uncategorized-monster.jpg",
null,
"https://carriageandhorses.net/wp-content/uploads/2021/07/uncategorized-fractions_lowest_terms_easy_001_pin-freets-for-4th-grade-language-arts-grammar-math-multiplication-science.jpg",
null,
"https://carriageandhorses.net/wp-content/uploads/2021/07/grade-maths-resources-step-word-problems-printable-worksheets-lets-share-knowledge-problem-free-math-for.jpg",
null,
"https://carriageandhorses.net/wp-content/uploads/2021/07/uncategorized-fantastic-free-math-worksheets-for-4th-grade-word-problems-image-inspirationsing-pages-educational-worksheet-printableing4free-com_.gif",
null,
"https://carriageandhorses.net/wp-content/uploads/2021/07/uncategorized-4th-gradeath-worksheets-division-word-problems-free-wallpapers-for-fantastic.jpg",
null,
"https://carriageandhorses.net/wp-content/uploads/2021/07/free-math-worksheets-for-4th-grade-word-problems-uncategorized-fantastic-image-inspirations-problem-2nd-graders-with-no.jpg",
null,
"https://carriageandhorses.net/wp-content/uploads/2021/07/reading-worksheets-for-3rd-grade-free-scienceth-math-word-problems-common-core-standards-fun.JPG",
null,
"https://carriageandhorses.net/wp-content/uploads/2021/07/4th-grade-math-worksheets-with-answers-pdf-free-printable-for-fourth-word-problems-common-core.png",
null,
"https://carriageandhorses.net/wp-content/uploads/2021/07/4th-grade-number-math-wordms-multi-step-3rd-free-worksheets-for-fantastic.jpg",
null,
"https://carriageandhorses.net/wp-content/uploads/2021/07/free-worksheets-for-4the-language-arts-grammar-reading-comprehension-3rd-math.gif",
null,
"https://carriageandhorses.net/wp-content/uploads/2021/07/monstermath3-monster-math-word-problems-free-pdfs-for-halloween-fantastic-worksheets-4thde-image-inspirations-reading.jpg",
null,
"https://carriageandhorses.net/wp-content/uploads/2021/07/grade-money-wordblem-worksheets-k5-learning-uncategorizedblems-free-math-for-4th-fantastic.gif",
null
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https://discuss.interviewbit.com/t/stuck-with-couple-of-possible-scenarios-can-anyone-help/14365 | [
"",
null,
"# Stuck with couple of possible scenarios, can anyone help?\n\n#1\n\nQuestions:\n\nAre two arrays of same length or different lengths?\n\nand, same old question, Why can’t it be O(n*n)\n\nThis is what I observed\n\nlets say arr[i] = [2,3,4,5,6,7,8,9,10]; arr[j] = [3,4,5,6,7]\n\nfor i = 0; the inner loop iterates ‘n’ times because ( j < n && arr[i] < arr[j] ) succeeds for all j & i = 0; here time complexity is O(n) and the loop breaks when j equals to n\n\nNow, the outer loop still runs for ‘n’ times\n\nso why can’t it be O(n*n)\n\nThe arrays I considered aren’t the worst case scenario?\n\nCan someome help me in understanding this please ?\n\n#2\n\nLet’s get to the problem, we are interested in finding the worst case of this problem that is Big O. Consider the worst case that outer loop will run total ‘n’ times and inner loop will also run ‘n’ times and assume the array is in ascending order, so arr[i]<arr[j] is always true. So our main concern now is “j<n” condition in while loop.\nNormally we would say that the complexity of this problem is O(n^2) because they are nested loop, but here is the thing in this problem.\nIf you see closely that j is global initialized with 0 and it is never initialized with ‘i’ inside any of the loops. So for our very first iteration when ‘i=0’ our while loop will run ‘n’ times since we assumed array to be ascending order. For other values of i such that ‘i=1,2,3…n’ while loop will be always false because ‘j’ does not satisfy the condition it is already equal to ‘n’.\nSo our i loop will run ‘n’ times and so our complexity becomes O(2*n), therefore O(n).\nI hope you understood !\n\n#3\n\nI understood why you said our loop run 'ni times. But after that how did you say that Complexity is O(2*n) .Isn;t it directly O(n)?\n\n#4\n\ni=0, j=0;\narr<arr this condition never gets true therefore while loop runs for one time only for each value of i\ntherefore program runs N times\nhence O(N)\n\n#5\n\nShould they also mention what kind of an array it is? Ascending, descending, random. Please explain me the case when the array is random.\n\n#6\n\nOn what basis we are assuming arr to be in ascending order?\n\n#7\n\nHere O(2*N) is same as O(N+N) which comes to O(N)\n\n#8\n\nNo the order must not be mentioned as we always need to look for the worst case. Here the worst case is when the condition stands true anyhow at its maxm possible. here if we dry run on descending array then only we find that arr[i] < arr[j] till its end. hence we choose descending order"
]
| [
null,
"https://discuss-files.s3.dualstack.us-west-2.amazonaws.com/original/4X/9/6/6/966130bf03a57db2de73897c626d264b27cc950f.png",
null
]
| {"ft_lang_label":"__label__en","ft_lang_prob":0.7755757,"math_prob":0.9692361,"size":550,"snap":"2021-43-2021-49","text_gpt3_token_len":173,"char_repetition_ratio":0.08974359,"word_repetition_ratio":0.0,"special_character_ratio":0.33272728,"punctuation_ratio":0.15492958,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99638164,"pos_list":[0,1,2],"im_url_duplicate_count":[null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-10-26T21:26:19Z\",\"WARC-Record-ID\":\"<urn:uuid:77565657-2bab-4053-a849-6a06415cf980>\",\"Content-Length\":\"16187\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:5c947e8c-6200-48ed-8dde-4eb6fcb0ddee>\",\"WARC-Concurrent-To\":\"<urn:uuid:19dd672d-6cf5-4a17-9394-f66cc08a2700>\",\"WARC-IP-Address\":\"52.11.75.178\",\"WARC-Target-URI\":\"https://discuss.interviewbit.com/t/stuck-with-couple-of-possible-scenarios-can-anyone-help/14365\",\"WARC-Payload-Digest\":\"sha1:N7PVUBUWE7VXJLGESZ2LXIKSQDD56VVH\",\"WARC-Block-Digest\":\"sha1:Y4RRUH7WVP6OV3ME4UUCNKXGJHLFXIJM\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-43/CC-MAIN-2021-43_segments_1634323587926.9_warc_CC-MAIN-20211026200738-20211026230738-00195.warc.gz\"}"} |
https://www.latestinterviewquestions.com/php-multiple-choice-questions-answers | [
"# PHP Multiple choice Questions & Answers\n\nPosted On:January 13, 2019, Posted By: Latest Interview Questions, Views: 6611, Rating :",
null,
"",
null,
"",
null,
"",
null,
"",
null,
"## Best PHP Multiple choice Questions and Answers\n\nDear Readers, Welcome to PHP Objective Questions have been designed specially to get you acquainted with the nature of questions you may encounter during your Job interview for the subject of PHP. These Objective type PHP Questions are very important for campus placement test and job interviews. As per my experience good interviewers hardly plan to ask any particular question during your Job interview and these model questions are asked in the online technical test and interview of many Medical companies.\n\n### 1) Which of the following are not considered as Boolean false?\n\na. FALSE\n\nb. 0\n\nc. “0”\n\nd. “FALSE”\n\ne. 1\n\nf. NULL\n\nAns: d and e",
null,
"### 2) Consider the following class:\n\nClass Insurance\n\n{\n\nfunction clsName()\n\n{\n\necho get_class(\\$this);\n\n}\n\n}\n\n8. \\$cl = new Insurance();\n\n9. \\$cl -> clsName();\n\n10. Insurance::clsName();\n\nWhich of the following Lines should be commented to print the class name without errors?\n\na. Line 8 and 9\n\nb. Line 10\n\nc. Line 9 and 10\n\nd. All the three lines 8,9, and 10 should be left as it is.\n\nAns: d\n\n### 3) Variables/functions in PHP don't work directly with:\n\na. echo()\n\nb. isset()\n\nc. print()\n\nd. All of the above\n\nAns: b\n\n### 4) What is the output of the following code?\n\na. Array ([x]=>9 [y]=>3 [z]=>-7)\n\nb. Array ([x]=>3 [y]=>2 [z]=>5)\n\nc. Array ([x]=>12 [y]=>5 [z]=>-2)\n\nd. Error\n\ne. None of the above\n\nAns: c\n\n### 5) Which of the following multithreaded servers allow PHP as a plug-in?\n\na. Netscape FastTrack\n\nb. Microsoft's Internet Information Server\n\nc. O'Reilly's WebSite Pro\n\nd. All of the above\n\nAns: d\n\n### 6) Which of the following statements is incorrect with regard to interfaces?\n\na. A class can implement multiple interfaces\n\nb. An abstract class cannot implement multiple interfaces\n\nc. An interface can extend multiple interfaces\n\nd. Methods with same name, arguments and sequences can exist in the different interfaces implmented by a class\n\nAns: d\n\na. real\n\nb. double\n\nc. decimal\n\nd. boolean\n\nAns: c\n\n### 8) What will be the output of the following code?\n\n\\$var = 10;\n\nfunction fn ()\n\n{\n\n\\$var = 20;\n\nreturn \\$var;\n\n}\n\nfn ();\n\necho \\$var;\n\na. 10\n\nb. 20\n\nc. Undefined Variable\n\nd. Syntax Error\n\nAns: a\n\n### 9) which of the following are the valid PHP data types?\n\na. resource\n\nb. null\n\nc. boolean\n\nd. string\n\ne. Both a and c\n\nf. Both b, c and d\n\ng. All of the above\n\nAns: g\n\n### 10) which of the following characters are taken care of by htmlspecialchars?\n\na. <\n\nb. >\n\nc. single quote\n\nd. double quote\n\ne. &\n\nf. All of the above\n\nAns: f\n\n### 11) What do you infer from the following code?\n\na. Only first character will be recognised and new line will be inserted.\n\nb. Last will not be recognised and only first two parts will come in new lines.\n\nc. All the will work and text will be printed on respective new lines.\n\nd. All will be printed on one line irrespective of the\n\nAns: d\n\n### 12) which of the following is a correct declaration?\n\na. static \\$varb = array(1,'val',3);\n\nb. static \\$varb = 1+(2*90);\n\nc. static \\$varb = sqrt(81);\n\nd. static \\$varb = new Object;\n\nAns: a\n\na. True\n\nb. False\n\nAns: b\n\n### 14) what is true regarding \\$a + \\$b where both of them are arrays?\n\na. Duplicated keys are NOT overwritten\n\nb. \\$b is appended to \\$a\n\nc. The + operator is overloaded\n\nd. This produces a syntax error\n\nAns: a,b\n\na. &\n\nb. =\n\nc. :?\n\nd. ?:\n\ne. +=\n\nf. &&\n\nAns: d\n\na. Integer\n\nb. String\n\nc. Boolean\n\nd. Array\n\nAns: b,d\n\na. +,-\n\nb. ==, !=\n\nc.\n\nd. &=, |=\n\nAns: b\n\n### 18) Which of the following attribute is needed for file upload via form?\n\na. enctype='multipart/form-data'\n\nb. enctype='singlepart/data'\n\nc. enctype='file'\n\nd. enctype='form-data/file'\n\nAns: a\n\na. echo()\n\nb. print()\n\nc. println()\n\nd. display()\n\nAns: a,b\n\n### 20) The inbuilt function to get the number of parameters passed is:\n\na. arg_num()\n\nb. func_args_count()\n\nc. func_num_args()\n\nd. None of the above\n\nAns: c\n\n### 21) Consider the following two statements:\n\nI while (expr) statement\n\nII while (expr): statement... endwhile;\n\nWhich of the following are true in context of the given statements?\n\na. I is correct and II is wrong\n\nb. I is wrong and II is correct\n\nc. Both I & II are wrong\n\nd. Both I & II are correct\n\nAns: d\n\n### 22) Multiple select/load is possible with:\n\na. Checkbox\n\nb. Select\n\nc. File\n\nd. All of the above\n\nAns: b\n\n### 23) Which of the following statement is not correct for PHP?\n\na. It is a server side scripting language\n\nb. A PHP file may contain text, html tags or scripts\n\nc. It can run on windows and Linux systems only\n\nd. It is compatible with most of the common servers used today\n\nAns: c\n\n### 24) Which of the following printing construct/function accpets multiple parameters?\n\na. echo\n\nb. print\n\nc. printf\n\nd. All of the above\n\nAns: a\n\na. Yes\n\nb. No\n\nAns: b\n\n### 26) We have two variable definitions:\n\n1. 023\n\nb. x23\n\nChoose the correct options:\n\na. 1 is octal\n\nc. 2 is octal\n\nAns: a,b\n\na. TRUE\n\nb. FALSE\n\nc. NULL\n\nd. _FILE_\n\ne. CONSTANT\n\nAns: d\n\n### 28) What will be the output of the following code?\n\n\\$a = 10;\n\nif (\\$a > 5 OR < 15)\n\necho 'true';\n\nelse\n\necho 'false';\n\na. true\n\nb. false\n\nc. No output\n\nd. Parse Error\n\nAns: d\n\n### 29) You need to check the size of a file in PHP function.\n\n\\$size = X(filename);\n\nWhich function will suitably replace 'X'?\n\na. filesize\n\nb. size\n\nc. sizeofFile\n\nd. getSize\n\nAns: a\n\n### 30) Which of the following vaiables is not related to file uploads?\n\na. max_file_size\n\nb. max_execution_time\n\nc. post_max_size\n\nd. max_input_time\n\nAns: d\n\na. \\$var_1\n\nb. \\$var1\n\nc. \\$var-1\n\nd. \\$var/1\n\ne. \\$v1\n\nAns: c,d\n\n### 32) What will be the output of the following code?\n\nfunction fn (&\\$var)\n\n{\n\n\\$var = \\$var - (\\$var/10*5);\n\nreturn \\$var;\n\n}\n\necho fn(100);\n\na. 100\n\nb. 50\n\nc. 98\n\nd. Error message\n\ne. None of the above\n\nAns: d\n\n### 33) What will be the output of following code?\n\n\\$a = 10;\n\necho 'Value of a = \\$a';\n\na. Value of a = 10\n\nb. Value of a = \\$a\n\nc. Undefined\n\nd. Syntax Error\n\nAns: a\n\n### 34) What will be the output of the following code?\n\n\\$Rent = 250;\n\nfunction Expenses(\\$Other)\n\n{\n\n\\$Rent = 250 + \\$Other;\n\nreturn \\$Rent;\n\n}\n\nExpenses(50);\n\necho \\$Rent;\n\na. 300\n\nb. 250\n\nc. 200\n\nd. Program will not compile\n\nAns: b\n\n### 35) Which of the following variable declarations within a class is invalid in PHP5?\n\na. private \\$type = 'moderate';\n\nb. internal \\$term= 3;\n\nc. public \\$amnt = '500';\n\nd. protected \\$name = 'Quantas Private Limited';\n\nAns: b\n\n### 36) Which of the following is used to maintain the value of a variable over different pages?\n\na. static\n\nb. global\n\nc. session_register\n\nd. None of the above\n\nAns: c\n\nAns: a\n\n### 38) The following php variables are declared:\n\n\\$company = 'ABS Ltd';\n\n\\$\\$company = ',Sydney';\n\nWhich of the following is not a correct way of printing 'ABS Ltd,Sydney'?\n\na. echo '\\$company \\$\\$company';\n\nb. echo '\\$company \\${\\$company}';\n\nc. echo '\\$company \\${'ABS Ltd'}';\n\nd. echo '\\$company {\\$\\$company}';\n\nAns: a\n\n### 39) Which of the following regular expressions can be used to check the validity of an e-mail addresses?\n\na. ^[^@ ]+@[^@ ]+.[^@ ]+\\$\n\nb. ^[^@ ]+@[^@ ]+.[^@ ]+\\$\n\nc. \\$[^@ ]+@[^@ ]+.[^@ ]+^\n\nd. \\$[^@ ]+@[^@ ]+.[^@ ]+^\n\nAns: b\n\n### 40) You wrote the following script to check of the right category:\n\nCorrect category!\n\nIncorrect category!\n\nWhat will be the output of the program if value of 'cate' remains 5?\n\na. Correct category!\n\nb. Incorrect category!\n\nc. Error due to use of invalid operator in line 6:'if (\\$cate==5)'\n\nd. Error due to incorrect syntax at line 8, 10, 12 and 14\n\nAns: a\n\na. True\n\nb. False\n\nAns: a\n\n### 42) Which of the following is not supported in PHP5?\n\na. Type Hinting\n\nb. Reflection\n\nc. Magic Methods\n\nd. Multiple Inheritance\n\ne. Object Cloning\n\nAns: d\n\n### 43) What will be the output of the following code?\n\necho 30*5 . 7;\n\na. 150.7\n\nb. 1507\n\nc. 150.7\n\nd. you can't concatenate integers\n\ne. error will occur\n\nAns: b\n\n### 44) In your PHP application you need to open a file. You want the application to issue a warning and continue execution, in case the file is not found. The ideal function to be used is:\n\na. include()\n\nb. require()\n\nc. nowarn()\n\nd. getFile(false)\n\nAns: a\n\n### 45) What will be the output of the following code?\n\na. int(3*4)\n\nb. int(12)\n\nc. 3*4\n\nd. 12\n\ne. None of the above\n\nAns: b\n\n### 46) You need to count the number of parameters given in the URL by a POST operation. The correct way is:\n\na. count(\\$POST_VARS);\n\nb. count(\\$POST_VARS_PARAM);\n\nc. count(\\$_POST);\n\nd. count(\\$HTTP_POST_PARAM);\n\nAns: c\n\n### 47) Which of the following is correct with regard to echo and print?\n\na. echo is a construct and print is a function\n\nb. echo is a function and print is a construct\n\nc. Both are functions\n\nd. Both are constructs\n\nAns: d\n\na. echo()\n\nb. print()\n\nc. println()\n\nd. display()\n\nAns: a,b\n\n### 49) How would you start a session?\n\na. session(start);\n\nb. session();\n\nc. session_start();\n\nd. begin_session();\n\nAns: c\n\n### 50) What will be the output of following code?\n\n\\$a = 10;\n\necho 'Value of a = \\$a';\n\na. Value of a = 10\n\nb. Value of a = \\$a\n\nc. Undefined\n\nd. Syntax Error\n\nAns: a\n\n### 51) Which of the following is not true regarding XForms?\n\na. PHP provides support for XForm\n\nb. It can be used on PDF documents\n\nc. The data is sent in XML format\n\nd. The action and method parameters are defined in the body\n\nAns: d\n\n### 52) You have defined three variables \\$to, \\$subject and \\$body to send an email. Which of the following methods would you use for sending an email?\n\na. mail(\\$to,\\$subject,\\$body)\n\nb. sendmail(\\$to,\\$subject,\\$body)\n\nc. mail(to,subject,body)\n\nd. sendmail(to,subject,body)\n\nAns: a\n\na. __call, __get, __set\n\nAns: a\n\n### 54) Which of the following is a PHP resource?\n\na. Domxml document\n\nc. File\n\nd. All of the above\n\nAns: d\n\na. Yes\n\nb. No\n\nAns: a"
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https://iris.sissa.it/handle/20.500.11767/124369 | [
"Over many years, our current understanding of the Universe has been extremely relied on perturbation theory (PT) both theoretically and experimentally. There are, however, many situations in cosmology in which the analysis beyond PT is required. In this thesis we study three examples: the resonant decay of gravitational waves (GWs), the dark energy (DE) instabilities induced by GWs, and the tail of the primordial field distribution function. The first two cases are within the context of the Effective Field Theory (EFT) of DE, whereas the last one is within inflation. We first review the construction of the EFT of DE, which is the most general Lagrangian for the scalar and tensor perturbations around the flat FLRW metric. Specifically, this EFT can be mapped to the covariant theories, known as Horndeski and Beyond Horndeski theories. We then study the implications on the dark energy theories coming from the fact that GWs travel with the speed \\$c_T = 1\\$ at LIGO/Virgo frequencies. After that, we consider the perturbative decay of GWs into DE fluctuations (\\$gamma ightarrow pipi\\$) due to the \\$ ilde{m}_4^2\\$ operator. This process is kinematically allowed by the spontaneous breaking of Lorentz invariance. Therefore, having no perturbative decay of gravitons together with \\$c_T = 1\\$ at LIGO/Virgo, rules out all quartic and quintic beyond Horndeski theories. As the first non-perturbative regime in this thesis, we study the decay of GWs into DE fluctuations \\$pi\\$, taking into account the large occupation numbers of gravitons. When the \\$m_3^3\\$ (cubic Horndeski) and \\$ ilde{m}_4^2\\$ (beyond Horndeski) operators are present, the GW acts as a classical background for \\$pi\\$ and modifies its dynamics. In particular, \\$pi\\$ fluctuations are described by a Mathieu equation and feature instability bands that grow exponentially. In the regime of small GW amplitude which corresponds to narrow resonance, we calculate analytically the produced \\$pi\\$, its energy and the change of the GW signal. Eventually, the resonance is affected by \\$pi\\$ self-interactions in a way that we cannot describe analytically. The fact that \\$pi\\$ self-couplings coming from the \\$m_3^3\\$ operator become quickly comparable with the resonant term affects the growth of \\$pi\\$ so that the bound on \\$alpha_{ m B}\\$ remains inconclusive. However, in the case of the \\$ ilde{m}_4^2\\$ operator self-interactions can be neglected at least in some regimes. Therefore, our resonant analysis improves the perturbative bounds on \\$alpha_{ m H}\\$, ruling out quartic Beyond Horndeski operators. In the second non-perturbative regime we show that \\$pi\\$ may become unstable in the presence of a GW background with sufficiently large amplitude. We find that dark-energy fluctuations feature ghost and/or gradient instabilities for GW amplitudes that are produced by typical binary systems. Taking into account the populations of binary systems, we conclude that the instability is triggered in the whole Universe for \\$|alpha_{ m B}| gtrsim 10^{-2}\\$, i.e. when the modification of gravity is sizable. The fate of the instability and the subsequent time-evolution of the system depend on the UV completion, so that the theory may end up in a state very different from the original one. In conclusion, the only dark-energy theories with sizable cosmological effects that avoid these problems are \\$k\\$-essence models, with a possible conformal coupling with matter. In the second part of the thesis we consider physics of inflation. Inflationary perturbations are approximately Gaussian and deviations from Gaussianity are usually calculated using in-in perturbation theory. This method, however, fails for unlikely events on the tail of the probability distribution: in this regime non-Gaussianities are important and perturbation theory breaks down for \\$|zeta| gtrsim |f_{ m NL}|^{-1}\\$. We then show that this regime is amenable to a semiclassical treatment, \\$hbar ightarrow 0\\$. In this limit the wavefunction of the Universe can be calculated in saddle-point, corresponding to a resummation of all the tree-level Witten diagrams. The saddle can be found by solving numerically the classical (Euclidean) non-linear equations of motion, with prescribed boundary conditions. We apply these ideas to a model with an inflaton self-interaction \\$propto lambda dot{zeta}^4\\$. Numerical and analytical methods show that the tail of the probability distribution of \\$zeta\\$ goes as \\$exp(-lambda^{1/4}zeta^{3/2})\\$, with a clear non-perturbative dependence on the coupling. Our results are relevant for the calculation of the abundance of primordial black holes.\n\nBeyond Perturbation Theory in Cosmology / Yingcharoenrat, Vicharit. - (2021 Sep 16).\n\n### Beyond Perturbation Theory in Cosmology\n\n#### Abstract\n\nOver many years, our current understanding of the Universe has been extremely relied on perturbation theory (PT) both theoretically and experimentally. There are, however, many situations in cosmology in which the analysis beyond PT is required. In this thesis we study three examples: the resonant decay of gravitational waves (GWs), the dark energy (DE) instabilities induced by GWs, and the tail of the primordial field distribution function. The first two cases are within the context of the Effective Field Theory (EFT) of DE, whereas the last one is within inflation. We first review the construction of the EFT of DE, which is the most general Lagrangian for the scalar and tensor perturbations around the flat FLRW metric. Specifically, this EFT can be mapped to the covariant theories, known as Horndeski and Beyond Horndeski theories. We then study the implications on the dark energy theories coming from the fact that GWs travel with the speed \\$c_T = 1\\$ at LIGO/Virgo frequencies. After that, we consider the perturbative decay of GWs into DE fluctuations (\\$gamma ightarrow pipi\\$) due to the \\$ ilde{m}_4^2\\$ operator. This process is kinematically allowed by the spontaneous breaking of Lorentz invariance. Therefore, having no perturbative decay of gravitons together with \\$c_T = 1\\$ at LIGO/Virgo, rules out all quartic and quintic beyond Horndeski theories. As the first non-perturbative regime in this thesis, we study the decay of GWs into DE fluctuations \\$pi\\$, taking into account the large occupation numbers of gravitons. When the \\$m_3^3\\$ (cubic Horndeski) and \\$ ilde{m}_4^2\\$ (beyond Horndeski) operators are present, the GW acts as a classical background for \\$pi\\$ and modifies its dynamics. In particular, \\$pi\\$ fluctuations are described by a Mathieu equation and feature instability bands that grow exponentially. In the regime of small GW amplitude which corresponds to narrow resonance, we calculate analytically the produced \\$pi\\$, its energy and the change of the GW signal. Eventually, the resonance is affected by \\$pi\\$ self-interactions in a way that we cannot describe analytically. The fact that \\$pi\\$ self-couplings coming from the \\$m_3^3\\$ operator become quickly comparable with the resonant term affects the growth of \\$pi\\$ so that the bound on \\$alpha_{ m B}\\$ remains inconclusive. However, in the case of the \\$ ilde{m}_4^2\\$ operator self-interactions can be neglected at least in some regimes. Therefore, our resonant analysis improves the perturbative bounds on \\$alpha_{ m H}\\$, ruling out quartic Beyond Horndeski operators. In the second non-perturbative regime we show that \\$pi\\$ may become unstable in the presence of a GW background with sufficiently large amplitude. We find that dark-energy fluctuations feature ghost and/or gradient instabilities for GW amplitudes that are produced by typical binary systems. Taking into account the populations of binary systems, we conclude that the instability is triggered in the whole Universe for \\$|alpha_{ m B}| gtrsim 10^{-2}\\$, i.e. when the modification of gravity is sizable. The fate of the instability and the subsequent time-evolution of the system depend on the UV completion, so that the theory may end up in a state very different from the original one. In conclusion, the only dark-energy theories with sizable cosmological effects that avoid these problems are \\$k\\$-essence models, with a possible conformal coupling with matter. In the second part of the thesis we consider physics of inflation. Inflationary perturbations are approximately Gaussian and deviations from Gaussianity are usually calculated using in-in perturbation theory. This method, however, fails for unlikely events on the tail of the probability distribution: in this regime non-Gaussianities are important and perturbation theory breaks down for \\$|zeta| gtrsim |f_{ m NL}|^{-1}\\$. We then show that this regime is amenable to a semiclassical treatment, \\$hbar ightarrow 0\\$. In this limit the wavefunction of the Universe can be calculated in saddle-point, corresponding to a resummation of all the tree-level Witten diagrams. The saddle can be found by solving numerically the classical (Euclidean) non-linear equations of motion, with prescribed boundary conditions. We apply these ideas to a model with an inflaton self-interaction \\$propto lambda dot{zeta}^4\\$. Numerical and analytical methods show that the tail of the probability distribution of \\$zeta\\$ goes as \\$exp(-lambda^{1/4}zeta^{3/2})\\$, with a clear non-perturbative dependence on the coupling. Our results are relevant for the calculation of the abundance of primordial black holes.\n##### Scheda breve Scheda completa Scheda completa (DC)\n16-set-2021\nCreminelli, Paolo\nYingcharoenrat, Vicharit\nFile in questo prodotto:\nFile\nThesis_Vicharit_Yingcharoenrat.pdf\n\naccesso aperto\n\nDescrizione: Theoretical Cosmology\nTipologia: Tesi\nLicenza: Non specificato\nDimensione 11.92 MB\nUtilizza questo identificativo per citare o creare un link a questo documento: `https://hdl.handle.net/20.500.11767/124369`\n•",
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https://research.tilburguniversity.edu/en/publications/regression-anytime-with-brute-force-svd-truncation | [
"# `Regression anytime' with brute-force SVD truncation\n\nChristian Bender, Nikolaus Schweizer\n\nResearch output: Contribution to journalArticleScientificpeer-review\n\n## Abstract\n\nWe propose a new least-squares Monte Carlo algorithm for the approximation of conditional expectations in the presence of stochastic derivative weights. The algorithm can serve as a building block for solving dynamic programming equations, which arise, e.g., in non-linear option pricing problems or in probabilistic discretization schemes for fully non-linear parabolic partial differential equations. Our algorithm can be generically applied when the underlying dynamics stem from an Euler approximation to a stochastic differential equation. A built-in variance reduction ensures that the convergence in the number of samples to the true regression function takes place at an arbitrarily fast polynomial rate, if the problem under consideration is smooth enough.\nOriginal language English Annals of Applied Probability Accepted/In press - 2020"
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https://maxnet80.com/qa/quick-answer-is-it-k-or-k-for-thousands.html | [
"",
null,
"# Quick Answer: Is It K Or K For Thousands?\n\n## Is it a capital K for thousands?\n\nThe capital letter K is sometimes used informally to represent one thousand (dollars), especially in newspaper headlines.\n\nThere is no space between the numeral and the letter K , as in 75 K ..\n\n## What does the K mean in 1k?\n\nYes, it does. k stands for kilo, so when you write any number followed by ‘k’ it means kilo or thousand. … K HERE MEANS A THOUSAND. THEREFORE, 1K = 1,000, 2K = 2000.IT COMES FROM THE INTERNATIONAL SYSTEM OF UNITS (SI) “KILO” MEANING A THOUSAND AND BEARING THE SYMBOL “K”.\n\n## What does K and M mean in numbers?\n\nWhen we write “30 K” for 30,000, we are not using Roman numerals but pseudo-metric, where K stands for “kilo-“. For the same reason, “30 M” means 30,000,000, using M for “mega-” (or just “million”). It would be wrong to use Roman numerals this way; M here can’t stand for thousand.\n\n## What is K in 10k salary?\n\n1k= one thousand (1,000) 10k = ten thousand (10,000) 100k= 100 thousand (100,000) it means thousand!\n\n## Is K bigger than M?\n\nThe unit symbol of Megabyte is MB. 1 KB (Kilobyte) is equal to 0.001 MB in decimal and 0.0009765625 MB in binary. It also means that 1 megabyte is equal to 1000 kilobytes in decimal and 1024 kilobytes in binary. … So as you can see, a Megabyte is one thousand times bigger than a Kilobyte.\n\n## Is million more than K?\n\nWhen the view reach 1,000k, it becomes one million (1mn). K is notation for 1000. 1000,000 makes 10 lacs, 1000k which makes 1 million. … 1000k is a million.\n\n## Why do we write K for thousand?\n\nIt actually references to an uncertain amount of time. Now, if this confusion was not enough, the French took the Greek word “Chilioi” and shortened it to “Kilo.” They then started the metric system and introduced kilo as 1000.\n\n## Is K or M bigger on Youtube?\n\nWhat does “K” or “M” behind numbers mean? On the Internet, 1K is used to represent 1 thousand and 10K is used to represent 10 thousand, similarly, 1M is used to represent 1 million. You must have understood this thing easily because “M” means Million, so “M” would be used for million.\n\n## How much is K money?\n\n“K” in money means a thousand. In Mathematics, Kilo means thousand, thus, the letter K. For example, 5K money basically just means five thousand (5,000). When used with currencies, 10K money is \\$10,000.\n\n## What is the meaning of 2.6 K?\n\nIt stands for thousand. It stands for thousand. See a translation.\n\n## What is K in US dollars?\n\n‘\\$’ is a symbol for representing United States Dollar aka ‘USD’. ‘k’ comes from Greek Kilo (or) Greek word chilioi or khilioi which means thousand. Its written in lower case ‘k’ when representing money. Since its prefixed by ‘\\$’ symbol it is 1800 dollars.\n\n## What’s bigger K or M on Instagram?\n\nK here mean kilo which equals to 1000. like M which stands for Mega or million. 1k followers = 1,000 followers. 1k is mean 1000."
]
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null,
"https://mc.yandex.ru/watch/68552596",
null
]
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https://search.r-project.org/CRAN/refmans/bestNormalize/html/exp_x.html | [
"exp_x {bestNormalize} R Documentation\n\n## exp(x) Transformation\n\n### Description\n\nPerform a exp(x) transformation\n\n### Usage\n\n```exp_x(x, standardize = TRUE, warn = TRUE, ...)\n\n## S3 method for class 'exp_x'\npredict(object, newdata = NULL, inverse = FALSE, ...)\n\n## S3 method for class 'exp_x'\nprint(x, ...)\n```\n\n### Arguments\n\n `x` A vector to normalize with with x `standardize` If TRUE, the transformed values are also centered and scaled, such that the transformation attempts a standard normal `warn` Should a warning result from infinite values? `...` additional arguments `object` an object of class 'exp_x' `newdata` a vector of data to be (potentially reverse) transformed `inverse` if TRUE, performs reverse transformation\n\n### Details\n\n`exp_x` performs a simple exponential transformation in the context of bestNormalize, such that it creates a transformation that can be estimated and applied to new data via the `predict` function.\n\n### Value\n\nA list of class `exp_x` with elements\n\n `x.t` transformed original data `x` original data `mean` mean after transformation but prior to standardization `sd` sd after transformation but prior to standardization `n` number of nonmissing observations `norm_stat` Pearson's P / degrees of freedom `standardize` was the transformation standardized\n\nThe `predict` function returns the numeric value of the transformation performed on new data, and allows for the inverse transformation as well.\n\n### Examples\n\n```x <- rgamma(100, 1, 1)\n\nexp_x_obj <- exp_x(x)\nexp_x_obj\np <- predict(exp_x_obj)\nx2 <- predict(exp_x_obj, newdata = p, inverse = TRUE)\n\nall.equal(x2, x)\n\n```\n\n[Package bestNormalize version 1.8.0 Index]"
]
| [
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https://solvedlib.com/suppose-y131-u2-u3-341-n-214-where-1-1-1-1421-find,454160 | [
"# Suppose y=131 U2 U3 3/41 ~ N, (μ, σ214) where μ=[ μ1 μ1 μ1 142]1. Find...\n\n###### Question:",
null,
"Suppose y=131 U2 U3 3/41 ~ N, (μ, σ214) where μ=[ μ1 μ1 μ1 142]1. Find the distribution of tto)\n\n#### Similar Solved Questions\n\n##### (b) Solve the following PDE using coefficient matching82u + 8u = 0 x € [0, 2],y € [0, 3] u(w,0) = 0 x € [0,2] u(1,3) = sin( Zx) + 2sin(4x) x € [0, 2] u(0, y) = ~ sin( ?3y) y € [0, 3] u(2,y) = 0 y € [0, 3]\n(b) Solve the following PDE using coefficient matching 82u + 8u = 0 x € [0, 2],y € [0, 3] u(w,0) = 0 x € [0,2] u(1,3) = sin( Zx) + 2sin(4x) x € [0, 2] u(0, y) = ~ sin( ?3y) y € [0, 3] u(2,y) = 0 y € [0, 3]...\n##### (10 points) Problem 5: A cylindrical container with diameter D\" is filled with water to height of \"H\" Now we connect it to another identical but empty container with a tube with small diameter \"d\" at the lowest part of the container. Calculate how long it takes until water stops flowing from the first container t0 the second;. Assume that D>> dand flow of water through the tube is slow and gradual;\n(10 points) Problem 5: A cylindrical container with diameter D\" is filled with water to height of \"H\" Now we connect it to another identical but empty container with a tube with small diameter \"d\" at the lowest part of the container. Calculate how long it takes until water s...\n##### Because there are 3 feet in every yard, the formula $F=3 \\cdot Y$ will convert $Y$ yards into $F$ feet. find $F$. $Y=2 \\frac{2}{3} \\text { yards }$\nBecause there are 3 feet in every yard, the formula $F=3 \\cdot Y$ will convert $Y$ yards into $F$ feet. find $F$. $Y=2 \\frac{2}{3} \\text { yards }$...\n##### If f(4)=-24 and F(4)=18 then Vxfx) )l dx X=424363018ubmit Answer\nIf f(4)=-24 and F(4)=18 then Vxfx) )l dx X=4 24 36 30 18 ubmit Answer...\n##### CoumtJan {0 scutIl = unilsandUnes . then what (s the volure ol tho cylinder shown ahove?cylinder392 Cubic Units210w Cubic Unlts56 cuDIc units1967 cubic unilsNori Q ueblonAebel\nCoumt Jan {0 scut Il = unilsand Unes . then what (s the volure ol tho cylinder shown ahove? cylinder 392 Cubic Units 210w Cubic Unlts 56 cuDIc units 1967 cubic unils Nori Q ueblon Aebel...\n##### Describe three techniques, methods, or recommendations you can use to promote retention in a study\nDescribe three techniques, methods, or recommendations you can use to promote retention in a study...\n##### An object with a mass of 24 kg, temperature of 150 ^oC, and a specific heat of 7 J/(kg*K) is dropped into a container with 48 L of water at 0^oC . Does the water evaporate? If not, by how much does the water's temperature change?\nAn object with a mass of 24 kg, temperature of 150 ^oC, and a specific heat of 7 J/(kg*K) is dropped into a container with 48 L of water at 0^oC . Does the water evaporate? If not, by how much does the water's temperature change?...\n##### The systcm the figure below equilibrium, und its free body diagram drawn on the right: The distance d i5 0,82 and each of the identical spring relaxed length Io =0.41 m The 96 kg brings the point down to height 17cm, The Mjs; the springs are negligible, Calculate the following quantities:(a) The angle deg (b) The force exerted on by the right spring The force exerted on by the left spring Fz The total spring length The stretch length 4/ = /-1o (f) The stiffness constant of the springsNlm\nThe systcm the figure below equilibrium, und its free body diagram drawn on the right: The distance d i5 0,82 and each of the identical spring relaxed length Io =0.41 m The 96 kg brings the point down to height 17cm, The Mjs; the springs are negligible, Calculate the following quantities: (a) The an...\n##### 6+10+10_Z6pts ) Let 4 =7]Find the eigenvalues A).by using the characteristic polynomial PA(A) det( A -Approximate the dominant eigenvalue and the corresponding eigenvector iterations us1ne Dower method with initial approximationFind theTelativ erTOT Erel. 123 There dominant eigenvalue obtained pan (a) and Aa the approximation the dominant eigenvalue after two iterations 0f cne powr methooApproximate the smallest Elgenialuecorresponding Eiwenrectoriterations ulnginferse WUTEL mechod with initial\n6+10+10_Z6pts ) Let 4 = 7] Find the eigenvalues A). by using the characteristic polynomial PA(A) det( A - Approximate the dominant eigenvalue and the corresponding eigenvector iterations us1ne Dower method with initial approximation Find the Telativ erTOT Erel. 123 There dominant eigenvalue obtained...\n2 2 Parabolic Answers; XXI zW-6x kx WW...\n##### 1. Below is production levels of lumber and oll for the United States and Canada. Country...\n1. Below is production levels of lumber and oll for the United States and Canada. Country Lumber Production (using 50 worker hours) 100 200 Oil Production (using 50 worker hours) 150 50 United States Canada a. Calculate the opportunity cost of producing 100 units of lumber for U.S. b. Calculate the ...\n##### Wikhol u 'oloatt hatnthom 12 Inerrto}I0)I6) 2J6)fif)\nwikhol u 'oloatt hatnthom 12 Inerrto} I0) I6) 2 J6) fif)...\n##### Comapare the attitude of the speaker to the reader Dulce et Decorum Est and In flanders...\ncomapare the attitude of the speaker to the reader Dulce et Decorum Est and In flanders field...\n##### Given the following information about a 2nd order transient circuit, solve for the coefficient az in...\nGiven the following information about a 2nd order transient circuit, solve for the coefficient az in the capacitor voltage ve (t) = e-mat (a essed 2-1 + aze Wi/ style-1) +ve (o)for t>0. C = 6 mF Vcl0+) = 9V Vc(infinity) = 8 V ic(0+) = 2 A Wn = 15 rad/sec zeta 1.46...\n##### Aadid; %mpperature changetempcrature change bctwcen 4975 Ano 2000. Fas Fame Using ' graph below: cakulatc the average the average emperature will rise betwoon 2010 and 20607Increase continues_ how much doyou 2stImatec Aerr Da TAlutAeulnal Aiaen6orgices €\nAadid; %mpperature change tempcrature change bctwcen 4975 Ano 2000. Fas Fame Using ' graph below: cakulatc the average the average emperature will rise betwoon 2010 and 20607 Increase continues_ how much doyou 2stImate c Aerr Da TAlut Aeulnal Aiaen 6orgices €...\n##### Question 35 (1 point) The inward chloride shift occurs whenBicarbonate ions enter the RBCBicarbonate ions leave the RBCHydrogen ions leave the RBCHydrogen ions enter the RBCNex Palge\nQuestion 35 (1 point) The inward chloride shift occurs when Bicarbonate ions enter the RBC Bicarbonate ions leave the RBC Hydrogen ions leave the RBC Hydrogen ions enter the RBC Nex Palge...\n##### Find the exact value using a double-angle identity\nfind the exact value using a double-angle identity...\n##### At a horizontal distance of 31 m from the bottom of a tree, the angle of...\nAt a horizontal distance of 31 m from the bottom of a tree, the angle of elevation to the top of the tree is 22°. How tall is the tree?...\n##### How do you find the first and second derivative of ln((x^2)(e^x))?\nHow do you find the first and second derivative of ln((x^2)(e^x))?...\n##### A barbell is mounted on a nearly frictionless axle through its center. The low-mass rod has...\nA barbell is mounted on a nearly frictionless axle through its center. The low-mass rod has a length d= 0.16 m, and each ball has a mass m = 0.1 kg. At this instant, there are two forces of equal magnitude F applied to the system as shown, with the directions indicated, and at this instant the angul...\n##### 1 order hom Kcesinomntnx 0 1 1342 L 1 1 lnaen Ein (@ \" 8\n1 order hom Kcesino mntnx 0 1 1 342 L 1 1 lnaen Ein (@ \" 8...\n##### QuestionEvaluateffI IertRtr where E is the portion of the unit ball 2+4 + 22 < that lies in the first octant. (Usc the spherical coordinate)\nQuestion Evaluate ffI IertRtr where E is the portion of the unit ball 2+4 + 22 < that lies in the first octant. (Usc the spherical coordinate)...\n##### Match each description with a component of blood in the key. Answers can be used more than once.a. red blood cellsb. white blood cellsc. red and white blood cells and plateletsd. plasmaFormed elements\nMatch each description with a component of blood in the key. Answers can be used more than once. a. red blood cells b. white blood cells c. red and white blood cells and platelets d. plasma Formed elements...\n##### The National Football League (NFL) records a variety of performance data for individuals and teams. To...\nThe National Football League (NFL) records a variety of performance data for individuals and teams. To investigate the importance of passing on the percentage of games won by a team, the data in the Excel Online file below show the conference (Conf), average number of passing yards per attempt (Yds/...\n##### WordEdit View inser Format ools Table Window Help AutoSavc 08 6-08 66 13 Ccll Division Meiosis Online Lab Insert Dravi Design Layout Refercnces Mailings Review ShareHomoTmas NeviED \"StylotstletOrganize the following statements the proper oracr of Interphase and _ first step with the nutnber (1) and continue Mciosis: Identify thc numbering ' the conect sequencc Homol logous chromiosones the center the cell; DNA replicationSpindle fibers pull homologous chromosomes thceu of the cell hap\nWord Edit View inser Format ools Table Window Help AutoSavc 08 6-08 66 13 Ccll Division Meiosis Online Lab Insert Dravi Design Layout Refercnces Mailings Review Share Homo Tmas Nevi ED \" Stylot stlet Organize the following statements the proper oracr of Interphase and _ first step with the nutn..."
]
| [
null,
"https://img.homeworklib.com/questions/75c61100-5cf7-11eb-8317-cde8e3a2ccab.png",
null
]
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