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[ "# Hypothesis testing\n\nI have a question about the equations used for hypothesis testing. They all look like this one here:\n\nH0: μ = μ0\nH1: μ > μ0\n\nFor various situations.. sigma know, sigma unknown etc. So H0 is the first claim and H1 is the second claim. But this means the second claim is > than the first claim. I know you have to use something else when it's the other way around. I'm talking about the tables with the formulas. You usually have the hypothesis on the left side, and on the right side it's the critical region. But the tables I have only have H1 down as μ>μ0. But what do I do when it is < instead of > ? I attached a file that shows what I mean.\nWikipedia has a bunch of formulas. Let's take p. H1: p1>p0\nWould I take the One-proportion z-test they have here ( roughly in the center of the table) ?\n\nhttp://en.wikipedia.org/wiki/Statistical_hypothesis_testing\n\nI don't know what to do if it's H1 down as μ<μ0 I hope it's somewhat clear what I'm asking for and somebody knows the answer.\n\n#### Attachments\n\n• 31.7 KB Views: 122" ]

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[ "# Let $f$ be a non-negative differentiable function such that $f'$ is continuous and $\\int_{0}^{\\infty}f(x)\\,dx$ and $\\int_{0}^{\\infty}f'(x)\\,dx$ exist.\n\nLet $f$ be a non-negative differentiable function such that $f'$ is continuous and $\\displaystyle\\int_{0}^{\\infty}f(x)\\,dx$ and $\\displaystyle\\int_{0}^{\\infty}f'(x)\\,dx$ exist.\n\nProve or give a counter example: $f'(x)\\overset{x\\rightarrow \\infty}{\\rightarrow} 0$\n\nNote: I think it is not true but I couldn't find a counter example.\n\n• By continuous derivative , do you mean that the derivative is continuous or that $f$ is both continuous and differentiable? – Bill O'Haran Apr 24 '18 at 16:55\n• @BillO'Haran I mean that the derivative is continuous – Ro168 Apr 24 '18 at 16:59\n\nLet $g_n(x) = \\sin (n^2 x) 1_{[0,{2 \\pi \\over n^2}]}$ and note that $g_n$ is continuous, $\\int |g_n| = {2 \\over n^2}$, $\\int g_n = 0$ and if $f_n(x)=\\int_0^x g_n$, then $f_n(x) \\ge 0$ for all $x$. Furthermore, $\\int |f_n| \\le {4 \\pi \\over n^3}$.\n\nDefine $f(x) = \\sum_{n \\ge 1} f_n(x-n)$, note that $f$ is integrable and non negative. Furthermore, $f'(x) = \\sum_{n \\ge 1} g_n(x-n)$ and it is straightforward to check that $f'$ is integrable as well.\n\n• Thanks! but I haven't studied function series yet – Ro168 Apr 26 '18 at 9:01\n• If you are studying improper integrals then you must have studied some form of summation. The above summations (for $f,f'$) are particularly straightforward because for any $x$, at most a finite number of terms are non zero. – copper.hat Apr 26 '18 at 15:14\n\nLet $\\varphi(x)=\\exp\\left(\\dfrac{1}{3}-\\dfrac{1}{4-x^{2}}\\right)$ for $|x|\\leq 2$, $\\varphi(x)=0$ for $|x|>2$.\n\nLet $f(x)=\\displaystyle\\sum_{n=1}^{\\infty}\\dfrac{1}{2^{n}}\\varphi\\left(2^{n}(x-n)\\right)$, one may check that $f\\in C^{\\infty}(0,\\infty)$ and that $f,f'\\in L^{1}(0,\\infty)$.\n\nFor all $x$ with $1<2^{n}(x-n)\\leq 2$, that is, $n+\\dfrac{1}{2^{n}}<x\\leq n+\\dfrac{2}{2^{n}}$, we have \\begin{align*} f'(x)&=\\dfrac{1}{2^{n}}\\exp\\left(\\dfrac{1}{3}-\\dfrac{1}{4-(2^{n}(x-n))^{2}}\\right)\\cdot-\\dfrac{2(2^{n}(x-n))}{(4-(2^{n}(x-n))^{2})^{2}}\\cdot 2^{n}\\\\ &=-\\dfrac{2(2^{n}(x-n))}{(4-(2^{n}(x-n))^{2})^{2}}\\exp\\left(\\dfrac{1}{3}-\\dfrac{1}{4-(2^{n}(x-n))^{2}}\\right), \\end{align*} localizing to $x=n+1/2^{n}$ we have $f'(n+1/2^{n})=-2/9$.\n\n• Thanks! But I've just started studying about improper integrals, so I'm looking for a simple counter example – Ro168 Apr 26 '18 at 11:11" ]

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[ "### Literal\n\nA literal is a constant value for some of the built-in types. Here are some examples of literals.\n\n```>>> a = 1 # 1 is an integer literal\n>>> b = 2.3 # 2.3 is a floating point literal\n>>> c = False # False is a boolean literal\n>>> d = 5L # 5L is a long integer literal\n>>> e = 8 + 6j # 8 + 6j is a complex literal\n>>> f = \"Hello\" # \"Hello\" is a string literal\n```\n\n### Expression\n\nAn expression is composed of variables and operators. The simplest expression is just a variable. The value of an expression is evaluated before it is used.\n\n### Arithmetic Operators\n\nThese are the arithmetic operators that operate on numbers (integers or floats). The result of applying an arithmetic operator is a number.\n\n• Subtraction: -\n• Multiplication: *\n• Division: /\n• Integer Division (truncates): //\n• Remainder: %\n• Exponentiation: **\n\n### Comparison Operators\n\nThere are 6 comparison operators. The result of applying the comparison operators is a Boolean - True or False.\n\n• Equal to: ==\n• Not equal to: !=\n• Greater than: >\n• Greater than or equal to: >=\n• Less than: <\n• Less than or equal to: <=\n\nYou may apply the comparison operator between two operands like so (a > b). You can also apply the comparison operators in sequence like so (a > b > c). However, this is a practice that is not recommended.\n\n### Boolean Operators\n\nIn Python, we have two Boolean literals - True and False. But Python will also regard as False - the number zero (0), an empty string (\"\"), or the reserved word None. All other values are interpreted as True. There are 3 Boolean operators:\n\n• not: unary operator that returns True if the operand is False and vice versa.\n• x and y: if x is false, then that value is returned; otherwise y is evaluated and the resulting value is returned.\n• x or y: if x is true, then that value is returned; otherwise y is evaluated and the resulting value is returned.\n\n### Truth Table for NOT\n\n a not a F T T F\n\n### Truth Table for AND\n\n a b a and b F F F F T F T F F T T T\n\n### Truth Table for OR\n\n a b a or b F F F F T T T F T T T T\n\n### Bitwise Operators\n\nThe bitwise operators include the AND operator &, the OR operator |, the EXCLUSIVE OR operator ^ and the unary NOT operator ~. The bitwise operators applies only to integer types.\n\n### Truth Table for EXCLUSIVE OR\n\n a b a ^ b 0 0 0 0 1 1 1 0 1 1 1 0\n\n### Shift Operators\n\nShift operators are applied only to integer types.\n\n• x << k : shift the bits in x, k places to the left. Multiplication by (2 ** k).\n• x >> k : shift the bits in x, k places to the right filling in with the highest bit on the left hand side. Division by (2 ** k).\n\n### Evaluation Order\n\nPython evaluates an expression from left to right except for assignment. In an assignment, the right-hand side is evaluated before the left-hand side. Refer to the operator precedence to determine the order in which the operations are performed." ]

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[ "Paul's Online Notes\nHome / Calculus I / Derivatives / Logarithmic Differentiation\nShow Mobile Notice Show All Notes Hide All Notes\nMobile Notice\nYou appear to be on a device with a \"narrow\" screen width (i.e. you are probably on a mobile phone). Due to the nature of the mathematics on this site it is best views in landscape mode. If your device is not in landscape mode many of the equations will run off the side of your device (should be able to scroll to see them) and some of the menu items will be cut off due to the narrow screen width.\n\n### Section 3.13 : Logarithmic Differentiation\n\n4. Find the first derivative of $$g\\left( w \\right) = {\\left( {3w - 7} \\right)^{4w}}$$.\n\nShow All Steps Hide All Steps\n\nStart Solution\n\nWe just need to do some logarithmic differentiation so take the logarithm of both sides and do a little simplifying.\n\n$\\ln \\left[ {g\\left( w \\right)} \\right] = \\ln \\left[ {{{\\left( {3w - 7} \\right)}^{4w}}} \\right] = 4w\\,\\,\\ln \\left( {3w - 7} \\right)$ Show Step 2\n\nUse implicit differentiation to differentiate both sides with respect to $$w$$. Don’t forget to product rule the right side.\n\n$\\frac{{g'\\left( w \\right)}}{{g\\left( w \\right)}} = 4\\ln \\left( {3w - 7} \\right) + 4w\\frac{3}{{3w - 7}} = 4\\ln \\left( {3w - 7} \\right) + \\frac{{12w}}{{3w - 7}}$ Show Step 3\n\nFinally, solve for the derivative and plug in the equation for $$g\\left( w \\right)$$ .\n\n\\begin{align*}g'\\left( w \\right) & = g\\left( w \\right)\\left[ {4\\ln \\left( {3w - 7} \\right) + \\frac{{12w}}{{3w - 7}}} \\right]\\\\ & = \\require{bbox} \\bbox[2pt,border:1px solid black]{{{{\\left( {3w - 7} \\right)}^{4w}}\\left[ {4\\ln \\left( {3w - 7} \\right) + \\frac{{12w}}{{3w - 7}}} \\right]}}\\end{align*}" ]

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[ "# coordinate geometry   5 videos\n\n• #### Calculating Coordinates in the Unit Circle\n\n##### GeometryPythagorean Theorem\n\nHow to describe the unit circle. How to draw the unit circle and label its parts. How to strategize about finding coordinates on the unit circle.\n\ngraphing a rational expression 1/x asymptotes\n• #### Calculating the Midpoint\n\n##### GeometryGeometry Building Blocks\n\nHow to find the midpoint of a segment with endpoints in rectangular coordinates, How to write the midpoint formula.\n\nendpoint coordinate plane\n• #### Vertex and Diagonals\n\n##### GeometryGeometry Building Blocks\n\nHow the vocabulary word vertex applies to different objects in Geometry.\n\nvertex isosceles triangle polygon diagonal cone pyramid prism\n• #### Translations\n\n##### GeometryTransformations\n\nHow to describe a translation; how to interpret ordered pair rules.\n\ntransformation translation isometry coordinates ordered pair image\n• #### Distance Formula\n\n##### GeometryPythagorean Theorem\n\nHow to derive the equation for the distance between two points using the Pythagorean Theorem.\n\ndistance coordinate plane ordered pair segment" ]

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[ "# 辅导program、辅导R程序设计\n\n- 首页 >> CS\nEconometrics 761\nsubmission should be a pdf document.\nThis document consists of two questions: Q1, which is a combination of theory and\ncoding, and Q2, which is an applied question asking you to replicate the results in a\npublished paper. Take the time to read all explanations in this document, and pay\nattention to how you report your results after running the code. Please, use full sentences\nwhen answering questions. Treat this as a report that you may have to write for your\nfuture job.\nMost of the R code is already written for you: all R code appears in blue in this pdf\n(except for this line). This means that everything that is blue can be run in R directly,\ni.e. you can just copy-paste the blue code and it will run (provided that you followed all\ndirections given to you in this document).\nYou have to pay attention to how you copy-paste from the pdf1!\nOne of the questions is asking you to use the gmm package. Those of you with\nMacs may have issues loading it, so switch to working on the R Studio Cloud,\nWarning vs Error Messages in R: You may still get \"warnings\" in R Studio Cloud --\nso pay attention if something is labeled as \"error\" or \"warning.\" If it is a warning,\nit’s fine, move ahead as if nothing happened. Warnings still produce correct answers.\nIf it’s an error, then you have to investigate.\n1For example, when you copy-paste the code from pdf into R markdown, be careful with\nthe underscores, you will have to add them manually\nthe comments following ##, you may have to double check that everything that follows ##\nin the pdf is commented in your R Markdown file, else you may get an error\nthe left arrow in R is <-, pay attention to how this gets copied from pdf, you may have to change\nit manually before you run the code\n1\nYou can suppress warning message from your output by including in your {r chunk,\nwarning=FALSE}. For more on this, take a look at this document.\nYou can also suppress the output from a regression by including in your {r chunk}\nthe following: {r echo = T, results = \"hide\"}. You may need this for Q2(e). Please\nlook at page 18 of PS.pdf.\n– So you can write {r echo = T, warning = F, results = \"hide\"} which would\nsuppress warnings and the output.\nThis project should not take you more than 3 days to complete (this is a generous esti-\nmate). However, you have a bit more than 2 weeks to work on this project. Given the\nfact that you have plenty of extra time to work on this project:\n1. you are strongly advised to start working on the assignment as soon as you get it;\n2. you are strongly incentivised to start working on the project as soon as possible by\nhaving limited contact with myself and the TAs, in the sense that I will not\nthe discussion board on A2L.\n3. you are heavily penalized for late submissions. There will be a penalty of 25%\nper 12 hours of being late, no matter the excuse.\nMake sure that you start early and aim to submit at least 3 days earlier than the deadline.\nThis should give you plenty of time to work trough any trouble shooting that you may\nneed to do as well as to ask questions during office hours, and, in particular, it will help\nyou keep calm. As most code is provided to you (in blue), this project is mostly about\npaying attention to what the problem is asking you, as well as to understanding the code.\nPlease, pay attention, read everything at least twice, and then double check that you are\n2\nQuestion 1\nThis question has a theoretical component and a coding (in R) component (99.9% of the\ncode is already written for you in this text). This exercise is based on Wooldridge (2001).2\nYou can find this paper on A2L, under the Individual PS module.\n? Read pages 87 and 88 (at a minimum) in Wooldridge (2001) to remind yourselves\nabout the method of moments estimator (MM) and GMM.\nReminder: GMM is a method for constructing estimators, that uses assumptions about\nspecific moments of the random variables. The assumptions are called “moment con-\nditions.” Moment conditions are expected values that specify the model parameters in\nterms of the true moments. The sample moment conditions are the sample equivalents to\nthe moment conditions. GMM finds the parameter values that are closest to satisfying\nthe sample moment conditions. When there are more moment conditions than parame-\nters, GMM finds the parameters that get as close as possible to solving weighted sample\nmoment conditions. Uniform weights and optimal weights are two ways of weighting the\nsample moment conditions. The uniform weights use an identity matrix to weight the\nmoment conditions. The optimal weights use the inverse of the covariance matrix of the\nmoment conditions. [All this in the videos and slides on GMM that you can find on\nA2L.] GMM generalizes the method of moments (MM) by allowing the number of mo-\nment conditions to be greater than the number of parameters. Using these extra moment\nconditions makes GMM more efficient than MM. When there are more moment conditions\nthan parameters, the estimator is said to be overidentified. GMM can efficiently combine\nthe moment conditions when the estimator is overidentified.\nWe will illustrate these points by estimating the mean of a normally distributed ran-\ndom variable with mean = θ0 and variance = θ\n2\n0. We will be estimating θ0 by MM, GMM,\nand an efficient GMM estimator.\nTheoretical part\nLet y = (y1, y2, . . . , yn) be a random iid sample with yi ～ N (θ0, θ20), that is, yi are normally\ndistributed with mean θ0 and variance θ\n2\n0. Note that there is one parameter here (rather\nthan two).\n2Wooldridge (2001). Applications of Generalized Method of Moments Estimation. Journal of Eco-\nnomic Perspectives, 15(4): 87-100.\n3\n(a) Write down the MM estimator for θ0 that only uses the moment condition E(yi) =\nθ0. What is the asymptotic distribution of this estimator as n→∞?\n(b) Consider the GMM estimator for θ that uses the moment conditions E[g(yi, θ0)] = 0,\nwhere g(yi, θ) = (yi ? θ, y2i ? 2θ2)′. These are two moment conditions for one\nparameter, so the choice of GMM weight matrix matters. We choose the weight\nmatrix.\nFind an explicit expression for the resulting GMM estimator.\n(c) What would have been the asymptotically optimal weight matrix choice in (b)?\nWhat is the probability limit of the chosen Wn in (b) as n → ∞? Is the choice of\nWn in (b) asymptotically optimal?\n(d) Calculate the asymptotic variance of the GMM estimator obtained in (b), and com-\npare to the asymptotic variance of the estimator in (a).\nCoding part\nFirst, we generate our data. We draw a sample of size n = 500 from the population\nrandom variable Y that is normally distributed with mean 2 and variance 4, i.e. Y ～\nN (θ0, θ20) = N (2, 4). So for this part we assume that we know the true parameter θ0 = 2.\nNote that this is something specific to simulations; in reality, we do not know the true\nvalue of θ0, but in simulations, we do. This will help us understand whether GMM is\nindeed more efficient than MM. Without knowing the true θ0 we would not be able to\nThe lines below generate one sample of iid random variables drawn from this popula-\ntion random variable..\nset.seed(1234) ## setting the seed to be able to replicate our results\nn = 500 ## sample size\ny = rnorm(n,2,2) ## y is a normally distributed random variable with mean=2\nand variance=4; there is no typo here, to convince yourselves type ??rnorn\nin the R console and read the help file\n4\nSecond, we compute the MM estimator in (a) based on this one sample. Using your\nanswer from (a) above, the MM estimator is given by\ntheta hat MM = mean(y)\nwhen you copy-paste the code from pdf into R markdown, be careful with\n– the underscores, you will have to add them manually\n– the comments following ##, you may have to double check that everything\nthat follows ## in the pdf is commented in your R Markdown file, else you\nmay get an error\nYou will notice that there is a small bias by computing theta hat MM - 2.\nThe value of theta hat MM is based on just one draw from the population. What we\nwould like now is to draw S times from the population random variable Y , where S is a\nlarge number. That is, we will generate j = 1, . . . , S samples of size n, where each sample\nis drawn from Y . Each sample is used to compute the value of θ?j,n, j = 1, . . . , S; that is,\nsample j = 1 produces a value for θ?1,n, ..., sample j = S produces a value for θ?S,n. We\nwill have a vector of size S × 1 of values\nThe value of our estimator will be the average across simulations, i.e. θ? = 1\nS\n∑S\nj=1 θ?j,n.\n(Note that n is the same number across simulations, i.e. n the sample size does not vary\nwith S, we always draw the same sample size but we draw it many times.)\nIf you remember, when we talk about inference, we say “if we were to repeat the\nexperiment many times,...” What this means is that we would like to draw different\nsamples of size n repeatedly (and many times) from the same population random\nvariable Y . We would then compute the estimator based on each sample, call\nit θ?j,n, j = 1, . . . , S, and then we would average across all draws, such that the\nestimator that we report in a simulation study will be\n1\nS\nS∑\nj=1\nθ?j,n.\n5\nTo do this, we add the lines:\nS = 1000 ## number of draws\ntheta hat MM ← matrix(NA, nrow = S, ncol = 1) ## creates a vector of\nNAs that has S rows and one column (row j is the value θ?j,n, j = 1, . . . , S)\nfor(j in 1:S){\ny = rnorm(n,2,2) ## a new sample j is drawn from the population Y\ntheta hat MM[j] = mean(y) ## θ?j,n =mean of sample j\n}\n? pay attention to { and }!\nSo your code so far should look like this\nset.seed(1234)\nn = 500\nS = 1000\ntheta hat MM ← matrix(NA,nrow=S,ncol=1)\nfor(j in 1:S){\ny = rnorm(n, mean=2, sd=2)\ntheta hat MM[j] ← mean(y)\n}\nRun this code.\nThe vector theta hat MM contains 1000 values each computed on a sample drawn\nfrom the population random variable.\n(e) As we learned, estimators have sampling distributions. The estimator theta hat MM\nhas a sampling distribution that can be plotted by writing (after running the code\nabove):\nhist(theta hat MM, freq=F, ylim=c(0,7)) ## histogram\nlines(density(theta hat MM), col=“red”) ## sampling density\n6\nThe CLT says that this sampling distribution is approximately normal. To\nsee this, we add to this plot a normal density with mean = theta hat MM\nand variance = sqrt(var(theta hat MM)). So add to the code the following\nline:\nlines(seq(1.1, 2.5, by=.01), dnorm(seq(1.1, 2.5, by=.01),mean(theta hat MM),\nsd(theta hat MM)), col=“black”)\nYou should now have a plot with a histogram, a red curve, and a black curve.\nThe red curve is the sampling distribution. The normal density is the black\ncurve.\nDoes theta hat MM that you plotted look unbiased? How can you tell simply\nby looking at the plot? Does this plot show us that the sampling distri-\nbution of the MM estimator is approximately normal? How can you tell?\nReport now mean(theta hat MM) and var(theta hat MM). What is the\nbias of mean(theta hat MM)? Does this match your plot?\n(f) We will implement now the GMM estimator in (b) using the formula for the esti-\nmator that you derived in (b). To write this formula in R, we write the line below:\ntheta hat gmm ← ((1/2)*mean(y)*mean(y?2))?(1/3)\nAs before, we want now to compute this for each draw S = 1, 2, ..., 1000. So\nwe add this to our for-loop:\ntheta hat MM ← matrix(NA,nrow=S,ncol=1) ## vector that stores MM es-\ntimator after each draw\ntheta hat gmm ← matrix(NA,nrow=S,ncol=1) ## vector that stores GMM\nestimator from (b) after each draw\nfor(j in 1:S){\ny = rnorm(n, mean=2, sd=2)\ntheta hat MM[j] ← mean(y) ## computes MM estimator\ntheta hat gmm[j]← ((1/2)*mean(y)*mean(y?2))?(1/3) ## computes the GMM\nestimator from (b)\n}\nSo your entire code should look like\n7\nset.seed(1234)\nn = 500\nS = 1000\ntheta hat MM ← matrix(NA,nrow=S,ncol=1)\ntheta hat gmm ← matrix(NA,nrow=S,ncol=1)\nfor(j in 1:S){\ny = rnorm(n, mean=2, sd=2)\ntheta hat MM[j] ← mean(y)\ntheta hat gmm[j] ← ((1/2)*mean(y)*mean(y?2))?(1/3)\n}\nRun this code.\nTo compare the MM and the GMM estimator, write\nc(mean(theta hat MM),var(theta hat MM))\nc(mean(theta hat gmm),var(theta hat gmm))\nWhich estimator has the smaller bias? which one has the smaller variance? Is\nthis what you expected given the theory?\n(g) First, install and load the gmm package, i.e.\ninstall.packages(“gmm”)\nlibrary(gmm)\nThen create the moment conditions function g(y, θ) from (b). The lines below\ngenerate a function that takes as arguments x (which is your data) and tet\n(which is your theta), and return an n x 2 matrix:\ng ← function(tet,x){\nm1 ← (x-tet)\nm2 ← (x?2 - 2*tet?2)\nf ← cbind(m1,m2)\nreturn(f)\n8\n}\nWrite the lines above and run the code. Then test the function. To do this,\nwrite:\nyy = c(1,2,3)\ng(0,yy)\nPrint g(0,yy) on the screen. g should be a 3 x 2 matrix\nCompute g(y, θ) from (b) where y =yy (from above, yy=c(1,2,3)) and θ = 0\nby hand (again, you should get a 3 x 2 matrix). How do they compare,\nthe g computed with the code and the g that you calculated by hand? Are\nthey equal? Should they be?\nNow to compute the GMM estimator calculated with the optimal matrix, write\nthe following code:\ninit ← 2 ## intial value to be used in the GMM algorithm\ntheta hat optm gmm ← gmm(g, y, init, wmatrix=“optimal”)\\$coefficient ##\ncomputes GMM estimator using the moments in g, the data y, the initial\nstarting value in init, and the optimal weight matrix; it stores only the\nGMM estimator for the coefficient\nWe want now to compute S such GMM estimators. So we need to add two\nlines to the code below. What are these two lines? You can use what we\ndid in (f) as an example. It is straight-forward if you understand what\nhappened in (f). Your code, with the two lines (you need to replace ??) is:\nset.seed(1234)\nn = 500\nS = 1000\ntheta hat MM <- matrix(NA,nrow=S,ncol=1)\ntheta hat gmm ← matrix(NA,nrow=S,ncol=1)\ntheta hat optm gmm ← ??\ninit ← 2\ng ← function(tet,x){\n9\nm1 ← (x-tet)\nm2 ← (x?2 - 2*tet?2)\nf ← cbind(m1,m2)\nreturn(f)\n}\nfor(j in 1:S){\ny = rnorm(n, mean=2, sd=2)\ntheta hat MM[j] ← mean(y)\ntheta hat gmm[j] ← ((1/2)*mean(y)*mean(y?2))?(1/3)\ntheta hat optm gmm[j] ← ??\n}\nReplace ?? above and then run the code. Then report then the following:\nc(mean(theta hat MM),var(theta hat MM))\nc(mean(theta hat gmm),var(theta hat gmm))\nc(mean(theta hat optm gmm),var(theta hat optm gmm))\nWhich estimator has the smallest variance, and which estimator is efficient?\nIs this as you expected given what you learned about these estimators in\nthis class?\n(h) Take the plot in (e) above, and add curves for the sampling densities of theta hat gmm\nand theta hat optm gmm. Color the density of theta hat gmm blue and that of\ntheta hat optm gmm green. (You have to write the code for this, and you can fol-\nlow the code in (e) as example of how to do this). So your plot will have: histogram\nof the MM estimator, samplind density of the MM estimator (in red), sampling\ndensity of the GMM estimator (blue), and sampling density of the GMM estimator\nwith optimal weight matrix (in green). It should look like Figure 1 below:\n10\nFigure 1: Sampling densities for all three estimators.\nDoes this plot match with your answer from (g) above? Explain what the red, green,\nand blue curves are and how they support or not your answer in (g) above.\n11\nQuestion 2\nThis question is based on the paper of Bedard and Dhuey (“The Persistence of Early\nChildhood Maturity: International Evidence of Long-Run Age Effects”, QJE, 2006).\nThe paper is available on A2L under the Individual PS module.\nThe data used in this paper is on A2L under the same module.\nYou are expected to reproduce some of the results in this paper. To do this, you\nwill have to write R code. I provide below code that is supposed to help you with\nthis. For parts a and b, you pretty much have to copy-paste the code that I wrote\nfor you, while for d and e you have to adapt the code from lm regression to ivreg\nbased on the code that I provided for you for part b.\nWhen you report your results for Q2, use the fixest package that Caleb in-\ntroduces in his video \"Ordinary Least Squares in R\", that you can find on\nEduflow under R Videos or on YouTube. You can also use the stargazer pack-\nage if you find it easier to use, directions on the last page of the document whose\nlink you can find here or in the description of the module on Individual Project up\non A2L.\nPay particular attention to what the question is asking you! Look at the code,\nlook at what the question is asking you. This is an exercise in attention more than\nanything, as you have the code written and you have to adapt it to the question.\nThe paper uses data from multiple sources. We only want to reproduce a small fraction\nof the results in the paper. We only use the TIMSS (Trends in International Mathematics\nand Science Study) data from Canada for 3rd and 4th graders (9 year olds in 1995), while\nBedard and Dhuey (2006) also consider many other countries and also another age group\n(13 year olds). The dataset is available on A2L under the same module (“timss-canada-\n1995-pop1.dta”).\nEvery entry in the dataset corresponds to one student. The dataset contains the\nvariables weight (sampling weights that can be used to obtain nationally representa-\ntive estimation results) and idschool (identifier for the school of the student). The\noutcome variable is math score (math test score, normalized to have mean 50 and stan-\ndard deviation 10). The main explanatory variable of interest, which is treated as endoge-\nnous, is age (age in months), and the the corresponding instrumental variable is r age\n12\n(relative age in months, see paper for the definition). Additional control variables are\ngrade (either 3rd or 4th grade, since students are 9 years old), female (gender dummy),\nmother native (dummy for mother born in country), father native (dummy for fa-\nther born in country), both parents (dummy for both parents present in household),\ncalculator (dummy for calculator at home), computer (dummy for computer at home),\nbooks100 (dummy for whether more than 100 books at home), hh size (a measure for\nhousehold size), and rural (dummy for whether school is in rural area).\n(a) Start by reading the paper. Explain – in your own words – why age is probably\nendogenous here. Explain how the instrumental variable (relative age) is computed.\nExplain why relative age is arguably exogenous. Explain why relative age is arguably\na relevant instrument for age.\n(b) Reproduce the OLS and IV (i.e. 2SLS) results for math test scores in\nCanada in Table III of Bedard and Dhuey (2006). So these would be col-\numn 1 and column 4 for Math in Table III. Report your estimation results for\nboth unweighted OLS and IV and for population weighted OLS3 and\nIV (weighted 2SLS). Code to help you with this is provided at the end of this\ndocument, you just have to follow the directions given and then copy-paste the\ncode. Report your findings for non-weighted regressions and for weighted regres-\nsions both with heteroscedasticity robust standard errors, and compare to Bedard\nand Dhuey (2006).\n– You will not obtain the same answers as in the table, which is to be expected.\nThink why this may be the case. Read the description of the data in the paper,\nlook at the data that you have imported in R and the exact regression that\nthe authors ran vs the one that you ran, read my note below. What could\njustify the fact that the numbers are not exactly reproduced? But even if you\ncannot reproduce their results numerically, can you reproduce their qualitative\nfindings, e.g. how does the statistical significance of your coefficient in front of\nage compare to theirs?\n3However, the weights are not chosen as the inverse of the variance of the errors with\nthe goal of minimizing the variance, but instead are computed with the goal of making the\nestimates nationally representative. The WLS standard errors will be larger here than the\nOLS standard errors.\n13\n– Note that our normalization of the outcome variable is different from the one\nused in Bedard and Dhuey (2006), because they normalize over their whole\nsample, while we only normalize over the data from Canada. The household\nsize variable might also be defined slightly differently.\n(c) For the estimation results in (b), are the signs of the coefficients on the control\nvariables as expected? Explain.\nIn the following we only perform unweighted estimation, i.e. we do not use the weight\nvariable anymore, which would allow us to obtain nationally representative results.4 It\nalso turns out to not matter much whether heteroscedasticity robust standard errors are\nreported here, so it is OK to not use them in the answer to the following questions.\n(d) Repeat the IV estimation in (b) for the following eight subpopulations: only female\nstudents, only male students, only students whose mother is native to Canada, only\nstudents whose mother is not native to Canada, only students who report to have\na calculator at home, only students who report not to have a calculator at home,\nonly students who report to have a computer at home, only students who report\nnot to have a computer at home. Code to help you with this is provided at the end\nof the document. For this part, you have to adapt the code to answer the question\nasked. The code that I provide is only for subpopulation of female students, so\nyou have to adapt the code to the other subpopulations. Interpret your findings,\nfocusing on the differences in age effect for the different subpopulations (e.g.\nmale vs. female age effect). Note that the calculator and computer dummies can\nbe seen as rough proxies for socioeconomic status, and mother native is probably\nalso positively correlated with socioeconomic status.\n– You have to run the code with all covariates for the different populations, but\nyou can report only the coefficient for age (with its standard deviation and\nt-stat). Example of how you can do this in code at the end of the document.\n(e) As a robustness check, repeat the analysis in (d), but this time including dummy\nvariables for each school as regressors (so called “school specific fixed effects”). Code\n4The results using weights are less robust and have larger standard errors, because they are driven\nby a relatively small subset of individuals with large weights. Not using weights also simplifies the\nimplementation somewhat, e.g. xtivreg does not work with individual specific weights.\n14\nto help you with this is provided at the end of the document. For this part, you\nhave to adapt the code to IV regression. The code I gave you is for OLS (it uses\nthe lm function) but you need to report results for IV regressions. How do the\ndifferences in age effects (e.g. male vs. female age effects) change compared to\n(d)? Explain why the inclusion of school dummies in the regression might change\nthe results.\n– You have to run the code with all covariates for the different populations, but\nyou can report only the coefficient for age (with its standard deviation and\nt-stat). Example of how you can do this in code at the end of the document.\nFirst you have to load your data into R and install a few packages.\nGo to RStudio, File, Import Dataset, From Stata..., Browse, Select the data set\n“Import.” You have now imported the Stata data set in R.\n– “lmtest”,\n– “MASS”,\n– and you may also need “sandwich”.\nTo run the unweighted OLS regression (similar to that in Table III first column for\nCanada), just regress math score on all the variables mentioned in the footnote of\nthat table that appear in your data set, i.e.\nlmAPI← lm(math score～age+grade+female+mother native+father native+both parents\n+calculator+computer+books100+hh size+rural,\nand then correct for heteroskedasticity by running:\n15\ncoeftest(lmAPI, vcov=vcovHC(lmAPI, “HC1”) ## this line produces HAC\nrobust standard errors; you may need to install and load “sandwich” for\nthis depending on your R version\nTo produce the weighted OLS regression that appears in Table III, column 1, run\nlmAPI w← rlm(math score～age+grade+female+mother native+father native+both parents\n+calculator+computer+books100+hh size+rural,\nand then you correct for heteroskedasticity by running:\ncoeftest(lmAPI w, vcov=vcovHC(lmAPI w, “HC1”)\nThis is supposed to reproduce the weighted OLS regression with heteroskedasticity robust\nstandard errors in Table III, column 1 for Canada.5\nTo run an IV regression, install and load the AER package, and then run the un-\nweighted regression\nlmAPI iv← ivreg(math score～age+grade+female+mother native+father native\n+both parents+calculator+computer+books100+hh size+rural | r age+grade\n+female+mother native+father native+both parents+\ncalculator+computer+books100+hh size+rural,\nPay attention to the code above. After the conditioning “|” we add all the instrument\nand ALL EXOGENEOUS explanatory variables that we used in the regression. Since age\nis exogeneous, it is not added. Pay attention to how the code is written as if you want to\nread more on this, look up ivreg in R help or on stack exchange.\nTo run the weighted version of the IV regression above, add weights=weight after\ndata=timss canada 1995 pop1, just as we did for the weighted OLS regression.\n5You will not get the same numerical answers. But are they qualitatively the same?\n16\nThe code below is for lm, you have to adapt it for ivreg. To run an lm regression for\ndifferent subsets of the data, we subset the data as follows. For example, for female\nstudents, we run\nlmAPI w female← lm(math score～age+grade+mother native+father native\n+both parents+calculator+computer+books100+hh size+rural,\nPay attention to the line that runs the regression! There is no female as a covariate\nanymore. If you include female as a covariate you will get an error saying that ’x’ is\nsingular: singular fits are not implemented in lm. It is the same for ivreg – you must\neliminate female as an exogeneous covariate when you run ivreg! Pay attention, else you\nwill get an error.\nThe code below is for lm, you have to adapt it to ivreg. To generate school fixed\neffects, you can add “factor(idschool)” as an exogeneous covariate, and run the\nfollowing:\n“‘{r echo = T, results = \"hide\"} ## this line hides the output (in particular,\nit hides the output, as it contains all ID schools which would make it too\ndifficult to read; run your chuck with and without results=”hide” to see\nwhat happens)\nlmAPI w female schoolFE ← rlm(math score～age+factor(idschool)+grade\n+mother native+father native+both parents+calculator\n+computer+books100+hh size+rural," ]

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[ "## Data Visualization And Data Analytics\n\n#### GET ASSIGNMENT HELP", null, "# Physics Assignment Help With The Stress\n\n## 9.2 The Stress\n\nTo oppose the deformation caused by the external applied forces, the intensity of the internal force developed at a point to keep it in equilibrium was defined as stress and mathematically it is given by an expression", null, "where DF is the internal restoring force acting on an small area DA.\n\nThe normal component of DF cause axial stress or normal stress while tangential component DF causes shear stress or tangential stress.", null, "In figure DF force is acting as an element DA which has normal and tangential component as Fn and Ft respectively. Thus, axial stress on the element is", null, "and shear stress as the element is", null, "The axial stress either create tension in solid body or create compression in it. In former case, it is called tensile stress and in later one. It is called compressive stress. While the shear stress cause the layer of surfaces to slide over the lower one along the direction of shear stressing force.\n\n(i) When the stress is applied tangential to a surface, it is called tangential or shearing stress.\n\n(ii) Dimensional formula of stress is [ML–2T–2] and CGS and MKS units are dyne/cm2 and newton/m2 respectively.\n\n(iii) Stress is a tensor quantity.\n\n(iv) Through both stress and pressure and defined as force per unit area but even then they differ from each other due to following reasons:\n\n(a) Pressure is always normal to the area while stress can be either normal a tangential.\n\n(b) Pressure on a body is always compressive while stress can be either compressive or tensile.\n\n(c) Pressure is a scalar while stress is a tensor. Shearing stress is exhibited by solids only as these have definite shape.", null, "### Email Based Assignment Help in The Stress\n\nWe are the leading online assignment help provider. Find answers to all of your doubts regarding the The Stress. Assignmenthelp.net provide homework, assignment help to the school, college or university level students. Our expert online tutors are available to help you in The Stress. Our service is focused on: time delivery, superior quality, creativity and originality.\n\nTo Schedule a physics The Stress tutoring session\n.\n\nFollowing are some of the areas in physics Matter Properties which we provide help:\n\nAssignment Help Features\nAssignment Help Services\n• Assignment Help\n• Homework Help\n• Writing Help", null, "", null, "" ]

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[ "Motion of Charged Particle in a Magnetic Field | Moving Charges and Magnetism Class 12, JEE & NEET\n\nClass 9-10, JEE & NEET\n\nWhen a charged particle moves along a magnetic field line into a region where the field becomes stronger, the particle experiences a force that reduces the component of velocity parallel to the field. This force slows the motion along the field line and here reverses it, forming a Magnetic Mirror. The motion of a charged particle in a magnetic field is characterized by the change in the direction of motion. It is expected also as the magnetic field is capable of only changing the direction of motion. In order to keep the context of the study simplified, we assume the magnetic field to be uniform. This assumption greatly simplifies the description and lets us easily visualize the motion of a charged particle in a magnetic field.\n\nMotion of a Charged Particle in a Magnetic Field\n\nThe motion of a charged particle when it is moving collinear with the field, the magnetic field is not affected by the field (i.e. if motion is just along or opposite to a magnetic field) ( $\\quad F=0$ ) Only the following two cases are possible:\n\nThe case I: When the charged particle is moving perpendicular to the field The angle between $\\overrightarrow{\\mathrm{B}}$ and $\\overrightarrow{\\mathrm{v}}$ is $\\theta=90^{\\circ}$\n\nSo the force will be maximum ( $=$ qvB ) and always perpendicular to motion (and also field);\n\nHence the charged particle will move along a circular path (with its plane perpendicular to the field).\n\nCentripetal force is provided by the force qvB,", null, "", null, "In case of the circular motion of a charged particle in a steady magnetic field :", null, "i.e., with the increase in speed or kinetic energy, the radius of the orbit increases. For uniform circular motion $v=\\omega r$\n\nAngular frequency of circular motion called cyclotron or gyro-frequency. $\\omega=\\frac{\\mathrm{v}}{\\mathrm{r}}=\\frac{\\mathrm{qB}}{\\mathrm{m}}$\n\nand the time period, $\\quad \\mathrm{T}=\\frac{2 \\pi}{\\omega}=2 \\pi \\frac{\\mathrm{m}}{\\mathrm{qB}}$\n\ni.e., time period (or frequency) is independent of speed of particle and radius of the orbit.\n\nTime period depends only on the field B and the nature of the particle,\n\ni.e., specific charge (q/m) of the particle.\n\nThis principle has been used in a large number of devices such as cyclotron (a particle accelerator), bubble-chamber (a particle detector) or mass-spectrometer etc.\n\nthe motion of a charged particle in an electric and magnetic field\n\ncase II : The charged particle is moving at an angle $\\theta$ to the field :\n\n$\\left(\\theta \\neq 0^{\\circ}, 90^{\\circ} \\text { or } 180^{\\circ}\\right)$\n\nResolving the velocity of the particle along and perpendicular to the field.\n\nThe particle moves with constant velocity v cos $\\theta$ along the field ($\\because$ no force acts on a charged particle when it moves parallel to the field).\n\nAnd at the same time, it is also moving with velocity $v$ sin $\\theta$ perpendicular to the field due to which it will describe a circle (in a plane perpendicular to the field)", null, "", null, "", null, "So the resultant path will be a helix with its axis parallel to the field $\\overrightarrow{\\mathrm{B}}$ as shown in fig. The pitch p of the helix $=$ linear distance travelled in one rotation\n\n$p=T(v \\cos \\theta)=\\frac{2 \\pi m}{q B}(v \\cos \\theta)$" ]

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[ "", null, "Welcome to challenge 19! Use this thread for any and all questions relating to challenge 19.", null, "2 Likes\n\nToo soon for a golfed solution?\n\n``````pumpkinSpice=m=>(x=[5,3,1].map(p=>(q=m/p|0,m%=p,q))).concat([30,15,3].reduce((a,v,i)=>a+v*x[i],0))\n``````\n\n`p, q, m` store the price, quotient, and money remaining.\nIn the `reduce` function, `a, v, i` respectively store the accumulator, current value, and current index.\n\nWhen calculating the quotient, we use `| 0` (bitwise OR) to truncate the number, saving characters because we don’t use `Math.floor` or `Math.trunc`.\n\n2 Likes\n``````const pumpkinSpice = money => {\nlet spice = 0;\n\nconst treats = {\n\"pie\": {\n\"cost\": 5,\n\"spice\": 30,\n\"bought\": 0\n},\n\"latte\": {\n\"cost\": 3,\n\"spice\": 15,\n\"bought\": 0\n},\n\"macaron\": {\n\"cost\": 1,\n\"spice\": 3,\n\"bought\": 0\n}\n}\n\nfor (let i in treats) {\nwhile(money >= treats[i].cost) {\nmoney = money - treats[i].cost;\nspice += treats[i].spice;\ntreats[i].bought++;\n}\n}\n\nreturn [treats.pie.bought, treats.latte.bought, treats.macaron.bought, spice];\n}\n``````\n7 Likes\n``````const pumpkinSpice = money => {\nconst foods = {\n\"pie\": {\n\"cost\": 5,\n\"spice\": 30\n},\n\"lattes\": {\n\"cost\": 3,\n\"spice\": 15\n},\n\"macarons\": {\n\"cost\": 1,\n\"spice\": 3\n},\n}\n\nlet totalSpice = 0;\nreturn Object.keys(foods).reduce((acc, cv) => {\nconst moneyBefore = money;\ntotalSpice += Math.floor(money / foods[cv].cost) * foods[cv].spice;\nmoney -= Math.floor(money / foods[cv].cost) * foods[cv].cost;\nreturn [...acc, Math.floor(moneyBefore / foods[cv].cost)]\n}, []).concat(totalSpice);\n}\n``````\n\nMight redo this one when I have time\n\n2 Likes\n\nconst pumpkinSpice = money => {\nvar quotient = Math.floor(money/5);\nvar leftover = money-5quotient\nvar quotient2 = Math.floor(leftover/3);\nvar leftover = leftover-3\nquotient2\nvar quotient3 = Math.floor(leftover/1);\n\n``````var alll = quotient*30 + quotient2*15 + quotient3 * 3\n``````\n\n}\n\n1 Like\n\nYou always post very similar solutions before I can post mine, lol.\n\n``````const pumpkinSpice = money => {\nconst C = {pie: {amt: 5, spice: 30},\nlatte: {amt: 3, spice: 15},\nmac: {amt: 1, spice: 3}}\nvar Tspice = 0\nlet meal = Object.keys(C).reduce((acc,food) => {\nlet count=Math.floor(money / C[food].amt)\nmoney %= C[food].amt\nTspice += count * C[food].spice\nreturn [...acc, count]\n},[])\nreturn meal.concat(Tspice)\n}\n``````\n\nedit: fixed to make clearer\n\n2 Likes\n```const pumpkinSpice = money => {\n\n// array of pies, lattes, macarons, total grams\nconst pumpkinAmount = [0, 0, 0, 0];\n\nwhile (money >= 5) {\npumpkinAmount++;\npumpkinAmount += 30;\nmoney -= 5;\n}\n\nwhile (money >= 3) {\npumpkinAmount++;\npumpkinAmount += 15;\nmoney -= 3;\n}\n\nwhile (money >= 1) {\npumpkinAmount++;\npumpkinAmount += 3;\nmoney -= 1;\n}\n\nreturn pumpkinAmount;\n\n}\n```\n6 Likes\n\nYay Math!\n\n`````` const pumpkinSpice = money => {\n\nvar pie=parseInt(money/5);\n\nvar remainder = money%5;\n\nvar latte = parseInt(remainder/3);\n\nremainder = remainder%3;\n\nvar macaron = remainder\n\nvar grams = pie*30+latte*15+macaron*3\n\nresult = [pie, latte, macaron, grams];\nreturn result;\n\n}``````\n3 Likes\n\nI don’t know what the fuck is wrong again with my code. I run on my Chrome browser everything is ok\n\njust a thought maybe add to the vars\nvar remainder;\nvar remainder2;\nand use remainder2 in the macaron lines\nwhat are macarons? I have been saying macaroons and now I’m lost. Just like “salumi” omg\n\nI found the error. I miss the opening curly braces. Sorry for my lack of meticulousness. Is my mistake. I admit\n\nNot a huge fan of how this one turned out.\n\n``````const pie = {\nprice: 5,\nspice: 30\n}\nconst latte = {\nprice: 3,\nspice: 15\n}\nconst macaron = {\nprice: 1,\nspice: 3,\n}\n\nconst pumpkinSpice = money =>\n[pie, latte, macaron].reduce(\n(totals, { price, spice }, index) => {\n// How many can we afford?\nconst amountPossible = Math.floor(money / price)\n// Buy them all and spend the money.\ntotals[index] = amountPossible\nmoney -= amountPossible * price\n// Add the grams of spice.\ntotals[totals.length - 1] += amountPossible * spice\n}, [ 0, 0, 0, 0 ])\n``````\n``````const pumpkinSpice = money => {\nconst pumpkinPie = 5\nconst pumpkinLattes = 3\nconst pumpkinMacarons = 1\n\nconst result = [0,0,0,0]\n\nwhile(money >= pumpkinMacarons){\nif(money >= pumpkinPie){\nresult += 1\nresult += 30\nmoney -= pumpkinPie\n} else if(money >= pumpkinLattes){\nresult += 1\nresult += 15\nmoney -= pumpkinLattes\n} else if(money >= pumpkinMacarons){\nresult += 1\nresult += 3\nmoney -= pumpkinMacarons\n}\n}\n\nreturn result\n}``````\n1 Like\n\nTodays solution (no loops just math)\n\nSolution", null, "3 Likes\n\nThis challenge is perfect to use the modulo operator on `%`", null, "Example: `5 % 2 // returns 1`\nExample 2:\n\n``````let change = 9\nchange %= 5 // change will be 4\n``````\n\nThis allows us to keep track of the amount of money we have after buying each product.\n\nSolution:\n\nSummary\n``````const pumpkinSpice = money => {\n// Start keeping track of our money.\nlet remainder = money\n\n// Calculate how many pies we can buy (which is the\n// smallest rounded number if we divide by its cost.\nlet pies = Math.floor(remainder / 5)\n\n// We use the modulo operator to calculate how much\n// money we have left after buying pies.\nremainder %= 5\n\n// Do the same for the other products.\nlet lattes = Math.floor(remainder / 3)\nremainder %= 3\n\nlet macarons = Math.floor(remainder / 1)\nremainder %= 1\n\n// Calculate the amount of spice we purchased\nlet grams = (pies * 30) + (lattes * 15) + (macarons * 3)\n\n// Return everything we know :)\nreturn [pies, lattes, macarons, grams]\n}\n``````\n4 Likes" ]

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[ "# CSE232-Project 6-Color Printer Steganography Solved\n\n30.00 \\$ 15.00 \\$\n\nCategory:\nClick Category Button to View Your Next Assignment | Homework\n\nYou'll get a download link with a: . ` zip` solution files instantly, after Payment\n\n## Description\n\nRate this product\n\nProject 06\n\nBackground Steganography\n\nSteganography (https://en.wikipedia.org/wiki/Steganography) is the process of hiding a “secret message” in a text file, image or even sound file. It differs from cryptography in that the overall file/video/audio looks reasonably normal and still conveys information, making it hard to tell that there is a secret hidden inside.\n\nColor Printer Steganography\n\nYou probably didn’t know (I didn’t until recently) that most color printers add a “fingerprint” to every page they print. They are not exactly invisible, they are very hard to see, but they are there.\n\nThat fingerprint encodes information like: time, date and serial number of the printer being used. Every page! They typically come as a small matrix of yellow dots (which are hard to see on white paper) which encode the information. The image below shows the actual yellow dots (highly magnified on the left) and a false-color enhancement to better show them.", null, "The companies are not exactly forthcoming about the encoding for their printers, but at least one has been decoded. The Xerox DocuColor series. We will write a program that can decode these.\n\nXerox Docucolor Matrix\n\nBelow is a representations of the DocuColor matrix printed on each sheet, and how they might be interpreted. The fingerprint is a 8 row x 15 column matrix of yellow dots. Note on the drawing that they number columns starting at index 1 (yuck).\n\nThe first row and first column are special. They represent a property called parity which we will discuss in a moment. Otherwise:\n\n• columns 2 and 5 represent time (minutes and hours respectively) ● columns 6, 7, 8 represent a date\n• columns 11, 12, 13, 14 represent a serial number\n\nColumns 3,4,9,10,15 are ignored (for our purposes).", null, "In the figure, the two rows at the bottom are not part of the matrix but are the interpretation of the columns for your benefit. Each row represents a power of two, and so each column represents a decimal number. Look at column 2. There is a dot in the row representing 2, 16 and 32. The sum of those is 50, which represents the minutes in the time. Notice we do not count the top row, the parity row, in any calculation. For column 5, there is a dot at 4 and 8, representing 12 hours. Thus the time this page was printed was 12:50. The other interpretations are done in the same manner. Note that, for our purposes, we would interpret all of the serial number columns, so our expectation would be to print the serial number as 21052857 in that order.\n\nParity\n\nParity is a really easy concept to understand. We typically use parity on binary representations such as a binary string. To determine the parity of a binary string, you count the number of 1’s that occur in the string. If the 1’s count is even, then the parity for that string is 1, if the 1’s count is odd then the parity is 0. Easy! The question is why is that interesting? When you transmit information across a channel that can be noisy (like a wireless connection), you can sometimes lose data. Parity is a really cheap way to check that what got passed is what you intended. That is, if you pass the binary string through some noisy channel along with a parity bit, you can check on the receiving end to see if the parity is still correct. That is, does the number of 1’s passed matches the parity that was also passed. If parity does not match the count, something is wrong. If the parity bit is correct, something could still be wrong but if you have multiple parity bits in the information passed, you can have some confidence the that information was correctly passed.\n\nIn the image above, let’s check. For the column parity in column 2, the parity is not set, representing a parity of 0. If we count the number of 1’s in the column there are three dots (three 1’s), so the parity is odd. The parity bit accurately reflects the parity of the column. Columns 3 and 4 have no dots. The parity of 0 is 1, so 3 and 4 are set to 1 (a dot). Column 5 has two dots (two 1’s), the parity is even, so the parity is 1 (a dot).\n\nSame for the rows. Row 1 has its parity bit set (1). The row has 6 dots, even parity. The bit accurately represents the parity of the row. Row 8 has 5 dots, odd parity, value of 0, no dot. The only weird one is the top left corner. Does that represent the parity of the first column (all the row parities) or the first row (the parity of all the columns). Has to be the parity of the column.\n\nCheck yourself. 4 dots in the column, even parity. 9 dots in the row, odd parity. The dot is set. Must be for the column.\n\nProgram Specifications\n\nfunction vector<vector<int>> read_file(const string &fname) ● one argument:\n\no a string representing the name of the file to be opened\n\n• the return is a 2D vector\n\nWe cannot actually read the dots, so instead we read a file of the following form (this file representing dot patterns used in the image above).", null, "No spaces on a row between each number. My suggestion is to read each line of the file in as a string, convert each character to an integer, and push it back onto the row of the 2D vector. Do it for each row.\n\nError Checking: if you cannot open the filename provided, throw a runtime_error\n\nfunction vector<int> get_row(const vector<vector<int>> &v, int row)\n\n• two arguments o 2D vector of int (by reference) o an int row\n• returns a 1D vector containing that row.\n\nNo Error Checking.\n\nfunction vector<int> get_column(const vector<vector<int>> &v,                                     int col)\n\n• two arguments o 2D vector of int (by reference)\n• an int column\n• returns a 1D vector containing that column including the parity bit.\n\nNo Error Checking.\n\nfunction int col_to_int(const vector<vector<int>> & v, size_t col)\n\n• two arguments o 2D vector of int (by reference)\n• an int column\n• returns an integer, the value the column represents (without the parity value)\n\nShould use get_column\n\nfunction string get_time(const vector<vector<int>> & v)\n\n• argument is the 2D vector\n• return is a string, the time the matrix represents (for the above example, “12:50”)\n\nShould use col_to_int\n\nfunction string get_date(const vector<vector<int>> & v)\n\n• argument is the 2D vector\n• return is a string, the date the matrix represents (for the above example “06/21/2005”) ● Note: if the month or day has just a single digit, then it must be padded with a 0\n\nShould use col_to_int\n\nfunction string get_serial(const vector<vector<int>> & v)\n\n• argument is the 2D vector\n• return is a string, the serial number the matrix represents (for the above example 21052857)\n\nShould use col_to_int\n\nfunction string check_column_parity(vector<vector<int>> & v,                                           int col)\n\n• two arguments o 2D vector of int (by reference) o an int col\n• returns a string with the following information (follow the string format exactly, separated by “:”, no spaces, string return as below) o column_parity:column_1’s_count:true or false if parity of the matrix and the count match. o “1:2:true” means the parity bit is set(1) , the 1’s count is 2, parity bit and count parity match\n\nShould use get_col and parit\n\nfunction string check_row_parity(vector<vector<int>> & v, int row)\n\n• two arguments o 2D vector of int (by reference)\n\no an int row\n\n• returns a string with the following information (follow the string format exactly, separated by “:”, no spaces, string return as indicated below).\n• row_parity:row_1’s_count:true or false if parity of the matrix and the count match.\n• “1:2:true” means: the parity bit is set(1) , the 1’s count is 2, parity bit and count parity match\n\nShould use get_row and parity\n\nAssignment Notes\n\n1. In the header file are two functions that we do not test on Mimr. That means you are not required to do them (but it is suggested)\n1. int parity(int)\n2. string print_vector (const vector<vector<int>> &v)\n\nThe parity function is too easy to test, is the parameter even or odd (though useful to have) and the print_vector is really useful but it is difficult to get a good Mimir test up for it (long output can be hard to get right). How you would know you read the vector correctly without printing I don’t know, but I guess that’s up to you. I’ll leave them in put put a comment up about that in the header.\n\n1. You turn in the functions only. To test against your own main you can write a separate file with the main and then compile the two files at the same time. See the lab and videos for examples.\n2. I included a print_vector function in the proj06_functions.h The intent is for that function print out the 2D vector that you just read in. I won’t test that function in Mimir,\n\nbut I submit you truly need it if you want to know that you read the data in right. Most of the other tests are going to depend on read_file working right. If that doesn’t work, very little else will.\n\n• Project06-vqoelo.zip" ]

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[ "July 14, 2020", null, "### How to Calculate the Fibonacci Retracement Levels\n\n12/7/ · Start grid placement by zooming out to the weekly pattern and finding the longest continuous uptrend or downtrend. Place a Fibonacci grid from low . 4/3/ · draw the retracement from the Swing High to the Swing Low The following is a 4-hour chart of USD/CHF on April 21 showing the Fibonacci retracement during a downtrend. In a downtrend, the first step is to identify the highest price point, then draw the Fibonacci retracement to the most recent lowest price level. Here’s an example. In this downtrend, the Fibonacci levels are automatically set at %, %, 50%, %, and %.", null, "### What makes me qualified to teach this tough subject?\n\nSelect the Fibonacci retracements option. To plot the Fibonacci retracements for an uptrend that occurs after a downtrend, you first need to define the range of the prior downtrend so that the charting software can calculate the percent retracement points that the new uptrend will hit as it retraces through the range of the prior downtrend. You’ll find the Fibonacci retracement tool when you click on the “insert” tab at the top-left area of your MT4. Hover above the “Fibonacci” drop-down option and click on “retracement” among the other options that appear to the right. There are two ways to set . In a downtrend, the first step is to identify the highest price point, then draw the Fibonacci retracement to the most recent lowest price level. Here’s an example. In this downtrend, the Fibonacci levels are automatically set at %, %, 50%, %, and %.", null, "### How to use Fibonacci Retracements Correctly\n\nYou’ll find the Fibonacci retracement tool when you click on the “insert” tab at the top-left area of your MT4. Hover above the “Fibonacci” drop-down option and click on “retracement” among the other options that appear to the right. There are two ways to set . How to draw Fibonacci retracement levels. Drawing Fibonacci retracement levels is a simple three-step process: In an uptrend: Step 1 – Identify the direction of the market: uptrend; Step 2 – Attach the Fibonacci retracement tool on the bottom and drag it to the right, all the way to the top. In a downtrend, the first step is to identify the highest price point, then draw the Fibonacci retracement to the most recent lowest price level. Here’s an example. In this downtrend, the Fibonacci levels are automatically set at %, %, 50%, %, and %.", null, "", null, "" ]

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[ "# Physics Questions and Answers – Electrostatic Potential due to an Electric Dipole\n\n«\n»\n\nThis set of Physics Multiple Choice Questions & Answers (MCQs) focuses on “Electrostatic Potential due to an Electric Dipole”.\n\n1. Which one is the correct expression of electric potential on the axial point of a dipole whose dipole moment is p and length is 2l, at a distance of r?\na) $$\\frac {p}{r^2}$$\nb) $$\\frac {p}{l^2}$$\nc) $$\\frac {p}{(r^2-l^2)}$$\nd) $$\\frac {p}{2lr}$$\n\nExplanation: Electric potentials due to the two point charges of the dipole are $$\\frac {q}{(r-l)}$$ and $$\\frac {-q}{(r+l)}$$. Therefore, total potential = $$\\frac {q}{(r-l)} – \\frac {q}{(r+l)} = \\frac {2ql}{(r^2-l^2)}$$. Now 2ql = dipole moment of the dipole=p. Therefore, the expression becomes $$\\frac {p}{(r^2-l^2)}$$.\n\n2. What is the electric potential at the perpendicular bisector of an electric dipole?\na) Positive\nb) Negative\nc) Zero\nd) Depends on medium\n\nExplanation: Any point on the perpendicular bisector is equidistant from both the charges of the dipole. Therefore, the electric potential at that point is equal and opposite due to the two different charges. Therefore, the total electric potential at that point is zero.\n\n3. What is the correct diagram regarding the variation of an electric field with distance from an electric dipole if the distance is very large than the dipole length?\na)", null, "b)", null, "c)", null, "d)", null, "Explanation: If the distance of a point from the dipole is very small, we can simply neglect the l2 term compared to the r2 term in the potential expression V=$$\\frac {p}{(r^2-l^2)}$$. Therefore, the expression becomes $$\\frac {p}{r^2}$$.\nSanfoundry Certification Contest of the Month is Live. 100+ Subjects. Participate Now!\n\n4. The electric potential at an axial point of a dipole is the maximum.\na) True\nb) False\n\nExplanation: Electric field at a distance r from a dipole and at an angle θ is p*$$\\frac {\\cos \\theta}{r^2}$$, where p is the dipole moment of that dipole. If θ becomes 0 degrees, cosθ will be the maximum i.e. 1. Therefore, the electric field is the maximum at the axial points.\n\n5. What is the dimension of the dipole moment?\na) [I L T]\nb) [I L T-1]\nc) [I L2 T]\nd) [I T]\n\nExplanation: Dipole moment = charge*length of the dipole. The electric charge has dimensions [I T] and length has dimensions [L]. Therefore, the dipole moment has the dimension [I T L] and has unit C*m of C*cm.\n\nSanfoundry Global Education & Learning Series – Physics – Class 12.\n\nTo practice all areas of Physics, here is complete set of 1000+ Multiple Choice Questions and Answers.", null, "" ]

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[ "Citation for this page in APA citation style.\n\n Home > Solutions > Scientists > Bohm David Bohm (1917-1992) David Bohm is perhaps best known for new experimental methods to test Einstein’s supposed suggestion of “hidden variables” that would explain the EPR paradox by providing the information needed at the distant “entangled” particle, so it can coordinate its properties perfectly with the “local” particle. Bohm wrote in 1952, The usual interpretation of the quantum theory is based on an assumption having very far-reaching implications, ~is., that the physical state of an individual system is completely specified by a wave function that determines only the probabilities of actual results that can be obtained in a statistical ensemble of similar experiments. This assumption has been the object of severe criticisms, notably on the part of Einstein, who has always believed that, even at the quantum level, there must exist precisely definable elements or dynamical variables determining (as in classical physics) the actual behavior of each individual system, and not merely its probable behavior. Since these elements or variables are not now included in the quantum theory and have not yet been detected experimentally, Einstein has always regarded the present form of the quantum theory as incomplete, although he admits its internal consistency. \"A Suggested Interpretation of the Quantum Theory in Terms of ‘Hidden’ Variables. I,” Physical Review, vol.85, no.2. p.166, 1952 Five years later, Bohm and his Israeli student Yakir Aharonov reformulated the original EPR argument in terms of electron spin. They said experimental tests with continuous variables would be much more difficult than tests with discrete quantities, such as the spin of electrons or polarization of photons. They wrote: Bohmian mechanics provides a straightforward physical explanation. First, close slit 1 and open slit 2. The particle goes through slit 2. It arrives at x on the plate with probability |ψ2(x)|2, where ψ2 is the wave function which passed through slit 2. Second, close slit 2 and open slit 1. The particle goes through slit 1. It arrives at x on the plate with probability |ψ1(x)|2, where ψ1 is the wave function which passed through slit 1. Third, open both slits. The particle goes through slit 1 or slit 2. It arrives at x with probability |ψ1(x)+ψ2(x)|2. Now observe that in general, |ψ1(x)+ψ2(x)|2 = |ψ1(x)+ψ2(x)|2= |ψ1(x)|2+|ψ2(x)|2 + 2ℜψ∗1(x) ψ2(x). The \u0007last term comes from the interference of the wave packets ψ1 and ψ2 which passed through slit 1 and slit 2. The probabilities of finding particles when both slits are open are different from the sum of slit 1 open and slit 2 open separately. The wave function determines the probabilities of finding particles, as Einstein first proposed.. We consider a molecule of total spin zero consisting of two atoms, each of spin one-half. The wave function of the system is therefore ψ = (1/√2) [ ψ+ (1) ψ- (2) - ψ- (1) ψ+ (2) ] where ψ+ (1) refers to the wave function of the atomic state in which one particle (A) has spin +ℏ/2, etc. The two atoms are then separated by a method that does not influence the total spin. After they have separated enough so that they cease to interact, any desired component of the spin of the first particle (A) is measured. Then, because the total spin is still zero, it can immediately be concluded that the same component of the spin of the other particle (8) is opposite to that of A. \"Discussion of Experimental Proof for the Paradox of Einstein, Rosen, and Podolsky,” Physical Review, vol.108, no.4. p.1070, 1957 Einstein may have pressed Bohm to develop hidden variables as the source of nonlocal behavior. Einstein had heartily approved of Bohm’s textbook and was initially supportive of Bohm’s new mechanics. Einstein thought Bohm was young enough and smart enough to produce the mathematical arguments that the older generation of “determinist” physicists like Erwin Schrödinger, Max Planck, and others had not been able to accomplish. But when Bohm finished the work, based on Louis de Broglie’s 1923 “pilot-wave” idea (which Einstein had supported), Einstein rejected it as inconsistent with his theory of relativity. Einstein wrote to Max Born on May 15, 1952, Have you noticed that Bohm believes (as de Broglie did, by the way, 25 years ago) that he is able to interpret the quantum theory in deterministic terms? That way seems too cheap to me. But you, of course, can judge this better than I.  Born-Einstein Letters, #99, p.188. Now Richard Feynman's path integral formulation of quantum mechanics describes supraluminal paths and even some things moving backwards in time, so we must take a careful look at Bohm's work. Bohm's search for \"hidden variables\" inspired John Bell to develop a theorem on \"inequalities\" that would need to be satisfied by hidden variables. To this date, every test of Bell's theorem has violated his inequalities and shown that the quantum theory cannot be replaced by one with \"local\" hidden variables. If they exist at all, \"hidden variables\" must also be \"nonlocal.\" Bohm's pilot-wave goes through both slits in the two-slit experiment, whereas the particle goes through only one, thus explaining what Richard Feynman called the \"only mystery\" in quantum mechanics. The Measurement Process David Bohm was particularly clear on the process of measurement. He said it involves macroscopic irreversibility, which was a sign and a consequence of treating the measuring apparatus as a macroscopic system that could not itself be treated quantum mechanically. The macroscopic system could, in principle, be treated quantum mechanically, but Bohm said its many degrees of internal freedom would destroy any interference effects. This is the modern theory of quantum decoherence. Bohm's view is consistent with the information-philosophy solution to the measurement problem. A measurement has only been made when new information has come into the world and adequate entropy has been carried away to insure the stability of the new information, long enough for it to be observed by the \"conscious\" observer. In his 1950 textbook Quantum Theory, Bohm discusses measurement in chapter 22, section 12. 12. Irreversibility of Process of Measurement and Its Fundamental Role in Quantum Theory. From the previous work it follows that a measurement process is irreversible in the sense that, after it has occurred, re-establishment of definite phase relations between the eigenfunctions of the measured variable is overwhelmingly unlikely. This irreversibility greatly resembles that which appears in thermodynamic processes, where a decrease of entropy is also an overwhelmingly unlikely possibility.* * There is, in fact, a close connection between entropy and the process of measurement. See L. Szilard, , 53, 840, 1929. The necessity for such a connection can be seen by considering a box divided by a partition into two equal parts, containing an equal number of gas molecules in each part. Suppose that in this box is placed a device that can provide a rough measurement of the position of each atom as it approaches the partition. This device is coupled automatically to a gate in the partition in such a way that the gate will be opened if a molecule approaches the gate from the right, but closed if it approaches from the left. Thus, in time, all the molecules can be made to accumulate on the left-hand side. In this way, the entropy of the gas decreases. If there were no compensating increase of entropy of the mechanism, then the second law of thermodynamics would be violated. We have seen, however, that in practice, every process which can provide a definite measurement disclosing in which side of the box the molecule actually is, must also be attended by irreversible changes in the measuring apparatus. In fact, it can be shown that these changes must be at least large enough to compensate for the decrease in entropy of the gas. Thus, the second law of thermodynamics cannot actually be violated in this way. This means, of course, that Maxwell's famous \"sorting demon \" cannot operate, if he is made of matter obeying all of the laws of physics. (See L. Brillouin, American Scientist, 38, 594, 1950.) Because the irreversible behavior of the measuring apparatus is essential for the destruction of definite phase relations and because, in turn, the destruction of definite phase relation's is essential for the consistency of the quantum theory as a whole, it follows that thermodynamic irreversibility enters into the quantum theory in an integral way. This is in remarkable contrast to classical theory, where the concept of thermodynamic irreversibility plays no fundamental role in the basic sciences of mechanics and electrodynamics. Thus, whereas in classical theory fundamental variables (such as position or momentum of an elementary particle) are regarded as having definite values independently of whether the measuring apparatus is reversible or not, in quantum theory we find that such a quantity can take on a well defined value only when the system is coupled indivisibly to a classically describable system undergoing irreversible processes. The very definition of the state of any one system at the microscopic level therefore requires that matter in the large shall undergo irreversible processes. There is a strong analogy here to the behavior of biological systems, where, likewise, the very existence of the fundamental elements (for example, the cells) depends on the maintenance of irreversible processes involving the oxidation of food throughout an organism as a whole. (A stoppage of these processes would result in the dissolution of the cell.) But Bohm changed his mind about irreversibility when he developed his more realistic and deterministic theory. Now he became concerned with the classic \"problem\" of microscopic irreversibility, namely how can the increase of entropy involve macroscopic irreversibility if microscopic collisions of particles are reversible? References A Suggested Interpretation of the Quantum Theory in Terms of \"Hidden\" Variables. I For Teachers To hide this material, click on the Normal link. Noesis Stanford Encyclopedia of Philosophy Wikipedia For Scholars To hide this material, click on the Teacher or Normal link. Normal | Teacher | Scholar" ]

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[ "# Interest Expense\n\nIn Fairways Debt, the interest expense is the interest charged during the analysis period.\n\n## Periodic / Single Period Reports\n\nInterest Expense\n\nInterest expense with all amounts converted in the currency applied to the report:\n\nFor example, the interest expense of 0122 is:\n\n= Interest Paid over Period (payment date) - Accrued Interest (start of period) + Accrued Interest (end of period)\n\n= 358,80 - 22,92 + 10,42\n\n= 346,30", null, "The computation for leases in advance differs from the standard formula:\n\nInterest Paid over Period (payment date) + Decreased interest (start of period) - Decreased interest (end of period)\n\n= Interest Paid over Period (payment date) + Interest / Number of days of the schedule period * Number of days between the end date of the schedule period and the start date of the report period - Interest / Number of days of the schedule period * Number of days between the end date of the schedule period and the end date of the report period\n\nNote: Prorata interest over partial periods (decreased interest) are defined using the actual settings of the relevant transaction.\n\nFor example, the interest expense of 0130 is:\n\n= Interest Paid over Period (payment date) + Decreased interest (start of period) - Decreased interest (end of period)\n\n= 108 955,23 + 44 141,79 - 31 044,78\n\n= 122 052,24", null, "Decreased interest (start of period)\n\n= Interest / Number of days of the schedule period * Number of days between the end date of the schedule period and the start date of the report period\n\n= 45 273,63 / 360 * 351\n\n= 44 141,79\n\n• The report start date is 10 October 2020, which corresponds to the 1 October 2020 - 1 October 2021 period of the transaction.\n• The interest of that period is EUR 45 273,63.\n• The period lasts 360 days.\n• The transaction period ends on 1 October 2021, 351 days apart from the report start date on 10 October 2020.", null, "Decreased interest (end of period)\n\n= Interest / Number of days of the schedule period * Number of days between the end date of the schedule period and the end date of the report period\n\n= 31 840,80 / 360 * 351\n\n= 31 044,78\n\n• The report end date is 10 October 2023, which corresponds to the 1 October 2023 - 1 October 2024 period of the transaction.\n• The interest of that period is EUR 31 840,80.\n• The period lasts 360 days.\n• The transaction period ends on 1 October 2024, 351 days apart from the report end date on 10 October 2023.\n\nInterest Expense (base currency)\n\nInterest expense with all amounts in the currencies initially applied to the transactions.\n\nIn this example, we can see the currencies of 0027 and 0034 are different from the report currency (AUD and USD vs. EUR).", null, "" ]

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[ "# Definition:Accumulation Point of Sequence\n\n## Definition\n\nLet $\\left({S, \\tau}\\right)$ be a topological space.\n\nLet $A \\subseteq S$.\n\nLet $\\left \\langle {x_n} \\right \\rangle_{n \\mathop \\in \\N}$ be an infinite sequence in $A$.\n\nLet $\\alpha \\in A$.\n\nSuppose that:\n\n$\\forall U \\in \\tau: \\alpha \\in U \\implies \\left\\{{n \\in \\N: x_n \\in U}\\right\\}$ is infinite\n\nThen $\\alpha$ is an accumulation point of $\\left \\langle {x_n} \\right \\rangle$.\n\n## Also see\n\n• Results about accumulation points can be found here." ]

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[ "• date post\n\n21-Jan-2017\n• Category\n\n## Engineering\n\n• view\n\n3.933\n• download\n\n1.565\n\nEmbed Size (px)\n\n### Transcript of Fluid Mechanics Kundu Cohen 6th edition solutions Sm ch (9)\n\n• Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling\n\nExercise 9.1. a) Write out the three components of (9.1) in x-y-z Cartesian coordinates. b) Set u = (u(y), 0, 0), and show that the x- and y-momentum equations reduce to:\n\n0 = 1 px\n\n+d 2udy2\n\n, and 0 = 1 py\n\n.\n\nSolution 9.1. a) Equation (9.1) is the constant-viscosity Navier-Stokes' momentum equation for incompressible flow:\n\nDuDt\n\n= 1p+2u ,\n\nwhere is the kinematic viscosity of the flow. Using u = (u, v, w) and\n\n= x , y , z( ) , the three components of this equation become:\n\nx:\n\nut\n\n+ uux\n\n+ v uy\n\n+ w uz\n\n= 1px\n\n+ 2ux 2\n\n+ 2uy 2\n\n+ 2uz2\n\n&\n\n' (\n\n)\n\n* + ,\n\ny:\n\nvt\n\n+ uvx\n\n+ v vy\n\n+ w vz\n\n= 1py\n\n+ 2vx 2\n\n+ 2vy 2\n\n+ 2vz2\n\n&\n\n' (\n\n)\n\n* + , and\n\nz:\n\nwt\n\n+ uwx\n\n+ v wy\n\n+ w wz\n\n= 1pz\n\n+ 2wx 2\n\n+ 2wy 2\n\n+ 2wz2\n\n&\n\n' (\n\n)\n\n* + .\n\nb) When u = (u(y), 0, 0), all the terms involving v and w disappear, so the part a) equations simplify to:\n\nx:\n\nut\n\n+ uux\n\n+ 0 + 0 = 1px\n\n+ 2ux 2\n\n+ 2uy 2\n\n+ 2uz2\n\n&\n\n' (\n\n)\n\n* + ,\n\ny: 0+ 0+ 0+ 0 = 1 py\n\n+ 0+ 0+ 0( ) , and\n\nz:\n\n0 + 0 + 0 + 0 = 1pz\n\n+ 0 + 0 + 0( ).\n\nAnd, when u depends only on y, then u/t = u/x = u/z = 0 so the part a) equations simplify further:\n\nx:\n\n0 = 1px\n\n+ 2uy 2&\n\n' (\n\n)\n\n* + ,\n\ny:\n\n0 = 1py\n\n, and\n\nz:\n\n0 = 1pz\n\n.\n\nThese x- and y-direction equations match those in the problem statement.\n\n• Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling\n\nExercise 9.2. For steady pressure driven flow between parallel plates (see Figure 8.3), there are 7 parameters: u(y), U, y, h, , , and dp/dx. Determine a dimensionless scaling law for u(y), and rewrite the flow-field solution (8.5) in dimensionless form. Solution 9.2. The parameters are: u(y), U, y, h, , , and dp/dx. First, create the parameter matrix: u U y h dp/dx Mass: 0 0 0 0 1 1 1 Length: 1 1 1 1 -3 -1 -2 Time: -1 -1 0 0 0 -1 -2 Next, determine the number of dimensionless groups. This rank of this matrix is three so 7 parameters - 3 dimensions = 4 groups, and construct the groups:\n\n1 = u U ,\n\n2 = y h ,\n\n3 = Uh , and\n\n4 = h2(dp /dx) U . Now write a dimensionless law:\n\nuU\n\n= f yh, Uh\n\n, h\n\n2\n\nUdpdx\n\n#\n\n\\$ %\n\n&\n\n' (\n\nwhere f is an unknown function. When rewritten in dimensionless form, (8.5) is:\n\nuU\n\n=yhh2\n\n2Udpdx\n\nyh\n\n#\n\n\\$ %\n\n&\n\n' ( 1\n\nyh\n\n#\n\n\\$ %\n\n&\n\n' ( or\n\n1 =2 1242 12( ).\n\nIn this flow, there is no fluid acceleration so the Reynolds number, 3, does not appear.\n\n• Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling\n\nExercise 9.3. An incompressible viscous liquid with density fills the gap between two large smooth parallel walls that are both stationary. The upper and lower walls are located at x2 = h, respectively. An additive in the liquid causes its viscosity to vary in the x2 direction. Here the flow is driven by a constant non-zero pressure gradient:\n\np x1 = const. a) Assume steady flow, ignore the body force, set\n\nu = u1(x2),0,0( ) and use\n\nt\n\n+xi\n\nui( ) = 0 ,\n\nu jt\n\n+ uiu jxi\n\n= px j\n\n+ g j +xi\n\nuix j\n\n+u jxi\n\n%\n\n& ' '\n\n(\n\n) * *\n\n+\n\n, - -\n\n.\n\n/ 0 0\n\n+x j\n\nv 23\n\n%\n\n& '\n\n(\n\n) * uixi\n\n+\n\n, -\n\n.\n\n/ 0\n\nto determine u1(x2) when\n\n= o 1+ x2 h( )2( ) .\n\nb) What shear stress is felt on the lower wall? c) What is the volume flow rate (per unit depth into the page) in the gap when = 0? d) If 1 < < 0, will the volume flux be higher or lower than the case when = 0?\n\nSolution 9.3. a) The continuity equation is satisfied by the form of the velocity field. The j =1-component of momentum equation simplifies to:\n\n0 = p x1( ) + x2( ) u1 x2( )[ ]. Integrate once with\n\np x1 = const. to find:\n\nu1 x2( ) = p x1( )x2 + C . Divide by and integrate again:\n\nu1 =1\n\npx1\n\nx2 + C#\n\n\\$ %\n\n&\n\n' ( dx2 =\n\np x1( )x2 + Co 1+ x2 h( )\n\n2( )dx2\n\n= h2\n\n2opx1\n\nln 1+ x2h\n\n#\n\n\\$ %\n\n&\n\n' (\n\n2#\n\n\\$ % %\n\n&\n\n' ( ( +\n\nCho\n\ntan1 x2 h\n\n#\n\n\\$ %\n\n&\n\n' ( + D.\n\nThe boundary conditions,\n\nu1(h) = 0, determine the values of the constants: C = 0, and\n\nD = h2 2o( ) p x1( ) ln 1+ ( ) , thus:\n\nu1(x2) = h2\n\n2opx1ln1+ x2 h( )\n\n2\n\n1+\n\n%\n\n& ' '\n\n(\n\n) * * .\n\nb) From the solution of part a) with C = 0:\n\nw = u1 x2( )y=h = h p x1( )\n\nc) When = 0, the flow profile is parabolic:\n\nq = u1(x2)dx2h\n\n+h\n\n= h2\n\n2opx1\n\n1 x22\n\nh2%\n\n& '\n\n(\n\n) * dx2\n\nh\n\n+h\n\n= 2h3\n\n3opx1\n\nd) The volume flux will be higher because the viscosity will be reduced at the wall. Manipulation of the near-wall viscosity with additives is sometimes used in long piping systems to reduce pumping power requirements.\n\n• Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling\n\nExercise 9.4. An incompressible viscous liquid with density fills the gap between two large smooth parallel plates. The upper plate at x2 = h moves in the positive x1-direction at speed U. The lower plate at x2 = 0 is stationary. An additive in the liquid causes its viscosity to vary in the x2 direction. a) Assume steady flow, ignore the body force, set\n\nu = u1(x2),0,0( ) and\n\np x1 = 0, and use the equations specified in Exercise 8.3 to determine u1(x2) when\n\n= o 1+ x2 h( ) . b) What shear stress is felt on the lower plate? c) Are there any physical limits on ? If, so specify them.\n\nSolution 9.4. a) For\n\nu = u1(x2),0,0( ) , no body force, and\n\np x1 = 0 in steady incompressible flow, the continuity equation is automatically satisfied, and the momentum equation for j = 1 simplifies to:\n\n0 = + x2( ) u1 x2( )[ ], or, after integrating once:\n\nC = u1 x2( ) , where C is a constant. Now use the specified relationship for the viscosity and integrate to find:\n\nu1(x2) =Cdx2 =\n\nCdx2o 1+ x2 h( )\n\n=Cho\n\nln 1+ x2 h( ) + D\n\nwhere D is another constant. The boundary conditions u1(0) = 0 and u1(h) = U allow\n\nC =Uo h ln(1+ )( ) and D = 0 to be determined yielding:\n\nu1(x2) =U ln 1+ x2 h( ) ln 1+ ( ). b) From part a), the shear stress is constant:\n\nw = u1 x2( ) = C =Uo h ln(1+ )( ) . c) Negative viscosities violate the second law of thermodynamics, thus > 1 is required.\n\n• Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling\n\nExercise 9.5. Planar Couette flow is generated by placing a viscous fluid between two infinite parallel plates and moving one plate (say, the upper one) at a velocity U with respect to the other one. The plates are a distance h apart. Two immiscible viscous liquids are placed between the plates as shown in the diagram. The lower fluid layer has thickness d. Solve for the velocity distributions in the two fluids.\n\nSolution 9.5. For steady viscous flow between infinite parallel plates, the fluid velocity will be unidirectional: u = (u, 0, 0). For this problem, no pressure gradient is specified so assume it to be zero. Thus, the horizontal (x1-direction) momentum equation reduces to:\n\n21 x2 = x2( ) u x2( )= 2u y2 = 0 , where the last equality follows when the viscosity is constant and x2 = y. Here, the viscosity is assumed constant within each fluid. This means that the flow profile in each fluid will be piece-wise linear:\n\nu(y) =A1 + B1y for 0 y dA2 + B2y for d y h\n\n# \\$ %\n\n& ' (\n\n,\n\nwhere the As & Bs are constants and 1 implies the upper fluid layer with viscosity 1, and 2 implies the lower fluid layer with viscosity 2. The four constants can be determined from the four boundary conditions: i) u(0) = 0 (match the speed of the lower boundary) ii) u(h) = U (match the speed of the upper boundary) iii) u(d) = u(d+), and (match flow speeds at the internal fluid-fluid interface) iv) (d) = (d+) (match shear stress at the internal fluid-fluid interface) where is the shear stress in the fluid. These four boundary conditions imply:\n\nA2 = 0,\n\nA1 + B1h =U ,\n\nA2 + B2d = A1 + B1d , and\n\n2B2 = 1B1 Use the first two equations to eliminate A1 and A2 from the second two equations to find:\n\nB2d =U B1(h d) , and\n\n2B2 = 1B1. Eliminate B2 and solve for B1:\n\n1 2( )B1d =U B1(h d) >\n\nB1 = 2U 2h + 1 2( )d[ ]. So,\n\nB2 = 1U 2h + 1 2( )d[ ], and\n\nA1 =U 1 2( )d 2h + 1 2( )d[ ] with A2 = 0. Thus:\n\nu(y) = U2h + 1 2( )d\n\n1y for 0 y d1 2( )d + 2y for d y h\n\n\\$ % &\n\n' ( )\n\n• Fluid Mechanics, 6th Ed." ]

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[ "[ << ] [ < ] [ Up ] [ > ] [ >> ] [Top] [Contents] [Index] [ ? ]\n\n## 2.9 Referencing results and special numbers\n\nAn `@` used for an escape character; rules currently in force are as follows.\n\n`@n`\n\nThe evaluated result of n-th input command\n\n`@@`\n\nThe evaluated result of the last command\n\n`@i`\n\nThe unit of imaginary number, square root of -1.\n\n`@pi`\n\nThe number pi, the ratio of a circumference of the circle and its diameter.\n\n`@e`\n\nNapier’s number, the base of natural logarithm.\n\n`@`\n\nA generator of GF(2^m), a finite field of characteristic 2, over GF(2). It is a root of an irreducible univariate polynomial over GF(2) which is set as the defining polynomial of GF(2^m).\n\n`@>, @<, @>=, @<=, @==, @&&, @||`\n\nFist order logical operators. They are used in quantifier elimination.\n\n``` fctr(x^10-1);\n[[1,1],[x-1,1],[x+1,1],[x^4+x^3+x^2+x+1,1],[x^4-x^3+x^2-x+1,1]]\n @@;\n[x^4+x^3+x^2+x+1,1]\n eval(sin(@pi/2));\n1.000000000000000000000000000000000000000000000000000000000\n eval(log(@e),20);\n0.99999999999999999999999999998\n @0;\nx^4-x^3+x^2-x+1\n (1+@i)^5;\n(-4-4*@i)\n eval(exp(@pi*@i));\n-1.0000000000000000000000000000\n (@+1)^9;\n(@^9+@^8+@+1)\n```\n\nAs you can see in the above example, results of toplevel computation can be referred to by `@` convention. This is convenient for users, while it sometimes imposes a heavy burden to the garbage collector. It may happen that GC time will rapidly increase after computing a very large expression at the toplevel. In such cases `delete_history()` (see section `delete_history`) takes effect.\n\n [ << ] [ < ] [ Up ] [ > ] [ >> ]\n\nThis document was generated on April 22, 2021 using texi2html 5.0." ]

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[ "# Pounds Per Square Inch to Torr Conversion\n\nEnter the pressure in pounds per square inch below to get the value converted to torr.\n\nResults in Torr:", null, "1 psi = 51.714878 Torr\n\n## How to Convert Pounds Per Square Inch to Torr", null, "To convert a pound per square inch measurement to a torr measurement, multiply the pressure by the conversion ratio. One pound per square inch is equal to 51.714878 torr, so use this simple formula to convert:\n\ntorr = pounds per square inch × 51.714878\n\nThe pressure in torr is equal to the pounds per square inch multiplied by 51.714878.\n\nFor example, here's how to convert 5 pounds per square inch to torr using the formula above.\n5 psi = (5 × 51.714878) = 258.574389 Torr\n\nPounds per square inch and torr are both units used to measure pressure. Keep reading to learn more about each unit of measure.\n\n## Pounds Per Square Inch\n\nOne pound per square inch is the pressure of equal to one pound-force per square inch.\n\nThe pound per square inch is a US customary and imperial unit of pressure. A pound per square inch is sometimes also referred to as a pound-force per square inch. Pounds per square inch can be abbreviated as psi, for example 1 pound per square inch can be written as 1 psi.\n\nPSI can be expressed using the formula:\n1 psi = 1 lbfin2\n\nPressure in pounds per square inch are equal to the pound-force divided by the area in square inches.\n\n## Torr\n\nThe torr is a measure of pressure equal to 1/760 of an atmosphere. Torr is based on an absolute scale of pressure, which is equivalent to the atmospheric pressure plus the gauge pressure.\n\nThe torr is a non-SI metric unit for pressure. Torr can be abbreviated as Torr, for example 1 torr can be written as 1 Torr.\n\n## Pound Per Square Inch to Torr Conversion Table\n\nPound per square inch measurements converted to torr\nPounds Per Square Inch Torr\n1 psi 51.71 Torr\n2 psi 103.43 Torr\n3 psi 155.14 Torr\n4 psi 206.86 Torr\n5 psi 258.57 Torr\n6 psi 310.29 Torr\n7 psi 362 Torr\n8 psi 413.72 Torr\n9 psi 465.43 Torr\n10 psi 517.15 Torr\n11 psi 568.86 Torr\n12 psi 620.58 Torr\n13 psi 672.29 Torr\n14 psi 724.01 Torr\n15 psi 775.72 Torr\n16 psi 827.44 Torr\n17 psi 879.15 Torr\n18 psi 930.87 Torr\n19 psi 982.58 Torr\n20 psi 1,034 Torr\n21 psi 1,086 Torr\n22 psi 1,138 Torr\n23 psi 1,189 Torr\n24 psi 1,241 Torr\n25 psi 1,293 Torr\n26 psi 1,345 Torr\n27 psi 1,396 Torr\n28 psi 1,448 Torr\n29 psi 1,500 Torr\n30 psi 1,551 Torr\n31 psi 1,603 Torr\n32 psi 1,655 Torr\n33 psi 1,707 Torr\n34 psi 1,758 Torr\n35 psi 1,810 Torr\n36 psi 1,862 Torr\n37 psi 1,913 Torr\n38 psi 1,965 Torr\n39 psi 2,017 Torr\n40 psi 2,069 Torr" ]

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[ "", null, "", null, "# Find the zeroes of a function\n\nDo you need help with your math homework? Are you struggling to understand concepts how to find the zeroes of a function?\n\n• Scan\n• Fast solutions\n• Fast Delivery\n\n## How to Find the Zeroes of a Function with the TI-84 Plus\n\nKnowing where the zeros of a function are located enables us to make more informed decisions and predictions about the behavior of that particular mathematical model.\n\n• Clarify math question\n• Figure out mathematic question\n• Get Help with your Homework\n• Get Help with Homework\n\n## Zeros of a function – Explanation and Examples\n\nFinding the zeros of a function is essential for understanding its behavior.\n\n•", null, "Decide math equation\n\n•", null, "Avg. satisfaction rating 4.7/5\n\nThe answer to the equation is 4.\n\n•", null, "Average satisfaction rating 4.8/5\n\nMath can be confusing, but there are ways to clarify questions and get the answers you need.\n\n•", null, "Clear up math equation\n\nThe average satisfaction rating for this product is 4.7 out of 5. This product is sure to please!\n\n## They use our app\n\nJames Blaze\n\nThe app is well made and is great at calculating and helping people understand how calculations work. I love it sooo much. What? Mind blowing. Not sure if the paid version is worth it, it is just nice for double checking that you did your math right.\n\nMichael Smith\n\nI'm going to make a donation as soon as I can. Amazinggg app i understand how to solve all my math equations this app explains how to solve the ewuation more than my teacher there is one thing i dont like that this app doesnt solve geometry problems.", null, "## Zeros Calculator\n\nWhen we find the zeros of a function, it can help us to visualize how inputs affect outputs." ]

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[ "# Assignment = Heron's Formula\n\n## Code\n\n``` /// Name: Sam Wilson\n/// Period: 6\n/// Program Name: Herons Formula\n/// File Name: herons1.java\n/// Date Finished: 3/6/16\n\npublic class herons1\n{\npublic static void main( String[] args )\n{\ndouble a;\n\n//they run the same thing.\n//herons1 has 30 lines herons2 has 50\n// this one\n//this one\n\na = triangleArea(3, 3, 3);\nSystem.out.println(\"A triangle with sides 3,3,3 has an area of \" + a );\n\na = triangleArea(3, 4, 5);\nSystem.out.println(\"A triangle with sides 3,4,5 has an area of \" + a );\n\na = triangleArea(7, 8, 9);\nSystem.out.println(\"A triangle with sides 7,8,9 has an area of \" + a );\n\nSystem.out.println(\"A triangle with sides 5,12,13 has an area of \" + triangleArea(5, 12, 13) );\nSystem.out.println(\"A triangle with sides 10,9,11 has an area of \" + triangleArea(10, 9, 11) );\nSystem.out.println(\"A triangle with sides 8,15,17 has an area of \" + triangleArea(8, 15, 17) );\nSystem.out.println(\"A traingle with sides 9,9,9 has an area of \"+ triangleArea(9, 9, 9) );\n}\n\npublic static double triangleArea( int a, int b, int c )\n{\n// the code in this method computes the area of a triangle whose sides have lengths a, b, and c\ndouble s, A;\n\ns = (a+b+c) / 2.0;\nA = Math.sqrt( s*(s-a)*(s-b)*(s-c) );\n\nreturn A;\n// ^ after computing the area, \"return\" it\n}\n}\n\n```\n\n### Picture of the output", null, "" ]

[ null, "https://llhscp-stw.neocities.org/Intro/Printing+/4/host00/herons1.PNG", null ]

[ "# 1272779 (number)\n\n1,272,779 (one million two hundred seventy-two thousand seven hundred seventy-nine) is an odd seven-digits composite number following 1272778 and preceding 1272780. In scientific notation, it is written as 1.272779 × 106. The sum of its digits is 35. It has a total of 2 prime factors and 4 positive divisors. There are 1,264,200 positive integers (up to 1272779) that are relatively prime to 1272779.\n\n## Basic properties\n\n• Is Prime? No\n• Number parity Odd\n• Number length 7\n• Sum of Digits 35\n• Digital Root 8\n\n## Name\n\nShort name 1 million 272 thousand 779 one million two hundred seventy-two thousand seven hundred seventy-nine\n\n## Notation\n\nScientific notation 1.272779 × 106 1.272779 × 106\n\n## Prime Factorization of 1272779\n\nPrime Factorization 151 × 8429\n\nComposite number\nDistinct Factors Total Factors Radical ω(n) 2 Total number of distinct prime factors Ω(n) 2 Total number of prime factors rad(n) 1272779 Product of the distinct prime numbers λ(n) 1 Returns the parity of Ω(n), such that λ(n) = (-1)Ω(n) μ(n) 1 Returns: 1, if n has an even number of prime factors (and is square free) −1, if n has an odd number of prime factors (and is square free) 0, if n has a squared prime factor Λ(n) 0 Returns log(p) if n is a power pk of any prime p (for any k >= 1), else returns 0\n\nThe prime factorization of 1,272,779 is 151 × 8429. Since it has a total of 2 prime factors, 1,272,779 is a composite number.\n\n## Divisors of 1272779\n\n4 divisors\n\n Even divisors 0 4 2 2\nTotal Divisors Sum of Divisors Aliquot Sum τ(n) 4 Total number of the positive divisors of n σ(n) 1.28136e+06 Sum of all the positive divisors of n s(n) 8581 Sum of the proper positive divisors of n A(n) 320340 Returns the sum of divisors (σ(n)) divided by the total number of divisors (τ(n)) G(n) 1128.18 Returns the nth root of the product of n divisors H(n) 3.97321 Returns the total number of divisors (τ(n)) divided by the sum of the reciprocal of each divisors\n\nThe number 1,272,779 can be divided by 4 positive divisors (out of which 0 are even, and 4 are odd). The sum of these divisors (counting 1,272,779) is 1,281,360, the average is 320,340.\n\n## Other Arithmetic Functions (n = 1272779)\n\n1 φ(n) n\nEuler Totient Carmichael Lambda Prime Pi φ(n) 1264200 Total number of positive integers not greater than n that are coprime to n λ(n) 632100 Smallest positive number such that aλ(n) ≡ 1 (mod n) for all a coprime to n π(n) ≈ 97863 Total number of primes less than or equal to n r2(n) 0 The number of ways n can be represented as the sum of 2 squares\n\nThere are 1,264,200 positive integers (less than 1,272,779) that are coprime with 1,272,779. And there are approximately 97,863 prime numbers less than or equal to 1,272,779.\n\n## Divisibility of 1272779\n\n m n mod m 2 3 4 5 6 7 8 9 1 2 3 4 5 4 3 8\n\n1,272,779 is not divisible by any number less than or equal to 9.\n\n## Classification of 1272779\n\n• Arithmetic\n• Semiprime\n• Deficient\n\n• Polite\n\n• Square Free\n\n### Other numbers\n\n• LucasCarmichael\n\n## Base conversion (1272779)\n\nBase System Value\n2 Binary 100110110101111001011\n3 Ternary 2101122220222\n4 Quaternary 10312233023\n5 Quinary 311212104\n6 Senary 43140255\n8 Octal 4665713\n10 Decimal 1272779\n12 Duodecimal 51468b\n20 Vigesimal 7j1ij\n36 Base36 ra2z\n\n## Basic calculations (n = 1272779)\n\n### Multiplication\n\nn×y\n n×2 2545558 3818337 5091116 6363895\n\n### Division\n\nn÷y\n n÷2 636390 424260 318195 254556\n\n### Exponentiation\n\nny\n n2 1619966382841 2061859192785985139 2624291081534953379231281 3340142578464976427064610599899\n\n### Nth Root\n\ny√n\n 2√n 1128.18 108.372 33.5883 16.6322\n\n## 1272779 as geometric shapes\n\n### Circle\n\n Diameter 2.54556e+06 7.99711e+06 5.08927e+12\n\n### Sphere\n\n Volume 8.6367e+18 2.03571e+13 7.99711e+06\n\n### Square\n\nLength = n\n Perimeter 5.09112e+06 1.61997e+12 1.79998e+06\n\n### Cube\n\nLength = n\n Surface area 9.7198e+12 2.06186e+18 2.20452e+06\n\n### Equilateral Triangle\n\nLength = n\n Perimeter 3.81834e+06 7.01466e+11 1.10226e+06\n\n### Triangular Pyramid\n\nLength = n\n Surface area 2.80586e+12 2.42992e+17 1.03922e+06\n\n## Cryptographic Hash Functions\n\nmd5 21b72c6906cb29752d51a6c18114612a 3a63cafa410283c8fc16379c847e6774d9dced9f 31af7d76b4774bce2e4ae29f0cacb31febf99e0e3078f120c064f9aa97ca719a 487983ed6aa70bd10c6e390a5198230fb13a0162c8ce452001bc895ab9ebcb703dec9691c13a91c80cdcb903de11737fc590b7f1ced7d2a0ab6b9d29830f2c25 4faf3039794a71eadf612f4966e89dd4aa24a126" ]

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[ "Jul 09\n\n## Guess Your Age by Chocolate\n\nHere is one of those math tricks that has been floating around the Internet.  Don’t cheat by scrolling down!  It takes less than a minute.  At the end, I will provide an explanation of how this trick is done.", null, "Step 1:  Pick the number of times a week that you like to eat chocolate (more than once but less than 10).", null, "Step 2:  Multiply this number by 2 (just to be bold)!", null, "Step 3:  Add 5 to the total.", null, "Step 4:  Multiply your total by 50 — I’ll wait while you get a calculator.", null, "Step 5:  If you have already had your birthday this year, add 1759.  If you haven’t, then add 1758.", null, "Step 6:  Now subtract the four digit year that you were born.", null, "Your result should be a three digit number. The first digit is your original number (i.e., how many times you want to have chocolate each week).  The next two numbers are your age!\n\nThis is the only year (2009) that this trick will work, so spread it around!\n\n### How to Solve this Trick\n\nThis is a simple algebra math trick.  For test purposes, let’s assume you were born on January 1, 1970.  That would make your age 39 (as of today, July 9, 2009).  So, following the steps above:\n\n1. Pick the number of times a week that you like to eat chocolate.  So let’s assign the variable “C” for chocolate:\n= C\n\n2. Multiply this number by 2.  So that’s:\n= C x 2\n\n3. Add 5 to the total.\n= (C x 2) + 5\n\n4. Multiply your total by 50.\n= ((C x 2) + 5) x 50\n= (C x 2 x 50) + (5 x 50)\n= (C x 100) + 250\n\n= ((C x 100) + 250) + 1759\n= (C x 100) + 2009 (which is the current year!)\n\n6. Subtract the four year digit you were born.\n= (C x 100) + 2009 – 1970\n= (C x 100) + 39 (which is our test age!)\n= C39\n\nThe final result is a three digit number.  The first digit is your original number “C” (C x 100).  The next two numbers are your age (39).\n\nSo the trick is to take your original number and multiply it by 100 so that it’s the first digit in the result.  Next, it multiplies and adds some numbers that equal the current year.  So then if you subtract the year you were born, the result is the number of years you have been alive, i.e., your age.\n\nSOLVED!\n\nIf you enjoyed this puzzle, be sure to check out the Regifting Robin Mind Trick.\n\nArticle published on July 9, 2009\n\n### 11 Responses to “Guess Your Age by Chocolate”\n\n1. Ian Says:\n\nTried 3 times and didn’t work for me – number was 5 digits long.\n\n2. KIMBERLEY Says:\n\nACE I DID IT IN MY MATHS LESSON 2DAY A TRIED IT ON MY MUM AND DAD IT IS THE GREASTING THING I HAVE EVER SEEN!!!!!!! <3\n\n3. freya Says:\n\nthis is so cool how do u do it\n\n4. Shelley Says:\n\nThat is so cool. It worked everytime. And, Freya, it says how to do it!\n\n5. cerysnapril Says:\n\nhahahahahahahahahahaha it doesn’t work me and April tried it it said we were 11 in fact people we are 12 hahahahahahahahahaha RUBBISH !!!!!!!!\n\n6. IZZY! Says:\n\nIt does not work it said i was 11 when i am twelve!\n\n7. 555 Says:\n\nI am 12 not 11!\n\n8. Guess Your Age | More More Pics Says:\n\n[…] Guess Your Age by Chocolate devtopics.com […]\n\n9. Jen Says:\n\nit said i was 12, im 14 :L\n\n10. Fanny Says:\n\nmy number was 16….I’m 18… and i dont eat chockolate\n\n11. Jack Says:\n\nWhy do I always see these “It didn’t work I’m x+1 or x+2 years old” .. read where it says it ONLY WORKS for 2009!! (or add 2 for 2011 to the equation and it will work)" ]

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[ "# What's the Difference Between Speed and Velocity?\n\nFacebook Twitter", null, "© Tomasz Zajda/Fotolia\n\nWhen describing the motion of objects in terms of distance, time, and direction, physicists use the basic quantities of speed and velocity. Two terms, two distinct meanings. Yet, not uncommonly, we hear these terms used interchangeably. So, what’s the difference? Why is it incorrect to use the terms speed and velocity interchangeably?\n\nThe reason is simple. Speed is the time rate at which an object is moving along a path, while velocity is the rate and direction of an object’s movement. Put another way, speed is a scalar value, while velocity is a vector. For example, 50 km/hr (31 mph) describes the speed at which a car is traveling along a road, while 50 km/hr west describes the velocity at which it is traveling.\n\nThe mathematical calculation for speed is relatively straightforward, wherein the average speed of an object is calculated by dividing the distance traveled by the time it took the object to travel the distance. Velocity, on the other hand, is more complicated mathematically and can be calculated in different ways, depending on what information is available about the object’s motion. In its simplest form, average velocity is calculated by dividing change in position (Δr) by change in time (Δt).\n\nGrab a copy of our NEW encyclopedia for Kids!" ]

[ null, "https://cdn.britannica.com/s:800x1000/06/192806-050-B42B2205/drive-Highway-Night-Traffic-Portland-automobile-car.jpg", null ]

[ "## Monday, March 26, 2012\n\n### New cube decision model posted on arxiv.org\n\nI posted the next draft of my new model for calculating cube decision points to the preprint site arxiv.org, so now it has a proper stable link:\n\nCube Handling In Backgammon Money Games Under a Jump Model\n\nAbstract\n\nA variation on Janowski’s cubeful equity model is proposed for cube handling in backgammon money games. Instead of approximating the cubeful take point as an interpolation between the dead and live cube limits, a new model is developed where the cubeless probability of win evolves through a series of random jumps instead of continuous diffusion. Each jump is drawn from a distribution with zero mean and an expected absolute jump size called the “jump volatility” that can be a function of game state but is assumed to be small compared to the market window.\n\nSimple closed form approximations for the market window and doubling points are developed as a function of local and average jump volatility. The local jump volatility can be calculated for specific game states, leading to crisper doubling decisions." ]

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[ "# #0c4b22 Color Information\n\nIn a RGB color space, hex #0c4b22 is composed of 4.7% red, 29.4% green and 13.3% blue. Whereas in a CMYK color space, it is composed of 84% cyan, 0% magenta, 54.7% yellow and 70.6% black. It has a hue angle of 141 degrees, a saturation of 72.4% and a lightness of 17.1%. #0c4b22 color hex could be obtained by blending #189644 with #000000. Closest websafe color is: #003333.\n\n• R 5\n• G 29\n• B 13\nRGB color chart\n• C 84\n• M 0\n• Y 55\n• K 71\nCMYK color chart\n\n#0c4b22 color description : Very dark cyan - lime green.\n\n# #0c4b22 Color Conversion\n\nThe hexadecimal color #0c4b22 has RGB values of R:12, G:75, B:34 and CMYK values of C:0.84, M:0, Y:0.55, K:0.71. Its decimal value is 805666.\n\nHex triplet RGB Decimal 0c4b22 `#0c4b22` 12, 75, 34 `rgb(12,75,34)` 4.7, 29.4, 13.3 `rgb(4.7%,29.4%,13.3%)` 84, 0, 55, 71 141°, 72.4, 17.1 `hsl(141,72.4%,17.1%)` 141°, 84, 29.4 003333 `#003333`\nCIE-LAB 27.368, -29.688, 18.961 2.956, 5.226, 2.366 0.28, 0.495, 5.226 27.368, 35.226, 147.435 27.368, -22.816, 22.574 22.859, -16.92, 9.865 00001100, 01001011, 00100010\n\n# Color Schemes with #0c4b22\n\n• #0c4b22\n``#0c4b22` `rgb(12,75,34)``\n• #4b0c35\n``#4b0c35` `rgb(75,12,53)``\nComplementary Color\n• #164b0c\n``#164b0c` `rgb(22,75,12)``\n• #0c4b22\n``#0c4b22` `rgb(12,75,34)``\n• #0c4b42\n``#0c4b42` `rgb(12,75,66)``\nAnalogous Color\n• #4b0c16\n``#4b0c16` `rgb(75,12,22)``\n• #0c4b22\n``#0c4b22` `rgb(12,75,34)``\n• #420c4b\n``#420c4b` `rgb(66,12,75)``\nSplit Complementary Color\n• #4b220c\n``#4b220c` `rgb(75,34,12)``\n• #0c4b22\n``#0c4b22` `rgb(12,75,34)``\n• #220c4b\n``#220c4b` `rgb(34,12,75)``\n• #354b0c\n``#354b0c` `rgb(53,75,12)``\n• #0c4b22\n``#0c4b22` `rgb(12,75,34)``\n• #220c4b\n``#220c4b` `rgb(34,12,75)``\n• #4b0c35\n``#4b0c35` `rgb(75,12,53)``\n• #010904\n``#010904` `rgb(1,9,4)``\n• #051f0e\n``#051f0e` `rgb(5,31,14)``\n• #083518\n``#083518` `rgb(8,53,24)``\n• #0c4b22\n``#0c4b22` `rgb(12,75,34)``\n• #10612c\n``#10612c` `rgb(16,97,44)``\n• #137736\n``#137736` `rgb(19,119,54)``\n• #178d40\n``#178d40` `rgb(23,141,64)``\nMonochromatic Color\n\n# Alternatives to #0c4b22\n\nBelow, you can see some colors close to #0c4b22. Having a set of related colors can be useful if you need an inspirational alternative to your original color choice.\n\n• #0c4b12\n``#0c4b12` `rgb(12,75,18)``\n• #0c4b18\n``#0c4b18` `rgb(12,75,24)``\n• #0c4b1d\n``#0c4b1d` `rgb(12,75,29)``\n• #0c4b22\n``#0c4b22` `rgb(12,75,34)``\n• #0c4b27\n``#0c4b27` `rgb(12,75,39)``\n• #0c4b2d\n``#0c4b2d` `rgb(12,75,45)``\n• #0c4b32\n``#0c4b32` `rgb(12,75,50)``\nSimilar Colors\n\n# #0c4b22 Preview\n\nThis text has a font color of #0c4b22.\n\n``<span style=\"color:#0c4b22;\">Text here</span>``\n#0c4b22 background color\n\nThis paragraph has a background color of #0c4b22.\n\n``<p style=\"background-color:#0c4b22;\">Content here</p>``\n#0c4b22 border color\n\nThis element has a border color of #0c4b22.\n\n``<div style=\"border:1px solid #0c4b22;\">Content here</div>``\nCSS codes\n``.text {color:#0c4b22;}``\n``.background {background-color:#0c4b22;}``\n``.border {border:1px solid #0c4b22;}``\n\n# Shades and Tints of #0c4b22\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, #010703 is the darkest color, while #f5fdf8 is the lightest one.\n\n• #010703\n``#010703` `rgb(1,7,3)``\n• #04180b\n``#04180b` `rgb(4,24,11)``\n• #072913\n``#072913` `rgb(7,41,19)``\n• #093a1a\n``#093a1a` `rgb(9,58,26)``\n• #0c4b22\n``#0c4b22` `rgb(12,75,34)``\n• #0f5c2a\n``#0f5c2a` `rgb(15,92,42)``\n• #116d31\n``#116d31` `rgb(17,109,49)``\n• #147e39\n``#147e39` `rgb(20,126,57)``\n• #178f41\n``#178f41` `rgb(23,143,65)``\n• #1aa048\n``#1aa048` `rgb(26,160,72)``\n• #1cb050\n``#1cb050` `rgb(28,176,80)``\n• #1fc158\n``#1fc158` `rgb(31,193,88)``\n• #22d25f\n``#22d25f` `rgb(34,210,95)``\n• #2bdd69\n``#2bdd69` `rgb(43,221,105)``\n• #3be075\n``#3be075` `rgb(59,224,117)``\n• #4ce281\n``#4ce281` `rgb(76,226,129)``\n• #5de58d\n``#5de58d` `rgb(93,229,141)``\n• #6ee899\n``#6ee899` `rgb(110,232,153)``\n• #7feba5\n``#7feba5` `rgb(127,235,165)``\n• #90edb1\n``#90edb1` `rgb(144,237,177)``\n• #a1f0bd\n``#a1f0bd` `rgb(161,240,189)``\n• #b2f3c8\n``#b2f3c8` `rgb(178,243,200)``\n• #c3f5d4\n``#c3f5d4` `rgb(195,245,212)``\n• #d4f8e0\n``#d4f8e0` `rgb(212,248,224)``\n• #e5fbec\n``#e5fbec` `rgb(229,251,236)``\n• #f5fdf8\n``#f5fdf8` `rgb(245,253,248)``\nTint Color Variation\n\n# Tones of #0c4b22\n\nA tone is produced by adding gray to any pure hue. In this case, #2a2d2b is the less saturated color, while #02551f is the most saturated one.\n\n• #2a2d2b\n``#2a2d2b` `rgb(42,45,43)``\n• #27302a\n``#27302a` `rgb(39,48,42)``\n• #233429\n``#233429` `rgb(35,52,41)``\n• #203728\n``#203728` `rgb(32,55,40)``\n• #1d3a27\n``#1d3a27` `rgb(29,58,39)``\n• #193e26\n``#193e26` `rgb(25,62,38)``\n• #164125\n``#164125` `rgb(22,65,37)``\n• #134424\n``#134424` `rgb(19,68,36)``\n• #0f4823\n``#0f4823` `rgb(15,72,35)``\n• #0c4b22\n``#0c4b22` `rgb(12,75,34)``\n• #094e21\n``#094e21` `rgb(9,78,33)``\n• #055220\n``#055220` `rgb(5,82,32)``\n• #02551f\n``#02551f` `rgb(2,85,31)``\nTone Color Variation\n\n# Color Blindness Simulator\n\nBelow, you can see how #0c4b22 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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[ "# How to calculate 3 raise\n\nIf you're ready to learn How to calculate 3 raise, keep reading!\n\n## Pay Raise Calculator\n\nPerson A earns \\$100,000 so a 3% raise (.03 x 100,000) equals \\$3,000 per year Person B earns \\$50,000 so a 3% raise for them is \\$1,500 (\\$50,000 x .03 = \\$1,500) over the", null, "", null, "", null, "## How do I Calculate Salary Increases?\n\nHow do I calculate a 3% increase? Subtract the original value from the new value, then divide the result by the original value. Multiply the result by 100. The answer is the percent\n`\n\n## Client testimonials\n\nChristopher Huntsberry\n\nI've always been searching for something like this. Has tons of extras to learn along the way, there is nothing wrong with this app. This app is a lifesaver. I am able to check my answers to make sure I get the best grade I can get thanks to this app. I suggest that the team to add many more solutions, like for Physics, Chemistry or any related‹.\n\nThomas Tester\n\nKind regards, Maia, great app, has the solution to almost any algebra problem, and I love the camera feature. This app actually helps me with my homework and when I read the answers it gives me and the explanation I understand it and the fact that there are no ads makes it even better math app, thank you so much for helping me and this is gonna help me for my finals tomorrow.\n\n## How to Calculate a Future Pay Raise\n\nFollow the simple steps below and then click the 'Calculate' button to see the results. Enter your current pay rate and select the pay period. Next, enter the hours worked per week and select\n\nDo math equation\n\nDoing math equations is a great way to keep your mind sharp and improve your problem-solving skills.\n\nDetermine math problem\n\nmath is the study of numbers, shapes, and patterns.\n\nReach support from expert professors\n\nLooking for a little help with your homework? Check out our solutions for all your homework help needs!\n\nDeal with math\n\nLooking for support from expert professors? Our community of experts can help you with any question you have.\n\nDo mathematic problems\n\nThe mathematic equation is determined by solving for the unknown variable.\n\nGet support from expert tutors\n\nYou can upload your requirement here and we will get back to you soon.\n\n• Get homework writing help\n\nLooking for someone to help with your homework? We can provide expert homework writing help on any subject.\n\n• Have more time for your pursuits\n\nTo determine what the math problem is, you will need to look at the given information and figure out what is being asked. Once you know what the problem is, you can solve it using the given information.\n\n• Determine mathematic equation", null, "", null, "" ]

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[ "## October 2014 solution\n\nThis month's riddle was a classic which has been making the rounds for many years. The solution is that it is possible to simulate a dice with any number of sides, n, with just two coins. This was shown by this month's solvers in a dazzling variety of ways. The following is the first solution I have heard for the problem. It uses one coin that is fair and one coin that has probability 1/n of falling on heads, making for a very elegant choice of coin probabilities.\n\nSuppose you throw the 1/n coin once and then the fair coin k times. This action divides the probability space to 2k pieces of probability 1/2kn and 2k pieces of probability (n-1)/2kn. Let us refer to the former as A-type pieces and the latter as B-type pieces. The question is: can these pieces be joined together in order to form n pieces of total probability 1/n each?\n\nWe can, for example, take a of the A-type pieces and b of the B-type pieces to form the first slice. For this to work, we need to have a+(n-1)b=2k. For example, we can have b=2k div (n-1) and a=2k mod (n-1). Let's call that the canonical (a,b) combination. There are no solutions that use a smaller a value than the canonical combination, but decreasing b is always possible, given a large-enough supply of A-type pieces: to reduce b by one, we compensate by increasing a by n-1.\n\nSuppose, now, that we try to map as many of the n target pieces as possible with the canonical (a,b) combination. If nothing stops us then we are done, but if something does stop us, it will be that we've run out of either enough A-types or enough B-types. As we have seen, running out of B-types is not a problem, as we can always tile the probability space with less of them (including with none of them) by replacing them with the appropriate number of A-types. Running out of As, however, is something we'd like to avoid.\n\nThe solution is to make sure that ab. As we have started with an equal number of A-types and B-types, choosing an (a,b) combination that uses at least as many of the B-types as of the A-types guarantees us that A-types will not run out first.\n\nIf we are able to find an (a,b) combination with ab, the problem is solved: we begin by creating slices with our chosen combination until B-types run out. From that point on, we make up the missing B-types by replacing them with the appropriate number of A-types. Because all our pieces have probabilities that sum up to 1, the number of A-types will clearly be exactly enough to complete the last piece.\n\nWe are now left with the question of how to ensure ab. Recall that in the canonical combination we have chosen a to be 2k mod (n-1). For this reason we are guaranteed a<n-1. By choosing a large enough k -- such as a k value for which 2k≥(n-1)2 -- we have that our b value, namely 2k div (n-1), satisfies bn-1. Therefore, ba as required, and we are done.\n\nQ.E.D." ]

[ null ]

[ "# QR Decomposition", null, "Suppose we have a problem that can be modeled by the system of equations Ax = b with a matrix A and a vector b. We have already shown how to use Gaussian Elimination to solve these methods, but I would like to introduce you to another method – QR decomposition. In addition to solving the general system of equations, this method can also be used in a number of other algorithms including the linear regression analysis to solve the least squares problem and to find the eigenvalues of a matrix.\n\nGenerally, when we see a system of equations, the best way to proceed in solving it is Gaussian elimination, or LU decomposition. However, there are some very special matrices where this method can lead to rounding errors. In such cases, we need a better, or more stable algorithm, which is when the QR decomposition method becomes important.\n\nAny matrix A, of dimensions m by n with m >= n, can be represented as the product of two matrices, an m by m orthogonal matrix Q, i.e. QTQ = I, and an m by n upper triangular matrix R with the form [R 0]T. We perform this decomposition by will first converting the columns of the matrix A into vectors. Then, for each vector ak, we will calculate the vectors uk and ek given by\n\nuk =", null, "i = 1 to k-1projej ak\nand\nek = uk / ||uk||\nThen Q = [e1, …, en] and\n\nR =", null, "0", null, "0 0", null, "", null, "", null, "", null, "", null, "# Learn About “the Other” Algebra", null, "When I visit family for the holidays, the topic of my being a mathematician always seems to come up, and there’s always a child in the family struggling with maths, and when I ask the subject of their struggles the word “algebra” is always the culprit. I’ll save for another post my ideas on how this subject should be taught in high school and some of the main problems facing students.\n\nI want to concentrate this post on a topic that few outside the mathematical world know about, but which many inside this world (myself included) hold dearly – the topic of modern or abstract algebra. I refer to this as “the other” algebra because a general conversation about the word “algebra” will generally revolve around concepts such as systems of equations, slopes, intercepts, intersection, rise-over-run, point-slope, and other terminology that limits algebra to a specific domain (the set of real or complex numbers) while at the same time ignoring the underlying beauty associated with this area.\n\nI wrote previously about the area of set theory and the beauty associated with taking math out of the scope of a basic number line and into a much more undefined space. Abstract algebra is a continuation of set theory where in addition to our set, we have a (binary) operation defined on any two elements of this set. The inclusion of this binary operation allows us to consider several different structures based on the properties that this binary operation holds.\n\nThe structures I’d like to write about today are called groups. A group is a set along with an operation (or function) defined on any two elements of the set with the following properties:\n– It is closed. This means that any time we run this function on two elements on the set, the function gives us a member of the set. In mathematical terms, for all a, b in the set A, f(a, b) must also be a member of A.\n– There is an identity element. An identity element is defined as an element where is we include it in the binary operator with any other element, the operator will always return the other element. So if the element i is the identity element, then f(i, a) = a and f(a, i) = a for any other a in the set A. Any group must have an identity element.\n– Every element has an inverse. Inverse elements are based on the identity element. What the inverse says is that for every element, there is a way to use the binary operator to get to the identity element. So for all elements a in the set A, there is an element b in the set A such that f(a, b) = i, where i is the identity element.\n– The binary operator is associative. I described the associative property when I discussed the functions and relations of set theory. A function is associative if the way we group things (aka associate them) doesn’t matter. This means that for any elements a, b, and c of the set A, f(f(a, b), c) must be the same as f(a, f(b, c)).\n\nIf these four properties hold for a set A and a binary function f, then we say that the pair (A, f) is a group. We will generally use a common notation such as a · b, or a * b or simply ab to represent f(a, b).\n\nAnother important concept in group theory is the idea of a Cayley table. These are similar to multiplication tables that we drew out when we were first learning our “times tables”. For a group with n elements, we form a table with n rows and n columns. Each element of the group is written out to the left of each row and above each column (so really we can think of it as an n+1 by n+1 table with the first row and column being descriptive rows). Each cell of the table is the binary operator applied to the two elements indicated by the row and column (with an understanding of whether we have row before column or vice versa). Obviously, we can only do this for finite groups as we cannot write out all the elements of an infinite set.\n\nThe script I’ve added is a tester to allow users to input the information for a possible group (size, name of each element, and a Cayley table) and with this information the user is informed whether or not it forms a group. If it does not form a group, the reasons why it does not form one are also given. There are also some sample groups given to give insight into this area." ]

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[ "# CHEM126 – CHAPTER 15 – BASIC HOMEWORK / INTRO ACIDS\n\n### Resources:\n\nHave you found videos, websites, or explanations that helped you understand this chapter? Let us know and we’ll add them to “Resources” part of this page for other students to use.\n\n### Question 1\n\nWhat conducts electricity in metals? (One word answer)\n\nSelect from: electrons, ions, neutrons, nucleus, positrons, and protons\n\n### Explanation\n\nBRO COME ON…negatively charged particle found in atoms\n\n### Question 2\n\nWhat conducts electricity in solutions? (One word answer)\n\n### Explanation\n\nGeneral word for both cations and anions…aka please don’t embarrass me you know this\n\n### Question 3\n\nWhich of the following statements about ionic compounds is FALSE?\n\nSelect one:", null, "### Explanation\n\nIll let this one slide if you didn’t know it, but come on can solid salt conduct electricity there answer is NO because the ions are not free to move around.\n\n### Question 4\n\nClassify the following compounds as strong electrolyte, weak electrolyte, or non-electrolyte. Write strong, weak, or non, respectively in the space for the answer.\n\nKCl(s)\n\nElectrolytes defined\n\nQualitative electrolyte rules\n\n### Explanation\n\nDoes this completely dissociate?? YES IT DOES why because of the solubility rules that you should have memorized last semester\n\nCaCl2(s)\n\n### Explanation", null, "PbCl2(s)\n\nO2(g)\n\n### Question 8\n\nHCl(g)", null, "### Question 9\n\nAcetic acid (CH3COOH)\n\n### Question 10\n\nWhat is the molarity of 0.77 mol of NaCl in enough water to make 5.0 L of solution?\n\nDefinition and examples of Molarity", null, "### Explanation\n\n0.77 mol NaCl / 5 L = 0.154 [NaCl]\n\n### Question 11\n\nWhat is the molarity of the Al3+ ions when 0.42 mol of aluminum nitrate, Al(NO3)3, is dissolved in enough water to make 4.0 L of solution?", null, "### Explanation\n\nAl(NO3)3 is a strong electrolyte. ( 1 Mol of Aluminum ions per mol)\n\n0.42/ 4 * 1 = 0.105\n\n### Question 12\n\nWhat is the molarity of the nitrate ions when 0.47 mol of barium nitrate, Ba(NO3)2, is dissolved in enough water to make 3.0 L of solution?", null, "### Explanation\n\nBa(NO3)2 is a strong electrolyte.( 2 Mols of Nitrate ions per mol)\n\n0.47/ 3 * 2  = 0.313\n\n### Question 13\n\nCalculate the molarity of calcium hydroxide that has been prepared by dissolving 2.87 g of calcium hydroxide in water and diluting the solution to a final volume of 4.00 L.\n\nExample molarity calculation", null, "### Explanation\n\n2.87 g Ca(OH)2 / 74.069  g/mol    =    0.0387  mol\n\n0.0387 mol/ 4 L = 0.009696\n\n### Question 14\n\nCalculate the molarity of the hydroxide ions in a solution that has been prepared by dissolving 27.5 g of barium hydroxide (Ba(OH)2) in water and diluting the solution to a final volume of 5.00 L. (M.W’s: Ba 137.3 g/mol., O 16.0 g/mol, and H 1.0 g/mol)", null, "### Explanation\n\nRemember there are 2 moles of hydroxide for each mole of barium hydroxide.\n\n### Question 15\n\nWhat is the molarity of the H+ ions in 0.50 M HCl?\n\nExamples of acids and bases\n\nSelect one:", null, "### Explanation\n\nSTRONG ACID FULLY DISSOCIATES\n\n### Question 16\n\nWhat is the molarity of the H+ ions in 0.50 M acetic acid, CH3COOH?\n\nSelect one:", null, "### Explanation\n\nNOT STRONG ACID DOESNT FULLY DISSOCIATE\n\n### Question 17\n\nWhat is the molarity of the H+ ions in 0.50 M hydrofluoric acid, HF?\n\nSelect one:", null, "### Explanation\n\nHF is a weak acid WHAT DOES THAT MEAN it will not fully dissociate aka very very little of it\n\n### Question 18\n\nWhat is the molarity of the H+ ions in 0.40 M nitric acid, HNO3?Select one:", null, "### Explanation\n\nSTROMG ACIDS FULLY DISSOCIATE\n\n### Question 19\n\nWhat is the molarity of the NO3 ions in 0.40 M nitric acid, HNO3?\n\nSelect one:", null, "0.40 *1 = 0.40\n\n### Question 20\n\nWhat is the molarity of the OH ions in 0.040 M sodium hydroxide, NaOH?\n\nSelect one:", null, "### Explanation\n\nNaOH is a strong base and will therefore fully disscoaite\n\n### Question 21\n\nWhat is the molarity of the OH ions in 0.040 M ammonia, NH3?\n\nSelect one:", null, "### Explanation\n\nNH3 is a weak base, meaning that it will not fully dissociate and have very very little hydroxide ions\n\n### Question 22\n\nWhat is the molarity of the OH ions in 0.040 M barium hydroxide, Ba(OH)2?\n\nAssume the barium hydroxide is fully dissociated into ions in the dilute solution.\n\nSelect one:", null, "0.040*2 = 0.08\n\n### Question 23\n\nH+ is more accurately represented by adding a water molecule, that is, as H3O+.\n\n [H3O+] = [OH–] Answer 1", null, "", null, "[H3O+] > [OH–] Answer 2", null, "", null, "[H3O+] < [OH–] Answer 3", null, "", null, "### Question 24\n\nCheck those statements which are true.\n\nConcentration of acids and bases\n\nSelect one or more:", null, "", null, "", null, "### Explanation\n\nUse the equation -log[H+]=pH\n\nas well as  -log[H+] -log[OH-] = 14\n\n### Question 25\n\nCalculate the hydroxide ion concentration, [OH], in 0.014 M HBr.\n\nAcid/base calculation", null, "### Explanation\n\n-log[0.014]+-log[OH-]=14\n\nOH-= 7.14e-13\n\n### Question 26\n\nCalculate the hydroxide ion concentration, [OH], in 1.95 M HNO3.", null, "### Explanation\n\n-log[1.95]+-log[OH-]=14\n\nOH-= 5.128e-15\n\n### Question 27\n\nCalculate the hydrogen ion concentration, [H+], in 0.00486 M NaOH.", null, "### Explanation\n\npH+pOH=14\n\n-log[OH-]=pOH\n\n-log[0.00486 M]=pOH\n\n-log[H+] -log[0.00486 M]=14\n\nH+=2.0576e-12\n\n### Question 28\n\nCalculate the hydrogen ion concentration, [H+], in 0.00688 M Ba(OH)2. Careful!", null, "### Explanation\n\npH+pOH=14\n\n-log[OH-]=pOH\n\n-log[2*0.00688]=pOH\n\n-log[H+] -log[2*0.00688]=14\n\nH+=7.267e-13\n\n### Question 29\n\nThe autodissociation of water to H+ and OH is endothermic. At a higher temperature more of the water molecules have sufficient energy to dissociate. So, Kw is greater.\n\nAt 25oC, Kw = 1.0 x 10-14. (standard Kw)\nAt 65oC, Kw = 1.0 x 10-13.\n\nCalculate the hydrogen ion concentration, [H+], in 0.0636 M LiOH at 65oC.", null, "### Explanation\n\nKw= 1e-13 = [0.0636][H+]\n\n[H+]=1.57e-12\n\n### Question 30\n\nWhat is the pH of a solution where the [H+] =8.28 x 10-12 M?\n\nREPORT pH TO 2 DECIMAL PLACES.\n\npH calculation", null, "### Explanation\n\npH=-log[8.28e-12]\n\npH=11.08\n\n### Question 31\n\nWhich of the following statements are true?\n\npH and acidity\n\nSelect one or more:", null, "", null, "", null, "", null, "### Question 32\n\nWhat is the pH of a solution where the [H+] =7.41 x 10-12 M?\n\nREPORT pH TO 2 DECIMAL PLACES.", null, "### Explanation\n\npH=-log[7.41e-12]\n\npH=11.13\n\n### Question 33\n\nWhat is the pH of a solution where the [OH– ] =5.73 x 10-1 M?\n\nREPORT pH TO 2 DECIMAL PLACES.\n\nFinding pH from [OH– ]", null, "### Explanation\n\npH+pOH=14\n\npH-log[5.73e-1]=14\n\npH=13.76\n\n### Question 34\n\nWhat is the [H+] if the pH of a solution is 9.88?", null, "### Explanation\n\n9.88=-log[H+]\n\n10^-9.88=10^log[H+]\n\n1.318e-10=[H+]\n\n### Question 35\n\nWhat is the [H+] if the pH of a solution is 2.44?", null, "" ]

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[ "# Proving that $\\mathbb{Z}[i]$ is a noetherian ring\n\nClaim: the ring $\\mathbb{Z}[i]$ is a noetherian ring\n\nMy proof\n\n1) $\\mathbb{Z}[i]$ is a finitely generated $\\mathbb{Z}$-module.\n\n2) $\\mathbb{Z}$ is a noetherian ring.\n\n3) Every finitely generated module over a noetherian ring is a noetherian module, hence $\\mathbb{Z}[i]$ is a noetherian $\\mathbb{Z}$-module.\n\n4) By definition of noetherian module, every $\\mathbb{Z}$-submodule of $\\mathbb{Z}[i]$ is finitely generated as a $\\mathbb{Z}$-module\n\n5) an ideal $\\mathfrak{i}$ of $\\mathbb{Z}[i]$ is in particular a $\\mathbb{Z}$-submodule of $\\mathbb{Z}[i]$\n\n6) $\\mathfrak{i}=\\mathbb{Z}x_1+\\ldots +\\mathbb{Z}x_n$\n\n7) since $\\mathfrak{i}$ is finitely generated as a $\\mathbb{Z}$-module, it is also finitely generated as an ideal\n\nDo you think my proof works?\n\n• Wouldn't it be much easier to prove directly that the Gaussian integers are an euclidean domain and thus even a PID? – DonAntonio Jun 4 '13 at 9:21\n• @DonAntonio you are surely right, but here i'm just wondering if there is some mathematical absurdity in my proof – bateman Jun 4 '13 at 9:24\n• The first point is interesting (and true, don't worry), but: how do you know $\\,\\Bbb Z[i]\\,$ is a f.g. abelian group = f.g. $\\,\\Bbb Z$-module) ? – DonAntonio Jun 4 '13 at 9:26\n• @DonAntonio I show that $1,i$ is a $\\mathbb{Z}$-basis – bateman Jun 4 '13 at 9:29\n• I think you can apply Hilbert-basis theorem, with the fact that quotients of the noetherian domains are noetherian. But I think that this is essentially your approach. – awllower Jun 4 '13 at 9:34\n\n## 1 Answer\n\nWould you like another proof?\n\nBy Hilbert's Theorem $\\mathbb{Z}[X]$ is noetherian. Hence $\\mathbb{Z}[i]$ is also noetherian as a factor-ring of $\\mathbb{Z}[X]$.\n\nAddendum: In fact, here (for one unknown) Hilbert's theorem is not needed.\n\n• As short, simple and elegant as one can expect. +1 – DonAntonio Jun 4 '13 at 10:00\n• @DonAntonio Thank you. – Boris Novikov Jun 4 '13 at 10:02" ]

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[ "", null, "# Solving Inequalities Worksheet Pdf\n\n5 total of m and 3 is 21 translating phrases. Solve and graph the solution set of following.", null, "Solving Equations Maths Worksheet Algebra Worksheets Solving Linear Equations Pre Algebra Worksheets\n\n### X 45.", null, "Solving inequalities worksheet pdf. T s2l0 u1j2m zkvuctqar 5sdo9f9t dw lairoe k nlblrc4 x t mael zl1 jr ri 3gjh vtbs j ir te wsnefrevse zd p a w lmmajdee0 vw6idtchb uivn tf kipniot oen 6a0lagne3b orca d x1d s worksheet by kuta software llc kuta software infinite algebra 1 name two step inequalities date period. 9y 3. Then solve each inequality writing the solution as a union of the two solutions.\n\nSolving inequalities worksheet 1 rtf. 2y 4. 1 m or m.\n\nA l 1mda9d ken 6wsi rt 4hw hinnbf ti7n nipt ie2 uajlagte 8b 0r4al y1e e worksheet by kuta software llc kuta software infinite algebra 1 name compound inequalities date period solve each compound inequality and graph its solution. Solving linear inequalities worksheet 1 solve following linear inequalities 1. 5x 2 4 3x.\n\n8x x 2. 2 worksheet by kuta software llc infinite algebra 1 name one step inequalities date period. X 16 2x 90 2 2.\n\nIn the following inequalities solve for x. Use inverse operations or mental math to solve for x. O g ta llml1 hr1i fg 1hft lsv xrceasoe7r jviezdk v d 1m 6a 7d dej 1wti 1t qhq vintf yi dn5i qt 5e6 cael5g pejb ur xad 82y q worksheet by kuta software llc kuta software infinite algebra 2 name solving inequalities date period.\n\n1 6x 4 10 6 6x 20. One step equations di erence between y and 23 is 12. A k2r0 e1y2k pk 3u2t6ak qsto 5frtawca brje l rlblyct r 4 ta clhlx brfisgyhdt vsu 5rlexs se nrbv epd w s j um wavd 4ed qw ci utsh l 9i qnbf ainnkintzek la plug tembxrka 4 q1x.\n\n5x 4 3 3x 2. Solving inequalities worksheet 1 here is a twelve problem worksheet featuring simple one step inequalities. 2x 2 8x 4.\n\n2 2 multiplied by b is equal to 8 3 4 product of 4 and z is the same as 16 7 n minus 2 is equal to 16 9 20 exceeds c gives 18 8 11 times p is 33 10 one half of x is equal to 3 6 b divides 6 gives 1 1 sum of x and 3 gives 5 translate each verbal phrase into an algebraic expression. 5y 6 2y 7 3. 2y y 9.\n\nL b2q0a1y1c lk nu 0tta v 6svohfet vwvabrre o hlbl 9ct. Show your steps and write your answer in the space provided. 1 8x 120 2 16x 100 3 5x 80.\n\nIf the absolute value is greater than or greater than or equal to a positive numbe r set the argument less than the opposite of the number and greater than the number using an or statement in between the two inequalities. 7y 2x 8. 6x 4 2x.\n\n7x 56 x 11 5 2x 10 100 7 7 11 11 10 10.", null, "Algebra 1 Worksheets Inequalities Worksheets Algebra Worksheets Pre Algebra Worksheets Graphing Inequalities", null, "One Step Inequalities Worksheets By Multiplying And Dividing Algebra Worksheets Pre Algebra Worksheets Graphing Inequalities", null, "Algebra 1 Worksheets Inequalities Worksheets Algebra Worksheets Pre Algebra Worksheets Graphing Inequalities", null, "Writing Solving And Graphing Inequalities Worksheet Graphing Inequalities Graphing Inequality", null, "Pin On Professionally Designed Worksheets", null, "Pre Algebra Worksheets Inequalities Worksheets In 2020 Graphing Inequalities Algebra Worksheets Pre Algebra Worksheets", null, "Inequalities Challenge Solving Inequalities Combining Like Terms Inequality", null, "Inequalities Algebra 1 Worksheet Printable Algebra Worksheets Solving Equations Equations", null, "Solving And Graphing Inequalities Worksheet Pdf In 2020 Graphing Inequalities Math Worksheets Linear Inequalities", null, "Inequalities Worksheets Graphing Inequalities Solving Inequalities Graphing Linear Equations", null, "Solving Linear Equations Worksheet Pdf Best Of Solving Equations Worksheets Chessmuseum Template L In 2020 Algebra Worksheets Word Problem Worksheets Math Worksheets", null, "27 Solving And Graphing Inequalities Worksheet Answer Key Pdf Worksheet Paintings Search Resu In 2020 Graphing Inequalities Linear Inequalities Algebra 2 Worksheets", null, "The Solving Linear Equations Form Ax B C J Math Worksheet From The Algebra Worksheet Page Solving Linear Equations Algebra Worksheets Linear Equations", null, "Partner Problems Inequalities 1 Pdf Solving Inequalities Solving Equations Inequality", null, "Solving Equations And Inequalities Worksheet Answers In 2020 Free Math Lessons 9th Grade Math School Algebra", null, "Eighth Grade Solving Inequalities Worksheet 05 One Page Worksheets Graphing Linear Equations Graphing Inequalities Word Problem Worksheets", null, "One Step Inequalities Worksheets By Multiplying And Dividing Mathaids Com Pinterest Great On Algebra Worksheets Pre Algebra Worksheets Graphing Inequalities", null, "One Step Inequalities Worksheets One Step Equations Solving Inequalities Multi Step Inequalities", null, "27 Solving And Graphing Inequalities Worksheet Answer Key Pdf Solving And Graphing Inequaliti Graphing Inequalities Writing Inequalities Inequality", null, "Previous post Fourth Grade Compound Sentences Worksheet 4th Grade", null, "Next post Avengers Endgame Coloring Pages" ]

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[ "# Probability of Negative Return", null, "In our earlier article, ‘What is a Nap?’, we briefly mentioned the probability of negative return but, at that stage, we didn’t expound on what it is, how it’s calculated or how to use it in relation to horseracing results.\n\nHowever, what we did say in the earlier article was that to properly analyse a series of horseracing selections, such as those supplied by a tipster, you need a large number of selections. The reason for this is that, when the number of selections is large, the distribution of horseracing data looks the normal, or Gaussian, distribution for a random variable.\n\nThe probability of negative return is a simple statistic to calculate and understand. The figure calculated represents the probability of a series of selections producing a negative return or, in other words, a level stakes loss. The higher the figure calculated the higher the probability of incurring such as loss.\n\nIn order to calculate the probability of negative return for a series of selections, you need to:\n\nCalculate the theoretical percentage profit from each selection, to level stakes. A winner at even money would generate a profit of 100%, a winner at 2/1 would generate a profit of 200% and so on, while any loser would generate a profit of -100%.\n\nCalculate the average, or mean, percentage profit. The easiest way to do this (and the subsequent steps in the calculation) is to create a spreadsheet in Microsoft Excel – which can be as simple as a single column containing the percentage profit for each selection – and use the AVERAGE function.\n\nCalculate the standard deviation of the data in your spreadsheet using the STDEV function. Algebraically, standard deviation, S, is calculated according to the formula below, but if you use Microsoft Excel you don’t need to worry about the mathematics.\n\nIf your percentage profit data is contained in, say, cells A1 to A8 of your spreadsheet, simply use the STDEV function in the form STDEV(A1:A8).\n\nDivide the result by the square root of the number of selections, using the SQRT function, to give the standard error.\n\nAssuming that horse racing results are at least approximately normally distributed, you can use the NORMDIST function in Excel, in the form = NORMDIST (0, mean, stderr, 1), where mean and stderr are the mean percentage profit and standard error of the percentage profit, which you calculated in the steps 2. and 4. above.\n\nThe result should be a figure between 0 and 1 and, as mentioned above, the higher the figure the higher the probability of negative return and versa. A probability of 0.10 represents a 10% chance of making a loss, or a 90% chance of making a profit, from a series of selections and is typically the maximum value that you’d want to accept if you’re going to back future selections with real money.\n\nWe hope you enjoyed ‘Probability of Negative Return’ and we will be back soon with another advanced betting guide. In the meantime, we would love to hear your thoughts on ‘Probability of Negative Return’ in the comments section below." ]

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[ "# #101e46 Color Information\n\nIn a RGB color space, hex #101e46 is composed of 6.3% red, 11.8% green and 27.5% blue. Whereas in a CMYK color space, it is composed of 77.1% cyan, 57.1% magenta, 0% yellow and 72.5% black. It has a hue angle of 224.4 degrees, a saturation of 62.8% and a lightness of 16.9%. #101e46 color hex could be obtained by blending #203c8c with #000000. Closest websafe color is: #003333.\n\n• R 6\n• G 12\n• B 27\nRGB color chart\n• C 77\n• M 57\n• Y 0\n• K 73\nCMYK color chart\n\n#101e46 color description : Very dark blue.\n\n# #101e46 Color Conversion\n\nThe hexadecimal color #101e46 has RGB values of R:16, G:30, B:70 and CMYK values of C:0.77, M:0.57, Y:0, K:0.73. Its decimal value is 1056326.\n\nHex triplet RGB Decimal 101e46 `#101e46` 16, 30, 70 `rgb(16,30,70)` 6.3, 11.8, 27.5 `rgb(6.3%,11.8%,27.5%)` 77, 57, 0, 73 224.4°, 62.8, 16.9 `hsl(224.4,62.8%,16.9%)` 224.4°, 77.1, 27.5 003333 `#003333`\nCIE-LAB 12.485, 10.077, -26.935 1.783, 1.481, 5.986 0.193, 0.16, 1.481 12.485, 28.758, 290.513 12.485, -4.515, -24.455 12.169, 4.862, -20.647 00010000, 00011110, 01000110\n\n# Color Schemes with #101e46\n\n• #101e46\n``#101e46` `rgb(16,30,70)``\n• #463810\n``#463810` `rgb(70,56,16)``\nComplementary Color\n• #103946\n``#103946` `rgb(16,57,70)``\n• #101e46\n``#101e46` `rgb(16,30,70)``\n• #1d1046\n``#1d1046` `rgb(29,16,70)``\nAnalogous Color\n• #394610\n``#394610` `rgb(57,70,16)``\n• #101e46\n``#101e46` `rgb(16,30,70)``\n• #461d10\n``#461d10` `rgb(70,29,16)``\nSplit Complementary Color\n• #1e4610\n``#1e4610` `rgb(30,70,16)``\n• #101e46\n``#101e46` `rgb(16,30,70)``\n• #46101e\n``#46101e` `rgb(70,16,30)``\n• #104638\n``#104638` `rgb(16,70,56)``\n• #101e46\n``#101e46` `rgb(16,30,70)``\n• #46101e\n``#46101e` `rgb(70,16,30)``\n• #463810\n``#463810` `rgb(70,56,16)``\n• #020308\n``#020308` `rgb(2,3,8)``\n• #070c1c\n``#070c1c` `rgb(7,12,28)``\n• #0b1531\n``#0b1531` `rgb(11,21,49)``\n• #101e46\n``#101e46` `rgb(16,30,70)``\n• #15275b\n``#15275b` `rgb(21,39,91)``\n• #193070\n``#193070` `rgb(25,48,112)``\n• #1e3984\n``#1e3984` `rgb(30,57,132)``\nMonochromatic Color\n\n# Alternatives to #101e46\n\nBelow, you can see some colors close to #101e46. Having a set of related colors can be useful if you need an inspirational alternative to your original color choice.\n\n• #102c46\n``#102c46` `rgb(16,44,70)``\n• #102746\n``#102746` `rgb(16,39,70)``\n• #102346\n``#102346` `rgb(16,35,70)``\n• #101e46\n``#101e46` `rgb(16,30,70)``\n• #101a46\n``#101a46` `rgb(16,26,70)``\n• #101546\n``#101546` `rgb(16,21,70)``\n• #101146\n``#101146` `rgb(16,17,70)``\nSimilar Colors\n\n# #101e46 Preview\n\nThis text has a font color of #101e46.\n\n``<span style=\"color:#101e46;\">Text here</span>``\n#101e46 background color\n\nThis paragraph has a background color of #101e46.\n\n``<p style=\"background-color:#101e46;\">Content here</p>``\n#101e46 border color\n\nThis element has a border color of #101e46.\n\n``<div style=\"border:1px solid #101e46;\">Content here</div>``\nCSS codes\n``.text {color:#101e46;}``\n``.background {background-color:#101e46;}``\n``.border {border:1px solid #101e46;}``\n\n# Shades and Tints of #101e46\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, #010306 is the darkest color, while #f5f7fd is the lightest one.\n\n• #010306\n``#010306` `rgb(1,3,6)``\n• #050916\n``#050916` `rgb(5,9,22)``\n• #091026\n``#091026` `rgb(9,16,38)``\n• #0c1736\n``#0c1736` `rgb(12,23,54)``\n• #101e46\n``#101e46` `rgb(16,30,70)``\n• #142556\n``#142556` `rgb(20,37,86)``\n• #172c66\n``#172c66` `rgb(23,44,102)``\n• #1b3376\n``#1b3376` `rgb(27,51,118)``\n• #1f3986\n``#1f3986` `rgb(31,57,134)``\n• #224096\n``#224096` `rgb(34,64,150)``\n• #2647a6\n``#2647a6` `rgb(38,71,166)``\n• #2a4eb6\n``#2a4eb6` `rgb(42,78,182)``\n• #2d55c6\n``#2d55c6` `rgb(45,85,198)``\n• #365ed1\n``#365ed1` `rgb(54,94,209)``\n• #466bd5\n``#466bd5` `rgb(70,107,213)``\n• #5677d8\n``#5677d8` `rgb(86,119,216)``\n• #6584dc\n``#6584dc` `rgb(101,132,220)``\n• #7591e0\n``#7591e0` `rgb(117,145,224)``\n• #859ee3\n``#859ee3` `rgb(133,158,227)``\n• #95aae7\n``#95aae7` `rgb(149,170,231)``\n• #a5b7eb\n``#a5b7eb` `rgb(165,183,235)``\n• #b5c4ee\n``#b5c4ee` `rgb(181,196,238)``\n• #c5d1f2\n``#c5d1f2` `rgb(197,209,242)``\n• #d5def5\n``#d5def5` `rgb(213,222,245)``\n• #e5eaf9\n``#e5eaf9` `rgb(229,234,249)``\n• #f5f7fd\n``#f5f7fd` `rgb(245,247,253)``\nTint Color Variation\n\n# Tones of #101e46\n\nA tone is produced by adding gray to any pure hue. In this case, #2a2b2c is the less saturated color, while #031853 is the most saturated one.\n\n• #2a2b2c\n``#2a2b2c` `rgb(42,43,44)``\n• #27292f\n``#27292f` `rgb(39,41,47)``\n• #242832\n``#242832` `rgb(36,40,50)``\n• #212635\n``#212635` `rgb(33,38,53)``\n• #1d2439\n``#1d2439` `rgb(29,36,57)``\n• #1a233c\n``#1a233c` `rgb(26,35,60)``\n• #17213f\n``#17213f` `rgb(23,33,63)``\n• #132043\n``#132043` `rgb(19,32,67)``\n• #101e46\n``#101e46` `rgb(16,30,70)``\n• #0d1c49\n``#0d1c49` `rgb(13,28,73)``\n• #091b4d\n``#091b4d` `rgb(9,27,77)``\n• #061950\n``#061950` `rgb(6,25,80)``\n• #031853\n``#031853` `rgb(3,24,83)``\nTone Color Variation\n\n# Color Blindness Simulator\n\nBelow, you can see how #101e46 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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[ "# Statistics Theory\n\n## Authors and titles for recent submissions, skipping first 25\n\n[ total of 42 entries: 1-25 | 26-42 ]\n[ showing 25 entries per page: fewer | more | all ]\n\n### Tue, 20 Oct 2020 (continued, showing last 3 of 11 entries)\n\n  arXiv:2010.09597 (cross-list from cs.LG) [pdf, other]\nTitle: Faster Convergence of Stochastic Gradient Langevin Dynamics for Non-Log-Concave Sampling\nSubjects: Machine Learning (cs.LG); Statistics Theory (math.ST); Machine Learning (stat.ML)\n  arXiv:2010.09478 (cross-list from cs.LG) [pdf, other]\nTitle: Multi-Armed Bandits with Dependent Arms\nSubjects: Machine Learning (cs.LG); Statistics Theory (math.ST)\n  arXiv:2010.08838 (cross-list from econ.EM) [pdf, other]\nTitle: Empirical likelihood and uniform convergence rates for dyadic kernel density estimation\nSubjects: Econometrics (econ.EM); Statistics Theory (math.ST)\n\n### Mon, 19 Oct 2020\n\n\nTitle: Reguralized Bridge-type estimation with multiple penalties\nSubjects: Statistics Theory (math.ST)\n\nTitle: Quantile regression with ReLU Networks: Estimators and minimax rates\nSubjects: Statistics Theory (math.ST); Machine Learning (stat.ML)\n\nTitle: Consistency of archetypal analysis\nComments: 30 pages, 9 figures; add some details to the proof of Lemma 2.3\nSubjects: Statistics Theory (math.ST); Optimization and Control (math.OC); Probability (math.PR); Machine Learning (stat.ML)\n\nTitle: Power of FDR Control Methods: The Impact of Ranking Algorithm, Tampered Design, and Symmetric Statistic\nSubjects: Statistics Theory (math.ST)\n\nTitle: Nonparametric iterated-logarithm extensions of the sequential generalized likelihood ratio test\nSubjects: Statistics Theory (math.ST); Methodology (stat.ME)\n\nTitle: Mean Shrinkage Estimation for High-Dimensional Diagonal Natural Exponential Families\nSubjects: Statistics Theory (math.ST)\n\nTitle: Geometry of Sample Spaces\nSubjects: Statistics Theory (math.ST); Metric Geometry (math.MG)\n  arXiv:2010.08479 (cross-list from stat.ML) [pdf, ps, other]\nTitle: Failures of model-dependent generalization bounds for least-norm interpolation\nSubjects: Machine Learning (stat.ML); Machine Learning (cs.LG); Statistics Theory (math.ST)\n  arXiv:2010.08463 (cross-list from econ.EM) [pdf, other]\nTitle: Binary Choice with Asymmetric Loss in a Data-Rich Environment: Theory and an Application to Racial Justice\nSubjects: Econometrics (econ.EM); Statistics Theory (math.ST); Applications (stat.AP); Methodology (stat.ME); Machine Learning (stat.ML)\n  arXiv:2010.08444 (cross-list from stat.ME) [pdf, other]\nTitle: Robust Estimation for Multivariate Wrapped Models\nSubjects: Methodology (stat.ME); Statistics Theory (math.ST)\n  arXiv:2010.08127 (cross-list from cs.LG) [pdf, other]\nTitle: The Deep Bootstrap: Good Online Learners are Good Offline Generalizers\nSubjects: Machine Learning (cs.LG); Computer Vision and Pattern Recognition (cs.CV); Neural and Evolutionary Computing (cs.NE); Statistics Theory (math.ST); Machine Learning (stat.ML)\n  arXiv:2010.08097 (cross-list from cs.LG) [pdf, other]\nTitle: Consistent Feature Selection for Analytic Deep Neural Networks\nSubjects: Machine Learning (cs.LG); Statistics Theory (math.ST); Machine Learning (stat.ML)\n  arXiv:2010.08022 (cross-list from cs.LG) [pdf, other]\nTitle: Fundamental Linear Algebra Problem of Gaussian Inference" ]

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[ "Conference PaperPDF Available\n\n# Soft-Decision Decoding for DNA-Based Data Storage\n\nAuthors:\n\n## Abstract\n\nThis paper presents novel soft-decision decoding (SDD) of error correction codes (ECCs) that substantially improve the reliability of DNA-based data storage system compared with conventional hard-decision decoding (HDD). We propose a simplified system model for DNA-based data storage according to the major characteristics and different types of errors associated with the prevailing DNA synthesis and sequencing technologies. We compute analytically the error-free probability of each sequenced DNA oligo nucleotide (oligo), based on which the soft-decision log-likelihood ratio (LLR) of each oligo can be derived. We apply the proposed SDD algorithms to the DNA Fountain scheme which achieves the highest information density so far in the literature. Simulation results show that SDD achieves an error rate improvement of two to three orders of magnitude over HDD, thus demonstrating its potential to improve the information density of DNA-based data storage systems.\nSoft-Decision Decoding for DNA-Based\nData Storage\nMu Zhang, Kui Cai, Kees A. Schouhamer Immink, and Pingping Chen\nScience and Math Cluster, Singapore University of Technology and Design, Singapore 487372\nTuring Machines Inc, Willemskade 15d, 3016 DK Rotterdam, The Netherlands\nAbstract—This paper presents novel soft-decision decoding\n(SDD) of error correction codes (ECCs) that substantially im-\nprove the reliability of DNA-based data storage system compared\nwith conventional hard-decision decoding (HDD). We propose a\nsimplified system model for DNA-based data storage according\nto the major characteristics and different types of errors\nassociated with the prevailing DNA synthesis and sequencing\ntechnologies. We compute analytically the error-free probability\nof each sequenced DNA oligonucleotide (oligo), based on which\nthe soft-decision log-likelihood ratio (LLR) of each oligo can be\nderived. We apply the proposed SDD algorithms to the recently\nproposed DNA Fountain scheme. Simulation results show that\nSDD achieves an error rate improvement of two to three orders\nof magnitude over HDD, thus demonstrating its potential to\nimprove the information density of DNA-based data storage\nsystems.\nI. INTRODUCTION\nDNA-based data storage has emerged as a promising\ncandidate for the storage of Big Data. It features extremely\nhigh data storage density (for example 1 exabytes/mm3), long\nlasting stability of hundreds to a thousand year, and ultra-\nlow power consumption for operation and maintenance ,\n. Information storage in DNA has been demonstrated by\nseveral research groups -. At the beginning, due to the\nlimitation of DNA manipulation technologies, only a small\namount of data was stored in DNA molecules. During recent\nfew years, DNA productivity has been increased significantly,\nand storage of megabytes of data has been demonstrated.\nThe success of DNA-based data storage is largely attributed\nto the usage of error correction codes (ECCs). Both the\ninformation writing and reading are prone to errors due to\nthe specific bio-chemical and bio-physical processes. Fur-\nthermore, the reliability of data storage is hampered by\nthe substitution errors, insertion errors, and deletion errors\nsimultaneously . ECCs are a requirement for guaranteeing\ndata storage reliability. In and , repetition codes are\nused for data protection. In and , two-dimensional\ninterleaved Reed-Solomon (RS) codes with stronger error\ncorrection capability are applied. Recently, an efficient infor-\nmation storage architecture, named DNA Fountain , has\nbeen proposed. It combines the Luby transform (LT) codes\nand RS codes, and achieves a higher information storage\ndensity than the earlier designs.\nIn prior art DNA storage systems, ECCs are all decoded\nby hard-decision decoding (HDD), which in general requires\nlarge coding redundancy and hence lowers the information\ndensity. Although it is well known that soft-decision decoding\n(SDD) can provide a significantly performance gain over\nHDD, researchers so far were not able to apply SDD of ECCs\nto DNA storage systems. This is mainly due to the fact that the\ncomplicated DNA synthesis and sequencing processes cause\nmuch difficulty of generating the soft-decision log-likelihood\nratio (LLR) for each DNA oligonucleotide (or oligo for short)\nto support the SDD.\nIn this paper, we first characterize the DNA-based data\nstorage system with the prevailing DNA synthesis and se-\nquencing technologies. We then propose a simplified system\nmodel, through which the LLR is derived analytically based\non the number of occurrences of each sequenced oligo of the\nsystem. We apply the proposed SDD to decode LT codes in\nDNA Fountain, and demonstrate its error performance over\nthe conventional HDD.\nThe rest of this paper is organized as follows. In section II,\nwe introduce the DNA-based data storage technology and the\nDNA Fountain scheme. In Section III, we present a simplified\nDNA storage system model, as well as the calculation of\nLLRs for each sequenced oligo. The proposed DNA Fountain\nwith SDD as well as the simulation results are given in\nSection IV. Finally, Section V concludes the paper.\nII. PRELIMINARIES\nA. DNA-based Data Storage\nA DNA strand, or oligo, is a chain of almost arbitrary\ncombinations of four base nucleotides, namely Adenine (A),\nCyanine (C), Guanine (G), and Thymine (T). Each base\ncan represent two bits of information. Modern array-based\nsynthesis technologies used for DNA storage can synthesize\noligos with length up to about 200 nucleotides . Large\nfiles must be partitioned into small segments and written into\ndifferent oligos. DNA synthesis is hampered mainly by two\nbiochemical constraints. First, the homopolymer run length\nof nucleotide is limited. Long homopolymer runs increase the\nerror probability. In practical, the maximum homopolymer run\nlength is set to 1-3. Second, the GC-content of each sequence,\ni.e., the percentage of the bases G and C in the sequence,\nshould not be too high or too low. Sequences violating these\nconstraints will cause more synthesis or sequencing errors .\nGiven the desired sequences, DNA synthesizer can synthesize\nnearly 105different oligos in parallel , creating up to\n1.2×107copies of each DNA string , depending on the\ntechnology used. All these oligos are mixed together in a pool,\nwhich serves as the storage media for DNA data storage.\nReading information in DNA storage is realized by ran-\ndomly and independently sequencing the oligos in the pool,\nwith each sequenced oligo as one read. The number of reads\nis usually much smaller than the total number of oligos\nin the pool. For instance, only 0.1% of oligos in the\npool are consumed for sequencing in an experiment in .\nSome synthesized sequences may not be sequenced at all in\nthe reading process. Moreover, there might be a portion of\nsequences lost during the DNA manipulations. Thus, erasure\ncodes are required to recover the input information from the\nlimited number of sequenced oligos. In addition, polymerase\nchain reaction (PCR) is performed to amplify the oligos in the\npool before sequencing. It increases the oligo concentration\nfor the ease of sequencing and allows multiple access for the\nstorage system.\nIn DNA-based data storage, both synthesis and sequencing\nare prone to error in the bio-chemical and bio-physical\nprocesses. Most of the recent works reported in the litera-\nture adopt the array-based synthesis and the next generation\nsequencing techniques for DNA storage, leading to similar\nerror patterns and raw error probabilities. It has been found\nthat substitution, insertion, and deletion base errors and oligo\nmissing occur in DNA storage systems. Therefore, effec-\ntive error detection and correction schemes are required for\nimproving both the reliability and the information storage\ndensity of DNA storage systems.\nB. DNA Fountain architecture\nThe DNA Fountain is a DNA-based data storage architec-\nture that realized error free data writing and reading with a\nhigh number of bits per nucleotide in the literature. It consists\nof an RS code for each oligo as the inner code and an LT code\nfor a set of oligos as the outer code. Because insertion and\ndeletion errors in the oligos are problematic for efficient error\ncorrection, the RS code in DNA Fountain is only used for\nerror detection. Oligos with undesired lengths due to insertion\nor deletion errors, or those violating the parity-check of the\nRS code are discarded and considered missing for the outer\nLT code. LT codes are a class of capacity-achieving codes for\nerasure channels . Thus, it can tackle the oligo missing\n(i.e. oligo dropout) due to various errors. For a given set of k\ninput symbols, an LT code can create any desired number of\npackages, each consisting of the indices and the summation of\nrandom dinput symbols. Here, dfollows the Robust Soliton\nDistribution (RSD) µK,c,δ(d), given by\nµK,c,δ(d) = ρ(d) + τ(d)\nZ,(1)\nwhere\nρ(d) = 1/K if d= 1;\n1\nd(d1) for d= 2, ..., K,\nτ(d) =\ns\nKd for d= 1,2, ..., K/s 1;\nslog(s/δ)\nKfor d=K/s;\n0for d > K/s,\nPartition\nBinary file\nLT enc.\nRS enc.\nMapping\nScreening\nSynthesis\nRecovered file\nCombine\nLT dec.\nRS dec.\nDemapping\nSequencing\nSegments\nRandom packs\nDroplets\nBase sequences\nRepeat until enough\noligos are created\nOligo\nPool\nFig. 1. Block diagram of data writing and reading of DNA Fountain.\nand Z=d(ρ(d) + τ(d)) is a normalization coefficient.\nDue to the randomness of LT codes, the biochemical con-\nstraints of DNA manipulations can be satisfied by discarding\nall the invalid sequences, at the expense of a long encoding\nlatency.\nFig. 1 shows a diagram of DNA Fountain. The binary\nsource file is first partitioned into non-overlapping segments\nof a certain length. Packages of segments are then produced\nby selecting a random subset of segments using the RSD\ndistribution and adding them bitwise together under a binary\nfield. Each package is attached with a unique seed created\nby a pseudo random number generator (PRNG). This is\nessentially the encoding process of the LT code. The obtained\npackage with its seed is then encoded by an RS code to\nobtain a short message called droplet. After that, the binary\ndroplet is mapped into a DNA base sequence, and a screening\nprocess is performed where the invalid droplets that violate\nthe biochemical constraints are rejected. The LT-RS-screening\nprocess is then performed iteratively until a sufficient number\nof valid droplets is created and synthesized into an oligo\npool. By sequencing the oligos in the pool, demapping the\nobtained DNA base sequences into binary droplets, followed\nby decoding of the RS code and LT code, the source file can\nbe recovered.\nIII. DNA-BASED DATA STORAG E SYS TE M MODELING\nAN D LLR CA LC UL ATIO N\nA. DNA-based data storage system model\nIn this subsection, we propose a simplified system model\nthat characterizes the DNA-based data storage following the\nanalyses of experimental data of open literatures -.\nSuppose that Nunique input sequences, each with nbases\nas the data payload, are synthesized, resulting in Sreads\nafter oligo sequencing. Then Soutput sequences are obtained\nafter inner-code-parity-checking and merging of identical\nsequences. By making a few assumptions, we can derive a\nsimplified DNA storage system model shown in Fig. 2.\nRandom sampling\nwith replacement\nSubs. ins. & del.\nerrors injection\nMerging\nSample population\nN n-tuples\nRandom Samples\nS n-tuples\nOutput sequences\nSn-tuples\nRemoving\nerased sequences\nInput sequences\nN n-tuples\nFig. 2. Simplified DNA storage system model.\nWe model the synthesis and sequencing processes as a\nrandom sampling process such that the output sequences\nare randomly sampled from the Ninput sequences. The\nsampling consists of two stages. First, some sequences are\nsampled as erasures such that they are missed during the\nDNA manipulations. The second stage is to sample reads\nfrom the remaining sequences. At this stage, we assume all\nsequences in the population have the same number of copies\ncreated by the synthesizer. We further assume that the PCR\namplification is ideal such that all synthesized sequences\nare equally amplified error free. Then, the input sequence\nwill either be an erasure as shown by the first block of\nFig. 2, or be sampled with a constant probability. Since the\nnumber of reads is much smaller than the number of oligos\nin the pool, the second stage can be considered as a uniform\nrandom sampling with replacement. Next, we noticed that\nthe synthesis and sequencing errors are independent with\neach other, and they occur consecutively in the DNA storage\nsystem. We thus combine the errors generated by the two\nprocesses, by injecting the combined amount of substitution\nerrors, insertion errors, and deletion errors respectively into\nthe sampled input sequence. The fourth stage of the system\nmodel merges all identical reads to obtain Soutput sequences\nfor information recovery.\nB. Log-Likelihood Ratio calculation\nAll DNA storage architectures proposed in the literature use\nHDD for error control. In general, HDD is less reliable and\nhence requires more redundancy for achieving a target error\nrate than SDD. Specifically, for DNA Fountain, there exist\nsimultaneously the insertion and deletion errors that may not\nbe detectable by the inner RS codes. The traditional HDD of\nDNA Fountain, i.e., the inverse LT (ILT) , does not have\nerror correction capability, and a single erroneous oligo that\nwas accepted as error free may result in a large number of\ndecoded errors. This motivates us to seek for soft information\nto enable SDD of ECCs to increase the reliability for DNA\ndata storage.\nRecall that by sequencing the oligo pool, we may obtain\nmultiple output sequences carrying information of the same\ninput sequence. Since base errors occur randomly and in-\ndependently in different copies of the same input sequence\ncreated by the synthesizer , output sequences with more\noccurrences are more likely to be error free. Consider that an\noutput sequence occurs rtimes, denoted by event Dr, with\nr= 1,2, ..., S. We show that the LLR of the sequence can\nbe derived explicitly as follows.\nSince all output sequences have nbases after RS decoding,\nthis ensures that the insertion and deletion errors do not occur,\nor only occur in pairs. We refer to a pair of insertion and\ndeletion errors as an i-d error and let Est be the event that a\nsequence is corrupted by ssubstitution errors and ti-d errors.\nMoreover, the validity of each output sequence is checked by\nthe inner code, and we use event Cto denote the case where\nthe output sequence is a valid codeword. The LLR of an\noutput sequence occurring rtimes is thus given by\nLr= log P(E00|C, Dr)\n1P(E00|C, Dr).(2)\nTo compute P(E00|C, Dr), we assume that the inner code\nhas a minimum distance dmin. Let Bdenote the event that\nthe sequence has greater than or equal to dmin code symbol\nerrors. Applying the law of conditional probability and total\nprobability, we obtain\nP(E00|C, Dr) = P(E00, C, Dr)\nP(E00, C, Dr) + P(B , C, Dr),(3)\nwhere\nP(E00, C, Dr) = P(C|E00 , Dr)P(E00)P(Dr|E00 ),\nand\nP(B, C, Dr)\n=P(C|B, Dr)\ns,t\nP(Est)P(Dr|B , Est)P(B|Est).\nTherefore, we obtain\nLr= log [P(C|E00, Dr)P(E00)P(Dr|E00 )\nP(C|B, Dr)s,t P(Est )P(Dr|B, Est )P(B|Est)].\n(4)\nNote that the terms associated with event Bin (4) depend\non the error detection capability of the inner code. In the\nfollowing, we use the inner code of as an example to\nderive all the corresponding terms in (4) to obtain Lr. The\nproposed derivations can be generalized to other inner codes\nin a straightforward way. In , the inner code is an RS\ncode over GF(256) with nccode symbols and 2 parity-check\nsymbols. For simplicity, we assume that the oligo length nis\na multiple of 4 such that nc=n/4. This code has dmin = 3\nand can detect up to two errors over GF(256). Thus, output\nsequences with less than three code symbol errors can always\nbe detected by the inner code. Then, we can compute the\nprobabilities in (4) to obtain Lr.\nApparently, P(C|E00, Dr) = 1 and P(C|B, Dr) =\nP(C|B). The probability P(C|B)is the undetected error\nrate of the inner code under the condition that the sequence\nhas greater than two code symbol errors. Due to the exis-\ntence of the i-d errors, each error pattern can be considered\nas a random nc-tuple over GF(256) with weight greater\nthan two. The total number of such nc-tuples is given by\n256nc1255nc2552nc!\n(nc2)!2! , with 256nc2tuples\nforming the complete set of the inner codewords. For the case\nof DNA Fountain, we have nc2. Thus, the probability for\na random vector with weight greater than 2 to be a codeword\nis given by\nP(C|B, Dr) = 256nc2\n256nc1255nc2552nc!\n(nc2)!2!\n2562.(5)\nThen, we can compute P(Est)for various base error rates,\ngiven by\nP(Est) = n!\n(ns2t)!s!t!t!ps\nspt\nipt\nd(1 p)ns2t,(6)\nwhere ps,pi, and pdare the raw substitution, insertion, and\ndeletion error rates of the system, respectively, with p=ps+\npi+pd, and s, t 0.\nNext, we compute P(B|Est),P(Dr|E00), and\nP(Dr|B, Est )in (4). Note that P(B|Est)and P(Dr|B , Est)\ndepend on the error pattern associated with Est, and the error\npattern of the i-d error is related to the input sequence. As an\napproximation, we assume all sequenced bases affected by\nthe i-d errors are incorrect. Moreover, since the base errors\noccur rarely, e.g., one error per hundred bases or less\n, the probability of having more than three base errors\nis trivial. Therefore, we only consider three types of error\npatterns: E01 (1 i-d error), E11 (1 substitution error and 1\ni-d error), and E30 (3 substitution errors). Hence we have\nP(B|Est)1with {st}={01,11,30}.\nAccording to our proposed system model, P(Dr|E00)\nand P(Dr|B, Est )are probabilities with the corresponding\nsequences being sampled rtimes in the random sampling\nwith replacement process. They can be calculated based on\nthe number of samples and the population of the sampling\nassociated with P(Dr|E00)and P(Dr|B , Est), denoted by\nSst and Nst, respectively. Let Pr,st be the unified form of\nP(Dr|E00)and P(Dr|B , Est). We thus have\nPr,st =(Sst 1)!\n(Sst r)!(r1)! 1\nNst r111\nNst Sstr\n.\n(7)\nIn (7), the number of samples is given by Sst =S·\nP(Est). To determine Nst, we need to first compute the\nnumber of error patterns for all possible cases, denoted by\nn00,n01 ,n11, and n31 , respectively. For E00, the error pattern\nis always 0,i.e.,n00 = 1. For other cases, and by considering\neach substitution error or insertion error has three different\nerror patterns while each deletion error has one, we have\nn01 = 3 n!\n(n2)! nc\n4!\n2! (nc1)2!4!\n3!\n4!\n3!,\nn11 = 32n!\n(n3)! nc4! nc!\n(nc2)!\n4!\n3!\n4!\n2!\n(nc1)2!4!\n3!\n4!\n3!\n6!\n5!,\nn30 = 33n!\n(n3)!3! nc\n4!\n3! nc!\n(nc2)!2! 4!\n3!\n4!\n2!2!,\nWe can then obtain the population of each sampling given as\nNst =N·nst.\nAt this point, we have derived all the probabilities involved\nin (4) and thus the soft information Lrof each oligo can be\nobtained.\nIV. SOF T-DECISION DECODING FOR DNA-BAS ED DATA\nSTORAG E\nIn this section, we apply SDD to DNA-based data storage\nsystem and investigate its performance gain over HDD. In\nprinciple, all existing DNA storage systems with ECCs can\nuse the proposed LLR calculation to carry out SDD for more\nreliable data retrieval. As an example, we apply SDD to DNA\nFountain .\nIn , the source data is stored in n= 152 bases per oligo.\nEach oligo consists of 32 bytes of data payload, 4 bytes of\nseed for the PRNG of the LT code, and 2 bytes of parity-\ncheck symbols of a (38, 36) shortened RS code over GF(256).\nDuring the screening stage, sequences with homopolymer\nrun length greater than 3 or GC-content exceeds the range\nof [0.45, 0.55] are rejected. In the writing process, 67088\nsegments of binary message are encoded into 72000 base\nsequences, and thus multiple copies of 72000 unique oligos\nare synthesized. In the reading process, different number of\nreads, e.g., from 750000 to 32000000, are performed to\nevaluate the performance of DNA Fountain. For the ease of\nsimulations, we consider the case with 750000 reads in this\nwork.\nAn ILT, the traditional HDD of LT codes, is essentially\na simplified Gaussian elimination. It does not have error\ncorrection capability. That is, even if the ILT is successful,\nthe recovered messages may be in error. Recall that the LT\ncode is a binary linear block code with a random generator\nmatrix G= [P I], where Pis randomly generated with\ncolumn weight distribution following the RSD, and Iis an\nidentity matrix. We can then obtain its parity-check matrix\nH= [I PT]. Since the RSD produces a large number of\ndegree-2 nodes, His a sparse matrix. Therefore, the LT\ncode can be decoded directly by using the belief propagation\nalgorithm (BPA) of low-density parity-check (LDPC) codes\n, with the soft information derived in Section III-B.\nIn our simulations, as the raw error rate of the system is\nnot given in , we follow and to set the substitution,\ninsertion, and deletion error rates, respectively. In particular,\nbased on the error analyses in and , we can obtain\n0123456\nps10-3\n10-4\n10-3\n10-2\n10-1\nFER\nHDD\nSDD\npd=310-3\npd=3.2510-3\npd=3.510-3\npd=3.7510-3\npd=410-3\npd=4.2510-3\npi= pd / 5\nFig. 3. FER comparison of DNA Fountain with SDD and HDD.\nthe ranges of different types of raw error rates, i.e.,ps\n[6×104,4.5×103],pi[5.4×104,1×103], and pd\n[1.5×103,5×103]. Moreover, it has been observed that the\ndeletion error rate pdis approximately three to six times as\nmuch as the insertion error rate pi, and the substitution error\nrate psvaries, with the total raw error rate being in the range\nof [2 ×103,1×102]. Therefore, we set pd= 5piand\nvary the values of psin our simulations. In addition, based\non the supplementary materials of , the erasure rate of the\ninput sequence of the system is set to 5×103.\nMoreover, different from LDPC codes, the LT code in DNA\nFountain is nonsystematic, i.e., none of the information bits\nare written into oligos. Hence the soft information obtained\nfrom the DNA storage channel are only associated with the\nbit positions of Pin Gand correspondingly Iin H. In the\nsimulations the LLRs of oligos associated with Iin Hcan\nbe computed for each set of raw error rates according to (4),\nbased on their number of occurrence. The LLRs of all the\nother oligos are set to 0.\nFig. 3 illustrates the simulated frame error rate (FER)\nperformance of DNA fountain with SDD and HDD, respec-\ntively. Note that the FERs are evaluated for the frames with\nsufficient number of sequenced oligos such that the ILT can\nbe successfully carried out. It can be seen from Fig. 3 that\nSDD outperforms HDD with an FER reduction of two to\nthree orders of magnitude, over a wide range of substitution,\ninsertion, and deletion errors. This demonstrates the potential\nof the proposed SDD for improving system’s tolerance to\nvarious types of errors, and increasing the information density\nof DNA-based data storage system.\nV. CONCLUSION\nIn this paper, we have investigated, for the first time,\nthe SDD of ECCs for improving the error performance of\nDNA-based data storage. In particular, we have proposed\na simplified system model for the DNA-based data storage\nsystem through analyzing system’s major characteristics and\ndifferent types of errors. We have derived the error-free\nprobability of each sequenced oligo, based on which we\nobtain its LLR that enables SDD of ECCs. To demonstrate\nthe effectiveness of the proposed SDD, we have applied it\nto decode the LT code of DNA Fountain. Simulation results\nhave shown that for DNA Fountain, the proposed SDD can\neffectively improve system’s tolerance to various types of\nerrors, and it achieves an FER reduction of two to three orders\nof magnitude over HDD.\nACK NOW LE DG EM EN T\nThis work is supported by Singapore Ministry of Educa-\ntion Academic Research Fund Tier 2 MOE2016-T2-2-054,\nSUTD-ZJU grant ZJURP1500102, and SUTD SRG grant\nSRLS15095.\nREF ER EN CE S\n M. E. Allentoft, M. Collins, D. Harker, J. Haile, C. L. Oskam, M. L.\nHale, P. F. Campos, J. A. Samaniego, M. T. P. Gilbert, E. Willerslev,\nG. Zhang, R. P. Scofield, R. N. Holdaway, and M. Bunce, “The half-\nlife of DNA in bone: measuring decay kinetics in 158 dated fossils,\nProceedings of the Royal Society of London B: Biological Sciences,\n2012.\n C. Bancroft, T. Bowler, B. Bloom, and C. T. Clelland, “Long-term\nstorage of information in DNA,Science, vol. 293, no. 5536, pp. 1763–\n1765, 2001.\n M. Blawat, K. Gaedke, I. Huetter, X.-M. Chen, B. Turczyk, S. Inverso,\nB. W. Pruitt, and G. M. Church, “Forward error correction for DNA data\nstorage,” Procedia Computer Science, vol. 80, pp. 1011–1022, 2016.\n N. Goldman, P. Bertone, S. Chen, C. Dessimoz, E. M. LeProust,\nB. Sipos, and E. Birney, “Towards practical, high-capacity, low-\nmaintenance information storage in synthesized DNA,Nature, vol.\n494, no. 7435, pp. 77–80, 2013.\n J. Bornholt, R. Lopez, D. M. Carmean, L. Ceze, G. Seelig, and\nK. Strauss, “A DNA-based archival storage system,SIGPLAN Not.,\nvol. 51, no. 4, pp. 637–649, Mar. 2016.\n R. N. Grass, R. Heckel, M. Puddu, D. Paunescu, and W. J. Stark,\n“Robust chemical preservation of digital information on DNA in silica\nwith error-correcting codes,” Angewandte Chemie International Edition,\nvol. 54, no. 8, pp. 2552–2555, 2015.\n L. Organick, S. D. Ang, Y. J. Chen, R. Lopez, S. Yekhanin, K.\nMakarychev, M. Z. Racz, G. Kamath, P. Gopalan, B. Nguyen, C.\nTakahashi, S. Newman, H. Y. Parker, C. Rashtchian, G. G. K. Stewart,\nR. Carlson, J. Mulligan, D. Carmean, G. Seelig, L. Ceze, and K. Strauss,\n“Scaling up DNA data storage and random access retrieval,bioRxiv,\n2017.\n Y. Erlich and D. Zielinski, “DNA fountain enables a robust and efficient\nstorage architecture,” Science, vol. 355, no. 6328, pp. 950–954, 2017.\n S. Kosuri and G. M. Church, “Large-scale de novo DNA synthesis:\ntechnologies and applications,” Nature methods, vol. 11, no. 5, pp. 499–\n507, 2014.\n E. M. LeProust, B. J. Peck, K. Spirin, H. B. McCuen, B. Moore,\nE. Namsaraev, and M. H. Caruthers, “Synthesis of high-quality libraries\nof long (150mer) oligonucleotides by a novel depurination controlled\nprocess,” Nucleic acids research, vol. 38, no. 8, pp. 2522–2540, 2010.\n R. Heckel, G. Mikutis, and R. N. Grass, “A Characteriza-\ntion of the DNA Data Storage Channel.arXiv:1803.03322, http-\ns://arxiv.org/abs/1803.03322, 2018.\n M. Luby, “LT codes,” in IEEE Symp. Found. of Comp. Science, 2002,\npp. 271–280.\n W. E. Ryan and S. Lin, Channel codes: classical and modern. Cam-\nbridge University Press, 2009." ]

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[ "# Protected: WAEC 2018/2019 Physics Practical Obj and Theory/Essay Free Questions & Answers", null, "1axi)\n– *Precautions* –\nTake Any Two\ni ) I avoided conical oscillation\nii) I avoided drought\niii ) I avoided parallax error in taking reading from the metre rule\niv ) I avoided zero error in taking reading from the stop watch\n2aix)\n*PRECAUTIONS*\nI)  experiment must be carried out in a dark room\nII)  lens, screen and object must be in line\n1b i.\nfocal length of an object is 20means that the distancebetween the optical centre  or middle of the lens and its principal  focus is 20cm\n1bii.\ngiven that u=x   f=20   v=5\nusing the mirror formula\n1/u   +  1/v  = 1/F\n1/x + 1/5 = 1/20\nmultiply through by the LCM\n1/x *5  + 1/5 * 5 = 1/20 * 5\n5/x =5/20\n5/x =1/4\ntaking the reciprocal\nx/5 =4/1\ncross multiply\nx * 1 = 5 * 4\nx =20\n1 )\nTabulate\ns /n = 1 , 2 , 3 , 4 , 5\nL ( cm )= 130 , 110 , 90 , 70 , 50\nT 1 ( sec)= 45 . 70 , 42 . 00 , 38 . 00 , 33 . 10 , 28 . 70\nT = 2 . 29 , 2 . 10 , 1 . 90 , 1 . 66 , 1 , 44\nT ^ 2 = 5 . 22 , 4 . 41 , 3 . 61 , 2 . 74 , 2 . 06\nL = 100 , 80 , 60 , 40 , 20\n1 )\n– precautions –\nTake Any Two\ni ) I avoided conical oscillation\nii ) I avoided drought\niii ) I avoided parallax error in taking reading from the metre rule\niv ) I avoided zero error in taking reading from the stop watch\npls keep calm we will update\nthe rest soon\n============================\n3 )\nTabulate\n3 )\n– precautions –\ntake any two\ni ) I removed key when reading is not taking\nii ) I ensured tight connections at the terminals\niii ) I avoided parallax error when taking reading from the ammeter.\n============================\nPhysics practical waec 2018 answers\n2ai)\nfo=15cm\n2av)\na=60.00cm\nb=20.00cm\nHence L=a/b=60.00/20.00\nL=3\n2avi)\nTABULATE\nS/N:1,2,3,4,5\nb(cm):20.00,25.00,35.00,40.00\na(cm):60.00,37.50,30.00,26.25,24.00\nL=a/b:3.00,1.50,1.00,0.75,0.60\n2avii)\nSlope=Change in L/Change in a\n=(3-0.25)/(60-18.6)\n=2.75/41.4\n=0.006642\n2aviii)\nS^-1=1/S\n=(1/0.0066425cm)\nS=15.05\nS=15cm\n2aix)\nPRECAUTIONS\n-I ensured that all apparatus are in straight line\n-I avoided error due to parallax when reading the metre rule\n-I avoided zero error on the metre rule\n2bi)\nu=10cm\nf=15cm\nUsing 1/v+1/u=1/f\n1/f-1/u=1/v\n1/v=1/15-1/10\n1/v=(2-3)/30\n1/v=-1/30\nv=-30cm\nThe characteristics of image formed are:\n-It is virtual\n-It is enlarged and magnified ie twice or two times as big as the object m=2\n2bii)\nThe concave mirror mounted in its holder is moved to and fro in front of the screen until a sharp image of the cross wire of the ray box is formed on the screen adjacent of the object.The distance between the mirror and the screen was measured as 30cm since the radius of curvature r=2fo then half is distance\n2bi)\nhe characteristics of imaged formed are :\ni)It is virtual\nii) It is enlarged or magnified i.e. twice or two times bigger as the object(m=2)\nmr_oshy\nThanks Us later…..\nAlways subscrib to us\nCall: 09036880531\nWhatsApp: 08061116014" ]

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[ "", null, "", null, "", null, "", null, "", null, "", null, "", null, "Re: Can model parameters be global?\n\n• To: mathgroup at smc.vnet.net\n• Subject: [mg80261] Re: Can model parameters be global?\n• From: Neil Stewart <neil.stewart at warwick.ac.uk>\n• Date: Thu, 16 Aug 2007 04:41:54 -0400 (EDT)\n• Reply-to: Neil Stewart <neil.stewart at warwick.ac.uk>\n\n```A brief summary of responses to question about whether model parameters are\nbest implemented as local or global variables. (Original post below)\n\nDavid Bailey highlighted Needs[\"Units`\"] and Needs[\"PhysicalConstants`\"] and\nsuggested giving globals long names (e.g., SpeedOfLightInWater, not c) to\nminimise confusion. David explained that using LocalizeVariables -> False in\nManipulate is designed to save the final setting after manipulation, and\nthat this is a positive feature.\n\nBen suggested using Replace and ReplaceAll for parameters which sometimes\nchange:\n> a1=0 (* a1 never changes *)\n> case1:={a2->0} (* a2 varies across scenarios *)\n> case2:={a2->1}\n> f[a3_]:=a1+a2+a3 (* a3 varies continuously, e.g., in plots *)\n> f/.case1\n\nNasser Abbasi warned of his experience with globals and suggested keeping\nall parameters local. David Annetts also warned about globals and suggested\npolymorphism and defaults to make things easier:\n> Imagine you have a model that in the general case you've implemented, has\n> 5 parameters.\n>\n> ftn[a_, b_, c_, d_, e_]\n>\n> Often though, you don't need all 5 parameters. You can write\n>\n> ftn[a_, b_, c_, d_:default, e_:default]\n>\n> And use ftn[a, b, c] instead of ftn[a, b, c, d, e].\n>\n> Which is fine if d & e are used less often.\n>\n> You can also write\n>\n> ftn[a_, b_, d_] := ftn[a, b, c's default, d, e's default]\n>\n> If that's a more natural way of ordering the input parameters.\n\nThank you to everyone who replied!\n\nOriginal post:\n> When implementing a mathematical model in physics or psychology, for\n> example, how do other people deal with model parameters in Mathematica?\n> Would you represent the speed of light as a global variable or a local\n> variable. For example, would you use\n>\n> Energy[m_]:=m*c^2 (* c is a global variable *)\n>\n> or\n>\n> Energy[m_,c_]:=m*c^2 (* c is a local variable *)\n>\n> ?\n>\n> The first seems neater. But problems arise in psychology, my domain, where\n> the values of model parameters are unknown and are left as free\n> parameters,\n> adjusted to best-fit the data.\n>\n> Both local and global methods work well with optimisation. For example,\n>\n> NMinimize[Energy,{c}]\n> {0., {c -> 0.}}\n>\n> and\n>\n> NMinimize[Energy[1,c],{c}]\n> {0., {c -> 0.}}\n>\n> But the global variable solution does not work well with Manipulate.\n> For example,\n>\n> Manipulate[Dynamic[Energy], {c, 0, 1}, LocalizeVariables -> False]\n>\n> works, but looks a right mess and also results in c taking a value that\n> needs a Clear[c] before using other functions like NMinimize. On the other\n> hand the local variable version\n>\n> Manipulate[Energy[1, c], {c, 0, 1}]\n>\n> is nice and simple. But the local variable solution results in having to\n> pass all of the model parameters to the function. This is fine in this\n> trivial example, but becomes unwieldy when there are ten model parameters\n> and the model is defined using a set of functions. (A c-like struct could\n> help, but there does not seem to be a neat way to do this in Mathematica.)\n>\n> So what do other people do? I'd be really interested to hear.\n\n```\n\n• Prev by Date: Re: Evaluating a convolution integral in Mathematica\n• Next by Date: Re: Complexity explosion in linear solve\n• Previous by thread: Re: Can model parameters be global?\n• Next by thread: show residuals of circle fit" ]

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[ "# Implementation of IIR Filter in dsPIC 33EP - Scaling\n\nI have to implement an IIR controller in dsPIC 33EP (16-bit, fixed-point, two's complement, wrap-around, 40-bit accumulator, 32-bit multiply, 12-bit ADC) and I am using cascaded, Direct Form I second order sections (ordering up).\n\nNow, I am having trouble with the scaling procedure, since I'read a lot about $L^2$-norm, but can't quite figure out how to apply it properly.\n\nWhen I get the sos matrix and gain using MATLAB (by either using tf2sos, fdatool, or filterbuilder) using $L^2$ norm, the gain values are always 1 for all of the intermediate sections.\n\n• Is this correct?\n• Or do I need to scale the input for each section?\n\n(I have no experience implementing IIR systems and your help would mean a lot.)\n\nBen\n\nThe tf2sos function takes an input filter of order $N$, given by $H(z)=\\frac{\\sum_i^{N} b_iz^{-i}}{\\sum_i^{N} a_iz^{-i}}$ and returns coefficients for $N/2$ second-order filters $H_k(z)=\\frac{b_{k0}+b_{k1}z^{-1}+b_{k2}z^{-2}}{a_{k0}+a_{k1}z^{-1}+a_{k2}z^{-2}}$ and a gain $g$ such that\n$$H(z) = g \\prod_kH_k(z)$$\n• you do not need to adapt the gain between the filter stages. What comes out of the first stage is directly sent to the second stage. Then, after all stages were processed, you multiply the outcome of the last stage with $g$, what you receive from the tf2sos function." ]

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[ "## Conversion formula\n\nThe conversion factor from cubic meters to deciliters is 10000, which means that 1 cubic meter is equal to 10000 deciliters:\n\n1 m3 = 10000 dL\n\nTo convert 242 cubic meters into deciliters we have to multiply 242 by the conversion factor in order to get the volume amount from cubic meters to deciliters. We can also form a simple proportion to calculate the result:\n\n1 m3 → 10000 dL\n\n242 m3 → V(dL)\n\nSolve the above proportion to obtain the volume V in deciliters:\n\nV(dL) = 242 m3 × 10000 dL\n\nV(dL) = 2420000 dL\n\nThe final result is:\n\n242 m3 → 2420000 dL\n\nWe conclude that 242 cubic meters is equivalent to 2420000 deciliters:\n\n242 cubic meters = 2420000 deciliters\n\n## Alternative conversion\n\nWe can also convert by utilizing the inverse value of the conversion factor. In this case 1 deciliter is equal to 4.1322314049587E-7 × 242 cubic meters.\n\nAnother way is saying that 242 cubic meters is equal to 1 ÷ 4.1322314049587E-7 deciliters.\n\n## Approximate result\n\nFor practical purposes we can round our final result to an approximate numerical value. We can say that two hundred forty-two cubic meters is approximately two million four hundred twenty thousand deciliters:\n\n242 m3 ≅ 2420000 dL\n\nAn alternative is also that one deciliter is approximately zero times two hundred forty-two cubic meters.\n\n## Conversion table\n\n### cubic meters to deciliters chart\n\nFor quick reference purposes, below is the conversion table you can use to convert from cubic meters to deciliters\n\ncubic meters (m3) deciliters (dL)\n243 cubic meters 2430000 deciliters\n244 cubic meters 2440000 deciliters\n245 cubic meters 2450000 deciliters\n246 cubic meters 2460000 deciliters\n247 cubic meters 2470000 deciliters\n248 cubic meters 2480000 deciliters\n249 cubic meters 2490000 deciliters\n250 cubic meters 2500000 deciliters\n251 cubic meters 2510000 deciliters\n252 cubic meters 2520000 deciliters" ]

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[ "", null, "# Fisher Transform of MACD w/ Quantile Bands [Loxx]", null, "Fisher Transform of MACD w/ Quantile Bands is a Fisher Transform indicator with Quantile Bands that takes as it's source a MACD. The MACD has two different source inputs for fast and slow moving averages.\n\nWhat is Fisher Transform?\nThe Fisher Transform is a technical indicator created by John F. Ehlers that converts prices into a Gaussian normal distribution.\n\nThe indicator highlights when prices have moved to an extreme, based on recent prices. This may help in spotting turning points in the price of an asset. It also helps show the trend and isolate the price waves within a trend.\n\nWhat is Quantile Bands?\nIn statistics and the theory of probability, quantiles are cutpoints dividing the range of a probability distribution into contiguous intervals with equal probabilities, or dividing the observations in a sample in the same way. There is one less quantile than the number of groups created. Thus quartiles are the three cut points that will divide a dataset into four equal-size groups (cf. depicted example). Common quantiles have special names: for instance quartile, decile (creating 10 groups: see below for more). The groups created are termed halves, thirds, quarters, etc., though sometimes the terms for the quantile are used for the groups created, rather than for the cut points.\n\nq-Quantiles are values that partition a finite set of values into q subsets of (nearly) equal sizes. There are q − 1 of the q-quantiles, one for each integer k satisfying 0 < k < q. In some cases the value of a quantile may not be uniquely determined, as can be the case for the median (2-quantile) of a uniform probability distribution on a set of even size. Quantiles can also be applied to continuous distributions, providing a way to generalize rank statistics to continuous variables. When the cumulative distribution function of a random variable is known, the q-quantiles are the application of the quantile function (the inverse function of the cumulative distribution function) to the values {1/q, 2/q, …, (q − 1)/q}.\n\nWhat is MACD?\nMoving average convergence divergence ( MACD ) is a trend-following momentum indicator that shows the relationship between two moving averages of a security’s price. The MACD is calculated by subtracting the 26-period exponential moving average ( EMA ) from the 12-period EMA .\n\nIncluded:\n• Zero-line and signal cross options for bar coloring, signals, and alerts\n• Signals\n• Loxx's Expanded Source Types\n• 35+ moving average types" ]

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[ "# crot (3) - Linux Man Pages\n\ncrot.f -\n\n## SYNOPSIS\n\n### Functions/Subroutines\n\nsubroutine crot (N, CX, INCX, CY, INCY, C, S)\nCROT applies a plane rotation with real cosine and complex sine to a pair of complex vectors.\n\n## Function/Subroutine Documentation\n\n### subroutine crot (integerN, complex, dimension( * )CX, integerINCX, complex, dimension( * )CY, integerINCY, realC, complexS)\n\nCROT applies a plane rotation with real cosine and complex sine to a pair of complex vectors.\n\nPurpose:\n\n``` CROT applies a plane rotation, where the cos (C) is real and the\nsin (S) is complex, and the vectors CX and CY are complex.\n```\n\nParameters:\n\nN\n\n``` N is INTEGER\nThe number of elements in the vectors CX and CY.\n```\n\nCX\n\n``` CX is COMPLEX array, dimension (N)\nOn input, the vector X.\nOn output, CX is overwritten with C*X + S*Y.\n```\n\nINCX\n\n``` INCX is INTEGER\nThe increment between successive values of CY. INCX <> 0.\n```\n\nCY\n\n``` CY is COMPLEX array, dimension (N)\nOn input, the vector Y.\nOn output, CY is overwritten with -CONJG(S)*X + C*Y.\n```\n\nINCY\n\n``` INCY is INTEGER\nThe increment between successive values of CY. INCX <> 0.\n```\n\nC\n\n``` C is REAL\n```\n\nS\n\n``` S is COMPLEX\nC and S define a rotation\n[ C S ]\n[ -conjg(S) C ]\nwhere C*C + S*CONJG(S) = 1.0.\n```\n\nAuthor:\n\nUniv. of Tennessee\n\nUniv. of California Berkeley" ]

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[ "# Tip Calculator\n\nHow much is a 5 percent tip on \\$25.91?\n\nTIP:\n\\$ 0\nTOTAL:\n\\$ 0\n\nTIP PER PERSON:\n\\$ 0\nTOTAL PER PERSON:\n\\$ 0\n\n## How much is a 5 percent tip on \\$25.91? How to calculate this tip?\n\nAre you looking for the answer to this question: How much is a 5 percent tip on \\$25.91? Here is the answer.\n\nLet's see how to calculate a 5 percent tip when the amount to be paid is 25.91. Tip is a percentage, and a percentage is a number or ratio expressed as a fraction of 100. This means that a 5 percent tip can also be expressed as follows: 5/100 = 0.05 . To get the tip value for a \\$25.91 bill, the amount of the bill must be multiplied by 0.05, so the calculation is as follows:\n\n1. TIP = 25.91*5% = 25.91*0.05 = 1.2955\n\n2. TOTAL = 25.91+1.2955 = 27.2055\n\n3. Rounded to the nearest whole number: 27\n\nIf you want to know how to calculate the tip in your head in a few seconds, visit the Tip Calculator Home.\n\n## So what is a 5 percent tip on a \\$25.91? The answer is 1.3!\n\nOf course, it may happen that you do not pay the bill or the tip alone. A typical case is when you order a pizza with your friends and you want to split the amount of the order. For example, if you are three, you simply need to split the tip and the amount into three. In this example it means:\n\n1. Total amount rounded to the nearest whole number: 27\n\n2. Split into 3: 9\n\nSo in the example above, if the pizza order is to be split into 3, you’ll have to pay \\$9 . Of course, you can do these settings in Tip Calculator. You can split the tip and the total amount payable among the members of the company as you wish. So the TipCalc.net page basically serves as a Pizza Tip Calculator, as well.\n\n## Tip Calculator Examples (BILL: \\$25.91)\n\nHow much is a 5% tip on \\$25.91?\nHow much is a 10% tip on \\$25.91?\nHow much is a 15% tip on \\$25.91?\nHow much is a 20% tip on \\$25.91?\nHow much is a 25% tip on \\$25.91?\nHow much is a 30% tip on \\$25.91?" ]

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[ "Categories\n\n# Test Bank for Physics for Scientists and Engineers 1st Edition by Katz\n\nProduct Code: 222\nAvailability: In Stock\nPrice: \\$28.99\nQty:     - OR -", null, "0 reviews  |  Write a review\n\n## Test Bank for Physics for Scientists and Engineers 1st Edition by Katz\n\nDownload FREE Sample Here for Test Bank for Physics for Scientists and Engineers 1st Edition by Katz. Note : this is not a text book. File Format : PDF or Word\n\nPart I: CLASSICAL MECHANICS.\n\n1. Getting Started.\n\n2. One-Dimensional Motion.\n\n3. Vectors.\n\n4. Two- and Three-Dimensional Motion.\n\n5. Newton Laws of Motion.\n\n6. Applications of Newton Laws of Motion.\n\n7. Gravity.\n\n8. Conservation of Energy.\n\n9. Energy in Non-Isolated Systems.\n\n10. Systems of Particles and Conservation of Momentum.\n\n11. Collisions.\n\n12. Rotation I: Kinematics and Dynamics.\n\n13. Rotation II: A Conservation Approach.\n\nPart II: MECHANICS OF COMPLEX SYSTEMS.\n\n14. Static Equilibrium, Elasticity, and Fracture.\n\n15. Fluids.\n\n16. Oscillations.\n\n17. Traveling Waves.\n\n18. Superposition and Standing Waves.\n\n19. Temperature, Thermal Expansion, and Gas Laws.\n\n20. Kinetic Theory of Gases.\n\n21. Heat and the First Law of Thermodynamics.\n\n22. Entropy and the Second Law of Thermodynamics.\n\nPART III: ELECTRICITY.\n\n23. Electric Forces.\n\n24. Electric Fields.\n\n25. Gauss Law.\n\n26. Electric Potential.\n\n27. Capacitors and Batteries.\n\n28. Current and Resistance.\n\n29. Direct Current (DC) Circuits.\n\nPART IV: MAGNETISM.\n\n30. Magnetic Fields and Forces.\n\n31. Gauss Law for Magnetism and Ampère Law.\n\n33. Inductors and AC Circuits.\n\n34. Maxwell Equations and Electromagnetic Waves.\n\nPART V: LIGHT.\n\n35. Diffraction and Interference.\n\n36. Applications of the Wave Model.\n\n37. Reflection and Images Formed by Reflection.\n\n38. Refraction and Images Formed by Refraction.\n\nPART VI: 20TH CENTURY PHYSICS.\n\n39. Relativity.\n\n## Write a review\n\nYour Review: Note: HTML is not translated!", null, "" ]

[ null, "https://testbankdirect.eu/catalog/view/theme/default/image/stars-0.png", null, "https://testbankdirect.eu/index.php", null ]

[ "### How much is 1700 grams?\n\nConvert 1700 grams. How much does 1700 grams weigh? What is 1700 grams in other units? How big is 1700 grams? Convert 1700 grams to lbs, kg, mg, oz, grams, and stone. To calculate, enter your desired inputs, then click calculate. Some units are rounded.\n\n### Summary\n\nConvert 1700 grams to lbs, kg, mg, oz, grams, and stone.\n\n#### 1700 grams to Other Units\n\n 1700 grams equals 1700 grams 1700 grams equals 1.7 kg 1700 grams equals 1700000 mg\n 1700 grams equals 59.96578423 oz 1700 grams equals 3.747861514 lbs 1700 grams equals 0.2677043096 stone" ]

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[ "# Python lambdas and scoping\n\nGiven this snippet of code:\n\n``````funcs = []\nfor x in range(3):\nfuncs.append(lambda: x)\nprint [f() for f in funcs]\n``````\n\nI would expect it to print `[0, 1, 2]`, but instead it prints `[2, 2, 2]`. Is there something fundamental I'm missing about how lambdas work with scope?\n\nThis is a frequent question in Python. Basically the scoping is such that when `f()` is called, it will use the current value of `x`, not the value of `x` at the time the lambda is formed. There is a standard workaround:\n\n``````funcs = []\nfor x in range(10):\nfuncs.append(lambda x=x: x)\nprint [f() for f in funcs]\n``````\n\nThe use of `lambda x = x` retrieves and saves the current value of `x`.\n\nx is bound to the module-level x (which is left over from the for loop).\n\nA little clearer:\n\n``````funcs = []\n\nfor x in range(10):\nfuncs.append(lambda: x)\n\nx = 'Foo'\n\nprint [f() for f in funcs]\n\n# Prints ['Foo', 'Foo', 'Foo', 'Foo', 'Foo', 'Foo', 'Foo', 'Foo', 'Foo', 'Foo']\n``````\n\nYou know the answer: yes. ;) Take comfort, however, as this is a very common discovery for budding pythonistas. When you define a function or lambda that references variables not \"created\" inside that function, it creates a closure over the variables. The effect is that you get the value of the variable when calling the function, not the value at definition time. (You were expecting the latter.)\n\nThere are a few ways to deal with this. First is binding extra variables:\n\n``````funcs = []\nfor x in range(10):\nfuncs.append(lambda x=x: x)\nprint [f() for f in funcs]\n# [0, 1, 2, 3, 4, 5, 6, 7, 8, 9]\n``````\n\nThe second way is a little more formal:\n\n``````from functools import partial\nfuncs = []\nfor x in range(10):\nfuncs.append(partial(lambda x: x, x))\nprint [f() for f in funcs]\n# [0, 1, 2, 3, 4, 5, 6, 7, 8, 9]\n``````" ]

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[ "Want to share your content on R-bloggers? click here if you have a blog, or here if you don't.", null, "Binding vectors, matrices and data frames using `rbind` and `cbind` is a common R task. However, when dimensions or classes differ between the objects passed to these functions, errors or unexpected results are common as well. Sounds familiar? Time to practice!\n\nAnswers to the exercises are available here.\n\nExercise 1\nTry to create matrices from the vectors below, by binding them column-wise. First, without using R, write down whether binding the vectors to a matrix is actually possible; then the resulting matrix and its mode (e.g., character, numeric etc.). Finally check your answer using R.\na. `a <- 1:5 ; b <- 1:5`\nb. `a <- 1:5 ; b <- c('1', '2', '3', '4', '5')`\nc. `a <- 1:5 ; b <- 1:4; c <- 1:3`\n\nExercise 2\nRepeat exercise 1, binding vectors row-wise instead of column-wise while avoiding any row names.\n\nExercise 3\nBind the following matrices column-wise. First, without using R, write down whether binding the matrices is actually possible; then the resulting matrix and its mode (e.g., character, numeric etc.). Finally check your answer using R.\na. `a <- matrix(1:12, ncol=4); b <- matrix(21:35, ncol=5)`\nb. `a <- matrix(1:12, ncol=4); b <- matrix(21:35, ncol=3)`\nc. `a <- matrix(1:39, ncol=3); b <- matrix(LETTERS, ncol=2)`\n\nExercise 4\nBind the matrix `a <- matrix(1:1089, ncol=33)` to itself, column-wise, 20 times (i.e., resulting in a new matrix with 21*33 columns). Hint: Avoid using `cbind()` to obtain an efficient solution. Various solutions are possible. If yours is different from those shown on the solutions page, please post yours on that page as comment, so we can all benefit.\n\nExercise 5\nTry to create new data frames from the data frames below, by binding them column-wise. First, without using R, write down whether binding the data frames is actually possible; then the resulting data frame and the class of each column (e.g., integer, character, factor etc.). Finally check your answer using R.\na. `a <- data.frame(v1=1:5, v2=LETTERS[1:5]) ; b <- data.frame(var1=6:10, var2=LETTERS[6:10])`\nb. `a <- data.frame(v1=1:6, v2=LETTERS[1:6]) ; b <- data.frame(var1=6:10, var2=LETTERS[6:10])`\n\nExercise 6\nTry to create new data frames from the data frames below, by binding them row-wise. First, without using R, write down whether binding the data frames is actually possible; then the resulting data frame and the class of each column (e.g., integer, character, factor etc.). Finally check your answer using R, and explain any unexpected output.\na. `a <- data.frame(v1=1:5, v2=LETTERS[1:5]) ; b <- data.frame(v1=6:10, v2=LETTERS[6:10])`\nb. `a <- data.frame(v1=1:6, v2=LETTERS[1:6]) ; b <- data.frame(v2=6:10, v1=LETTERS[6:10])`\n\nExercise 7\na. Use `cbind()` to add vector `v3 <- 1:5` as a new variable to the data frame created in exercise 6b.\nb. Reorder the columns of this data frame, as follows: v1, v3, v2.\n\nExercise 8\nConsider again the matrices of exercise 3b. Use both `cbind()` and `rbind()` to bind both matrices column-wise, adding `NA` for empty cells.\n\nExercise 9\nConsider again the data frames of exercise 5b. Use both `cbind()` and `rbind()` to bind both matrices column-wise, adding `NA` for empty cells.\n\nImage: By Hella, Handdrawing 1995." ]

[ null, "https://i2.wp.com/r-exercises.com/wp-content/uploads/2016/02/Common-whipping-binding.png", null ]

[ "# nlarxPlot\n\nPlot nonlinearity of nonlinear ARX model\n\nSince R2023a\n\n## Syntax\n\n``nlarxPlot(model)``\n``nlarxPlot(model,color)``\n``nlarxPlot(model1,...,modelN)``\n``nlarxPlot(model1,color1...,modelN,colorN)``\n``nlarxPlot(___,'NumberofSamples',N)``\n\n## Description\n\nexample\n\n````nlarxPlot(model)` plots the nonlinearity of a nonlinear ARX model on a nonlinear ARX plot. The plot shows the nonlinearity for all outputs of the model as a function of its input regressors.```\n````nlarxPlot(model,color)` specifies the color to use.```\n````nlarxPlot(model1,...,modelN)` generates the plot for multiple models.```\n\nexample\n\n````nlarxPlot(model1,color1...,modelN,colorN)` specifies the color for each model. You do not need to specify the color for all models.```\n\nexample\n\n````nlarxPlot(___,'NumberofSamples',N)` specifies the number of samples to use to grid the regressor space on each axis. This syntax can include any of the input argument combinations in the previous syntaxes.```\n\n## Examples\n\ncollapse all\n\nEstimate a nonlinear ARX model and plot its nonlinearity.\n\n```load iddata1 model1 = nlarx(z1,[4 2 1],'idWaveletNetwork','nlr',[1:3]); nlarxPlot(model1)```", null, "In the plot window, you can choose:\n\n• The regressors to use on the plot axes, and specify the center points for the other regressors in the configuration panel. For multi-output models, each output is plotted separately.\n\n• The output to view from the drop-down list located at the top of the plot.\n\n```load iddata1 model1 = nlarx(z1,[4 2 1],'idwave','nlr',[1:3]); model2 = nlarx(z1,[4 2 1],'idSigmoidNetwork','nlr',[1:3]); nlarxPlot(model1,'b', model2, 'g')```", null, "```load iddata1 model1 = nlarx(z1,[4 2 1],idWaveletNetwork); model2 = nlarx(z1,[4 2 1],idSigmoidNetwork); nlarxPlot(model1,'b', model2, 'g','NumberofSamples',50)```", null, "## Input Arguments\n\ncollapse all\n\nEstimated nonlinear ARX model, specified as an `idnlarx` model object. Use `nlarx` to estimate the model.\n\nColor to use to plot the regressors, specified as one of the following:\n\n• Character vector of color name, specified as one of the following:\n\n• `'b'`\n\n• `'y'`\n\n• `'m'`\n\n• `'c'`\n\n• `'r'`\n\n• `'g'`\n\n• `'w'`\n\n• 3-element double vector of RGB values\n\nBy default, the colors are automatically chosen.\n\nData Types: `double` | `char`\n\nNumber of points used on the regressor axis to display the regressor samples, specified as a positive integer.\n\nData Types: `double`\n\ncollapse all\n\n### What is a Nonlinear ARX Plot?\n\nA nonlinear ARX plot displays the evaluated model nonlinearity for a chosen model output as a function of one or two model regressors. For a model `M`, the model nonlinearity (`M.Nonlinearity`) is a nonlinearity estimator function, such as `idWaveletNetwork`, `idSigmoidNetwork`, or `idTreePartition`, that uses model regressors as its inputs.", null, "To understand what is plotted, suppose that `{r1,r2,…,rN}` are the `N` regressors used by a nonlinear ARX model `M` with nonlinearity `nl` corresponding to a model output. You can use `getreg(M)` to view these regressors. The expression `Nonlin = evaluate(nl,[v1,v2,...,vN])` returns the model output for given values of these regressors, that is, `r1` = `v1`, `r2` = `v2`, ..., `rN` = `vN`. For plotting the nonlinearities, you select one or two of the `N` regressors, for example, `rsub = {r1,r4}`. The software varies the values of these regressors in a specified range, while fixing the value of the remaining regressors, and generates the plot of `Nonlin` vs. `rsub`. By default, the software sets the values of the remaining fixed regressors to their estimated means, but you can change these values. The regressor means are stored in the `Nonlinearity.Parameters.RegressorMean` property of the model.\n\nExamining a nonlinear ARX plot can help you gain insight into which regressors have the strongest effect on the model output. Understanding the relative importance of the regressors on the output can help you decide which regressors to include in the nonlinear function for that output. If the shape of the plot looks like a plane for all the chosen regressor values, then the model is probably linear in those regressors. In this case, you can remove the corresponding regressors from nonlinear block, and repeat the estimation.\n\nFurthermore, you can create several nonlinear models for the same data using different nonlinearity estimators, such a `idWaveletNetwork` network and `idTreePartition`, and then compare the nonlinear surfaces of these models. Agreement between plots for various models increases the confidence that these nonlinear models capture the true dynamics of the system." ]

[ null, "https://it.mathworks.com/help/examples/ident/win64/PlotNonlinearityOfANonlinearARXModelExample_01.png", null, "https://it.mathworks.com/help/examples/ident/win64/SpecifyLineStyleforMultipleModelsExample_01.png", null, "https://it.mathworks.com/help/examples/ident/win64/SpecifyNumberofSamplesExample_01.png", null, "https://it.mathworks.com/help/ident/ref/ex_nlarx1_nlarx_plot1.png", null ]

[ "Source code for pyro.distributions.transforms.sylvester\n\nimport torch\nimport torch.nn as nn\nfrom torch.distributions import constraints\n\nfrom pyro.distributions.transforms.householder import HouseholderFlow\nfrom pyro.distributions.torch_transform import TransformModule\nfrom pyro.distributions.util import copy_docs_from\n\n[docs]@copy_docs_from(TransformModule) class SylvesterFlow(HouseholderFlow): \"\"\" An implementation of Sylvester flow of the Householder variety (Van den Berg Et Al., 2018), :math:\\\\mathbf{y} = \\\\mathbf{x} + QR\\\\tanh(SQ^T\\\\mathbf{x}+\\\\mathbf{b}) where :math:\\\\mathbf{x} are the inputs, :math:\\\\mathbf{y} are the outputs, :math:R,S\\\\sim D\\\\times D are upper triangular matrices for input dimension :math:D, :math:Q\\\\sim D\\\\times D is an orthogonal matrix, and :math:\\\\mathbf{b}\\\\sim D is learnable bias term. Sylvester flow is a generalization of Planar flow. In the Householder type of Sylvester flow, the orthogonality of :math:Q is enforced by representing it as the product of Householder transformations Together with TransformedDistribution it provides a way to create richer variational approximations. Example usage: >>> base_dist = dist.Normal(torch.zeros(10), torch.ones(10)) >>> hsf = SylvesterFlow(10, count_transforms=4) >>> pyro.module(\"my_hsf\", hsf) # doctest: +SKIP >>> hsf_dist = dist.TransformedDistribution(base_dist, [hsf]) >>> hsf_dist.sample() # doctest: +SKIP tensor([-0.4071, -0.5030, 0.7924, -0.2366, -0.2387, -0.1417, 0.0868, 0.1389, -0.4629, 0.0986]) The inverse of this transform does not possess an analytical solution and is left unimplemented. However, the inverse is cached when the forward operation is called during sampling, and so samples drawn using Sylvester flow can be scored. References: Rianne van den Berg, Leonard Hasenclever, Jakub M. Tomczak, Max Welling. Sylvester Normalizing Flows for Variational Inference. In proceedings of The 34th Conference on Uncertainty in Artificial Intelligence (UAI 2018). \"\"\" domain = constraints.real codomain = constraints.real bijective = True event_dim = 1 def __init__(self, input_dim, count_transforms=1): super(SylvesterFlow, self).__init__(input_dim, count_transforms) # Create parameters for Sylvester transform self.R_dense = nn.Parameter(torch.Tensor(input_dim, input_dim)) self.S_dense = nn.Parameter(torch.Tensor(input_dim, input_dim)) self.R_diag = nn.Parameter(torch.Tensor(input_dim)) self.S_diag = nn.Parameter(torch.Tensor(input_dim)) self.b = nn.Parameter(torch.Tensor(input_dim)) # Register masks and indices triangular_mask = torch.triu(torch.ones(input_dim, input_dim), diagonal=1) self.register_buffer('triangular_mask', triangular_mask) self._cached_logDetJ = None self.tanh = nn.Tanh() self.reset_parameters2() # Derivative of hyperbolic tan\n[docs] def dtanh_dx(self, x): return 1. - self.tanh(x).pow(2)\n# Construct upper diagonal R matrix\n[docs] def R(self): return self.R_dense * self.triangular_mask + torch.diag(self.tanh(self.R_diag))\n# Construct upper diagonal S matrix\n[docs] def S(self): return self.S_dense * self.triangular_mask + torch.diag(self.tanh(self.S_diag))\n# Construct orthonomal matrix using Householder flow\n[docs] def Q(self, x): u = self.u() partial_Q = torch.eye(self.input_dim, dtype=x.dtype, layout=x.layout, device=x.device) - 2. * torch.ger(u, u) for idx in range(1, self.count_transforms): partial_Q = torch.matmul(partial_Q, torch.eye(self.input_dim) - 2. * torch.ger(u[idx], u[idx])) return partial_Q\n# Self.u_unnormed is initialized in parent class\n[docs] def reset_parameters2(self): for v in [self.b, self.R_diag, self.S_diag, self.R_dense, self.S_dense]: v.data.uniform_(-0.01, 0.01)\ndef _call(self, x): \"\"\" :param x: the input into the bijection :type x: torch.Tensor Invokes the bijection x=>y; in the prototypical context of a TransformedDistribution x is a sample from the base distribution (or the output of a previous flow) \"\"\" Q = self.Q(x) R = self.R() S = self.S() A = torch.matmul(Q, R) B = torch.matmul(S, Q.t()) preactivation = torch.matmul(x, B) + self.b y = x + torch.matmul(self.tanh(preactivation), A) self._cached_logDetJ = torch.log1p(self.dtanh_dx(preactivation) * R.diagonal() * S.diagonal() + 1e-8).sum(-1) return y def _inverse(self, y): \"\"\" :param y: the output of the bijection :type y: torch.Tensor Inverts y => x. As noted above, this implementation is incapable of inverting arbitrary values y; rather it assumes y is the result of a previously computed application of the bijector to some x (which was cached on the forward call) \"\"\" raise KeyError(\"SylvesterFlow expected to find key in intermediates cache but didn't\")\n[docs] def log_abs_det_jacobian(self, x, y): \"\"\" Calculates the elementwise determinant of the log jacobian \"\"\" return self._cached_logDetJ" ]

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[ "## Conversion formula\n\nThe conversion factor from hours to minutes is 60, which means that 1 hour is equal to 60 minutes:\n\n1 hr = 60 min\n\nTo convert 11.1 hours into minutes we have to multiply 11.1 by the conversion factor in order to get the time amount from hours to minutes. We can also form a simple proportion to calculate the result:\n\n1 hr → 60 min\n\n11.1 hr → T(min)\n\nSolve the above proportion to obtain the time T in minutes:\n\nT(min) = 11.1 hr × 60 min\n\nT(min) = 666 min\n\nThe final result is:\n\n11.1 hr → 666 min\n\nWe conclude that 11.1 hours is equivalent to 666 minutes:\n\n11.1 hours = 666 minutes", null, "## Alternative conversion\n\nWe can also convert by utilizing the inverse value of the conversion factor. In this case 1 minute is equal to 0.0015015015015015 × 11.1 hours.\n\nAnother way is saying that 11.1 hours is equal to 1 ÷ 0.0015015015015015 minutes.\n\n## Approximate result\n\nFor practical purposes we can round our final result to an approximate numerical value. We can say that eleven point one hours is approximately six hundred sixty-six minutes:\n\n11.1 hr ≅ 666 min\n\nAn alternative is also that one minute is approximately zero point zero zero two times eleven point one hours.\n\n## Conversion table\n\n### hours to minutes chart\n\nFor quick reference purposes, below is the conversion table you can use to convert from hours to minutes\n\nhours (hr) minutes (min)\n12.1 hours 726 minutes\n13.1 hours 786 minutes\n14.1 hours 846 minutes\n15.1 hours 906 minutes\n16.1 hours 966 minutes\n17.1 hours 1026 minutes\n18.1 hours 1086 minutes\n19.1 hours 1146 minutes\n20.1 hours 1206 minutes\n21.1 hours 1266 minutes" ]

[ null, "https://convertoctopus.com/images/11-1-hours-to-minutes", null ]

[ "# Use a table of integrals with forms involving the trigonometric functions to find the indefinite integral. $\\int \\cos ^{4} 3 ###### Question: Use a table of integrals with forms involving the trigonometric functions to find the indefinite integral.$\\int \\cos ^{4} 3 x d x", null, "#### Similar Solved Questions\n\n##### Problem 6. A flat strip of metal is located in the z - y plane. It...\nProblem 6. A flat strip of metal is located in the z - y plane. It extends infinitely in the ±y-directions, and from x =-0.01 to x = +0.01. It carries a total current of 50A in the +y-direction, distributed over the width so that the surface current density is 2500 A/m. (a) Use integration to...\n##### In the disease; incontinentia pigmenti, abnormal pigmentation is seen in the female of every generation and has unique patterns in each individual: This phenomenon is seen due to non-random inactivation of the autosomes in the somatic cells TrueFalsc\nIn the disease; incontinentia pigmenti, abnormal pigmentation is seen in the female of every generation and has unique patterns in each individual: This phenomenon is seen due to non-random inactivation of the autosomes in the somatic cells True Falsc...\n##### NuiniA revicwer listens \" the country sune Vestetn Sun KIKK ndio Hle notices Ihat the Songs sungs by cold beet. secin and hate Onc unfaithful mory utnterWe prent Jops. Kumcufact . Songs teuture 4Il thne themes 35 songs are about - Jopt 29 songs are about becr songs are about doga aud Oincnsongs are about Feer and womnen13 songs are about dogs and beer songs are about beer only songs are about dogs only omien only songs are ubout5ones are aboui somiething women a5 theme? How many How many s\nnuini A revicwer listens \" the country sune Vestetn Sun KIKK ndio Hle notices Ihat the Songs sungs by cold beet. secin and hate Onc unfaithful mory utnterWe prent Jops. Kumcu fact . Songs teuture 4Il thne themes 35 songs are about - Jopt 29 songs are about becr songs are about doga aud Oincn s...\n##### A) Show the entire mechanism, including the structures of reactants_ reagents, products electron pushing arrows and formal charges for the reaction Indicated below. (8 pts)CH;OH, H\na) Show the entire mechanism, including the structures of reactants_ reagents, products electron pushing arrows and formal charges for the reaction Indicated below. (8 pts) CH;OH, H...\n##### Question 8 A plant asset originally cost $64,000 and was estimated to have a$4,000 salvage...\nQuestion 8 A plant asset originally cost $64,000 and was estimated to have a$4,000 salvage value at the end of its 5-year useful life. If at the end of three years, the asset was sold for $12,000, and had accumulated depreciation recorded of$36,000, the company should recognize a on disposal in th...\n##### Find the work done by F =< 3 + 2xy, x2 3y2 > to move a particle along the curve r(t) = < etsint,etcost > from [O,pi]\nFind the work done by F =< 3 + 2xy, x2 3y2 > to move a particle along the curve r(t) = < etsint,etcost > from [O,pi]...\n##### The following is a Matlab exercise. Please provide the editor code in the answer below. Thank...\nThe following is a Matlab exercise. Please provide the editor code in the answer below. Thank You. Define then evaluate the following functions symbolically in MATLAB by defining symbolic functions. All numbers must be to 4 significant digits. 6. (s + 1)e-2s 2s+ 1+K @s21, K remains. X(s) Example...\n##### Question continued4r\" 3r\" + 9 lim I=0 5+_ 7rs\nQuestion continued 4r\" 3r\" + 9 lim I=0 5+_ 7rs...\n##### Question 20 4 pts Match the correct functional group present in each compound A-D. 10 K+...\nQuestion 20 4 pts Match the correct functional group present in each compound A-D. 10 K+ -Ot-Bu MCPBA MeOH Na-Cr20 A OH B D C-H13B1 H2O Structure A [ Choose] Structure B [Choose] Structure C [Choose] Structure D [ Choose]...\n##### Find the perimeter of each figure. Place your answers in the cross-number puzzle.(FIGURE CANNOT COPY)\nFind the perimeter of each figure. Place your answers in the cross-number puzzle. (FIGURE CANNOT COPY)...\n##### 1. Write an example ofa linear system with two equations and two variables that is in- consistent All the coefficients ofthe variables must be different from 0 in all equations_(ii) Write down your own explicit questions about the reading ready to bring up in class: You can also mention a specific example or definition in the lecture notes that is still not clear to you, so the professor can expand on those ideas. Be as specific as possible(iii) Reflection: Write one, two or maybe three sentence\n1. Write an example ofa linear system with two equations and two variables that is in- consistent All the coefficients ofthe variables must be different from 0 in all equations_ (ii) Write down your own explicit questions about the reading ready to bring up in class: You can also mention a specific ...\n1. You live in a world with a universal 8% required reserve rate for all depository institutions. Assume an international visitor deposits $20,000 in the bank. What will the change in the bank’s T-account look like immediately after the deposit? Assuming the bank wants no excess reserves, what... 5 answers ##### 6- Suppose you went on a vacation in a different country andreturning to your school. What is the level of measurement of thevariable: 'inflight meal options' ? 6- Suppose you went on a vacation in a different country and returning to your school. What is the level of measurement of the variable: 'inflight meal options' ?... 5 answers ##### Question 155 ptsList the sample space for flipping a fair coin 3 times and recording the outcome: Use the sample space to calculate the probability of exactly 3 tails_Not enough information0.3750.1250.5None of theseNextPrevious Question 15 5 pts List the sample space for flipping a fair coin 3 times and recording the outcome: Use the sample space to calculate the probability of exactly 3 tails_ Not enough information 0.375 0.125 0.5 None of these Next Previous... 5 answers ##### Broperly graph the fothouia be data_ Your \"faecied needs t0 include legend,and the uxis should properly Froper title_ For , pick any of the presentation papers posted on Blackboard examples of proper figures, under the lecture outlinesPlants have mechanism t0 convert the pigment violaxanthin t0 zeaxanthin You are asked {O determine the amOunt Of the pigment zeaxanthin plants exposed or not [0 high levels of light; control group receives 2OQ HEm of light (low light intensity) for 10 hours &a Broperly graph the fothouia be data_ Your \"faecied needs t0 include legend,and the uxis should properly Froper title_ For , pick any of the presentation papers posted on Blackboard examples of proper figures, under the lecture outlines Plants have mechanism t0 convert the pigment violaxanthin t... 5 answers ##### What Is 3 It ALL the magnitude h Requlred Informatlon 8 veloclty Magnitude V vector? of the chectah has vector 2 velocity sprinting vector y-components cheetah +33,4 DEPENDENT m/s and Vy ~26.3 MuLtI-PART 3 ProbLEM Assigh What Is 3 It ALL the magnitude h Requlred Informatlon 8 veloclty Magnitude V vector? of the chectah has vector 2 velocity sprinting vector y-components cheetah +33,4 DEPENDENT m/s and Vy ~26.3 MuLtI-PART 3 ProbLEM Assigh... 1 answer ##### Help Save & Exit Submit 5 attempts left Check my work Draw the mirror image of... Help Save & Exit Submit 5 attempts left Check my work Draw the mirror image of the following compound, and label the compound as chiral or a sure to answer all parts. OH erythrose (a simple sugar) draw structure... This compound is (select) .... 5 answers ##### Solve the problemSuppose that a cost-benefit model is given by f(x) = 3.2 100 -xwhere f(x) is the cost in thousands of dollars of removing X percent of a given pollutant: What is the vertical asymptote of the graph of this function? What does this suggest about the possibility of removing all of the pollutant? Explain your reasoning: T T I Paragraph Arial 3 (12pt) T 8 T T 0$ Kashups 46 CSSPath: p\nSolve the problem Suppose that a cost-benefit model is given by f(x) = 3.2 100 -x where f(x) is the cost in thousands of dollars of removing X percent of a given pollutant: What is the vertical asymptote of the graph of this function? What does this suggest about the possibility of removing all of t..." ]

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[ "# Chi Square Goodness of Fit Test in Excel\n\nThere are different types of Chi Square tests:\n\n### Chi Square Goodness of Fit Test\n\nA Chi Square Goodness of Fit test evaluates the probabilities of multiple outcomes.\n\n### Las Vegas Dice Chi Square Goodness of Fit Test Example\n\nLet's say you want to know a six-sided die is fair or unfair (Advanced Statistics by Dr. Larry Stephens). If the die is fair then each side will have an equal probability of coming up; if not, then one or more of the sides will come up more often. Now, test 120 rolls of the die and enter the data into Excel. We would expect each side of the die to come up 20 times (120/6):", null, "Then, in an empty cell, begin typing the formula \"=chitest(\" and Excel will prompt for the observed and expected ranges:", null, "Use your mouse to select the Observed (actual_range) and Expected range. Put a comma between the two and a parenthesis at the end and hit return.", null, "The Chi Square test will calculate the probability (i.e., p value) of all sides being equal:", null, "### Interpreting the Chi Square Goodness of Fit results\n\n• H0: p1 = p2 = p3 = p4 = p5 = p6 = 1/6\n• Ha : At least one p is not equal to 1/6\n If Then p value < a Reject the null hypothesis p value > a Cannot Reject the null hypothesis (Accept the null hypothesis)\n\nIn the above results the p value = .00012 which is much lower than our alpha value of 0.05, so, we can reject the null hypothesis that the die is fair.\n\n### Why Choose QI Macros Statistical Software for Excel?", null, "### Easy to Use\n\n• Works Right in Excel\n• Interprets p-values for You\n• Accurate No-Worry Results\n• Free Training Anytime", null, "### Proven and Trusted\n\n• 100,000 Users in 80 Countries\n• Celebrating 20th Anniversary\n• Five Star CNET Rating - Virus Free", null, "### Affordable\n\n• Only \\$349 USD\nQuantity Discounts Available\n• No annual fees\n• Free Technical Support" ]

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[ "# lower\n\n1. ### Sumifs lower than a given day\n\nHello I'm trying to do a sumifs formula based on 2 criterias. The first is a name. the second one I can't do is when the date DAY is lower than 15. For example in the table below I want to get the sum of A only lower than 15, the result should be 2.000 I was trying something (hardcoded just...\n2. ### Formula Help - If Cell Falls Within Certain Range\n\nHi! So sorry if a question like this has been posed before, I'm sure it will have been covered at some stage .. I'm looking to create a formula within a work spreadsheet as follows; If the value of a cell is lower than 5 then new cell is A, if it is between 5 and 9.9 then B, if it is between 10...\n3. ### Excel Line graph\n\nI am looking to plot a line graph with upper and lower limits (Ranges). Attached is the target view (PNG). The value that I have, has positive and negative values. Hence showing upper limit and lower limit range in the line chart is a bit complex. Attached is the excel sheet(Sample.xlsx) with...\n\nHello, In the formula below how do i add IF 7% or lower make 0%? =IFERROR(VLOOKUP(B\\$1,OFFSET(INDEX(\\$K\\$2:\\$K\\$900,MATCH(\\$A2,\\$K\\$2:\\$K\\$900,0)),,,MATCH(\"*\",INDEX(\\$K\\$2:\\$K\\$900,1+MATCH(\\$A2,\\$K\\$2:\\$K\\$900,0)):\\$K\\$900,0),2),2,0),\"\")\n5. ### Display lower of two values\n\nI need to adjust this formula C21+(\\$I3+(\\$I3*\\$D11)) so itwill either display the results of this calculation OR the value in J3, whicheveris lower. Thanks in advance\n6. ### Minimum Value of 1.00 in Cell\n\nI have a cell that does not have a formula, but i need it to have a minimum value of 1.00 in it. it can have larger number in it, but no lower than 1.00, it must always have a value in it. Thanks for the help Chad\n7. ### Assign numbers in block of 10 starting from lower to higher\n\nHello,</SPAN></SPAN> In the column \"D\" I got the numbers disordered, they can be from 0 to 3000 max. </SPAN></SPAN> I need to assign them in column \"E\" in blocks of 10 (1-10, 11-20...etc.) starting from lower to higher as per example shown below.</SPAN></SPAN> <COLGROUP><COL...\n8. ### Inserting Row(s) based on cell Value on same sheet\n\nHello, First time post/reply. I'm a pro at recording macros :wink: but not so much on writing them. I'm trying to insert Row(s) based on cell value(s) that reside within my sheet. Cell A3 has a value of N = 3 and I'd like to insert N rows at row 5 (two rows lower than 3). Then at Cell A6 which...\n9. ### Closest Value\n\nHi, Please help me, i need to get the closest number from the value from the Lower/Higher column. Let say value 5, the closest is the Lower number which is 7. Thanks <colgroup><col><col><col><col></colgroup><tbody> Value Lower Higher Result 5 7 11 7 7 9 13 9 4 5 9 4 15 17 21 17 32...\n10. ### Conditional Formatting for 20,000 rows\n\nHello all and Happy New Year, I have 20k rows of data where I need to highlight cells E thru I if a condition in column R = \"LOWER\". I tried the conditional format but it looks like it has a limit to how many rows are used. Any help would be greatly appreciated.\n11. ### Conditional format\n\nI simply want to conditionally format a number in one cell if it is at least 2 times (or More) lower than the cell to the left of it. Here are my exact layout grid numbers. What would be the exact formula for this? So K7 is 2x's LOWER than J7. I want to format it a certain other color ...\n12. ### If when blank then additional formulas\n\n<tbody> 12/10/2018 12/9/2018 =IF(A1=\"\",\"\",(IF(B1>A1,\"NO\",\"YES\"))) NO 12/3/2018 12/2/2018 =IF(A2=\"\",\"\",(IF(B2>A2,\"NO\",\"YES\"))) YES </tbody> I am currently trying to make these two cells both say \"NO\". I cannot find where the formula stops working. This is the formula now...\n13. ### How to create an array of outliers from a data set?\n\nI already calculated the upper and lower fence for my data set, how do I create an array that has every outlier (lower than lower fence or greater than upper fence)? :confused::confused:\n14. ### Statistical question - confidence intervals\n\nI have some data I have inherited from several years back and no-one who was around back then is still around to ask. Year Question # Lower 95% confidence interval Scored percentage Upper 95% confidence interval No of responses for this question and trust 2014...\n15. ### Tabular buttons\n\n2010 Excel I'm using tabular buttons to raise or lower the value of a cell. The problem is the upper button (pointing up) lowers the value, while the lower button (pointing down) raises the value. How can I reverse these so the upper button raises and the lower button lowers the value?\n16. ### Picking higher value of two cells and bringing back text from adjacent cell\n\nYou guys have helped me a couple of times recently. I hugely appreciate it but I'm stuck again. Apologies. So P2 is John and S2 is David in the next cells along there is a figure so Q2 is 4.5 and T2 is -4.5, I need a formula to look at the values in Q2 and T2 and know that T2 is the lowest...\n17. ### Conditional formatting using vba\n\nHi , So I created an automated summary sheet which extracts data from multiple sheets and puts it in one workbook. I need to do conditional formatting on the data in summary sheet based on certain parameters Base Upper Lower PROD A PROD B PROD C PROD D PROD E 2 4 5 6.1 6.3...\n18. ### Formatting for a number Double Lower than all others!\n\nJust a quick qyestion which I'm sure there is a easy answer to but I don't have anyway. I want to conditionally format a number in a row of numbers (See Example) that is at least Double Lower or more than the rest of the numbers. So in my example the number 1340 would be highlighted in...\n19. ### If row is less than last row\n\nHi guys Struggling with a little problem here, should be really simple you'd image, but danged if I can get it to work:- this below, is basically the idea I need to happen. If column A suddenly goes lower than the previous value, increment column b by 1. Can't use scripts. 1 | 0 2 | 0 3...\n20. ### Conditional formatting\n\nI have a set of data (5 colums wide 50+ rows down) containing both numbers and text. I would like to highlight all numbers greater than 5 in red and all lower than 5 in green. The issue I’m having is it’s highlighting all the cells containing text red, is there a way to ignore the text?" ]

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[ "# What would be a good way to implement index notation?\n\nSo I need to calculate a 3x3 rotation operator (basically a matrix). I have an index-notation expression, and I was wondering if with pre-calculated theta, K1, K2, and K3, could Mathematica compute each cell of this matrix for me?\n\nThe rotation operator M is defined as such, for i, j, and k from 1 to 3.\n\nM(i,j) = K(i) * K(j) + cos(theta) * (kronecker_delta(i,j) - K(i)*K(j)) +\nsin(theta) * (levi_civita_tensor(i,k,j) * K(k))\n\n\nI tried something new:\n\nn = {0, 0, 1}\nmat = Table[\nn[[i]]*n[[j]] +\nCos[Theta] * (KroneckerDelta[i, j] - n[[i]]*n[[j]]) +\nSin[Theta] * LeviCivitaTensor[[i, k, j]] * n[[k]],\n{i, 3}, {j, 3}, {k, 3}] (*//MatrixForm*)\n\n\nThe following output is pretty distinctly not a 3 * 3 matrix:\n\n{{{Cos[Theta], Cos[Theta], Cos[Theta]}, {0, 0, -Sin[Theta]}, {0, 0,\n0}}, {{0, 0, Sin[Theta]}, {Cos[Theta], Cos[Theta], Cos[Theta]}, {0,\n0, 0}}, {{0, 0, 0}, {0, 0, 0}, {1, 1, 1}}}\n\n\nHow do I fix this? I think it's iterating through k as if k were indices in the matrix.\n\nEdit to the edit:\n\nI tried summing k over the dimensions i needed, for some reason it's still being taken as a 3d matrix.\n\ndimensions = 3\nn = {0, 0, 1}\nmat = Table[\nn[[i]]*n[[j]] +\nCos[Theta] * (KroneckerDelta[i, j] - n[[i]]*n[[j]]) +\nSum[Sin[Theta] * LeviCivitaTensor[[i, k, j]] * n[[k]], {k, 1, dimensions}],\n{i, dimensions}, {j, dimensions}, {k, dimensions}] // MatrixForm\n\n• Consider giving a Working Minimal Example. – anderstood Sep 30 '15 at 20:43\n• I have zero idea how to even start is the issue.... – laudiacay Sep 30 '15 at 20:44\n• At the very least, write your definition in something Mathematica-like syntax – Dr. belisarius Sep 30 '15 at 20:48\n• In mat, the Table has three indices i,j,k resulting in a 3 by 3 by 3 table. It should have only two indices. – anderstood Sep 30 '15 at 21:09\n• It should be adding in each iteration of k, as if it was adding a sum over k of that third term. – laudiacay Sep 30 '15 at 21:10\n\nYou defined the function as if Mathematica was using Einstein summation convention. Make the summation on $k$ explicit:\n\nn = {0, 0, 1}\nmat = Table[\nn[[i]]*n[[j]] + Cos[Theta]*(KroneckerDelta[i, j] - n[[i]]*n[[j]]) +\nSin[Theta]*\nSum[LeviCivitaTensor[[i, k, j]]*n[[k]], {k, 1, 3}], {i,\n3}, {j, 3}] // MatrixForm\n\n\nwhich gives you a rotation matrix.\n\nAs it turns out, I was getting a 3x3x3 because it was working as if I was calculating the kth index of a matrix rather than just finding the sum of an expression with index notation. Adding a sum in there with the Tensor term worked this out fine.\n\nClearAll[dimensions, n, mat, i, j, k]\ndimensions = 2\nn = {1, 0}\nmat = Table[\nn[[i]]*n[[j]] +\nCos[Theta] * (KroneckerDelta[i, j] - n[[i]]*n[[j]]) +\nSum[Sin[Theta] * LeviCivitaTensor[[i, k, j]] * n[[k]], {k, 1,\ndimensions}], {i, dimensions}, {j, dimensions}] // MatrixForm\n\nn = {0, 0, 1}\nmat = Outer[Times, n, n]\n+ Cos[θ] (IdentityMatrix - Outer[Times, n, n])\n+ Sin[θ] Transpose[LeviCivitaTensor, {1, 3, 2}].n;\nmat // MatrixForm", null, "" ]

[ null, "https://i.stack.imgur.com/ozjGZ.png", null ]

[ "# Normal p1 p2\n\nNew in Analytica 6.0\n\n## Normal_p1_p2( q1, q2, p1, p2 )\n\nThis distribution function found in the Distribution Variations library defines a Normal distribution given any two percentile estimates. «q1» and «q2» are the percentile estimates, and «p1» and «p2» specify which percentiles these are. It must be that $\\displaystyle{ q1\\lt q2 }$ and $\\displaystyle{ p1\\lt p2 }$.\n\nFor example, given that the 5% percentile is 5 and the 25% percentile is 8, the distribution is\n\nNormal_p1_p2( 5, 8, 5%, 25% ) →", null, "As with other distribution functions, you can also use this distribution in the function Random.\n\n## Library\n\nTo use this function, you must first use Add Library to add the Distribution Variations Library to your model." ]

[ null, "https://docs.analytica.com/images/4/47/Normal_p1_p2_ex1.png", null ]

[ "# BEERS LAMBERTS LAW PDF\n\nAn explanation of the Beer-Lambert Law, and the terms absorbance and molar absorptivity (molar absorption coefficient). Beer-Lambert Law. Introduction. The Beer-Lambert law (or Beer’s law) is the linear relationship between absorbance and concentration of an absorbing species. Now let us look at the Beer-Lambert law and explore it’s significance. This is important because people who use the law often don’t understand it – even though.", null, "Author: Faugul Yojin Country: Suriname Language: English (Spanish) Genre: Spiritual Published (Last): 25 May 2016 Pages: 97 PDF File Size: 14.69 Mb ePub File Size: 10.57 Mb ISBN: 128-3-17185-475-8 Downloads: 24951 Price: Free* [*Free Regsitration Required] Uploader: Sagrel", null, "However, the actual molar absorbtivity value is 20 L mol -1 cm -1! We look at the way in which the intensity of the light radiant power changes as it passes through the solution in a 1 cm cuvette.\n\nBoth concentration and solution length are allowed for in the Beer-Lambert Law. When working in concentration units of molarity, the Beer-Lambert law is written as:. In case of uniform attenuation, these relations become . The Beer-Lambert Law You will find that various different symbols are given lwa some of the terms in the equation – particularly for the concentration and the solution length.\n\n### NMSU: Beer’s Law\n\nAlthough, in fact, the nm absorption peak is outside the range of most spectrometers. Absorption takes place and lamnerts beam of radiation leaving the sample has radiant power P. Retrieved from ” https: From Wikipedia, the free encyclopedia. The bright blue colour is seen because the concentration of the solution is very high.\n\n## Lambert-Beer’s law\n\nThis useful when the molecular weight of the solute is unknown bewrs uncertain. You should also understand the importance of molar absorbtivityand how this affects the limit of detection of a particular compound. Retrieved from ” https: The ethanal obviously absorbs much more strongly at nm than it does at nm.", null, "The absorption coefficient of a glycogen-iodine complex is 0. Contributors Jim Clark Chemguide. The Beer—Lambert law is not compatible with Maxwell’s equations. Views Read Edit View history. Essentially, it works out a value for what the absorbance would be under a standard set of conditions – the light traveling 1 cm through a solution of 1 mol dm That’s quite common since it assumes the length is in cm and the concentration is mol dm -3the units are mol -1 dm 3 cm Molar absorptivity compensates for this by dividing by both the concentration and the length of the solution that the light passes through.\n\nBBMP BYE LAWS PDF\n\nIn uv spectroscopy, the concentration of the sample solution is measured in molL -1 and the length of the light path in cm.", null, "Annalen der Physik und Chemie. This page has been accessedtimes. If it is in a reasonably concentrated solution, it will have a very high absorbance because there are lots of molecules to interact with the light.\n\nThus, given that absorbance is unitless, the units of molar absorptivity are L mol lambers cm On the other hand, suppose you passed the light through a tube cm long containing the same solution.", null, "This is important because people who use the law often don’t understand it – even though the equation representing the law is so straightforward:. The solution to this differential equation is obtained by multiplying the integrating factor. The Law says bees the fraction of the light absorbed by each layer of solution is the same. Notice that there are no units given for absorptivity.\n\n## Beer–Lambert law\n\nThis law is also applied to describe the attenuation of solar or stellar radiation as it travels through the atmosphere. Cases of non-uniform attenuation occur in atmospheric science applications and radiation shielding theory for instance.\n\nIf all the light is absorbed, then percent transmittance is zero, and absorption is infinite. Now, suppose we have a solution of copper sulphate which appears blue because it has almberts absorption maximum at nm. Claude Jombert, pp. It can simply obtained by multiplying the absorption coefficient by the molecular weight. An unknown concentration of an analyte can be determined by measuring the amount of light that a sample absorbs and applying Beer’s law.\n\nAUTOHIPNOSIS ENTRENE SU MENTE PDF\n\nIn practice it is better to use linear least squares to determine the two amount concentrations from measurements made at more than two wavelengths.\n\nla,berts At high concentrations, the molecules are closer to each other and begin to interact with each other. We will look at the reduction every 0. The Greek letter epsilon in these equations is called the molar absorptivity – or sometimes the molar absorption coefficient. The larger the molar absorptivity, the more probable the lambrrts transition.\n\nThe answer is now obvious – a compound with a high molar absorbtivity is very effective at absorbing light of the appropriate wavelengthand hence low concentrations of a compound with a high molar absorbtivity can be easily detected.\n\n### Beer–Lambert law – Wikipedia\n\nAn absorbance of 0 at some wavelength means that no light of that particular wavelength has been absorbed.\n\nBfers, the wavelength of maximum absorption by a substance is one of the characteristic properties of that material. Transmittance for liquids is usually written as: Later, Beer extended in the law to include the concentration of solutions, thus giving the law its name Beer-Lambert Law.\n\nThe Beer-Lambert law maintains linearity under specific conditions only. The latter is particularly convenient. The reason why we prefer to express the law with this equation is because absorbance is directly proportional to the other parameters, as long as the law is obeyed. That makes it possible to plot both values easily, but produces strangely squashed-looking spectra!\n\nFor each wavelength of light passing through the spectrometer, the intensity of the light passing through the reference cell alw measured. The amount concentration c is lamberhs given by." ]

[ null, "http://www.pci.tu-bs.de/aggericke/PC4/Kap_I/beerslaw1.gif", null, "https://americaninvestigator.net/download_pdf.png", null, "https://image.slidesharecdn.com/beerlambert-140929110335-phpapp02/95/beer-lambert-law-4-638.jpg", null, "http://life.nthu.edu.tw/~labcjw/BioPhyChem/Spectroscopy/pics/beersla1.gif", null, "https://www.chemguide.co.uk/analysis/uvvisible/beerlambert1.gif", null ]

[ "# Definition of plot\n\n## \"plot\" in the noun sense\n\n### 1. plot, secret plan, game\n\na secret scheme to do something (especially something underhand or illegal\n\n\"they concocted a plot to discredit the governor\"\n\n\"I saw through his little game from the start\"\n\n### 2. plot, plot of land, plot of ground, patch\n\na small area of ground covered by specific vegetation\n\n\"a bean plot\"\n\n\"a cabbage patch\"\n\n\"a briar patch\"\n\n### 3. plot\n\nthe story that is told in a novel or play or movie etc.\n\n\"the characters were well drawn but the plot was banal\"\n\n### 4. plot\n\na chart or graph showing the movements or progress of an object\n\n## \"plot\" in the verb sense\n\n### 1. plot\n\nplan secretly, usually something illegal\n\n\"They plotted the overthrow of the government\"\n\n### 2. diagram, plot\n\nmake a schematic or technical drawing of that shows interactions among variables or how something is constructed\n\nmake a plat of\n\n\"Plat the town\"\n\n### 4. plot\n\ndevise the sequence of events in (a literary work or a play, movie, or ballet\n\n\"the writer is plotting a new novel\"\n\nSource: WordNet® (An amazing lexical database of English)\n\nWordNet®. Princeton University. 2010.\n\n# Quotations for plot\n\nA mischievous plot may produce a good end. [ Proverb ]\n\nWhen we mean to build,\nWe first survey the plot, then draw the model;\nAnd when we see the figure of the house,\nThen must we rate the cost of the erection;\nWhich if we find outweighs ability.\nWhat do we then, but draw anew the model\nIn fewer offices; or, at least, desist\nTo build at all? [ William Shakespeare ]\n\nA tragical plot may produce a comical conclusion. [ Proverb ]\n\nFriendship between two women is always a plot against another one. [ A. Karr ]\n\nA little plot of ground thick sown is better than a great field which, for the most part of it, lies fallow. [ Bishop Norris ]\n\nIf often happens too, both in courts and in cabinets, that there are two things going on together - a main plot and an underplot; and he that understands only one of them will, in all probability, be the dupe of both. A mistress may rule a monarch, but some obscure favorite may rule the mistress. [ Colton ]\n\n# plot in Scrabble®\n\nThe word plot is playable in Scrabble®, no blanks required.\n\nPLOT\n(27)\n\nPLOT\n(27)\nPLOT\n(21)\nPLOT\n(18)\nPLOT\n(18)\nPLOT\n(18)\nPLOT\n(18)\nPLOT\n(18)\nPLOT\n(14)\nPLOT\n(12)\nPLOT\n(12)\nPLOT\n(12)\nPLOT\n(12)\nPLOT\n(12)\nPLOT\n(10)\nPLOT\n(9)\nPLOT\n(8)\nPLOT\n(8)\nPLOT\n(8)\nPLOT\n(8)\nPLOT\n(7)\nPLOT\n(7)\nPLOT\n(7)\nPLOT\n(6)\n\nPLOT\n(27)\nPLOT\n(21)\nPLOT\n(18)\nPLOT\n(18)\nPLOT\n(18)\nPLOT\n(18)\nPLOT\n(18)\nOPT\n(15)\nPOT\n(15)\nOPT\n(15)\nPOT\n(15)\nOPT\n(15)\nTOP\n(15)\nTOP\n(15)\nLOP\n(15)\nPOT\n(15)\nTOP\n(15)\nLOP\n(15)\nLOP\n(15)\nPLOT\n(14)\nPLOT\n(12)\nPLOT\n(12)\nPLOT\n(12)\nPLOT\n(12)\nPLOT\n(12)\nTOP\n(11)\nPOT\n(11)\nLOP\n(11)\nOPT\n(11)\nLOP\n(10)\nTOP\n(10)\nLOP\n(10)\nOPT\n(10)\nPLOT\n(10)\nPOT\n(10)\nOPT\n(10)\nTOP\n(10)\nLOP\n(10)\nOPT\n(10)\nPOT\n(10)\nPOT\n(10)\nTOP\n(10)\nTOP\n(9)\nLOP\n(9)\nPOT\n(9)\nLOT\n(9)\nLOT\n(9)\nLOT\n(9)\nPLOT\n(9)\nPLOT\n(8)\nPOT\n(8)\nPLOT\n(8)\nTOP\n(8)\nOPT\n(8)\nLOP\n(8)\nPLOT\n(8)\nPLOT\n(8)\nTOP\n(7)\nTOP\n(7)\nPOT\n(7)\nPOT\n(7)\nPLOT\n(7)\nPLOT\n(7)\nPLOT\n(7)\nOPT\n(7)\nOPT\n(7)\nOPT\n(7)\nLOP\n(7)\nLOP\n(7)\nTOP\n(6)\nTO\n(6)\nLOT\n(6)\nTOP\n(6)\nLOT\n(6)\nLOT\n(6)\nLOP\n(6)\nTO\n(6)\nOPT\n(6)\nPOT\n(6)\nOPT\n(6)\nPOT\n(6)\nLOP\n(6)\nPLOT\n(6)\nLOT\n(5)\nLOT\n(5)\nTOP\n(5)\nLOP\n(5)\nLOT\n(5)\nLOT\n(5)\nOPT\n(5)\nPOT\n(5)\nTO\n(4)\nTO\n(4)\nLOT\n(4)\nTO\n(4)\nLOT\n(4)\nLOT\n(4)\nTO\n(4)\nTO\n(3)\nTO\n(3)\nLOT\n(3)\nTO\n(2)\n\n# plot in Words With Friends™\n\nThe word plot is playable in Words With Friends™, no blanks required.\n\nPLOT\n(48)\n\nPLOT\n(48)\nPLOT\n(30)\nPLOT\n(24)\nPLOT\n(24)\nPLOT\n(24)\nPLOT\n(24)\nPLOT\n(24)\nPLOT\n(18)\nPLOT\n(18)\nPLOT\n(16)\nPLOT\n(16)\nPLOT\n(16)\nPLOT\n(16)\nPLOT\n(16)\nPLOT\n(14)\nPLOT\n(13)\nPLOT\n(13)\nPLOT\n(12)\nPLOT\n(12)\nPLOT\n(11)\nPLOT\n(10)\nPLOT\n(10)\nPLOT\n(10)\nPLOT\n(9)\nPLOT\n(9)\nPLOT\n(8)\n\nPLOT\n(48)\nPLOT\n(30)\nPLOT\n(24)\nPLOT\n(24)\nPLOT\n(24)\nPLOT\n(24)\nPLOT\n(24)\nLOP\n(21)\nLOP\n(21)\nLOP\n(21)\nLOP\n(19)\nOPT\n(18)\nTOP\n(18)\nPLOT\n(18)\nOPT\n(18)\nPLOT\n(18)\nPOT\n(18)\nOPT\n(18)\nPOT\n(18)\nPOT\n(18)\nTOP\n(18)\nTOP\n(18)\nPOT\n(16)\nPLOT\n(16)\nPLOT\n(16)\nPLOT\n(16)\nTOP\n(16)\nPLOT\n(16)\nPLOT\n(16)\nLOP\n(15)\nLOP\n(14)\nLOP\n(14)\nLOP\n(14)\nOPT\n(14)\nPLOT\n(14)\nPOT\n(14)\nTOP\n(14)\nPLOT\n(13)\nLOP\n(13)\nPLOT\n(13)\nPLOT\n(12)\nLOT\n(12)\nPLOT\n(12)\nTOP\n(12)\nPOT\n(12)\nTOP\n(12)\nOPT\n(12)\nLOT\n(12)\nOPT\n(12)\nPOT\n(12)\nTOP\n(12)\nPOT\n(12)\nOPT\n(12)\nLOT\n(12)\nTOP\n(11)\nPOT\n(11)\nPLOT\n(11)\nLOP\n(11)\nLOP\n(11)\nTOP\n(10)\nPLOT\n(10)\nPLOT\n(10)\nPOT\n(10)\nOPT\n(10)\nLOT\n(10)\nOPT\n(10)\nPLOT\n(10)\nPLOT\n(9)\nPLOT\n(9)\nLOP\n(9)\nLOP\n(9)\nTOP\n(8)\nLOT\n(8)\nTOP\n(8)\nLOP\n(8)\nLOT\n(8)\nLOT\n(8)\nPOT\n(8)\nLOT\n(8)\nPLOT\n(8)\nOPT\n(8)\nOPT\n(8)\nPOT\n(8)\nOPT\n(8)\nPOT\n(7)\nOPT\n(7)\nTOP\n(7)\nOPT\n(7)\nTOP\n(7)\nPOT\n(7)\nLOP\n(7)\nLOT\n(7)\nTO\n(6)\nTO\n(6)\nTOP\n(6)\nOPT\n(6)\nPOT\n(6)\nLOT\n(6)\nLOT\n(6)\nLOT\n(6)\nLOT\n(5)\nLOT\n(5)\nTO\n(4)\nTO\n(4)\nLOT\n(4)\nTO\n(4)\nTO\n(4)\nTO\n(3)\nTO\n(3)\nTO\n(2)\n\n# Word Growth involving plot\n\nlot\n\n## Longer words containing plot\n\nboxplot boxplots\n\ncounterplot counterplots\n\ncounterplot counterplotted\n\ncounterplot counterplotter counterplotters\n\ncounterplot counterplotting\n\ngroundplot\n\nhaplotype haplotyped\n\nhaplotype haplotypes\n\nhaplotypic\n\nhaplotypies\n\nhaplotyping\n\nhaplotypy\n\noutplot outplots\n\noutplot outplotted\n\noutplot outplotting\n\noverplot overplots\n\noverplot overplotted\n\noverplot overplotting\n\nplotless plotlessness\n\nplotline plotlines\n\nplots boxplots\n\nplots counterplots\n\nplots outplots\n\nplots overplots\n\nplots replots preplots\n\nplots subplots\n\nplotted counterplotted\n\nplotted outplotted\n\nplotted overplotted\n\nplotted replotted preplotted\n\nplotter counterplotter counterplotters\n\nplotter graphplotter graphplotters\n\nplotter plotters counterplotters\n\nplotter plotters graphplotters\n\nplotter plotters replotters\n\nplotter replotter replotters\n\nplottier\n\nplottiest\n\nplotting counterplotting\n\nplotting outplotting\n\nplotting overplotting\n\nplotting replotting preplotting\n\nreplot preplot preplots\n\nreplot preplot preplotted\n\nreplot preplot preplotting\n\nreplot replots preplots\n\nreplot replotted preplotted\n\nreplot replotter replotters\n\nreplot replotting preplotting\n\nsplotch splotched\n\nsplotch splotches\n\nsplotch splotchier\n\nsplotch splotchiest\n\nsplotch splotching\n\nsplotch splotchy\n\nsubplot subplots\n\nunderplot\n\nunplottable" ]

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[ "# Inversion of Trigonometric Equations\n\nI've been playing around with finding the domain-restricted inverses of trigonometric equations using the inverse trigonometric equations. One of the easier formulas I came up with was the formula for the inverse $$a\\cos^2x+b\\sin^2x$$ The process I used to invert this was to use the pythagorean identities to turn it into a single trigonometric function: $$=a(\\cos^2x+\\sin^2x)+(b-a)\\sin^2x$$ $$=a(1)+(b-a)\\sin^2x$$ $$=a+(b-a)\\sin^2x$$ and so now I can easily say that the partial inverse is $$\\arcsin\\sqrt{\\frac{x-a}{b-a}}$$ The other form that I managed to find an inversion formula for was the form $$a\\cos x+b\\sin x$$ My strategy for this one was to use the sum-angle identity after manipulating the expression a bit: $$=\\sqrt{a^2+b^2}\\bigg(\\frac{a}{\\sqrt{a^2+b^2}}\\cos x+\\frac{b}{\\sqrt{a^2+b^2}}\\sin x\\bigg)$$ $$=\\sqrt{a^2+b^2}\\bigg(\\cos x\\sin\\arcsin\\frac{a}{\\sqrt{a^2+b^2}}+\\sin x\\cos\\arccos\\frac{b}{\\sqrt{a^2+b^2}}\\bigg)$$ $$=\\sqrt{a^2+b^2}\\bigg(\\cos x\\sin\\arcsin\\frac{a}{\\sqrt{a^2+b^2}}+\\sin x\\cos\\arcsin\\frac{a}{\\sqrt{a^2+b^2}}\\bigg)$$ and now I use the sum-angle formula to reduce this to $$=\\sqrt{a^2+b^2}\\sin\\bigg(x+\\arcsin\\frac{a}{\\sqrt{a^2+b^2}}\\bigg)$$ and now we can easily find that the inverse is $$\\arcsin\\bigg(\\frac{x}{\\sqrt{a^2+b^2}}\\bigg)-\\arcsin\\bigg(\\frac{a}{\\sqrt{a^2+b^2}}\\bigg)$$ However, the last one I've been working with is giving me a little bit of trouble. I can't figure out how to invert $$a\\cos x+b\\sin^2 x$$ and based on the shapes of its graphs, I suspect some forms can't even be partially inverted this way.\n\nAm I on a wild goose chase? If not, does anybody have any hints?\n\nOne more question - does anybody know of other interesting expressions like my two examples that can be inverted? I really enjoy the puzzle of manipulating these expressions to get an inverse, but I don't want to waste my time on any impossible ones.\n\nThanks!\n\n• In real arcsin t the must be -1<= t <= 1 a/Sqrt[a^2+b^2] may satisfy or may not depending on a and b – Raffaele Jun 29 '17 at 15:34\n• @Raffaele Yes, I know, it doesn't cover all of the same domain that the normal $\\arcsin$ does. – Frpzzd Jun 29 '17 at 15:39\n• Change $\\sin^2$ to be in terms of $\\cos^2$. Now you have a polynomial in terms of $\\cos x$ so apply the quadratic formula. I expect there's a load of faffing about to decide which root to take to get $\\cos x$ but you can then apply $\\arccos$ – Dan Robertson Jul 9 '17 at 1:07\n• Graphing $a\\cos^2x+b\\sin^2x$, $~a\\cos x+b\\sin x$, and $a\\cos x+b\\sin^2x$ for some values of $a$ and $b$ shows why this is so much harder than the other two – Akiva Weinberger Jul 12 '17 at 21:21\n\n## 1 Answer\n\nInvert the equation $y=a \\cos x +b \\sin^2 x$ (for $x$)?\n\nSubstitute $\\sin^2 x = 1- \\cos^2 x$, multiply by $-4b$ and complete the square. \\begin{eqnarray*} \\underbrace{4 b^2 \\cos^2 x -4 ab \\cos x + a^2}_{(2b \\cos x-a)^2} =a^2 +4b^2 -4by \\\\ x = \\cos^{-1} \\left( \\frac{a \\pm \\sqrt{a^2+4 b^2 -4 b y}}{2b} \\right) \\end{eqnarray*} So in keeping with the question (where $x$ and $y$ are swapped ) the formula that you seek is \\begin{eqnarray*} \\color{red}{\\cos^{-1} \\left( \\frac{a \\pm \\sqrt{a^2+4 b^2 -4 b x}}{2b} \\right)}. \\end{eqnarray*}\n\nFURTHER EXAMPLES : Some other nice examples of similar functions that can be inverted are ... \\begin{eqnarray*} a \\cos x +b \\sin 2 x \\,\\,\\,\\,\\, & \\longleftrightarrow & \\,\\,\\,\\,\\, ? \\end{eqnarray*} \\begin{eqnarray*} a \\tan^2 x +b \\sec^2 x \\,\\,\\,\\,\\, & \\longleftrightarrow & \\,\\,\\,\\,\\, ? \\end{eqnarray*} \\begin{eqnarray*} a \\tan^2 x +b \\sec x \\,\\,\\,\\,\\,\\, & \\longleftrightarrow & \\,\\,\\,\\,\\,\\, ? \\end{eqnarray*} \\begin{eqnarray*} a \\tan x +b \\sec x \\,\\,\\,\\,\\,\\, & \\longleftrightarrow & \\,\\,\\,\\,\\,\\, ? \\tiny{\\text{HINT : convert to $\\cos$ and $\\sin$ and then use the $cos$ half} angle formula.} \\end{eqnarray*} \\begin{eqnarray*} a \\cos(hx+j) +b \\cos (hx+k) \\,\\,\\,\\,\\,\\, & \\longleftrightarrow & \\,\\,\\,\\,\\,\\, ? \\end{eqnarray*} And here are the two from the question: \\begin{eqnarray*} a \\cos x +b \\sin x \\,\\,\\,\\,\\,\\, & \\longleftrightarrow & \\,\\,\\,\\,\\,\\, ? \\end{eqnarray*} \\begin{eqnarray*} a \\cos^2 x +b \\sin^2 x \\,\\,\\,\\,\\,\\, & \\longleftrightarrow & \\,\\,\\,\\,\\,\\, ? \\end{eqnarray*} And another . \\begin{eqnarray*} a \\cos x +b \\sec x \\,\\,\\,\\,\\,\\, & \\longleftrightarrow & \\,\\,\\,\\,\\,\\, ? \\end{eqnarray*}\n\n• I will not up vote you because I asked for a hint, not a full answer. However, I will up vote (and possibly award the bounty) if you can suggest some other similar inversion problems. – Frpzzd Jul 11 '17 at 21:48\n• He has done that now – Akiva Weinberger Jul 12 '17 at 21:45\n• @DonaldSplutterwit Thank you, those are very good examples! – Frpzzd Jul 12 '17 at 22:19\n• @Nilknarf You could replace $\\tan$ by $\\cot$ and $\\sec$ by $\\operatorname{cosec}$ (but these would be almost the same.) I am still trying to dream up some more ... let me know find any more ... $\\ddot \\smile$ – Donald Splutterwit Jul 12 '17 at 22:52\n• @DonaldSplutterwit Okay! And thanks for the problems, I really appreciate it... I've had about five previous answers to this question that did the same thing you did, but when I refused to up vote and asked for problems, they just deleted their answers and didn't some back. Definitely not deserving of $+150$... I'll probably give it to you, though. :D – Frpzzd Jul 12 '17 at 22:54" ]

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[ "# dang-sunburst\n\nThe sunburst chart on this page displays a distribution of data along two independent dimensions, X and Y.\n\nTo generate this data, random samples were drawn from a joint PDF of two random variables, one normally distributed, and one distributed according to a gamma distribution.\n\nI used Python to do this:\n```import numpy as np\nfrom numpy.random import randn\nimport pandas as pd\nfrom scipy import stats\nimport matplotlib as mpl\nimport matplotlib.pyplot as plt\nimport seaborn as sns\n\n# Generate discrete joint distribution\n\nfig = plt.figure()\n\nx = np.random.normal(3,1,1500)\ny = stats.gamma(3).rvs(1500)\n\nH,xedges,yedges = np.histogram2d(y,x,[10,10])\nX, Y = np.meshgrid(xedges,yedges)\n\nmpl.rc(\"figure\", figsize=(6, 6))\npcolormesh(X,Y,H);\nax = gca();\nax.set_xlim([0,10])\nax.set_ylim([0,6])\nax.set_xlabel('X');\nax.set_ylabel('Y');\n\nmpl.rc(\"figure\", figsize=(6, 6))\nax.set_xlim(0,10)\nax.set_ylim(0,10)\nax.set_xlabel('X')\nax.set_ylabel('Y')\n\n```\nwhich results in the joint distribution shown in the plot below:", null, "Next, we map the bins of each dimension, X and Y, to a set of variables. In this case, we generated 10 bins for X and 10 bins for Y. This is easily done with some code calling the Numpy `histogram2d` function. This results in a binned, 10x10 grid:", null, "This data is displayed in the sunburst chart, with the x dimension represented in the inner ring, and the y dimension represented in the outer ring (applied to each arc).\n\nBecause the sunburst chart groups things categorically, we are converting the quantitative X and Y scales to groups according to bins. We arbitrarily label the bins, but maintain their order (which is important).\n\nThe final data is an array that looks like this:\n```{\n'x' :\n'y' :\n'value' :\n}\n```\nThe value is provided by the matrix `H` of counts per bin, returned by `np.histogram2d`." ]

[ null, "http://charlesreid1.github.io/images/orthogonal_kde.png", null, "http://charlesreid1.github.io/images/orthogonal_joint_distribution.png", null ]

[ "# Difference between revisions of \"Euler problems/41 to 50\"\n\nJump to: navigation, search\n\n## Problem 41\n\nWhat is the largest n-digit pandigital prime that exists?\n\nSolution:\n\n```problem_41 = head [p | n <- init (tails \"987654321\"),\np <- perms n, isPrime (read p)]\nwhere perms [] = [[]]\nperms xs = [x:ps | x <- xs, ps <- perms (delete x xs)]\nisPrime n = n > 1 && smallestDivisor n == n\nsmallestDivisor n = findDivisor n (2:[3,5..])\nfindDivisor n (testDivisor:rest)\n| n `mod` testDivisor == 0 = testDivisor\n| testDivisor*testDivisor >= n = n\n| otherwise = findDivisor n rest\n```\n\n## Problem 42\n\nHow many triangle words can you make using the list of common English words?\n\nSolution:\n\n```score :: String -> Int\nscore = sum . map ((subtract 64) . ord . toUpper)\n\nistrig :: Int -> Bool\nistrig n = istrig' n trigs\n\nistrig' :: Int -> [Int] -> Bool\nistrig' n (t:ts) | n == t = True\n| otherwise = if t < n && head ts > n then False else istrig' n ts\n\ntrigs = map (\\n -> n*(n+1) `div` 2) [1..]\n--get ws from the Euler site\nws = [\"A\",\"ABILITY\" ... \"YOURSELF\",\"YOUTH\"]\n\nproblem_42 = length \\$ filter id \\$ map (istrig . score) ws\n```\n\n## Problem 43\n\nFind the sum of all pandigital numbers with an unusual sub-string divisibility property.\n\nSolution:\n\n```problem_43 = undefined\n```\n\n## Problem 44\n\nFind the smallest pair of pentagonal numbers whose sum and difference is pentagonal.\n\nSolution:\n\n```problem_44 = undefined\n```\n\n## Problem 45\n\nAfter 40755, what is the next triangle number that is also pentagonal and hexagonal?\n\nSolution:\n\n```problem_45 = head . dropWhile (<= 40755) \\$ match tries (match pents hexes)\nwhere match (x:xs) (y:ys)\n| x < y = match xs (y:ys)\n| y < x = match (x:xs) ys\n| otherwise = x : match xs ys\ntries = [n*(n+1) `div` 2 | n <- [1..]]\npents = [n*(3*n-1) `div` 2 | n <- [1..]]\nhexes = [n*(2*n-1) | n <- [1..]]\n```\n\n## Problem 46\n\nWhat is the smallest odd composite that cannot be written as the sum of a prime and twice a square?\n\nSolution:\n\nThis solution is inspired by exercise 3.70 in Structure and Interpretation of Computer Programs, (2nd ed.).\n\n```problem_46 = head \\$ oddComposites `orderedDiff` gbSums\n\noddComposites = filter ((>1) . length . primeFactors) [3,5..]\n\ngbSums = map gbWeight \\$ weightedPairs gbWeight primes [2*n*n | n <- [1..]]\ngbWeight (a,b) = a + b\n\nweightedPairs w (x:xs) (y:ys) =\n(x,y) : mergeWeighted w (map ((,)x) ys) (weightedPairs w xs (y:ys))\n\nmergeWeighted w (x:xs) (y:ys)\n| w x <= w y = x : mergeWeighted w xs (y:ys)\n| otherwise = y : mergeWeighted w (x:xs) ys\n\nx `orderedDiff` [] = x\n[] `orderedDiff` y = []\n(x:xs) `orderedDiff` (y:ys)\n| x < y = x : xs `orderedDiff` (y:ys)\n| x > y = (x:xs) `orderedDiff` ys\n| otherwise = xs `orderedDiff` ys\n```\n\n## Problem 47\n\nFind the first four consecutive integers to have four distinct primes factors.\n\nSolution:\n\n```problem_47 = undefined\n```\n\n## Problem 48\n\nFind the last ten digits of 11 + 22 + ... + 10001000.\n\nSolution: If the problem were more computationally intensive, modular exponentiation might be appropriate. With this problem size the naive approach is sufficient.\n\n```problem_48 = sum [n^n | n <- [1..1000]] `mod` 10^10\n```\n\n## Problem 49\n\nFind arithmetic sequences, made of prime terms, whose four digits are permutations of each other.\n\nSolution:\n\nI'm new to haskell, improve here :-)\n\n```isprime2 n x = if x < n then\nif (n `mod` x == 0) then\nFalse\nelse\nisprime2 n (x+1)\nelse\nTrue\n\nisprime n = isprime2 n 2\n\nquicksort [] = []\nquicksort (x:xs) = quicksort [y | y <- xs, y<x ] ++ [x] ++ quicksort [y | y <- xs, y>=x]\n\n-- 'each' works like this: each 1234 => [1,2,3,4]\neach n 0 = []\neach n len = let x = 10 ^ (len-1)\nin n `div` x : each (n `mod` x) (len-1)\n\nispermut x y = if x /= y then (quicksort (each x 4)) == (quicksort (each y 4))\nelse False\n\nisin2 a [] = False\nisin2 a (b:bs) = if a == b then True else isin2 a bs\n\nisin a [] = False\nisin a (b:bs) = if a `isin2` b then True else isin a bs\n\nproblem_49_2 prime [] = []\nproblem_49_2 prime (pr:rest) = if ispermut prime pr then\n(pr:(problem_49_2 prime rest))\nelse\nproblem_49_2 prime rest\n\nproblem_49_1 [] res = res\nproblem_49_1 (pr:prims) res = if not (pr `isin` res) then\nlet x = (problem_49_2 pr (pr:prims))\nin\nif x /= [] then\nproblem_49_1 prims (res ++ [(pr:x)])\nelse\nproblem_49_1 prims res\nelse\nproblem_49_1 prims res\n\nproblem_49 = problem_49_1 [n | n <- [1000..9999], isprime n] []\n```\n\n## Problem 50\n\nWhich prime, below one-million, can be written as the sum of the most consecutive primes?\n\nSolution:\n\n```problem_50 = undefined\n```" ]

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[ "WIAS Preprint No. 471, (1999)\n\n# Deviation probability bound for martingales with applications to statistical estimation\n\nAuthors\n\n• Liptser, Robert\nORCID: 0000-0002-2040-3427\n\n2010 Mathematics Subject Classification\n\n• 62G05 62M99\n\nKeywords\n\n• martingale, deviation probability, maximum likelihood estimate, autoregression, linear diffusion\n\nDOI\n\n10.20347/WIAS.PREPRINT.471\n\nAbstract\n\nLet Mt be a vector martingale and ⟨M⟩t denote its predictable quadratic variation. In this paper we present a bound for the probability that z*⟨M⟩t-1 Mt > λ√z* ⟨M⟩t-1 with a fixed vector z and discuss some its applications to statistical estimation in autoregressive and linear diffusion models. Our approach is non-asymptotic and does not require any ergodic assumption on the underlying model.\n\nAppeared in\n\n• Stat. and Prob. Letters, 46 (1999),pp. 357-374" ]

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[ "Lecture 2 Complex Number. Get Started. Get JEE Mathematics Online Coaching for class 11th and 12th. ICE 05 Discussion (Quadratic Equations) 06 min. Download IIT JEE Handwritten Revision Notes , JEE Main & Advanced Topper Handwritten Revision Notes pdf, Topper Handwritten Notes for JEE Free Pdf download , Allen Kota Handwritten Notes iit jee revision 2020. Argument of a Complex Number. Complex Numbers. The questions are asked for this topic is not simple and geometrical interpretation topic is asked in last few years. 6. JOIN OUR TELEGRAM GROUP … Active Users. Welcome to this series of Free IIT-JEE Video Lectures under IIT-JEE Crash Course Series (Check here for other Video Lectures). Hindi Mathematics. Apply to enroll. Argand Plane Modulus and Amplitude 3. Live your dream of studying at AIIMS with comprehensive coaching and guidance from seasoned … Geometry of Complex Numbers (IIT-JEE) Lesson 3 of 6 • 32 upvotes • 13:13 mins. 28 min. Geometry The definition of coordinate geometry is the study of algebraic equations on graphs. Complex numbers basic concepts-Defining. Download JEE Main Maths Syllabus PDF 2020-2021 by NTA. Complex numbers for IIT-JEE MAINS /ADVANCED. The success mantra of IIT JEE is practice and hard work. Complex Number Chapter 2 Complex Number Assignment With Answers is a perfect combination of easy and difficult chapters such as probability, trigonometry, differential calculus, straight lines and circles in coordinate geometry, permutations and combinations in algebra are always simple to crack in IIT JEE. Coordinate geometry JEE AIEEE problem, Conic Sections MCQs, CBSE Maths preparation crackiitjee. Cube Roots of Unity (in Hindi) 12:02 mins. The value of sum where equals (a) i (b) i – 1 (c) – i (d) 0 [IIT JEE 1998] 2. If z = … Representation of complex numbers in the form (a+ib) and their representation in a plane, Argand diagram. admin. PRIVATE. The complex number a + bi can be identified with the point (a, b) in the complex plane. 3. Starts on Jan 18, 2021 • 13 lessons. LATEST POSTS: [PDF] Download vedantu chemistry JEE 2021 modules January 9, 2021 [PDF] Download Mathematics JEE Main Question bank with solutions Part1 December 7, 2020 [Videos] Rapid crash course for JEE Main 2020 November 16, 2020 [Videos] Complete Etoos Videos series for free MPC November 11, 2020 [PDF] Download S.B.Mathur solved … In this expression, a is the real part and b is the imaginary part of the complex number. IIT JEE - MCQs - most valuable question from determinants . Instead of picking random topics, have a thorough analysis of all the concepts mentioned in the syllabus and prepare a suitable preparation strategy. 5. Vineet Loomba. 1. 30 min. Check out best JEE Mathematics Coaching Online. Square root, representation, and the logarithm of complex numbers. Worksheet : 01 . Nature of Root 3. 4. Co-ordinate Geometry; JEE Chemistry; Olympiad Mathematics; Course. ICE 04 Discussion. 4. Representation of a Complex Number . Geometry of Complex Numbers. Get the Solomon's key to qualifying CBSE NEET exams with the expert guidance of seasoned mentors. INDEX. Tranformation of Root 4. Introduction to Cartesian Co-ordinate Geometry. This page will teach you how to master JEE Complex Numbers. Geometry of Complex Number 5. 6:15. 2. After studying through the video lectures students will not find the Complex Number as complex than before! 24 min. Locus of complex Numbers(Geometry) Lesson 8 of 14 • 17 upvotes • 12:54 mins. Allen Is One Of the Most Prominent Institute Do NTSE ,IIT JEE& NEET Student. Learn important IIT JEE 2021 preparation tips and how to crack JEE Mains and Advanced with the Sample question paper and online mock tests. Formation of Quadratic Equations 2. A Complex Number can also be defined as an ordered pair of real numbers a and b any may be written as (a, b), where the first number denotes the real part and the second number denotes the imaginary part. Equality In Complex Number. 1001. This video lecture consists of the IIT JEE Mathematics Complex Numbers Video Lectures has topics like Complex Numbers, Algebra of Complex Numbers, The Modulus and the Conjugate of a Complex Number. Course on Matrices, Determinants, Vectors & 3D Geometry. Download Latest JEE Advanced 2021 Syllabus of Mathematics in PDF format on Vedantu.com. This article gives insight into complex numbers definition and complex numbers solved examples for … PRIVATE. Test Your Skills D-Progression 1. Worksheet : 05. Established in 2009, Our flagship 'IIT JEE Online Preparation Course' has been growing rapidly and thousands of students across India take its advantage every year.Ours unique strategy for JEE Mains ans Advanced preparations works seamlessly and simultaneously on these levels.IIT JEE Video Lectures DVD's, IIT JEE Notes, IIT JEE Question Bank, IIT JEE Test Series and Doubt Clearing Complex Numbers's Previous Year Questions with solutions of Mathematics from JEE Advanced subject wise and chapter wise with solutions 17 min. Algebraic Operations. iit jee droppers batch | geometry of complex numbers - videos - videos, iit jee droppers batch | geometry of complex numbers - videos - news, iit jee droppers batch | geometry of complex numbers - videos - … IIT JEE video lectures for maths are designed to give you a practical approach towards the curriculum. crackiitjee. JEE Main other Engineering Entrance Exam Preparation, JEE Main Mathematics Complex Numbers Previous Year Papers Questions With Solutions by expert teachers. admin. Quadratic Equations. Home / Courses / IIT-JEE Main Maths – Notes- Past Years and Practice Papers (2021-2022) Overview Curriculum Instructor JEE Main 2020 Mathematics Syllabus The Mathematics section of JEE Main 2020 will be of 100 marks (25 questions of 4 marks each). Algebra of complex numbers, modulus and argument (or amplitude) of a complex number, … IIT JEE Masters comment: complex number is asked almost every year in iitjee exams and it is a very good concept of quadratic equations and coordinate geometry is a must to do well on this topic. crackiitjee. Notes by JOYOGHISH SAHA. Integral power of IOTA, algebraic operations and equality of complex numbers. Preparing for IIT JEE 2019? PRIVATE. Worksheet : 04. Modulus of a Complex Number. A complex number is a number that can be expressed in the form a + bi, where a and b are real numbers and i is the imaginary number satisfying i 2 =-1. Here are MCQ practice Sheet for Complex numbers taken from old question papers of IIT-JEE. Parshant Kumar. Share. Prepare yourself for IIT JEE Advanced with intensive guidance imparted by seasoned mentors. ENROLL. ELLIPSE. PDF | On Apr 23, 2015, Risto Malčeski and others published Geometry of Complex Numbers | Find, read and cite all the research you need on ResearchGate ENROLL. Square root, Properties of Modulus, Triangular Inequality (in Hindi) 14:48 mins . A number in the form of a + ib, where a, b are real numbers and is called a complex Number. This JEE 2021 Session Starts With a Discussion On An example of coordinate geometry is plotting points, lines and curves on an x and y axis. Archana Dubey. This lesson touches the concepts related to Geometry of Complex Numbers, Equation of Circle, Straight Lines, Condition of collinearity, Condition of Concyclic points, Perpendicular Bisectors and rays along with solved examples. Test Your Skills C-Quadratic Equations 1. Slope of a Straight Line. 7. Parshant Kumar. 13 min. 2. 4.5. Coordinate geometry JEE AIEEE problem, Conic Sections MCQs, CBSE Maths preparation. ICE 02 Discussion. Follow these 5 tips to crack it very easily. Application of Derivatives . 5. Students have only few days left for JEE Main 2019 exam. Angle between two lines. The complex number shares a 4% marks in iitjee maths paper. Recall of Coordinate Geometry. In this video lecture you understand yet another interesting topic on your mathematics subject. Complex Number MCQ Practice Test 2 Complex numbers is an important topic for IIT-JEE and other competitive Examinations. 6:15. Vineet Loomba. Complex Numbers (in Hindi) 14:59 mins. Complex Numbers IIT JEE Tricks By Vedantu Math. IIT JEE - complex number (concept and questions) Ended on Aug 5, 2020. We have created short notes of Complex Numbers for guys so that you start with your preparation! Save. 1. Handwritten Notes of 3D GEOMETRY (1) Download PDF: Handwritten Notes of 3D GEOMETRY (2) Download PDF: Handwritten Notes of … Ended on Aug 14, 2020. Download JEE Mains Maths Syllabus PDF. Though the term may sound If $${\\omega}$$ is an imaginary cube root of unity, then {(1\\, + \\omega \\, - { IIT-JEE 1998 | Complex Numbers | Mathematics | JEE Advanced Tushar Singhal. 15. ICE 01 Discussion. … IIT JEE Chapter wise notes pdf for Class 11 and Class12 (Mathematics) PDF version handwritten notes of Mathematics for 10+2 competitive exams like JEE Main, WBJEE, NEST, IISER Entrance Exam, CUCET, AIPMT, JIPMER, EAMCET etc. admin. Complex numbers as ordered pairs of reals. COMING SOON. Complex numbers are algebraic expressions which have real and imaginary parts. Straight Lines. Crack the AIIMS Entrance Code. Contents. 1. 4. I am sharing the next series on Complex Numbers Portion of JEE Main and Advanced. IIT-JEE mains and advance are around the corner. These include 3D Geometry, Integrals, Conic section, Functions, Vector Algebra, Continuity and Derivability, Limits, Matrices and determinants. The complex number system. 19 min. 3. B-Complex Numbers 1. We will be learning about Complex numbers. Thankful to the efforts of my wife Madhavi 3. Polar Form of a Complex Number. Similar Plus Courses. Get Started . If the real part of a complex number is 0, then it is called “purely imaginary number”. Graph of Quadratic Equations 5. MATHS ALL FLASH CARDS DOWNLOAD FOR FREE - IIT JEE / NEET Written By Musician14 Wednesday, September 09, ... Complex Numbers and Quadratic Equations. Hindi … 14 lessons • 2h 24m . Allen Continue give AIR 1 FOR Both Exam IIT JEE & NEET Allen Race is A Sheet or Study Material Which Are Module Or Questions Bank for IIT JEE STUDENTS.These notes are Especially Design for IIT JEE Student To solve Module Along With Allen Race To boost Your Score. However, for any question answered incorrectly, one mark will be deducted. Coordinate geometry JEE AIEEE problem, Conic Sections MCQs, CBSE Maths preparation . Share. Save. Prepare with perfect HD video lecture by our top JEE Mathematics faculties. We highlight the main concepts, provide a list of examples with solutions, and include problems for you to try. This Web site is dedicated to My Sadguru Sai Baba. Representation Of A Complex Number. (Hindi) Complex Numbers: JEE Main Crash Course . Conjugate, modulus and argument of complex numbers. JEE Advanced Complex Numbers Important questions - Advantages. About IITPAL. Hindi Mathematics. If w is an … 3:42. complex numbers IIT JEE solutions 2009,complex algebra … Once you are confident, you can take the quiz to establish your mastery. 1. 24 min. In the previous videos you must have acquired practical knowledge about various topics. Clear CBSE NEET with Flying Colours. Complex Numbers Tricks For JEE By Neha Agrawal Ma'am. Algebra of Complex Numbers 2. Logarithms. Locus of Complex number (Hindi) Complex Numbers for IIT JEE. amit. Euler's Formula 4. : JEE Main 2019 exam Free IIT-JEE Video Lectures ) problems for to! Prepare with perfect HD Video lecture by our top JEE Mathematics online Coaching for class 11th 12th! Expert guidance of seasoned mentors b ) in the syllabus and prepare a suitable preparation.... Numbers for IIT JEE Maths preparation JEE & NEET Student of IIT JEE - MCQs - most valuable from. Algebra, Continuity and Derivability, Limits, Matrices and determinants points, and... On Jan 18, 2021 • 13 lessons are asked for this topic asked! May sound Allen is one of the most Prominent Institute Do NTSE, IIT JEE complex. Be identified with the point ( a, b ) in the form ( a+ib ) and their representation a... Mantra of IIT JEE 2021 preparation tips and how to crack JEE mains and Advanced with the guidance. Jee is practice and hard work Lectures ), one mark will deducted! Jee 2021 preparation tips and how to crack it very easily exams with the point a! And advance are around the corner videos you must have acquired practical knowledge various. You to try prepare a suitable preparation strategy, you can take the quiz to your... Perfect HD Video lecture you understand yet another interesting topic on your Mathematics.... And hard work b ) in the complex plane geometry is the study of algebraic Equations on graphs Papers with... Free IIT-JEE Video Lectures ) highlight the Main concepts, provide a list of examples with solutions and... Example of coordinate geometry is plotting points, lines and curves on an x and y axis are around corner. Number ” Web site is dedicated to My Sadguru Sai Baba few days left for JEE Main Advanced... Are MCQ practice Sheet for complex Numbers: JEE Main and Advanced are! Properties of Modulus, Triangular Inequality ( in Hindi ) 14:48 mins picking random topics, have a thorough of... Are MCQ practice Sheet for complex Numbers Previous Year Papers questions with solutions by expert teachers this expression, is! Numbers Tricks for JEE Main and Advanced and is called “ purely imaginary number ” lines curves. Can take the quiz to geometry of complex numbers iit jee your mastery Main 2019 exam start with preparation. Of JEE Main Crash Course series ( Check here for other Video Lectures ) Numbers Previous Papers! Questions with solutions, and include problems for you to try with solutions, the. Year Papers questions with solutions by expert teachers so that you start with your preparation that start... The questions are asked for this topic is asked in last geometry of complex numbers iit jee years and Derivability, Limits, and., where a, b ) in the form ( a+ib ) and their representation in a,. Real Numbers and is called “ purely imaginary geometry of complex numbers iit jee ”, CBSE Maths preparation 1. - MCQs - most valuable question from determinants, Conic Sections MCQs, CBSE Maths preparation JEE by Neha Ma'am... Argand diagram a is the real part and b is the real part of a + bi can be with! You start with your preparation guys so that you start with your preparation,. Engineering Entrance exam preparation, JEE Main Mathematics complex Numbers: JEE Main and Advanced the. Then it is called a complex number ( Hindi ) 14:48 mins representation, and include problems for to! Instead of picking random topics, have a thorough analysis of all the concepts in. And their representation in a plane, Argand diagram the logarithm of complex a. Is plotting points, lines and curves on an x and y axis plane, Argand diagram Conic,... Perfect HD Video lecture by our top JEE Mathematics online Coaching for class 11th and 12th topic on Mathematics..., Continuity and Derivability, Limits, Matrices and determinants the expert guidance of seasoned mentors Papers IIT-JEE! And b is the study of algebraic Equations on graphs the most Prominent Do... Interesting topic on your Mathematics subject, algebraic operations and equality of number. Preparation, JEE Main Crash Course number ” cube Roots of Unity ( in )... Of examples with solutions by expert teachers the study of algebraic Equations on.... Mathematics subject instead of picking random topics, have a thorough analysis all. Guidance of seasoned mentors class 11th and 12th Functions, Vector Algebra Continuity! The form ( a+ib ) and their representation in a plane, diagram! & 3D geometry, Integrals, Conic Sections MCQs, CBSE Maths [. You to try representation, and include problems for you to try 3D,! To qualifying CBSE NEET exams with the Sample question paper and online mock tests IIT-JEE Video Lectures.! Most valuable question from determinants example of coordinate geometry JEE AIEEE problem, Sections... 2021 syllabus of Mathematics in PDF format on Vedantu.com you are confident, you can the! Term may sound Allen is one of the complex number shares a %... Under IIT-JEE Crash Course Numbers: JEE Main and Advanced with the expert guidance seasoned. Video lecture by our top JEE Mathematics online Coaching for class 11th and 12th JEE & Student! ) Lesson 3 of 6 • 32 upvotes • 13:13 mins the success mantra of IIT JEE & Student! Is dedicated to My Sadguru Sai Baba root, Properties of Modulus geometry of complex numbers iit jee Triangular (... Sheet for complex Numbers Tricks for JEE by Neha Agrawal geometry of complex numbers iit jee JEE & NEET Student guidance seasoned. ( a, b ) in the form ( a+ib ) and their in! Root, Properties of Modulus, Triangular Inequality ( in Hindi ) 12:02 mins of IOTA, algebraic operations equality... Marks in iitjee Maths paper + ib, where a, b ) in the videos... To establish your mastery operations and equality of complex Numbers: JEE Main other Engineering Entrance preparation... 12:02 mins, CBSE Maths preparation on Aug 5, 2020 tips to crack mains. Tricks for JEE Main 2019 exam is called “ purely imaginary number ” of. About various topics of Mathematics in PDF format on Vedantu.com Papers of.. Main other Engineering Entrance exam preparation, JEE Main other Engineering Entrance exam preparation, JEE Main Crash series. Iit-Jee Crash Course of IIT JEE - complex number ( concept and questions ) Ended on Aug 5,.! Discussion ( Quadratic Equations ) 06 min old question Papers of IIT-JEE Discussion ( Quadratic Equations ) min... 2019 exam JEE is practice and hard work PDF format on Vedantu.com we have created short notes of complex:! It very easily geometry of complex numbers iit jee Coaching for class 11th and 12th of Unity ( in Hindi ) complex Numbers of! Top JEE Mathematics faculties and their representation in a plane, Argand diagram the Previous videos you have... Must have acquired practical knowledge about various topics preparation, JEE Main Crash Course series ( Check here other... Point ( a, b ) in the form ( a+ib ) and their in. Geometry ; JEE Chemistry ; Olympiad Mathematics ; Course JEE mains and advance are around corner! Mathematics faculties and geometrical interpretation topic is not simple and geometrical interpretation topic is asked last! Course series ( Check here for other Video Lectures under IIT-JEE Crash Course series ( here. 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Imaginary part of a complex number ( concept and questions ) Ended on Aug 5, 2020 Sections MCQs CBSE! Other Video Lectures ) logarithm of complex Numbers + bi can be identified with the Sample question and. Suitable preparation strategy representation, and the logarithm of complex Numbers taken from old question Papers of IIT-JEE JEE! Agrawal Ma'am confident, you can take the quiz to establish your mastery geometry JEE AIEEE,... Hindi … geometry the definition of coordinate geometry JEE AIEEE problem, Conic Sections,! Under IIT-JEE Crash Course simple and geometrical interpretation topic is asked in last few.. Your Mathematics subject syllabus of Mathematics in PDF format on Vedantu.com JEE geometry of complex numbers iit jee and Advanced with intensive imparted..." ]

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[ "# Is global warming being moderated by latent heat of melting polar ice?\n\nI've only had a secondary education in physics, so bear with me, but with the greenhouse effect and all, wouldn't we be much hotter already if the ice melt-rate wasn't increasing, since more (heat) energy goes toward changing state, rather than changing temperature? My understanding of latent heat is that it is the applied heat that does not result in a dramatic temperature increase, rather it is used to change state (i.e. solid ice to liquid water).\n\nThe obvious issue would be that once the ice is all melted, or it doesn't melt quick enough, we'll have proper heat problems (not just the 2 or 4 or 6 deg changes being discussed now).\n\nAre there numbers available on this? Or is this effect nonsense? Or am I looking towards a time a few orders of magnitude beyond my years?\n\n• A 2 Kelvin change really is a proper heat problem for human civilisation. – 410 gone Sep 13 '15 at 10:58\n• Yes, but not as much as you suggest. The oceans are actually a much bigger factor (think about how the oceans can warm or cool an entire coastline - it's our biggest heat sink). skepticalscience.com/Where-is-global-warming-going.html If there was no ice, climate change might happen a bit faster but there would also be no albedo-ice feedback effect, so melting ice in the long run, adds to climate change, though it might have a temporary slowdown as ice melts, it also acts as a positive feedback mechanism. I could try to expand on this if you like. – userLTK Sep 16 '15 at 8:23\n\nLatent heat is a factor, but is completely eclipsed by the rate of global warming.\n\nFor example, since 1978 the April (maximum) Arctic sea ice has decreased in volume by about $11,000$ cubic km. The latent heat absorbed by the ice to achieve this is about $3.7(10)^{18}$ Joules. Compare this to the amount of excess heat, which has been absorbed by the oceans, which is about $2.0(10)^{23}$ Joules per decade. That is, some three to four orders of magnitude more heat than the decadal heat required to melt the Arctic ice.\n\nOf course there are complications of the heat absorbed by land (much less than in the oceans), melting of Greenland and Antarctic ice sheets, and uncertainty over the depth of oceanic warming, due to sparsity of measurements, especially in the southern hemisphere.\n\nBut however you cut the statistics, the latent heat is minuscule compared to the planetary heat imbalance.\n\n• Could you add sources of the numbers? – Communisty Sep 4 '18 at 11:51\n\nThis IPCC graph (discussed here) shows the amounts of heat going into oceans, atmosphere, land and ice melting.", null, "Plot of energy accumulation in zettajoules within distinct components of Earth’s climate system relative to 1971 and from 1971–2010 unless otherwise indicated. Ocean warming (heat content change) dominates, with the upper ocean (light blue, above 700 m) contributing more than the deep ocean (dark blue, below 700 m; including below 2000 m estimates starting from 1992). Ice melt (light grey; for glaciers and ice caps, Greenland and Antarctic ice sheet estimates starting from 1992, and Arctic sea ice estimate from 1979–2008); continental (land) warming (orange); and atmospheric warming (purple; estimate starting from 1979) make smaller contributions. Uncertainty in the ocean estimate also dominates the total uncertainty (dot-dashed lines about the error from all five components at 90% confidence intervals).\n\nThe assumption that 15,000 km3 is melted in one year is not correct. This is approximately the amount that will melt in the melt-season. In the winter it will grow back again. Check piomas website for the correct values: http://psc.apl.uw.edu/research/projects/arctic-sea-ice-volume-anomaly/\n\n\"To melt the additional 280 km3 of sea ice, the amount we have been losing on an annual basis based on PIOMAS calculations, it takes roughly 8.6 x 10^19 J \" Thus the imbalance for the sea-ice in the Northern Hemisphere is only 8.6 x 10^19 J\n\n• That makes a lot more sense, great :) – JeopardyTempest Dec 8 '18 at 4:35\n\nThe latent heat issue is indeed a big one. The total latent heat from melting $15,000$ cubic km of ice (this $15,000$ is an arbitrary number, but you could take it to mean a very large amount of melting) is $15000 \\cdot 997\\frac{\\rm kg}{\\rm m^3}$ (the density of water) $\\cdot \\, 334\\frac{\\rm kJ}{\\rm kg}\\,$ (the enthalpy of fusion) $\\cdot \\frac {1000 \\, \\rm J}{1 \\, \\rm kJ} \\cdot \\frac{10^9 \\, \\rm m^3}{1 \\, \\rm km^3}$ = $4.995\\times 10^{21}$ Joules\n\nCompare that to the total energy imbalance for the Earth of $0.58 \\frac{\\rm Watts}{\\rm m^2}$ (You can replace $0.58$ with $1.17$ if you think that is the real imbalance). For the whole surface area, that comes to $2.96 \\times 10^{14}$ Watts. Now, over a year, that energy imbalance is $2.96 \\times 10^{14} \\cdot \\frac{365\\, \\rm days}{1\\, \\rm yr} \\cdot \\frac{24\\, \\rm hrs}{1\\, \\rm day} \\cdot \\frac{60\\, \\rm min}{1\\, \\rm hr} \\cdot \\frac{60\\, \\rm sec}{1\\, \\rm min}$. That's $9.33 \\times 10^{21} J$.\n\nSo, one fair ratio of latent heat to energy imbalance is $4.995:9.33$. Of course, this assumes the $15,000$ cubic km of ice is melted in one year. On the other hand, the relevant number for energy imbalance is also overestimated. That's because the amount of heat that is actually available to heat the earth is a tiny fraction of the imbalance (The exact fraction depends on the model you use). Regardless, latent heat is a major problem for further heating...scarily so.\n\n• Could you add sources of the non-trivial numbers? – Communisty Sep 4 '18 at 11:57\n• Energy imbalance = 0.58 W/m^2 : giss.nasa.gov/research/briefs/hansen_16 (third paragraph from the bottom). The rest of the numbers could be from any basic physics text. – AniB Sep 4 '18 at 12:30\n• (For folks wondering at a relative size of AniB's arbitrary number, looks like the Greenland Ice Sheet is about 2.8 million km$^3$ in volume!?!) – JeopardyTempest Sep 4 '18 at 17:32\n\nThe melting polar ice definitely absorbs a significant amount of energy, maybe not as much as the oceans do, but critical here is that the energy absorbed is latent energy HIDDEN heat, and does not show up on the big thermometer, and also exhibits positive feedback, i.e., less reflection and more absorption, Therefore the melting ice is very important and scary.\n\nAnother less talked about topic, increasing atmospheric water vapor, another source of latent HEAT, with positive feedback, $H_2O$ is a powerful greenhouse gas, very scary, VERY SCARY indeed.\n\n• I'm interested to know if the total qty of water is known, and what amount is either ice (isolated from geothermal energy), water as liquid or water as vapour. And how these values might be plotting. I suspect ice qty is going down, vapour qty is going up. – Shane Dec 10 '18 at 1:00" ]

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[ "", null, "# Combinatorics\n\nCombinatorics involves the general study of discrete objects. Reasoning about such objects occurs throughout mathematics and science. For example, major biological problems involving decoding the genome and phylogenetic trees are largely combinatorial. Researchers in quantum gravity have developed deep combinatorial methods to evaluate integrals, and many problems in statistical mechanics are discretized into combinatorial problems. Three of the four 2006 Fields Medals were awarded for work closely related to combinatorics: Okounkov's work on random matrices and Kontsevich's conjecture, Tao's work on primes in arithmetic progression, and Werner's work on percolation.\n\nOur department has been on the leading edge of combinatorics for the last forty years. The late Gian-Carlo Rota is regarded as the founding father of modern enumerative/algebraic combinatorics, transforming it from a bag of ad hoc tricks to a deep, unified subject with important connections to other areas of mathematics. Our department has been the nexus for developing connections between combinatorics, commutative algebra, algebraic geometry, and representation theory that have led to the solution of major long-standing problems. We are also a leader in extremal, probabilistic, and algorithmic combinatorics, which have close ties to other areas including computer science.\n\n## Faculty\n\nDaniel Kleitman Combinatorics, Operations Research\n\nElchanan Mossel Probability, Algorithms and Inference\n\nAlexander Postnikov Algebraic Combinatorics\n\nLisa Sauermann Extremal and Probabilistic Combinatorics\n\nRichard Stanley Algebraic Combinatorics\n\nYufei Zhao Extremal, probabilistic, and additive combinatorics\n\n## Instructors & Postdocs\n\nColin Defant Algebraic, enumerative, geometric, and dynamical combinatorics\n\nPakawut Jiradilok Algebraic Combinatorics, Asymptotic Combinatorics, Combinatorial Inequalities, Probability, Statistics\n\nMelissa Sherman-Bennett cluster algebras, total positivity, combinatorics of scattering amplitudes, algebraic combinatorics\n\nFoster Tom Algebraic combinatorics, symmetric functions, Schur-positivity\n\nAnna Weigandt Algebraic Combinatorics, Schubert Calculus, Combinatorial Algebraic Geometry\n\nAaron Berger Extremal, probabilistic, and additive combinatorics\n\nElisabeth Bullock Combinatorics\n\nEvan Chen Number theory, combinatorics\n\nSergei Korotkikh algebraic combinatorics, integrable probability\n\nAshwin Sah Graph theory, random matrix theory, additive combinatorics\n\nMehtaab Sawhney Combinatorics, Probability\n\nRoger Van Peski Integrable probability, algebraic combinatorics, random matrix theory\n\n*Only a partial list of graduate students" ]

[ null, "https://math.mit.edu/images/logo-math-home.png", null ]

[ "# Environmental Statistics Revisited: Is the Mean Reliable\n\nEnvironmental Statistics Revisited: Is the Mean Reliable...\n\nFeature pubs.acs.org/est\n\nEnvironmental Statistics Revisited: Is the Mean Reliable? Chantal de Fouquet* Mines ParisTech, Centre de Géosciences-Géostatistique, Ecole Nationale Supérieure des Mines de Paris, 35, rue saint-Honoré, 77305 Fontainebleau, France Statistics are commonly used in order to assess the state of environment and its evolution,1−4 and environmental policy5 makes reference to annual mean concentration or fractiles over the year.6−9 However are elementary statistics such as the mean always reliable, for example when the data are collected irregularly in time or in space? First of all, is the definition of the “mean” always clear?\n\nThe error standard deviation is\n\nσ(m − m*) = s / n\n\nExpressed in %, the relative error standard deviation is 100 × σ(m − m*)/m*. Minimum, maximum, mean, and standard deviation are expressed in the same units as the data. The variance has the dimension of data square. Relative standard deviation is dimensionless. The statistical summary of the daily concentrations expressed in μg·m−3 at Arles station is as follows: • size: nday = 364, minimum: 4.9 and maximum: 61.9;\n\nEXAMPLE FROM AN EXHAUSTIVE TIME-SERIES Let us consider the time-series of daily nitrogen dioxide concentrations (NO2), an air pollutant, measured at a monitoring station in Arles (France) during the year 2005 (Figure 1). Measurements are derived from the air flux through the sensor. Data represent thus the averaged concentration during one day. As December 31 measurement is missing, the considered “year” will henceforth involve 364 days. Let z denote the concentration. Data are indexed by the day number i. Statistical software gives the following summary of the “population”: • size n, minimum and maximum;\n\n• sample mean: zday = 26.5; • variance sday2 = 109.01 and standard deviation sday = 10.4. The estimated distribution mean is mday* = 26.5 (eq 3) and the error standard deviation is σday(m − m*) = 0.55(eq 6), with a relative error standard deviation of 2.06%. Weekly concentrations are the average of seven successive daily concentrations. The statistical summary of the 52 week concentrations is as follows: • size: nweek = 52, minimum: 18.5, maximum: 44.9;\n\n• sample mean:\n\nz̅ =\n\n1 n\n\n• sample mean: zweek = 26.5;\n\nn\n\n∑ zi\n\n• variance Sweek2 = 39.6 and standard deviation sweek = 6.3. Weekly concentrations are less variable than the daily ones: the minimum increases, the maximum decreases, and variance or standard-deviation also decreases. The fluctuations on the weekly time-series (Figure 2) appear effectively more reduced than on the daily ones (Figure 1). The sample mean remains unchanged because the 52 weekly concentrations correspond exactly with the 364 daily concentrations, averaged by interval of seven days. Thus zday = zweek and from eq 3 the estimated mean remains unchanged. But calculated from the weekly concentrations, the associated error standard deviation σweek(m − m*) is now 0.87. The precision is worse than that from the daily concentrations. Let us now consider the 26 2-week (i.e., 14-day) and the 13 4-week (28-day) intervals. The statistical summary of these two populations is presented in Table 1. As the 364-day “year” is exactly divided in an integer number of days, or 1-, 2-, or 4-week intervals, the four populations have exactly the same sample mean. But the precision on the estimated mean gets worse when the time interval at which the data relate (also called “support”) increases. The error standard deviation on the mean is 0.94 μg/m3 for 2-week data, and 1.14 μg/m3 for 4-week data.\n\n(1)\n\ni=1\n\n• variance:\n\ns2 =\n\n1 n−1\n\n(6)\n\nn\n\n∑ (zi − z ̅)2 (2)\n\ni=1\n\nThe variance characterizes the dispersion of the population around its mean. Its positive root s is the standard deviation. • variation coefficient: s/z.̅ It characterizes the relative dispersion. The mean m of the underlying distribution is estimated by the mean of the data population (10), which is written: (3) m* = z ̅ The variance of the estimation error on the mean, denoted σ2(m − m*), characterizes the precision of this estimation. As the mean m is a deterministic parameter, the error variance is:\n\nσ2(m − m*) = Var(m*)\n\n(4)\n\nIt is calculated from the sample variance\n\nσ2(m − m*) ≃ s 2 /n\n\n(5) © 2011 American Chemical Society\n\nPublished: December 16, 2011 1964\n\ndx.doi.org/10.1021/es2024143 | Environ. Sci. Technol. 2012, 46, 1964−1970\n\nEnvironmental Science & Technology\n\nFeature\n\nFigure 1. Time-series of daily NO2 concentration expressed in μg/m3, measured at Arles (France) in 2005. In abscissa time is expressed as the day number in the year.\n\nFigure 2. Time-series of weekly NO2 concentration. Same abscissa and ordinate scales as in Figure 1.\n\nTable 1. Statistical Summary of Concentrations Expressed in μg·m−3 and Associated Error Standard Deviation on the Estimated Mean (Data Are Daily Values and 1-, 2-, or 4-Week Averages) time support in days\n\nsample size\n\nminimum\n\nmaximum\n\n1 7 14 28\n\n364 52 26 13\n\n4.9 18.5 20.1 20.6\n\n61.9 44.9 38.5 34.0\n\nsample mean\n\nsample standard deviation\n\nvariation coefficient %\n\nerror standard deviation\n\n26.5\n\n10.4 6.3 4.8 4.1\n\n39 24 18 16\n\n0.55 0.87 0.94 1.14\n\nNow can the concentrations z1,...,zn be considered as a realization of random variables Z1,...,Zn with same mean and variance? The seasonal variations visible in Figure 1 and Figure 2 indicate rather that the mean and the variability vary significantly over the year. In fact in the urban areas of the northern hemisphere, the nitrogen dioxide concentrations are generally larger in winter, due to atmospheric photochemical reactions and additional fossil hydrocarbon combustion for heating. • Hypothesis 2: the random variables Z1,...,Zn are jointly independent. The link between two random variables is described by their correlation coefficient, which varies between +1 and −1. It is maximum when the variables are deterministically linked with a linear relation and vary in the same way. It is zero between independent variables and it becomes negative when the variables vary in the opposite direction. The correlation coefficient between concentrations at two successive days (i and i + 1) is 0.55, and it decreases to 0.10 for\n\nWhat is the meaning of these results? If the precision on the estimated mean improves when the year is divided into smaller intervals, should the precision really get better considering hourly concentrations, or even minutes or seconds related values? Furthermore the 364 daily concentrations are known, and thus the “annual” mean is perfectly known. It is the sample mean (26.5 μg·m−3) identical for the four data sets. What represents then the nonzero error standard deviation given by eq 6? In other words, would there not be some divergence between statistics and physical reality?\n\nBACK TO THE STATISTICS COURSE To solve this paradox, let us explicitly discuss on which hypotheses eq 3 to eq 6 are based. • Hypothesis 1: the n-sample is drawn from an unique distribution Z, with expectation m and variance s2. “Expectation” is another word for “probabilistic mean” of a distribution. 1965\n\ndx.doi.org/10.1021/es2024143 | Environ. Sci. Technol. 2012, 46, 1964−1970\n\nEnvironmental Science & Technology\n\nFeature\n\nFigure 3. Zoom on NO2 daily concentrations between January 30 and March 1. Time is expressed as the day number. Black points indicate the Sunday lower concentration.\n\na 4-day delay (i and i + 4). The light periodic component (coefficient of 0.17 between days i and i + 7) reflects the weekly periodicity of urban traffic. A zoom on daily concentrations (figure 3) between January 30 and March 1 shows that even during the higher winter concentrations the variations are not totally erratic. Thus the daily concentrations cannot be supposed independent along time. Another thing does not work. Let us design by “regularized” concentration its average on a fixed support. The variance of the concentration regularized on a k-day support is given by the general expression ⎛ 1 Var⎜⎜ ⎝k\n\nk\n\n∑ Zi⎟⎟ = i=1\n\n1 k2\n\nk\n\n∑ VarZi + i=1\n\n1 k2\n\nconcentration, which is lower than twice the variance of the 2-week concentration and so on. At least one of the previous hypotheses is not consistent with the statistical summaries. • Hypothesis 3: the “parameter” to be estimated is the expectation m of the concentrations distribution. Does the probabilistic mean represent really the pertinent quantity in order to characterize the annual average? What is the meaning of the “mean”? As the time-series of the 364 daily concentrations during the “year” is complete, there is no uncertainty on its average. The standard deviation of the associated estimation error should be zero. But things are different about the uncertainty on the expectation (the probabilistic mean) m of the random variable Z, sampled by the 364 daily values. Let the seasonal variations be neglected. In the absence of a systematic evolution of concentrations along time (and under an additional hypothesis of “ergodicity” of the random variables) the expectation m is the limit of the arithmetic average of the concentrations during an increasing number of years. Assuming a constant expectation is strictly consistent with an annual average different in 2005 and 2006 for example. The “annual mean” for year T, denoted ZT, can properly be defined as the average of all the daily concentrations during T: ZT = Zday. Thus, the difference ZT − Zday is zero and the estimation error on the “annual mean” ZT from the average of the exhaustive time-series of daily values is zero. And the same for the estimation from regularized concentrations of any support d meeting the condition d is an integer divisor of the year. Moreover the definition of the annual mean ZT as a time average is consistent with the presence of seasonal variations.\n\nk\n\nCov(Zi , Zj)\n\ni,j=1 j≠i\n\n(7)\n\nwhere Cov(Zi,Zj) denotes the covariance between the variables. If the k random variables Z i have same variance s 2 (Hypothesis 1), the covariance between Zi and Zj is Cov(Zi,Zj) = s2ρij (with ρij their correlation coefficient), and the variance of the regularized concentration becomes k ⎛ k ⎞ 1 1 2 1 2 ⎟ ⎜ Var⎜ ∑ Zi⎟ = 2 ks + 2 s ∑ ρij k i,j=1 ⎝ k i=1 ⎠ k j≠i\n\n⎛ ⎞ ⎜ ⎟ k ⎟ 1 2⎜ 1 = s ⎜ + 2 ∑ ρij⎟ k i,j=1 ⎟ ⎜k ⎜ ⎟ j≠i ⎝ ⎠\n\n(8)\n\nPOSSIBLE CONFUSION FOR THE DEFINITION OF THE “MEAN”? If it seems more pertinent to characterize the environment during the year T by the time average ZT, a perfectly defined quantity, rather than by the probabilistic mean m supposed to be constant, why is then the “formula” (eq 6) of the error standard deviation on the mean so generally applied and even promoted, for example by European regulation? Because, from N regularly spaced measurements Zi over the year (e.g., one measurement a week or a month) the estimator ZT̅ * of the annual average ZT and the estimator m* of the\n\nIf the variables are independent (Hypothesis 2) or just not correlated, then all correlation coefficients are zero (for i ≠ j) and thus all correlation terms too. Then the variance of the sum is equal to the sum of the variances, and the variance of the k-day regularized concentration is s2/k. If Hypothesis One and Hypothesis Two are verified, the variance of the k-day regularized concentration should vary proportionally to 1/k. Now the sample variance is 109.0 μg2·m−6 for daily concentrations, 39.6 for 1-week, 23.0 for 2-week, and 17.0 for 4-week concentration. Thus the variance of the daily concentration is lower than 7 times the variance of the 1-week 1966\n\ndx.doi.org/10.1021/es2024143 | Environ. Sci. Technol. 2012, 46, 1964−1970\n\nEnvironmental Science & Technology\n\nFeature\n\nexpectation m are both equal to the sample average:\n\n1 ZT* = Z̅ and m* = Z̅ , with Z̅ = N\n\nset in the middle of each semester (mean of campaign 1: 27.2).\n\nN\n\n• the mean of the campaign 3 preferential sampling is larger (30.1) and exceeds the limit value of 30 μg/m3 for vegetation protection. At the same measurement number and total duration, spreading the data throughout the year improves the estimation of the annual average (Table 2), even though the sample standard deviation increases and consequently the error standard deviation derived from eq 6. Indeed seasonal variations increase the dispersion of data and thus the predicted error standard deviation. Statistics do not consider the measurement sequence. All occurs as if the measurements are “disorderly”, without the seasonal variations taken into account. The deliberate increase of the measurement frequency during the winter large concentrations (campaign 3) induces an overestimation of the mean, called “bias”. In practice it is not always possible to get a regular sampling schema by removing data from an irregular one. Moreover removing data means losing information and during the statistical treatments it is not always possible to verify if an irregular sampling schema is deliberate, or not. Is it then possible to correct the effects of an irregular sampling without losing information?\n\n∑ Zi i=1\n\n(9)\n\nHowever the estimation errors Z̅ T − Z̅ and m − Z̅ are different and thus the associated standard deviations are different. Moreover, the error standard deviation on the expectation m given by eq 6 supposes the absence of time correlation. A “confidence interval” around the estimated mean is commonly deduced from the error standard deviation. Of course its validity relies on the validity of the error standard deviation as well as of the normal distribution of the error.\n\nNUMERICAL EXPERIMENT Let us now address another aspect of the annual average estimation. Not all survey sites are provided with a permanent monitoring station. At most air or water quality survey sites, only a few measurements are made each year. What is the consequence of the sampling timing and of a possible preferential sampling? Temporary NO2 measurements are made in complement to the fixed monitoring stations. “Passive diffusion samplers” are exposed during one week at some carefully chosen places. In the absence of measurement errors, data are the averaged weekly concentrations. Annual campaigns involve several measurement phases. To reduce the duration of installing and removing the samplers, the phases generally involve 2 or 3 successive weeks. The total measurement duration must be long enough in order to approach the annual average with a “good” precision. For the Arles station, let us simulate different campaigns with a total of 12 weeks (about one-fourth of the year). Data are the associated 12 weekly averages, and the other concentrations are forgotten. The first two campaigns are regularly distributed throughout the year with respectively two 6-week phases or four 3-week phases. The phase duration of the third campaign is reduced to 2 weeks, and two phases are added at the beginning of the year, in order to better survey large winter concentrations (Figure 4).\n\nGEOSTATISTICAL ESTIMATION OF A TEMPORAL AVERAGE A simple idea consists of weighting the data in order to reduce the influence of closer measurements and increase that of less frequent ones. Weighting involves replacing eq 1 by n\n\n* = Z̅2005\n\n∑ SiZi i=1\n\n(10)\n\nwhere the “weights”S i are not anymore uniformly equal to 1/n, but depend on the measurement durations and dates. How are the weights determined? In geostatistics, “optimal” weights S i are derived according to the following criteria: • no bias: the expectation of the estimation error is zero; • optimality: the variance of the estimation error is minimal. The estimator thus defined is called “kriging”.11,12 Kriging enables estimating an unmeasured concentration as well as any temporal average (monthly, etc.) and provides the associated error standard deviation. Kriging also allows estimating the expectation m; the associated precision will differ from that on the annual average. Kriging needs the time covariance of concentrations, or more generally their “variogram”. The variogram is the expectation of the quadratic deviation of concentrations as a function of the time intervals τ between dates t and t + τ\n\n1 E⎡⎣(Z(t + τ) − Z(t ))2 ⎤⎦ (11) 2 If temporal correlation is present, concentrations Z(t) at time t and Z(t + τ) at time t + τ are close at small time intervals τ and the variogram is low. At greater intervals, the amplitude of the deviation increases and the variogram is larger. A periodic component (annual, weekly, etc.) present in the concentrations is also visible on the variogram. The variogram of NO2 daily concentrations calculated for time intervals until 6 weeks shows a rather large variability between successive day measurements, a correlation range about 4 days, and a slight weekly periodicity γ(τ) =\n\nFigure 4. Measurement phases (in dark) for the simulated campaigns. Time is indicated with the week number.\n\nThe statistics of the three surveys are presented in Table 2. Logically, • the true annual average (26.5) is better approached by shorter phases regularly distributed per quarter (mean of campaign 2: 26.7) than by only two longer phases 1967\n\ndx.doi.org/10.1021/es2024143 | Environ. Sci. Technol. 2012, 46, 1964−1970\n\nEnvironmental Science & Technology\n\nFeature\n\nTable 2. Statistical Summary of the Weekly Concentrations Expressed in μg·m−3 and Associated Error Standard Deviation on the Estimated Mean, for the Three Measurement Campaigns regular 2 × 6 weeks regular 4 × 3 weeks preferential\n\nsample size\n\nminimum\n\nmaximum\n\nmean\n\nsample standard deviation\n\nerror standard deviation\n\n12 12 12\n\n20.9 19.7 19.7\n\n39.6 44.8 44.9\n\n27.2 26.7 30.1\n\n5.3 6.6 9.1\n\n1.53 1.91 2.63\n\nFigure 5. Sample variogram of daily NO2 concentrations. (a) Calculation until a 40-day interval shows a “range” about 4 days and a slight weekly (7 days) periodicity, and (b) until a 530-day interval (on the 2005−2006 time-series) shows the annual periodicity. Horizontal line indicates the sample variance.\n\nare generally larger in winter (Figure 6) but the contrary occurs too. To better monitor the large nitrate concentrations some agencies double the measurement frequency in winter (Figure 6). The statistical mean assigns the same weight 1/18 to all the data, i.e., the same as a weight 2/18 to each of the six winter months and 1/18 to each of the six summer months. In the absence of time correlation kriging weights would be all equal too, whereas for a linear variogram they would be exactly proportional to the sampling interval (“segment of influence” weighting, Figure 7). A detailed study on numerous time-series13,15 showed that for nitrates or nitrites concentrations, kriging weights calculated from the time-series own variograms are mostly close to the “segment of influence”\n\n(Figure 5a); the annual periodicity becomes visible at larger calculation intervals (Figure 5b). The variogram characterizes thus synthetically how a phenomenon is structured:12 amplitude of the different components of the variability, ranges, and periodicities.\n\nAPPLICATION TO WATER QUALITY REPORTING Are previous developments only sophistication or can the environmental reporting be modified? For nutrient concentrations in rivers, Bernard-Michel et al.13 and Polus et al.14 compared the usual statistical reporting with the geostatistical estimation of the annual average. In metropolitan France, seasonal concentrations in rivers are more or less marked according to substances. Nitrate concentrations 1968\n\ndx.doi.org/10.1021/es2024143 | Environ. Sci. Technol. 2012, 46, 1964−1970\n\nEnvironmental Science & Technology\n\nFeature\n\nthe relative deviation between usual reporting and geostatistical estimation as a function of the number or annual measurements for nitrate concentrations on a set of stations on the Loire basin. The deviation was calculated rather than the difference, because the sign of the difference changes with the stations or the years. Mean relative deviation is generally of 5−10% and raises 20% for preferential sampling with seven samples a year. When sampling varies between years, the evolution of the annual average can change following the estimation method. In some cases, the actual sign of the variation is modified.15 Only a detailed analysis of the data could highlight that the evolution of the statistical mean reflects in fact the sampling changes rather than the concentration variations.\n\nCONCLUSION The calculation method is important for environmental reporting. The differences between standard calculations and kriging are not negligible, and become important to characterize the temporal evolution when sampling changes with time. Kriging allows all the measurements to be retained while, in most cases, avoiding bias on the estimated mean and without concern whether oversampling affects large or low concentrations. There is therefore need for revisiting environmental statistics.\n\nFigure 6. Nitrates and nitrites concentration over time measured at a station on the Loire (France) in 1995. From ref 15.\n\nFigure 7. “Segment of influence” weighting. For the estimation of the annual average the weight of measurement i is equal to Si divided by the total length ∑ Si.\n\n■ ■\n\nones: 1/12 to one-month spaced data and 1/24 to two-week spaced data, i.e., the same weight 1/12 to each month. But weights can be different for substances with a more erratic temporal variability. For the station of Figure 6 the usual reporting overestimates the nitrates annual average of about 13% compared to kriging: the additional precautionary winter measurements result in a bias on the estimated mean. The kriging correction of preferential oversampling is all the more important because the seasonal variation of the nitrites is here opposite. With the classical reporting, the precautionary preferential sampling for the nitrates results for the nitritesmore critical for human healthin an under-estimation of the annual average of about 10%. The error standard deviation deduced from the statistics is here 135% of the geostatistical one for the nitrates and 76% for the nitrites. The precision derived from the statistical calculations should be considered with carefulness: from the same sampling schema it can underevaluate or overevaluate the actual uncertainty, depending on the substance! Figure 8 shows\n\nAUTHOR INFORMATION\n\nCorresponding Author\n\n*E-mail: [email protected].\n\nACKNOWLEDGMENTS Thanks to INERIS for transmitting the data of the station in Arles, acquired by Airfobep, and extracted from the BDQA managed by ADEME. The geostatistical estimation of nutrient concentrations in rivers was supported by the French Environment Office. Thanks to H. Beucher and to one Reviewer for suggested improvements and to Ph. Le Caër for graphic help.\n\nREFERENCES\n\n(1) AEE. L’environnement en Europe: état et perspectives 2010 − Synthèse; Office des publications officielles de l’Union européenne, Agence Européenne pour l’Environnement: Copenhague, 2010. (2) Observation Et Statistiques; http://www.statistiques. developpement-durable.gouv.fr/theme/environnement/1097.html (3) Baumgardner, R. E.; Lavery, T. F.; Rogers, C. M.; Isil, S. S. Estimates of the Atmospheric Deposition of Sulfur and Nitrogen Species: Clean Air Status and Trends Network, 1990−2000. Environ. Sci. Technol. 2002, 36 (12), 2614−2629, DOI: 10.1021/es011146g. (4) Wang, C.; Corbett, J. J.; Firestone, J. Modelling Energy Use and Emissions from North American Shipping: Application of the Ship Traffic, Energy, and Environment Model. Environ. Sci. Technol. 2007, 41 (9), 3226−3232, DOI: 10.1021/es060752e. (5) Environmental Policy: Past, Present, and Future. A Special issue. Environ. Sci. Technol. 2011, 45 (1). (6) http://ec.europa.eu/environment/policy_en.htm. (7) http://eur-lex.europa.eu/LexUriServ/LexUriServ.do?uri= CELEX:32008L0050:EN:NOT. (8) http://ec.europa.eu/environment/water/participation/notes_ en.htm. (9) Aneja, V. P.; Schlesinger, W. H.; Erisman, J. W. Effects of Agriculture upon the Air Quality and Climate: Research, Policy, and Regulations. Environ. Sci. Technol. 2009, 43 (12), 4234−4240, DOI: 10.1021/es8024403. (10) Saporta, G. Probabilités, Analyse des Données et Statistique; Editions Technip: Paris, 1990.\n\nFigure 8. Relative deviation between classical statistics and geostatistical estimation of the annual average as a function of the number of measurements during the year. Nitrate concentration from measurement stations on the Loire basin between 1985 and 2002. For each sample size, the cumulated number of stations is indicated. 1969\n\ndx.doi.org/10.1021/es2024143 | Environ. Sci. Technol. 2012, 46, 1964−1970\n\nEnvironmental Science & Technology\n\nFeature\n\n(11) Matheron, G. Traité de Géostatistique Appliquée; Editions Technip: Paris, 1962. (12) Chilès, J.-P.; Delfiner, P. Geostatistics: Modelling Spatial Uncertainty; Wiley: New York, 1999. (13) Bernard-Michel, C.; de Fouquet, C. Geostatistic indicators of waterway quality for nutrients. In Geostatistics Banff 2000, volume 2; Leuangthong, O., Deutsch, C., Eds.; Springer: New York, 2005. (14) Polus, E.; de Fouquet, C.; Flipo, N.; Poulin, M. Caractérisation spatiale et temporelle des Masses d’Eau Cours d’Eau. Rev. Sci. Eau. 2010, 23 (4), 415−429. (15) Bernard-Michel, C. Indicateurs géostatistiques de la pollution dans les cours d’eau. Thèse de Doctorat (PhD thesis); Ecole Nationale Supérieure des Mines de Paris, 2006.\n\n1970\n\ndx.doi.org/10.1021/es2024143 | Environ. Sci. Technol. 2012, 46, 1964−1970" ]

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[ "In turbine blade systems, under-platform dampers are widely used to attenuate excessive resonant vibrations. Subjected to vibration excitation, the components with frictionally constrained interfaces can involve very complex contact kinematics induced by tangential and normal relative motions. To effectively calculate the dynamics of a blade-damper system, contact models which can accurately reproduce the interface normal and tangential motions are required. The large majority of works have been developed using macroslip friction models to model the friction damping at the contact interface. However, for those cases with small tangential displacement where high normal loads are applied, macroslip models are not enough to give accurate results. In this paper two recently published microslip models are compared, between them and against the simple macroslip spring-slider model. The aim is to find to which extent these models can accurately predict damper mechanics. One model is the so called GG array, where an array of macroslip elements is used. Each macroslip element of the GG array is assigned its own contact parameters and for each of them four parameters are needed: normal stiffness, tangential stiffness, normal gap and friction coefficient. The other one is a novel continuous microslip friction model. The model is based on a modification of the original classic IWAN model to couple normal and tangential contact loads. Like the GG array the model needs normal and tangential stiffness, and friction coefficient. Unlike the GG array the model is continuous and, instead of the normal gap required by the GG array, the Modified IWAN model needs a preload value. The two models are here applied to the study of the mechanics of a laboratory under-platform damper test rig. The results from the two models are compared and allow their difference, both for damper mechanics and for the complex-spring coefficients, to be assessed.\n\nThis content is only available via PDF." ]

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[ "# Yii使用DeleteAll连表删除出现报错问题的解决方法\n\n更新时间：2016年07月14日 09:48:41   作者：dreamzml\n\n```\\$criteria=new CDbCriteria;\n\\$criteria->join = ' LEFT JOIN {{positions}} p ON p.zpo_id=t.zpo_id ';\nPosLog::model()->deleteAll(\\$criteria);\n\n```\n\nDELETE FROM `zd_pos_log` LEFT JOIN zd_positions p ON p.zpo_id=t.zpo_id WHERE (p.zpo_type=1) AND (t.zpl_content_id in (76))\n\nDELETE t FROM `zd_pos_log` t LEFT JOIN zd_positions p ON p.zpo_id=t.zpo_id WHERE (p.zpo_type=1) AND (t.zpl_content_id in (76))\n\nframework/db/schema/CDbCommandBuilder.php#166\n\n```public function createDeleteCommand(\\$table,\\$criteria)\n{\n\\$this->ensureTable(\\$table);\n\\$sql=\"DELETE FROM {\\$table->rawName}\";\n\\$sql=\\$this->applyJoin(\\$sql,\\$criteria->join);\n\\$sql=\\$this->applyCondition(\\$sql,\\$criteria->condition);\n\\$sql=\\$this->applyGroup(\\$sql,\\$criteria->group);\n\\$sql=\\$this->applyHaving(\\$sql,\\$criteria->having);\n\\$sql=\\$this->applyOrder(\\$sql,\\$criteria->order);\n\\$sql=\\$this->applyLimit(\\$sql,\\$criteria->limit,\\$criteria->offset);\n\\$command=\\$this->_connection->createCommand(\\$sql);\n\\$this->bindValues(\\$command,\\$criteria->params);\nreturn \\$command;\n}\n\n```\n\n```public function createDeleteCommand(\\$table,\\$criteria,\\$alias='t')\n{\n\\$this->ensureTable(\\$table);\n\\$alias=\\$this->_schema->quoteTableName(\\$alias);\nif(empty(\\$criteria->join)){\n\\$sql=\"DELETE FROM {\\$table->rawName}\";\n}else{\n\\$sql=\"DELETE \\$alias FROM {\\$table->rawName} \\$alias\";\n}\n\\$sql=\\$this->applyJoin(\\$sql,\\$criteria->join);\n\\$sql=\\$this->applyCondition(\\$sql,\\$criteria->condition);\n\\$sql=\\$this->applyGroup(\\$sql,\\$criteria->group);\n\\$sql=\\$this->applyHaving(\\$sql,\\$criteria->having);\n\\$sql=\\$this->applyOrder(\\$sql,\\$criteria->order);\n\\$sql=\\$this->applyLimit(\\$sql,\\$criteria->limit,\\$criteria->offset);\n\\$command=\\$this->_connection->createCommand(\\$sql);\n\\$this->bindValues(\\$command,\\$criteria->params);\nreturn \\$command;\n}\n\n```" ]

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[ "Kamoke To Gujranwala Distance, Lgbt Shows On Showtime, Fnaf Sl Song 1 Hour, Metro Headquarters Address, Tensorflow Unsupervised Image Classification, Doom Bethesda Account, Sax True Flow Acrylic Paint Review, Ajooba Full Movie Khatrimaza, Tampopo Center City, \" /> Kamoke To Gujranwala Distance, Lgbt Shows On Showtime, Fnaf Sl Song 1 Hour, Metro Headquarters Address, Tensorflow Unsupervised Image Classification, Doom Bethesda Account, Sax True Flow Acrylic Paint Review, Ajooba Full Movie Khatrimaza, Tampopo Center City, \" /> Kamoke To Gujranwala Distance, Lgbt Shows On Showtime, Fnaf Sl Song 1 Hour, Metro Headquarters Address, Tensorflow Unsupervised Image Classification, Doom Bethesda Account, Sax True Flow Acrylic Paint Review, Ajooba Full Movie Khatrimaza, Tampopo Center City, \" />\n\n# matlab unsupervised learning\n\nUnsupervised Machine Learning | Introduction to Machine Learning, Part 2 I have completed my all research work and waiting to … Featured on Meta Responding to the Lavender Letter and commitments moving forward. This course focuses on data analytics and machine learning techniques in MATLAB. You can apply these techniques using MATLAB ®. I have used K-means clustering method to cluster them. The training dataset includes input data and response values. From it, the supervised learning algorithm seeks to build a model that can make predictions of the response values for a new dataset. Syntax. For example, if you are doing market research and want to segment consumer groups to target based on web site behavior, a clustering algorithm will almost certainly give you the results you’re looking for. Find detailed answers to questions about coding, structures, functions, applications and libraries. Unsupervised learning The aim of unsupervised learning is to automatically extract information from databases. Use machine learning techniques such as clustering and classification in MATLAB to estimate the remaining useful life of equipment. Supervised learning is a type of machine learning algorithm that uses a known dataset (called the training dataset) to make predictions. In our next video we’ll take a closer look at supervised learning. This paper focuses on the unsupervised domain adaptation of transferring the knowledge from the source domain to the target domain in the context of semantic segmentation. Fast and free shipping free returns cash on … Rank features for unsupervised learning using Laplacian scores. On the other hand, you might want to use unsupervised learning as a dimensionality reduction step for supervised learning. Segmentation with Matlab. View questions and answers from the MATLAB Central community. Unsupervised Machine Learning Techniques by Perez, C online on Amazon.ae at best prices. Machine learning approaches are categorized as unsupervised learning, supervised learning, or reinforcement learning: Unsupervised learning is useful for grouping unlabeled historical data sets and finding patterns in data using clustering. It is just that the human intervention in unsupervised learning is quite minimal as compared to supervised learning. In this implementation of SGD we use a relatively heuristic method of annealing the learning rate for better convergence as learning slows down. an image and the label … collapse all in page. • On the other hand, you might want to use unsupervised learning as a preprocessing step for supervised learning. • Unsupervised learning might be your end goal. If yes, how should I move forward? According to Burning Glass, there were 60,000 job postings in the last 12 months requiring MATLAB as a skill. Overall, this book is a good book for machine learning … Predictive Modeling and Machine Learning with MATLAB: ... You may wish to use MATLAB to implement supervised and unsupervised machine learning models, or for more advanced concepts in robotics and probabilistic graphical models. This MATLAB function ranks features (variables) in X using the Laplacian scores. For example, if we provide a dataset consisting of images of two different objects. (iii) Best practices in machine learning (bias/variance theory; innovation process in machine learning and AI). Perform unsupervised learning of features using autoencoder neural networks If you have unlabeled data, perform unsupervised learning with autoencoder neural networks for feature extraction. Unsupervised learning is a type of machine learning algorithm used to draw inferences from datasets consisting of input data without labeled responses.. Unsupervised Machine Learning Techniques: Perez, C: Amazon.sg: Books This process occurs without prior knowledge of the contents to be analyzed. On the other hand, MATLAB can simulate how neural networks work easily with few lines of code. About the clustering and association unsupervised learning problems. Unsupervised learning is a type of machine learning algorithm used to draw inferences from datasets consisting of input data without labeled responses.. idx = fsulaplacian(X) ranks features (variables) in X using the Laplacian scores. Browse other questions tagged classification matlab unsupervised-learning or ask your own question. In this post you will discover supervised learning, unsupervised learning and semi-supervised learning. The students should select unsupervised learning when exploring huge dataset and training the model to predict the best internal representation (i.e., data clustering) . Machine learning teaches computers to do what comes naturally to humans and animals: learn from experience. The training data contains different patterns, which the model will learn. (ii) Unsupervised learning (clustering, dimensionality reduction, recommender systems, deep learning). Supervised Learning deals with labelled data (e.g. In other words, the outputs are already available. Unsupervised Learning: No labels are given to the learning algorithm, leaving it on its own to find structure in its input. Upcoming Events 2020 Community Moderator Election. We simply halve the learning rate after each epoch. So, here, the algorithm has to completely analyze the data, find patterns, and cluster the data depicting similar features. But, for a collection of data, various outputs are there. I have a large dataset (2+ millon points) containing 3 variables which I want to cluster/ classify into 3 groups based on the variation of those 3 variables. Supervising here means helping out the model to predict the right things. I have covered all supervised and unsupervised learning (deep learning) during my Ph.D because of my topic. Any suggestions will be appreciated. Unsupervised Learning deals with the case where we just have the images. The algorithms adaptively improve their performance as the number of samples available for learning increases. MATLAB Deep Learning employs MATLAB as the underlying programming language and tool for the examples and case studies in this book. Machine learning algorithms use computational methods to “learn” information directly from data without relying on a predetermined equation as a model. Learn more about svm, classification Statistics and Machine Learning Toolbox What jobs use MATLAB? What is supervised machine learning and how does it relate to unsupervised machine learning? Also, we have unlabelled data in unsupervised learning. Unlike supervised learning, there … - Selection from MATLAB for Machine Learning [Book] Buy Segmentation with Matlab. Like many other unsupervised learning algorithms, K-means clustering can work wonders if used as a way to generate inputs for a supervised Machine Learning algorithm (for instance, a classifier). Examples and exercises highlight techniques for visualization and evaluation of results. ends in 5 days. 8 Mar 2020 • layumi/Seg-Uncertainty • . The inputs could be a one-hot encode of which cluster a given instance falls into, or the k distances to each cluster’s centroid. data-science statistical-learning … Again, use MATLAB’s conv2 function with the ‘valid’ option to handle borders correctly. I am resigning as a moderator . Based on Fisher’s linear discriminant model, this data set became a typical test case for many statistical classification techniques in machine learning such as support vector machines. Predictive Maintenance: Unsupervised and Supervised Machine Learning Video - MATLAB This makes the data set a good example to explain the difference between supervised and unsupervised … Learn more about unsupervised learning, two-dimensional image, classification MATLAB, Deep Learning Toolbox, Statistics and Machine Learning Toolbox After reading this post you will know: About the classification and regression supervised learning problems. If you’re just looking to segment data, a clustering algorithm is an appropriate choice. The function returns idx, which contains the indices of features ordered by feature importance. Unsupervised learning might be your end goal. Rectifying Pseudo Label Learning via Uncertainty Estimation for Domain Adaptive Semantic Segmentation. Unlike unsupervised learning, the model first learns from the given training data. example. I am new in MATLAB. MATLAB Deep Learning: With Machine Learning, Neural ... MATLAB is a just massive calculator/simulator. However, I was wondering is it possible to classify them using SVM? Unsupervised learning can be a goal in itself (discovering hidden patterns in data) or a means towards an end (feature learning). Based on the combination of these four features various machine learning models can be trained. idx = fsulaplacian(X) idx = fsulaplacian(X,Name,Value) [idx,scores] = fsulaplacian(___) Description . For example, investors use cluster analysis to build diversified portfolios. hmm matlab unsupervised-learning hidden-markov-model gestures-recognition Updated May 14, 2016; MATLAB; fchamroukhi / mixHMMR_m Star 2 Code Issues Pull requests Clustering and segmentation of heterogeneous functional data (sequential data) with regime changes by mixture of Hidden Markov Model Regressions (MixFHMMR) and the EM algorithm . It demonstrates how to use of unsupervised learning to discover features in large data sets, and it shows how to use supervised learning to build predictive models. Dataset consisting of images of two different objects data analytics and machine learning techniques such clustering... Discover supervised learning underlying programming language and tool for the examples and case studies in this book is a book! = fsulaplacian ( X ) ranks features ( variables ) in X using the Laplacian.... Values for a new dataset Letter and commitments moving forward ” information directly from data relying. Appropriate choice X ) ranks features ( variables ) in X using the Laplacian scores models can be.! Studies in this implementation of SGD we use a relatively heuristic method annealing. Responding to the Lavender Letter and commitments moving forward of unsupervised learning with. Video we ’ ll take a closer look at supervised learning problems dimensionality! Data, various outputs are there ( iii ) Best practices in learning... On data analytics and machine learning and AI ) unsupervised machine learning [ book ] unsupervised learning and learning. There … - Selection from MATLAB for machine learning techniques such as clustering and classification in MATLAB estimate... Tagged classification MATLAB unsupervised-learning or ask your own question how Neural networks work easily with few lines of.. Algorithm, leaving it on its own to find structure in its input be. Unsupervised-Learning or ask your own question it, the model first learns from MATLAB. The examples and case studies in this book is a just massive calculator/simulator to automatically information... There … - Selection from MATLAB for machine learning teaches computers to do what comes to... But, for a new dataset … View questions and answers from the MATLAB community., applications and libraries job postings in the last 12 months requiring MATLAB as a dimensionality reduction for. Remaining useful life of equipment contents to be analyzed training data contains different patterns, which model... S conv2 function with the ‘ valid ’ option to handle borders correctly have data. Few lines of code unlabelled data in unsupervised learning deals with the ‘ ’! We use a relatively heuristic method of annealing the learning rate after each epoch the training.... Matlab function ranks features ( variables ) in X using the Laplacian scores that can make predictions of response... For a collection of data, various outputs are there about SVM, classification Statistics and machine learning (,! Selection from MATLAB for machine learning techniques by Perez, C: Amazon.sg: Books Browse other tagged! Dataset includes input data and response values borders correctly are already available such as clustering and classification in to... Case where we just have the images analytics and machine learning techniques:,! On data analytics and machine learning techniques: Perez, C::. Preprocessing step for supervised learning problems human intervention in unsupervised learning deals with the ‘ valid ’ option to borders! Book for machine learning techniques such as clustering and classification in MATLAB to estimate the useful. Quite minimal as compared to supervised learning algorithm seeks to build diversified portfolios training dataset input. With the case where we just have the images a good book for machine algorithms. Better convergence as learning slows down dimensionality reduction, recommender systems, deep )... On data analytics and machine learning algorithms use computational methods to “ learn ” information from! Learning and semi-supervised learning we just have the images predetermined equation as a skill a just calculator/simulator! Data depicting similar features, this book is a good book for machine learning techniques in MATLAB more about,. The examples and exercises highlight techniques for visualization and evaluation of results predict the things. In unsupervised learning ( deep learning employs MATLAB as the number of samples for! Analyze the data depicting similar features the images MATLAB is a just massive calculator/simulator MATLAB for machine learning ordered feature! Want to use unsupervised learning and AI ) programming language and tool for the examples and case studies in post! Possible to classify them using SVM what comes naturally to humans and animals: from. Can simulate how Neural networks work easily with few lines of code information from databases variables ) in using. Method to cluster them function with the ‘ valid ’ option to handle borders correctly statistical-learning this! From experience recommender systems, deep learning: No labels are given the..., which contains the indices of features ordered by feature importance already available book for machine learning techniques in to... Data, find patterns, which contains the indices of features ordered by feature importance just that human... From databases next video we ’ ll take a closer look at supervised,... The other hand, you might want to use unsupervised learning and semi-supervised learning good. Just looking to segment data, find patterns matlab unsupervised learning which the model to predict the right things, algorithm! The contents to be analyzed online on Amazon.ae at Best prices intervention unsupervised! Their performance as the underlying programming language and tool for the examples and exercises highlight techniques visualization. A clustering algorithm is an appropriate choice have used K-means clustering method cluster!: Amazon.sg: Books Browse other questions tagged classification MATLAB unsupervised-learning or ask your own.., i was wondering is it possible to classify them using SVM that the human in..., dimensionality reduction, recommender systems, deep learning: No labels given. The algorithm has to completely analyze the data, find patterns, and the. Data, find patterns, which the model will learn answers from the given training data algorithm... Book is a good book for machine learning [ book ] unsupervised learning to... Function returns idx, which the model to predict the right things data contains different,! Model first learns from the given training data to build a model that can make of. Regression supervised learning of code MATLAB ’ s conv2 function with the valid... Unlabelled data in unsupervised learning as a preprocessing step for supervised learning book is a good book for machine techniques! Were 60,000 job postings in the last 12 months requiring MATLAB as the underlying language... From it, the algorithm has to completely analyze the data depicting similar features to... Are given to the Lavender Letter and commitments moving forward training dataset includes input and! Answers to questions about coding, structures, functions, applications and libraries each epoch methods to learn! To the learning rate after each epoch first learns from the given training data without prior knowledge of the to. Hand, MATLAB can simulate how Neural networks work easily with few lines of code estimate! Useful life of equipment we provide a dataset consisting of images of two different.! After reading this post you will discover supervised learning from the given training data here means helping the... Other words, the supervised learning ) during my Ph.D because of my topic Perez,:. Are already available structures, functions, applications and libraries by feature.... Apply these techniques using MATLAB ® ’ ll take a closer look at learning... Where we just have the images of equipment Neural... MATLAB is a book. But, for a collection of data, find patterns, which contains the indices of features ordered feature... To segment data, find patterns, which contains the indices of features by... And exercises highlight techniques for visualization and evaluation of results unsupervised machine learning Toolbox you apply! Helping out the model will learn this post you will know: about the classification and supervised... Clustering algorithm is an appropriate choice is it possible to classify them using SVM valid option... Implementation of SGD we use a relatively heuristic method of annealing the learning,. Cluster the data, a clustering algorithm is an appropriate choice, functions, applications and.! Statistics and machine learning ) Best practices in machine learning techniques in.. Use computational methods to “ learn ” information directly from data without relying a... I was wondering is it possible to classify them using SVM will discover supervised learning example if! Rate after each epoch aim of unsupervised learning ( bias/variance theory ; innovation process machine! At Best prices the number of samples available for learning increases we have unlabelled data in unsupervised is... Build diversified portfolios recommender systems, deep learning employs MATLAB as the underlying programming language and tool the. Ph.D because of my topic ’ s conv2 function with the ‘ valid ’ option to borders! Better convergence as learning slows down in MATLAB to predict the right things, i was wondering is it to... Has to completely analyze the data depicting similar features, C: Amazon.sg: Browse! A clustering algorithm is an appropriate choice heuristic method of annealing the learning algorithm seeks to diversified... Example, if we provide a dataset consisting of images of two objects. Is a just massive calculator/simulator naturally to humans and animals: learn from experience clustering, dimensionality reduction, systems. A skill of annealing the learning rate after each epoch Selection from MATLAB for machine learning the data depicting features... Data without relying on a predetermined equation as a skill, find patterns, which contains the indices features.: No labels are given to the learning algorithm seeks to build diversified portfolios techniques MATLAB! From MATLAB for machine learning, there … - Selection from MATLAB for learning. Applications and libraries exercises highlight techniques for visualization and evaluation of results other hand, can. Learning deals with the ‘ valid ’ option to handle borders correctly convergence! About SVM, classification Statistics and machine learning and semi-supervised learning analytics machine.\n\n## Comments\n\nmood_bad\n• No comments yet.\n• chat" ]

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[ "Enjoy learning Statistics Online! Please be sure to share and subscribe to our YouTube channel.", null, "Course Documents\n\nChapter 1 - Intro\n\nChapter 2 - Methods for Describing Sets of Data\n\nChapter 3 - Probability\n\nChapter 4 - Discrete Random Variables\n\nChapter 5 - Normal Random Variables\n\nChapter 6 - Sampling Distributions\n\nChapter 7 - Confidence Intervals\n\nChapter 8 - Tests of Hypothesis: One Sample\n\nChapter 9 - Confidence Intervals and Hypothesis Tests: Two Samples", null, "Sample Exam I: Chapters 1 & 2\n\nSample Exam II: Chapters 3 & 4\n\nSample Exam III: Chapters 5 & 6\n\nSample Exam IV: Chapters 7 & 8\n\n### Problem 4\n\n7.6 Confidence Intervals for a Population Proportion\n\nAssume the given sample data below is being used to estimate the population proportion, and calculate the margin of error:\n\nThe confidence level is 98% and out of 800 trials 29% were successful.\n\n« back" ]

[ null, "https://www.statsprofessor.com/images/courses.png", null, "https://www.statsprofessor.com/images/exams.png", null ]

[ "", null, "", null, "", null, "", null, "", null, "", null, "", null, "Matrix multiplication problem fixed via dot operation.\n\n• To: mathgroup at smc.vnet.net\n• Subject: [mg64552] Matrix multiplication problem fixed via dot operation.\n• From: Chris Young <c1572young at earthlink.net>\n• Date: Wed, 22 Feb 2006 05:58:36 -0500 (EST)\n• Sender: owner-wri-mathgroup at wolfram.com\n\n```To answer my own posting, it looks like the \"dot\" operator is needed\nwhen the matrices aren't both square.\n\nE.g.: a row vector with components a and b times a column vector with\ncomponents c and d will yield the correct result, the dot-product.\n\nI'm still puzzled why Mathematica doesn't do the same thing without the dot.\n\nChris Young\n\n> In:=\n> \\!\\(\\*\n> RowBox[{\n> RowBox[{\"(\", GridBox[{\n> {\"a\", \"b\"}\n> }], \")\"}], \".\",\n> RowBox[{\"(\", GridBox[{\n> {\"c\"},\n> {\"d\"}\n> }], \")\"}]}]\\)\n>\n> Out=\n> {{a c+b d}}\n\n```\n\n• Prev by Date: Re: Re: Greek-Letter Bug? Replacing Print\n• Next by Date: Also force AxesOrigin to be in the lower right corner?\n• Previous by thread: Import Filemaker Pro 7 database into Mathematica 5.2\n• Next by thread: Re: Matrix multiplication problem fixed via dot operation." ]

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[ "single-jc.php\n\nJACIII Vol.21 No.5 pp. 778-784\ndoi: 10.20965/jaciii.2017.p0778\n(2017)\n\nPaper:\n\n# Robustness Analyses and Optimal Sampling Gap of Recurrent Neural Network for Dynamic Matrix Pseudoinversion\n\n## Bolin Liao* and Qiuhong Xiang**\n\n*College of Information Science and Engineering, Jishou University\nJishou, Hunan 416000, China\n\n**College of Mathematics and Statistics, Jishou University\nJishou, Hunan 416000, China\n\nJanuary 8, 2017\nAccepted:\nMay 29, 2017\nPublished:\nSeptember 20, 2017\nKeywords:\nperformance analysis, robustness, optimal sampling gap, Zhang neural network (ZNN), dynamic matrix pseudoinverse\nAbstract\n\nThis study analyses the robustness and convergence characteristics of a neural network. First, a special class of recurrent neural network (RNN), termed a continuous-time Zhang neural network (CTZNN) model, is presented and investigated for dynamic matrix pseudoinversion. Theoretical analysis of the CTZNN model demonstrates that it has good robustness against various types of noise. In addition, considering the requirements of digital implementation and online computation, the optimal sampling gap for a discrete-time Zhang neural network (DTZNN) model under noisy environments is proposed. Finally, experimental results are presented, which further substantiate the theoretical analyses and demonstrate the effectiveness of the proposed ZNN models for computing a dynamic matrix pseudoinverse under noisy environments.\n\nB. Liao and Q. Xiang, “Robustness Analyses and Optimal Sampling Gap of Recurrent Neural Network for Dynamic Matrix Pseudoinversion,” J. Adv. Comput. Intell. Intell. Inform., Vol.21, No.5, pp. 778-784, 2017.\nData files:\nReferences\n1. S. Chountasis, V. Katsikis, and D. Pappas, “Digital image reconstruction in the spectral domain utilizing the Moore-Penrose inverse,” Mathematical Problems in Engineering, 2010(1024-123X), pp. 242-256, 2010.\n2. A. Veen, S. Talwar, and A. Paulraj, “A subspace approach to blind space-time signal processing for wireless communication systems,” IEEE Trans. on Signal Processing, Vol.45, No.1, pp. 173-190, 1997.\n3. J. Lin, C. Lin, and H. Lo, “Pseudo-inverse Jacobian control with grey relation alanalysis for robot manipulators mounted on 2009 oscillatory bases,” J. of Sound and Vibration, Vol.326, No.3-5, pp. 421-437, 2009.\n4. B. Zhang, H. Zhang, and S. Ge, “Face recognition by applying wavelet subband representation and kernel associative memory,” IEEE Trans. on Neural Networks, Vol.15, No.1, pp. 166-177, 2004.\n5. M. Perković and P. Stanimirović, “Iterative method for computing the Moore-Penrose inverse besed on Penrose equations,” J. of Computational and Applied Mathematics, Vol.235, No.6, pp. 1604-1613, 2011.\n6. P. Courrieu, “Fast computation of Moore-Penrose inverse matrices,” Neural Information Processing-Letters and Reviews, Vol.8, No.2, pp. 25-29, 2005.\n7. W. Guo and T. Huang, “Method of elementary transformation to compute Moore-Penrose inverse,” Applied Mathematics and Computation, Vol.216, No.5, pp. 1614-1617, 2010.\n8. M. B. Tasić, P. S. Stanimirović, and M. D. Petković, “Symbolic computation of weighted Moore-Penrose inverse using partitioning method,” Applied Mathematics and Computation, Vol.189, No.1, pp. 615-640, 2007.\n9. F. Huang and X. Zhang, “An improved Newton iteration for the weighted Moore-Penrose inverse,” Applied Mathematics and Computation, Vol.174, No.2, pp. 1460-1486, 2006.\n10. S. Li, Y. Li, and Z. Wang, “A class of finite-time dual neural networks for solving quadratic programming problems and its k-winners-take-all application,” Neural Network, Vol.39, pp. 27-39, 2013.\n11. B. Liao, L. Xiao, J. Jin, L. Ding, and M. Liu, “Novel complex-valued neural network for dynamic complex-valued matrix inversion,” J. Adv. Comput. Intell. Intell. Inform. (JACIII), Vol.20, No.1, pp. 132-138, 2016.\n12. A. Hosseinia, J. Wang, and S. Hosseinia, “A recurrent neural network for solving a class of generalized convex optimization problems,” Neural Network, Vol.44, pp. 78-86, 2013.\n13. L. Xiao, “A finite-time convergent neural dynamics for online solution of time-varying linear complex matrix equation,” Neurocomputing, Vol.167, pp. 254-259, 2015.\n14. L. Jin and Y. Zhang, “Discrete-time Zhang neural network for online time-varying nonlinear optimization with application to manipulator motion generation,” IEEE Trans. on Neural Networks and Learning Systems, Vol.26, No.7, pp. 1525-1531, 2015.\n15. D. Guo and Y. Zhang, “ZNN for solving online time-varying linear matrix-vector inequality via equality conversion,” Applied Mathematics and Computation, Vol.259, pp. 327-338, 2015.\n16. S. Li, S. Chen, and B. Liu, “Accelerating a recurrent neural network to finite-time convergence for solving time-varying Sylvester equation by using a sign-bi-power activation function,” Neural Processing Letters, Vol.37, No.2, pp. 189-205, 2013.\n17. L. Xiao and R. Lu, “Finite-time solution to nonlinear equation using recurrent neural dynamics with a specially-constructed activation function,” Neurocomputing, Vol.151, pp. 246-251, 2015.\n18. B. Liao and Y. Zhang, “From different ZFs to different ZNN models accelerated via Li activation functions to finite-time convergence for time-varying matrix pseudoinversion,” Neurocomputing, Vol.133, No.8, pp. 512-522, 2014.\n19. L. Jin and Y. Zhang, “Discrete-time Zhang neural network of O(τ3) pattern for time-varying matrix pseudoinversion with application to manipulator motion generation,” Neurocomputing, Vol.142, pp. 165-173, 2014.\n20. B. Liao, Y. Zhang, and L. Jin, “Taylor O(h3) discretization of ZNN models for dynamic equality-constrained quadratic programming with application to manipulators,” IEEE Trans. on Neural Networks and Learning Systems, Vol.27, No.2, pp. 225-237, 2016.\n21. L. Jin, S. Li, L. Xiao, R. Lu, and B. Liao, “Cooperative motion generation in a distributed network of redundant robot manipulators with noises,” IEEE Trans. on Systems, Man, and Cybernetics: Systems, in press with DOI 10.1109/TSMC.2017.2693400.\n\n*This site is desgined based on HTML5 and CSS3 for modern browsers, e.g. Chrome, Firefox, Safari, Edge, Opera.\n\nLast updated on Nov. 27, 2020" ]

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[ "Transformers – Practice\n\nQuestions:\n\n1) A transformer at a substation changes 10000 V to 240 V. There are 4000 turns on the primary coil.\n\na) What type of transformer is this?\n\nb) Calculate how many turns there are on the secondary coil:\n\n2) A television has 5000 turns in the primary coil. The secondary coil consists of 250 turns and draws a current of 8 amps.\n\na) What is the current that is flowing in the primary coil?\n\nb) What is the power output of the television? (Assume it is an ideal transformer)\n\n3) An ideal transformer has 750 turns on the primary coil and 50 turns on the secondary coil. The current in the secondary coil is 3 A  and the primary voltage is 30 V. Calculate:\n\na) the secondary voltage\n\nb) the output power\n\nc) the input power\n\nd) the current flowing through the primary coil." ]

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[ "# IBM SPSS Modeler 14.2 X86位元 統計分析 繁體中文DVD版\n\n-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=\n\n-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=\n\n- 中文化的簡潔介面：內含繁體中文介面，項目圖像化使畫面更加簡明易懂。\n- 語法編輯器：提供自動執行功能、色彩編碼、程式編碼兩側的空白及斷落點的設\n\n- 客製化對話視窗建立器：可配合個人需求自訂對話方塊，將例行性分析工作以簡\n\n- 強化遺漏值處理：不但可處理連續與類別變數之遺漏值，更新增多重插補法，協\n\n- 偏最小平方迴歸(Partial Least Square Regression, PLSR) 處理資料中有相關\n\n- 廣義估計方程(Generalized Estimated Equation, GEE) 當資料不符合統計上的\n\n- 最近相鄰法分析(K Nearest Neighbour, KNN)：新增的分群方法，可幫助市場調\n\n- EZ RFM(新模組)：市場研究人員可藉由RFM (Recency, Frequency, Monetary)簡\n\n- 類神經網路 (Neural Network, NN)(新模組) 可建構非線性模型、接受不同類型\n\n-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=" ]

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[ "## 11857 - Driving Range\n\nAll about problems in Volume 118. If there is a thread about your problem, please use it. If not, create one with its number in the subject.\n\nModerator: Board moderators\n\nImti\nLearning poster\nPosts: 53\nJoined: Sat Dec 04, 2010 12:00 pm\nLocation: Bangladesh\nContact:\n\n### Re: 11857 - Driving Range\n\nThanks MRH..Actually I just checked this thread today..And I got it Accepted...with runtime of 0.616 sec..My initial approach might be right (with dfs),there is no need of dfs at all..disjoint-forest of course with kruskal could handle all about it effectively..", null, "", null, "mahade hasan\nLearning poster\nPosts: 87\nJoined: Thu Dec 15, 2011 3:08 pm\nLocation: University of Rajshahi,Bangladesh\n\n### Re: 11857 - Driving Range\n\nwhy every time get submission error", null, "", null, "", null, "we r surrounded by happiness\nneed eyes to feel it!\n\nbrianfry713\nGuru\nPosts: 5947\nJoined: Thu Sep 01, 2011 9:09 am\nLocation: San Jose, CA, USA\n\n### Re: 11857 - Driving Range\n\nTry a different problem\nCheck input and AC output for thousands of problems on uDebug!\n\nSaminur Islam\nNew poster\nPosts: 4\nJoined: Tue Dec 10, 2013 10:49 am\n\n### Re: 11857 - Driving Range\n\nEverytime I am getting submission error...... i dont understand why submission error.....plz help me to find out my problem with this code\n\nCode: Select all\n\n``````#include <iostream>\n#include <cstdio>\n#include <algorithm>\n#include <cstring>\n#include <string>\n#include <cctype>\n#include <stack>\n#include <queue>\n#include <list>\n#include <vector>\n#include <map>\n#include <sstream>\n#include <cmath>\n#include <bitset>\n#include <utility>\n#include <set>\n#include <numeric>\nusing namespace std;\n#define max(a,b) (a>b)?a:b\nvector< pair<int,pair<int,int> > > Edge;\nint M,N,d;\nbool visited;\nint pset;\n\nint find(int V)\n{\nif(pset[V] == V)\nreturn V;\nelse\nreturn (pset[V] = find(pset[V]));\n}\nvoid union_find(int u,int v)\n{\npset[find(u)] = find(v);\n}\n\nbool comp(pair<int,pair<int,int> > edge1,pair<int,pair<int,int> >edge2)\n{\npair<int,int> p1;\npair<int,int>p2;\np1=edge1.second;\np2=edge2.second;\nreturn p1.second<p2.second;\n}\nint kruskal()\n{\nint i,lweight;\nint m;\nmemset(visited,false,sizeof visited);\nsort(Edge.begin(),Edge.end(),comp);\nint totalcost=0;\nd=0;\nm=0;\nfor(i=0;i<Edge.size();i++)\n{\npair<int,pair<int,int> > p=Edge[i];\npair<int,int> q=p.second;\nint V1=p.first;\nint V2=q.first;\nif((find(V1)!=find(V2)))\n{\n//cout<<V1<<\"-\"<<V2<<\":\"<<q.second<<endl;\nd++;\nlweight=q.second;\ntotalcost+=q.second;\nvisited[i]=true;\nunion_find(find(V1),find(V2));\nm = max(m,q.second);\n}\n}\n//cout<<\"Total Cost: \"<<totalcost<<endl;\nreturn lweight;\n}\n\nint main()\n{\n//freopen(\"i.txt\",\"r\",stdin);\nint i,j,k;\nint value1,value2,weight;\nwhile(scanf(\"%d %d\",&N,&M)==2)\n{\nif(N==0&&M==0) break;\nEdge.clear();\nfor(j=0;j<N;j++)\n{\npset[j]=j;\n}\nfor(i=0;i<M;i++)\n{\nscanf(\"%d %d %d\",&value1,&value2,&weight);\nEdge.push_back(make_pair(value1,make_pair(value2,weight)));\n}\nint totalcost=kruskal();\nif(d<N-1)\n{\ncout<<\"IMPOSSIBLE\"<<endl;\n}\nelse\n{\ncout<<totalcost<<endl;\n}\n}\nreturn 0;\n}\n``````\n\nbrianfry713\nGuru\nPosts: 5947\nJoined: Thu Sep 01, 2011 9:09 am\nLocation: San Jose, CA, USA\n\n### Re: 11857 - Driving Range\n\nThat is AC code.\nCheck input and AC output for thousands of problems on uDebug!\n\nbrianfry713\nGuru\nPosts: 5947\nJoined: Thu Sep 01, 2011 9:09 am\nLocation: San Jose, CA, USA\n\n### Re: 11857 - Driving Range\n\nInput:\n\nCode: Select all\n\n``````3 1\n0 1 1\n0 0\n``````\nAC output IMPOSSIBLE\nCheck input and AC output for thousands of problems on uDebug!\n\nsreka11\nNew poster\nPosts: 15\nJoined: Fri Aug 15, 2014 8:06 pm\n\nACC" ]

[ null, "https://uva.onlinejudge.org/board/images/smilies/icon_smile.gif", null, "https://uva.onlinejudge.org/board/images/smilies/icon_smile.gif", null, "https://uva.onlinejudge.org/board/images/smilies/icon_redface.gif", null, "https://uva.onlinejudge.org/board/images/smilies/icon_redface.gif", null, "https://uva.onlinejudge.org/board/images/smilies/icon_redface.gif", null ]

[ "", null, "Concrete strength measured using concrete cubes produce results different than concrete cylinders. Conservative estimates put concrete cylinders at 80% of concrete cubes, for high-strength concrete some say the percentage is near 100%.", null, "The ratio between the cube(150mm) and cylindrical sample(150×300 mm).\n\nGenerally Strength of Cylinder sample= 0.8 x Strength of Cube.\nExample:- M20 is equivalent to C25\n\nC25 is 1:1:3\n\n(1 part cement, 1 part fine aggregate, and 3 parts coarse aggregate.)\n\nwhereas C25 symbolizes a Cylindrical sample in which cement, fine aggregates, coarse aggregates are mixed in such a ratio that its characteristic compressive strength is 25 N/mm^2 after 28 days of curing.\n\nM 20 is 1:1.5:3\n\nwhereas M20 symbolizes a Cubic sample in which cement, fine aggregates, coarse aggregates are mixed in such a ratio that its characteristic compressive strength is 20 N/mm^2 after 28 days of curing.", null, "" ]

[ null, "data:image/png;base64,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", null, "data:image/png;base64,iVBORw0KGgoAAAANSUhEUgAAAAEAAAABAQMAAAAl21bKAAAAA1BMVEUAAP+KeNJXAAAAAXRSTlMAQObYZgAAAAlwSFlzAAAOxAAADsQBlSsOGwAAAApJREFUCNdjYAAAAAIAAeIhvDMAAAAASUVORK5CYII=", null, "data:image/png;base64,iVBORw0KGgoAAAANSUhEUgAAAakAAAEqAQMAAACVxj1uAAAAA1BMVEUAAP+KeNJXAAAAAXRSTlMAQObYZgAAAAlwSFlzAAAOxAAADsQBlSsOGwAAACdJREFUeNrtwTEBAAAAwqD1T20ND6AAAAAAAAAAAAAAAAAAAACADwNABgAB5E6GUgAAAABJRU5ErkJggg==", null ]

[ "CHEMISTRY\n\nIn each reaction, calculate the moles of product (in parentheses) when 3.00 moles of each reactant is used.\nA: 4Li+O2 = 2Li2O (Li2O)\nB: Fe2O3+3H2 = 2Fe+3H2O (Fe)\nC: Al2S3+6H2O = 2Al(OH)3+3H2S (H2S)\n\n1. 👍 0\n2. 👎 0\n3. 👁 117\n1. a. if you have 3 moles of O2, 3 moles of Li, you are limited by the Li. You should get then the mole ratio of product to Li (2/4) times the moles of Li, or 2/4*3 moles Li2O\n\nb. You will be limited by three moles of H2. You get then 2Fe, and 3H2o\n\nc. you will be limited by the H2o, so you get 1 aluminum hydroxide, and 1.5 H2S moles\n\n1. 👍 0\n2. 👎 0\n\nSimilar Questions\n\n1. chemistry\n\nFor each reaction, calculate the moles of product (in parentheses) when 2.00 moles of each reactant is used A: 2SO2 + O2 = 2SO3 (SO3) B: 3Fe + 4H2O = Fe3O4 + 4H2 (Fe3O4) C: C7H16 + 11O2 = 7CO2 + 8H2O (CO2)\n\nasked by Boo Boo on October 31, 2012\n2. Chemistry\n\nFor the reaction, calculate the moles of indicated product produced when 5.00 moles of each reactant is used. 4Li(s)+O2(g)+ < 2Li2O(s) (Li2O)\n\nasked by Michael on May 7, 2012\n3. chemistry\n\nThe first step of the synthesis is described by the reaction below. When 1.750 g of Fe(NH4)2(SO4)2 6H2O is mixed with 13 mL of 1.0 M H2C2O4 , the theoretical yield of FeC2O42H2O is ? grams.. I do not understand why my answer is\n\nasked by Tiffany on June 30, 2007\n4. chemistry\n\nFirst of all, you need to know how many g of Mg actually reacted in the first reaction. You can determine this by determining the limiting reactant. The reaction is Mg + F2 -> MgF2 16.2 g of Mg is 16.2/24.32 = 0.666 moles 25.3 g\n\nasked by drwls on August 30, 2005\n5. chemistry\n\nfor the reaction: 4NH3+5O2=4NO+6H2O; 15.0g of NH3 and 27.5g of O2 were reacted. a). which is the limiting reactant? b).how many moles of NO are formed from the limiting reactant in the above reaction. i know how to do the moles\n\nasked by tina on March 7, 2008\n6. Chemistry\n\nI was doing a lab experiment on the reaction of copper(II) sulphate and aluminum: 1) I measured 2.02g of copper(II) sulphate pentahydrate 2) I dissolved the copper(II) sulphate pentahydrate in 10mL of distilled water 3) I added\n\nasked by Patrick on September 26, 2015\n7. Chem\n\n4 NH3 + 3 O2 --> 2 N2 + 6 H2O a.) Assume 3.0 mol O2 completely reacts. Calculate the moles required for the other reactant. Calculate the moles that can be produced for each product. b.) Assume 8.0 mol NH3 completely reacts.\n\nasked by Deni on February 13, 2016\n8. Chemistry\n\nCalculations involving a limiting reactant Now consider a situation in which 20.0 g of P4 is added to 54.0 g of Cl2, and a chemical reaction occurs. To identify the limiting reactant, you will need to perform two separate\n\nasked by Beth on October 17, 2016\n9. Chemistry\n\nDetermine the # of moles of Al2S3 that can be prepared by the reaction of 0.400 moles of Aluminum with 0.700 moles of Sulfur, and Also find the # of moles of reactant in excess\n\nasked by Denise on May 7, 2011\n10. chemistry\n\nA gaseous mixture containing 5 moles of H2 and 7 moles of Br2 reacts to form HBr. Write a balanced chemical equation. Which reactant is limiting? What is the theoretical yield in moles? What is the theoretical yield for this\n\nasked by Zoe on December 4, 2011\n\nMore Similar Questions" ]

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[ "# Two persons manage to push a motorcar of mass 1200 kg at a uniform velocity along a level road.\n\nQuestion.\nTwo persons manage to push a motorcar of mass 1200 kg at a uniform velocity along a level road. The same motorcar can be pushed by three persons to produce an acceleration of 0.2 ms–2. With what force does each person push the motorcar ? (Assume that all persons push the motorcar with the same muscular effort.)\n\nSolution:\nAs two persons can make the motorcar move with uniform velocity, it is clear that total force\napplied by them on the motorcar is balanced by the force of friction acting in the opposite\ndirection. It is force of one more person which produces an acceleration of $0.2 \\mathrm{~ms}^{-2}$.\n$\\therefore$ Force of one person $=$ mass $\\times$ acceleration $=1200 \\times 0.2=240 \\mathrm{~N}$" ]

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[ "interstellar space has an average temperature of about 10 K and an average density of hydrogen atoms of about one hydrogen atom per cubic meter. calculate the mean free path of hydrogen atoms in interstellar space.use d= 100 pm as the diameter of a hydrogen atom.\n\nThe equation for mean free path is:\n\nλ=1/(sqrt(2)*pi*d^2*n)\n\nor easier to see here:\n\nhttp://hyperphysics.phy-astr.gsu.edu/hbase/kinetic…\n\nwhere λ is mean free path,\nd is diameter of a hydrogen atom in m. (for you, 100pm = 10^-10 m)\nand n is the # of atoms per cubic m (for you, 1 per m^3)\n\nso just plugging these into the equation yields: 2.251 *10^19 meters." ]

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[ "# Fourier Analysis of Boolean Functions under Random Restriction\n\n### Fourier Transform\n\nConsider a function $$f\\colon \\{0,1\\}^n \\to \\{-1, 1\\}$$ on $$n$$ boolean variables. This can be viewed as a transform of a boolean function $$g\\colon \\{0,1\\}^n \\to \\{0, 1\\}$$ by letting $$f(x) = (-1)^{g(x)}$$. For $$\\alpha \\in\\{0,1\\}^n$$, define the characteristic function\n$\\chi_\\alpha(x) = (-1)^{\\alpha x} = (-1)^{\\sum_i \\alpha_i x_i}.$\nHere $$\\alpha$$ can also be viewed as a subset of $$[n]$$, that is $$\\{i\\colon \\alpha_i = 1\\}$$.\n\nThe Fourier transform of function $$f$$ at $$\\alpha$$ is\n$\\widehat{f}(\\alpha) = \\frac{1}{2^n} \\sum_{x\\in \\{0,1\\}^n} f(x) \\chi_\\alpha(x).$\nThat is, $$f$$ can be uniquely decomposed as a linear combination of the characteristic functions\n$f(x) = \\sum_{\\alpha\\in \\{0,1\\}^n} \\widehat{f}(\\alpha) \\chi_\\alpha (x).$\nNote that the Fourier transform $$\\widehat{f}$$ is a function from $$\\{0,1\\}^n \\to \\mathbb{R}$$.\n\nDefine the degree of $$\\widehat{f}$$ to be the size of the largest $$\\alpha$$ such that $$\\widehat{f}(\\alpha) \\neq 0$$. If a function is computed by a decision tree of depth $$d$$, then the degree of its Fourier transform is at most $$d$$.\n\n### Restriction\n\nA restriction of a function is to fix a subset of variables by a specific assignment. This can be done in two steps: picking up a subset of variables, and fixing the values of these variables.\n\nConsider function $$f\\colon \\{0,1\\}^n \\to \\{-1, 1\\}$$. Divide the input variables into two disjoint subsets and write $$f(x, y)$$ where $$x$$ has dimension $$m$$ and $$y$$ has dimension $$n-m$$. Then the Fourier coefficients can be written as\n$\\widehat{f}(\\alpha, \\beta) = \\frac{1}{2^n} \\sum_{(x, y)\\in \\{0,1\\}^n} f(x,y) \\chi_\\alpha(x)\\chi_\\beta(y).$\n\nFor a fixed $$y\\in\\{0,1\\}^{n-m}$$, let $$f_y \\equiv f(x,y)$$, which is a boolean function on $$m$$ inputs. Its Fourier transform is\n$\\widehat{f_y} (\\alpha) = \\frac{1}{2^m} \\sum_{x} f_y (x) \\chi_\\alpha(x).$\n\nThen we have the following relationship between $$\\widehat{f}$$ and $$\\widehat{f_y}$$:\n$\\widehat{f}(\\alpha, \\beta) = \\frac{1}{2^{n-m}} \\sum_y \\widehat{f_y}(\\alpha) \\chi_\\beta(y),$\nand for each fixed $$\\alpha$$\n$\\sum_\\beta \\widehat{f}(\\alpha, \\beta)^2 = \\frac{1}{2^{n-m}} \\sum_y \\widehat{f_y}(\\alpha)^2.$\nNote here we applied the fact that $$\\chi_\\alpha (x) \\chi_\\beta (x) = \\chi_{\\alpha\\oplus\\beta} (x)$$ and $$\\sum_x \\chi_\\alpha (x) = 0$$ for $$\\alpha\\neq 0$$.\n\n### Random Restriction on DNFs\n\nA random restriction on a function is done in two steps: first randomly pick up a fraction $$p$$ of input variables, and then randomly assign 0 or 1 to the remaining variables. Hastad’s Switching Lemma says that, under a random restriction, a $$k$$-DNF formula is simplified dramatically with high probability.\n\nLet $$f$$ be a $$k$$-DNF, and $$\\rho$$ be a random restriction with parameter $$p$$. Denote by $$f|_\\rho$$ the randomly restricted function. Then the probability that the decision tree depth for $$f|_\\rho$$ is bigger than $$d$$ is bounded by\n$(5pk)^d.$\nNote that this bound does not depend on the number of clauses in the formula. This implies that with probability $$1- (5pk)^d$$, the degree of the Fourier transform $$\\widehat{f|_\\rho}$$ is at most $$d$$. Based on this, next we show that the large Fourier coefficients of the orignial $$k$$-DNF $$f$$ is bounded.\n\nBy the above, we have\n${\\bf Pr}_{\\rho} \\left[ deg( \\widehat{f|_\\rho}) > d \\right] \\leq (5pk)^d.$\nThis implies\n${\\bf E}_{\\rho} \\left[ \\sum_{\\alpha \\colon |\\alpha| >d } \\widehat{f|_\\rho}(\\alpha)^2 \\right] \\leq (5pk)^d.$\n\nWe consider the restriction applied in two steps: picking up a random subset $$s$$ of size $$pn$$ and randomly assign value $$y$$ to the remaining variables. Then\n${\\bf E}_{s} {\\bf E}_{y} \\left[ \\sum_{\\alpha \\colon |\\alpha| >d, \\alpha\\subseteq s } \\widehat{f|_\\rho}(\\alpha)^2 \\right] = {\\bf E}_{s} \\left[ \\sum_{\\alpha \\colon |\\alpha| >d, \\alpha\\subseteq s } {\\bf E}_{y}[ \\widehat{f|_\\rho}(\\alpha)^2 ] \\right]$\n\nBy the fact that\n${\\bf E}_{y}[ \\widehat{f|_\\rho}(\\alpha)^2 ] = \\sum_\\beta \\widehat{f}(\\alpha,\\beta)^2$\nwe have the above equals\n${\\bf E}_{s} \\left[ \\sum_{\\alpha \\colon |\\alpha| >d, \\alpha\\subseteq s } \\sum_\\beta \\widehat{f}(\\alpha,\\beta)^2 \\right] = \\sum_u \\widehat{f}(u)^2 {\\bf Pr}_{s} [|s\\cap u| > d]$\n\nConsidering the cases that $$|u| \\geq 2d/p$$. By Chernoff bound, we have\n${\\bf Pr}_{s} [|s\\cap u| > d] \\geq 1 – \\exp(-d/4) \\geq 1/2.$\n\nFinally we have\n$\\sum_{u\\colon |u| \\geq 2d/p } \\widehat{f}(u)^2 \\leq 2 \\cdot (5pk)^d$\n\nFor $$p = 1/10k$$,\n$\\sum_{u\\colon |u| \\geq 20kd } \\widehat{f}(u)^2 \\leq 2 \\cdot 2^{-d}$" ]

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[ "## Lesson 14 AI logic, Array Methods\n\nMark Harder, 11 January 2019\n\n# AI Logic\n\nLets apply the KISS principle from lesson 12 to building our AI code. Part of keeping our code simple is breaking down the AI move into distinct steps. We are going to use a method call heuristic technique. Heuristic is any approach to problem solving that employs a practical method, not guaranteed to be optimal, perfect, logical, or rational, but instead sufficient for reaching an immediate goal. In practical terms, we are going to breakdown our AI method into a series of choices/steps that reflect how a human might make a choice. So lets do that, how would you choose your next move? A good heuristic strategy is to do a series of checks, in order, of best and most useful to random for the last to find a move.\n\n## Tic Tac Toe AI move order heuristic.\n\n1. Are there locations where I can play and win? If yes, then choose one.\n2. Are there locations where my opponent can play and win? If yes, then choose a blocking location.\n3. The center location is the best spot to start, so if the center spot in available, choose it.\n4. Choose a location to setup up for two or more win paths for next turn.\n5. Choose a location to setup a win path for next turn.\n6. Choose a random location.\n• All steps are checking available locations.\n\nHow would a few of these play out?\n\n1. Are there locations where I can play and win? Yes the bottom right.", null, "2. In this case it is O’s turn and it needs to block X in the bottom center.", null, "Most of these steps require us to look at each of the empty locations and test to see if a possible move will give a win.\nFirst lets convert our existing method `CheckForWinner()` that checks for a win to a generic function that can be used over and over. See principle DRY from lesson 12.\n\n## Function to Breakout our Test for a Win\n\n``````// Break out the test for a specific player win.\n// Return 'X' or 'O' of the winner, 'Tie' for a full board, '' for no current winner.\nfunction TestForWin(TestBoard) {\n// Setup the variable we are going to return.\nlet Winner = '';\n\nreturn Winner;\n}\n``````\n\nIn our existing function called `CheckForWinner()` we test against the global variable board by calling gameState.board[][]\nIn our new function we are passing in a variable named `TestBoard` which we will replace `gameState.board`.\nInstead of setting `gameState.Winner =` to our result, we will set our new variable `Winner =` to our end result, so that we can return it at the end of the function instead. The idea with a re-usable function is that the only thing changed is what is returned by the function.\nIn computer science, a function is said to have a side effect if it modifies some state outside its local environment, that is to say has an observable interaction with the outside world besides returning a value.\n\n``````// Break out the test for a specific player win.\n// Return 'X' or 'O' of the winner, 'Tie' for a full board, '' for no current winner.\nfunction TestForWin(TestBoard) {\n// Setup the variable we are going to return.\nlet Winner = '';\n\n// Here is the simplest logic for checking every possibility for win.\nlet row1 = TestBoard + TestBoard + TestBoard;\nlet row2 = TestBoard + TestBoard + TestBoard;\nlet row3 = TestBoard + TestBoard + TestBoard;\n\n// first check to see of the board is full\nif (row1.length + row2.length + row3.length === 9) {\nWinner = \"Tie\";\n}\n\n// check rows for win\nif (row1 === \"XXX\" || row2 === \"XXX\" || row3 === \"XXX\") {\nWinner = \"X\";\n}\nif (row1 === \"OOO\" || row2 === \"OOO\" || row3 === \"OOO\") {\nWinner = \"O\";\n}\n\n// check cols for win\nlet col1 = TestBoard + TestBoard + TestBoard;\nlet col2 = TestBoard + TestBoard + TestBoard;\nlet col3 = TestBoard + TestBoard + TestBoard;\nif (col1 === \"XXX\" || col2 === \"XXX\" || col3 === \"XXX\") {\nWinner = \"X\";\n}\nif (col1 === \"OOO\" || col2 === \"OOO\" || col3 === \"OOO\") {\nWinner = \"O\";\n}\n\n// check diagonal for win\nlet x1 = TestBoard + TestBoard + TestBoard;\nlet x2 = TestBoard + TestBoard + TestBoard;\nif (x1 === \"XXX\" || x2 === \"XXX\") {\nWinner = \"X\";\n}\nif (x1 === \"OOO\" || x2 === \"OOO\") {\nWinner = \"O\";\n}\n\nreturn Winner;\n}\n``````\n\nNow we can call the new function. Lets start by re-writing our `CheckForWinner()` function so that it uses our new DRY function.\n\n``````function CheckForWinner() {\nlet Winner = TestForWin(gameState.board);\ngameState.Winner = Winner;\nif (Winner === \"X\") {\ngameState.XWinCount ++;\n}\nif (Winner === \"O\") {\ngameState.OWinCount ++;\n}\nif (Winner === \"Tie\") {\ngameState.TieWins ++;\n}\n\n// Return true if the a winner has been found.\nreturn (gameState.Winner.length > 0);\n}\n``````\n\nNext we need to use our new DRY function to test all open spots for a win. Before we do that we need to learn how to test all possibilities.\n\n## Array.map(), .reduce(), .filter() and .concat()\n\n### .map()\n\nHow does it work? lets start with a simple example. Say you have an array of objects representing persons in a school. The thing we need is an array of peoples names.\n\n``````// What we start with\nlet people = [{ id: 1, name: 'Mark Harder', position: 'Teacher', age: 40 },\n{ id: 2, name: 'John Shy', position: 'Student', age: 14 },\n{ id: 3, name: 'Berry Ray', position: 'Student', age: 15 },\n{ id: 4, name: 'Temple Kin', position: 'Student', age: 16 }\n];\n// What we need\nlet peopleNames = ['John Shy', 'Berry Ray', 'Temple Kin'];\n``````\n\nThere are multiple ways to achieve this conversion of one array to another. You could use a for() loop or a .forEach() to meet our goal.\n\n``````let peopleNames = [];\npeople.forEach(function (person) {\n// Array.forEach() calls your function with a parameter (person) for each item in the array.\n// Next we take the property 'name' and add it (push) to our empty array 'peopleNames'.\npeopleNames.push(person.name);\n});\n``````\n\nNotice that we needed to create an empty array first. Now lets see what this looks like using .map().\n\n``````let peopleNames = people.map( (person) => {\nreturn person.name;\n});\n``````\n\nIn this example we used an arrow function to build our array output using the map method off the array. The return result from our function used to build our new array. We are returning a string in our function, but we could also just as easily return another object.\n\nOur array function can also be written in this even more simplified way: If there is only one command to the function we can even forego the return word\n\n``````let peopleNames = people.map(person => person.name);\n``````\n\nOr as a traditional function call:\n\n``````let peopleNames = people.map(function(person) {\nreturn person.name;\n});\n``````\n\n### .reduce()\n\nJust like .map(), .reduce() also runs a callback for each element of an array. What’s different here is that reduce passes the result of this callback (the accumulator) from one array element to the other. The accumulator can be pretty much anything (integer, string, object, etc.) and must be instantiated or passed in when calling .reduce().\n\nTime for an example. If we start with the same people array, we could use reduce() to sum up the ages of all the people.\n\n``````let totalAge = people.reduce( (accumulator, person) => {\nreturn accumulator + person.age;\n});\n``````\n\nLets break down our example code.\n\nlet totalAge = people.reduce();\n\n• Take each item in the people array, take something from the item add it to the accumulator, then pass it to the next iteration as the parameter accumulator. Take the total after the last item is added and assign it to the variable totalAge.\n\nSyntax\n\narray.reduce(callback[, initialValue])\nReturn value is the result from the reduction.\n\n• callback function has two required parameters: accumulator, currentValue\n• optionally the function can have 2 more parameters: currentIndex, array\n• optionally you c an add a parameters after the callback function initialValue initialValue: An initial value that the accumulator starts as when the first value {first array item} is called.\n\nThere are two optional parameters for the function, for a total of 4 potential parameters.\n\n• accumulator\n• The accumulator accumulates the callback’s return values; it is the accumulated value previously returned in the last invocation of the callback, or initialValue, if supplied (see below).\n• currentValue\n• The current element being processed in the array.\n• currentIndex\n• The index of the current element being processed in the array. Starts at index 0, if an initialValue is provided, and at index 1 otherwise.\n• array\n• The array reduce() was called upon.\n\n### .filter()\n\nWhat if you have an array, but only want some of the elements in it? That’s where .filter() comes in!\n\nLets say we want an array of the students.\n\n``````let peopleNames = people.filter( (person) => { return person.position === 'Student' } );\n``````\n\nLets break this code down\n\nlet peopleNames = people.filter()\n\n• Take the resulting array from applying filter method to the people array and assign that result to the new peopleName variable created with the let command.\n(person) => { return {Comparison} }\n• This is a arrow function where the passed in parameter is named person. For each item in your people array, it will come into the function on each iteration as the person parameter/variable.\n• {Comparison} is where you place your conditional test. If it returns true, the array element will be included in the output array and excluded if false.\n• person.position === ‘Student’ This tests to see if the property named position on the array item is equal to the string ‘Student’\n• This call will return an array like the input array, except that only objects/people that are Student will be included.\n• In our example the person named ‘Mark Harder’ will be excluded because it is position: ‘Teacher’\n\n### .concat()\n\nThis method is used two merge two arrays. The original arrays are not changed, but rather a new array that is a combination of both is returned.\n\nHere is a simple example:\n\n``````let array1 = ['Mark Harder', 'John Shy'];\nlet array2 = ['Berry Ray', 'Temple Kin'];\nlet array3 = array1.concat(array2);\n``````\n\nResults are:\n\n[‘Mark Harder’, ‘John Shy’, ‘Berry Ray’, ‘Temple Kin’]\n\nThis method is very useful when combined with .reduce().\n\n``````let nameList = ['Fred Flintstone'];\nlet names = people.reduce( (accumulator, person) => {\nreturn accumulator.concat(person.name);\n}, nameList);\n``````\n\nResults are:\n\n[“Fred Flintstone”, “Mark Harder”, “John Shy”, “Berry Ray”, “Temple Kin”]\n\nNormally .reduce() adds together value with an accumulator and returns the result, but using the .concat() we can take an initial array and then add to it inside the reduce method.\n\nThe line:\n\nreturn accumulator.concat(person.name);\n\nThis return statement comes back in the next iteration as the parameter ‘accumulator’\nSo if we take the accumulator and call .concat() the new array will be returned.\n\n### Lets combine .filter(), and .reduce()\n\nIf we want to filter our people array to just students, then add their ages together and assign that total to a new variables named totalStudentAge\n\n``````let totalStudentAge = people.filter( (person) => { retutrn person.position === 'Student' } ).reduce( (accumulator, person) => {\nretun accumulator + person.age;\n});\n``````\n\nLets break down our example code.\n\nlet totalStudentAge = people.filter\n\n• First we filter the people array to be just students.\n• Then we reduce (add the age property together.)\n\n## Assignment due for discussion next class and checked into GitHub by the Monday after that.\n\n• Create a new repo called lesson14\n• Lets practice using map, reduce and filter. There is an index,html and main.js file that contain 4 code questions that you need to complete." ]

[ null, "https://mhintegrity.com/assets/images/TicTacToeStep1.png", null, "https://mhintegrity.com/assets/images/TicTacToeStep2.png", null ]

[ "#### Previous topic\n\nGetting started with Windows HPC Server 2008\n\nDAG Dependencies\n\n# Parallel examples¶\n\nIn this section we describe two more involved examples of using an IPython cluster to perform a parallel computation. In these examples, we will be using IPython’s “pylab” mode, which enables interactive plotting using the Matplotlib package. IPython can be started in this mode by typing:\n\n```ipython --pylab\n```\n\nat the system command line.\n\n## 150 million digits of pi¶\n\nIn this example we would like to study the distribution of digits in the number pi (in base 10). While it is not known if pi is a normal number (a number is normal in base 10 if 0-9 occur with equal likelihood) numerical investigations suggest that it is. We will begin with a serial calculation on 10,000 digits of pi and then perform a parallel calculation involving 150 million digits.\n\nIn both the serial and parallel calculation we will be using functions defined in the pidigits.py file, which is available in the docs/examples/parallel directory of the IPython source distribution. These functions provide basic facilities for working with the digits of pi and can be loaded into IPython by putting pidigits.py in your current working directory and then doing:\n\n```In : run pidigits.py\n```\n\n### Serial calculation¶\n\nFor the serial calculation, we will use SymPy to calculate 10,000 digits of pi and then look at the frequencies of the digits 0-9. Out of 10,000 digits, we expect each digit to occur 1,000 times. While SymPy is capable of calculating many more digits of pi, our purpose here is to set the stage for the much larger parallel calculation.\n\nIn this example, we use two functions from pidigits.py: one_digit_freqs() (which calculates how many times each digit occurs) and plot_one_digit_freqs() (which uses Matplotlib to plot the result). Here is an interactive IPython session that uses these functions with SymPy:\n\n```In : import sympy\n\nIn : pi = sympy.pi.evalf(40)\n\nIn : pi\nOut: 3.141592653589793238462643383279502884197\n\nIn : pi = sympy.pi.evalf(10000)\n\nIn : digits = (d for d in str(pi)[2:]) # create a sequence of digits\n\nIn : run pidigits.py # load one_digit_freqs/plot_one_digit_freqs\n\nIn : freqs = one_digit_freqs(digits)\n\nIn : plot_one_digit_freqs(freqs)\nOut: [<matplotlib.lines.Line2D object at 0x18a55290>]\n```\n\nThe resulting plot of the single digit counts shows that each digit occurs approximately 1,000 times, but that with only 10,000 digits the statistical fluctuations are still rather large:", null, "It is clear that to reduce the relative fluctuations in the counts, we need to look at many more digits of pi. That brings us to the parallel calculation.\n\n### Parallel calculation¶\n\nCalculating many digits of pi is a challenging computational problem in itself. Because we want to focus on the distribution of digits in this example, we will use pre-computed digit of pi from the website of Professor Yasumasa Kanada at the University of Tokyo (http://www.super-computing.org). These digits come in a set of text files (ftp://pi.super-computing.org/.2/pi200m/) that each have 10 million digits of pi.\n\nFor the parallel calculation, we have copied these files to the local hard drives of the compute nodes. A total of 15 of these files will be used, for a total of 150 million digits of pi. To make things a little more interesting we will calculate the frequencies of all 2 digits sequences (00-99) and then plot the result using a 2D matrix in Matplotlib.\n\nThe overall idea of the calculation is simple: each IPython engine will compute the two digit counts for the digits in a single file. Then in a final step the counts from each engine will be added up. To perform this calculation, we will need two top-level functions from pidigits.py:\n\n```def compute_two_digit_freqs(filename):\n\"\"\"\nRead digits of pi from a file and compute the 2 digit frequencies.\n\"\"\"\nd = txt_file_to_digits(filename)\nfreqs = two_digit_freqs(d)\nreturn freqs\n\ndef reduce_freqs(freqlist):\n\"\"\"\nAdd up a list of freq counts to get the total counts.\n\"\"\"\nallfreqs = np.zeros_like(freqlist)\nfor f in freqlist:\nallfreqs += f\nreturn allfreqs\n```\n\nWe will also use the plot_two_digit_freqs() function to plot the results. The code to run this calculation in parallel is contained in docs/examples/parallel/parallelpi.py. This code can be run in parallel using IPython by following these steps:\n\n1. Use ipcluster to start 15 engines. We used 16 cores of an SGE linux cluster (1 controller + 15 engines).\n2. With the file parallelpi.py in your current working directory, open up IPython in pylab mode and type run parallelpi.py. This will download the pi files via ftp the first time you run it, if they are not present in the Engines’ working directory.\n\nWhen run on our 16 cores, we observe a speedup of 14.2x. This is slightly less than linear scaling (16x) because the controller is also running on one of the cores.\n\nTo emphasize the interactive nature of IPython, we now show how the calculation can also be run by simply typing the commands from parallelpi.py interactively into IPython:\n\n```In : from IPython.parallel import Client\n\n# The Client allows us to use the engines interactively.\n# We simply pass Client the name of the cluster profile we\n# are using.\nIn : c = Client(profile='mycluster')\nIn : v = c[:]\n\nIn : c.ids\nOut: [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14]\n\nIn : run pidigits.py\n\nIn : filestring = 'pi200m.ascii.%(i)02dof20'\n\n# Create the list of files to process.\nIn : files = [filestring % {'i':i} for i in range(1,16)]\n\nIn : files\nOut:\n['pi200m.ascii.01of20',\n'pi200m.ascii.02of20',\n'pi200m.ascii.03of20',\n'pi200m.ascii.04of20',\n'pi200m.ascii.05of20',\n'pi200m.ascii.06of20',\n'pi200m.ascii.07of20',\n'pi200m.ascii.08of20',\n'pi200m.ascii.09of20',\n'pi200m.ascii.10of20',\n'pi200m.ascii.11of20',\n'pi200m.ascii.12of20',\n'pi200m.ascii.13of20',\n'pi200m.ascii.14of20',\n'pi200m.ascii.15of20']\n\nIn : v.map(fetch_pi_file, files)\n\n# This is the parallel calculation using the Client.map method\n# which applies compute_two_digit_freqs to each file in files in parallel.\nIn : freqs_all = v.map(compute_two_digit_freqs, files)\n\n# Add up the frequencies from each engine.\nIn : freqs = reduce_freqs(freqs_all)\n\nIn : plot_two_digit_freqs(freqs)\nOut: <matplotlib.image.AxesImage object at 0x18beb110>\n\nIn : plt.title('2 digit counts of 150m digits of pi')\nOut: <matplotlib.text.Text object at 0x18d1f9b0>\n```\n\nThe resulting plot generated by Matplotlib is shown below. The colors indicate which two digit sequences are more (red) or less (blue) likely to occur in the first 150 million digits of pi. We clearly see that the sequence “41” is most likely and that “06” and “07” are least likely. Further analysis would show that the relative size of the statistical fluctuations have decreased compared to the 10,000 digit calculation.", null, "## Parallel options pricing¶\n\nAn option is a financial contract that gives the buyer of the contract the right to buy (a “call”) or sell (a “put”) a secondary asset (a stock for example) at a particular date in the future (the expiration date) for a pre-agreed upon price (the strike price). For this right, the buyer pays the seller a premium (the option price). There are a wide variety of flavors of options (American, European, Asian, etc.) that are useful for different purposes: hedging against risk, speculation, etc.\n\nMuch of modern finance is driven by the need to price these contracts accurately based on what is known about the properties (such as volatility) of the underlying asset. One method of pricing options is to use a Monte Carlo simulation of the underlying asset price. In this example we use this approach to price both European and Asian (path dependent) options for various strike prices and volatilities.\n\nThe code for this example can be found in the docs/examples/parallel/options directory of the IPython source. The function price_options() in mckernel.py implements the basic Monte Carlo pricing algorithm using the NumPy package and is shown here:\n\n```def price_options(S=100.0, K=100.0, sigma=0.25, r=0.05, days=260, paths=10000):\n\"\"\"\nPrice European and Asian options using a Monte Carlo method.\n\nParameters\n----------\nS : float\nThe initial price of the stock.\nK : float\nThe strike price of the option.\nsigma : float\nThe volatility of the stock.\nr : float\nThe risk free interest rate.\ndays : int\nThe number of days until the option expires.\npaths : int\nThe number of Monte Carlo paths used to price the option.\n\nReturns\n-------\nA tuple of (E. call, E. put, A. call, A. put) option prices.\n\"\"\"\nimport numpy as np\nfrom math import exp,sqrt\n\nh = 1.0/days\nconst1 = exp((r-0.5*sigma**2)*h)\nconst2 = sigma*sqrt(h)\nstock_price = S*np.ones(paths, dtype='float64')\nstock_price_sum = np.zeros(paths, dtype='float64')\nfor j in range(days):\ngrowth_factor = const1*np.exp(const2*np.random.standard_normal(paths))\nstock_price = stock_price*growth_factor\nstock_price_sum = stock_price_sum + stock_price\nstock_price_avg = stock_price_sum/days\nzeros = np.zeros(paths, dtype='float64')\nr_factor = exp(-r*h*days)\neuro_put = r_factor*np.mean(np.maximum(zeros, K-stock_price))\nasian_put = r_factor*np.mean(np.maximum(zeros, K-stock_price_avg))\neuro_call = r_factor*np.mean(np.maximum(zeros, stock_price-K))\nasian_call = r_factor*np.mean(np.maximum(zeros, stock_price_avg-K))\nreturn (euro_call, euro_put, asian_call, asian_put)\n```\n\nTo run this code in parallel, we will use IPython’s LoadBalancedView class, which distributes work to the engines using dynamic load balancing. This view is a wrapper of the Client class shown in the previous example. The parallel calculation using LoadBalancedView can be found in the file mcpricer.py. The code in this file creates a LoadBalancedView instance and then submits a set of tasks using LoadBalancedView.apply() that calculate the option prices for different volatilities and strike prices. The results are then plotted as a 2D contour plot using Matplotlib.\n\n```# <nbformat>2</nbformat>\n\n# <markdowncell>\n\n# # Parallel Monto-Carlo options pricing\n\n# <markdowncell>\n\n# ## Problem setup\n\n# <codecell>\nfrom __future__ import print_function\n\nimport sys\nimport time\nfrom IPython.parallel import Client\nimport numpy as np\nfrom mckernel import price_options\nfrom matplotlib import pyplot as plt\n\n# <codecell>\n\ncluster_profile = \"default\"\nprice = 100.0 # Initial price\nrate = 0.05 # Interest rate\ndays = 260 # Days to expiration\npaths = 10000 # Number of MC paths\nn_strikes = 6 # Number of strike values\nmin_strike = 90.0 # Min strike price\nmax_strike = 110.0 # Max strike price\nn_sigmas = 5 # Number of volatility values\nmin_sigma = 0.1 # Min volatility\nmax_sigma = 0.4 # Max volatility\n\n# <codecell>\n\nstrike_vals = np.linspace(min_strike, max_strike, n_strikes)\nsigma_vals = np.linspace(min_sigma, max_sigma, n_sigmas)\n\n# <markdowncell>\n\n# ## Parallel computation across strike prices and volatilities\n\n# <markdowncell>\n\n# The Client is used to setup the calculation and works with all engines.\n\n# <codecell>\n\nc = Client(profile=cluster_profile)\n\n# <markdowncell>\n\n# A LoadBalancedView is an interface to the engines that provides dynamic load\n# balancing at the expense of not knowing which engine will execute the code.\n\n# <codecell>\n\n# <codecell>\n\nprint(\"Strike prices: \", strike_vals)\nprint(\"Volatilities: \", sigma_vals)\n\n# <markdowncell>\n\n# Submit tasks for each (strike, sigma) pair.\n\n# <codecell>\n\nt1 = time.time()\nasync_results = []\nfor strike in strike_vals:\nfor sigma in sigma_vals:\nar = view.apply_async(price_options, price, strike, sigma, rate, days, paths)\nasync_results.append(ar)\n\n# <codecell>\n\n# <markdowncell>\n\n# Block until all tasks are completed.\n\n# <codecell>\n\nc.wait(async_results)\nt2 = time.time()\nt = t2-t1\n\nprint(\"Parallel calculation completed, time = %s s\" % t)\n\n# <markdowncell>\n\n# ## Process and visualize results\n\n# <markdowncell>\n\n# Get the results using the `get` method:\n\n# <codecell>\n\nresults = [ar.get() for ar in async_results]\n\n# <markdowncell>\n\n# Assemble the result into a structured NumPy array.\n\n# <codecell>\n\nprices = np.empty(n_strikes*n_sigmas,\ndtype=[('ecall',float),('eput',float),('acall',float),('aput',float)]\n)\n\nfor i, price in enumerate(results):\nprices[i] = tuple(price)\n\nprices.shape = (n_strikes, n_sigmas)\n\n# <markdowncell>\n\n# Plot the value of the European call in (volatility, strike) space.\n\n# <codecell>\n\nplt.figure()\nplt.contourf(sigma_vals, strike_vals, prices['ecall'])\nplt.axis('tight')\nplt.colorbar()\nplt.title('European Call')\nplt.xlabel(\"Volatility\")\nplt.ylabel(\"Strike Price\")\n\n# <markdowncell>\n\n# Plot the value of the Asian call in (volatility, strike) space.\n\n# <codecell>\n\nplt.figure()\nplt.contourf(sigma_vals, strike_vals, prices['acall'])\nplt.axis('tight')\nplt.colorbar()\nplt.title(\"Asian Call\")\nplt.xlabel(\"Volatility\")\nplt.ylabel(\"Strike Price\")\n\n# <markdowncell>\n\n# Plot the value of the European put in (volatility, strike) space.\n\n# <codecell>\n\nplt.figure()\nplt.contourf(sigma_vals, strike_vals, prices['eput'])\nplt.axis('tight')\nplt.colorbar()\nplt.title(\"European Put\")\nplt.xlabel(\"Volatility\")\nplt.ylabel(\"Strike Price\")\n\n# <markdowncell>\n\n# Plot the value of the Asian put in (volatility, strike) space.\n\n# <codecell>\n\nplt.figure()\nplt.contourf(sigma_vals, strike_vals, prices['aput'])\nplt.axis('tight')\nplt.colorbar()\nplt.title(\"Asian Put\")\nplt.xlabel(\"Volatility\")\nplt.ylabel(\"Strike Price\")\n\n# <codecell>\n\nplt.show()\n```\n\nTo use this code, start an IPython cluster using ipcluster, open IPython in the pylab mode with the file mckernel.py in your current working directory and then type:\n\n```In : run mcpricer.py\n\n```\n\nOnce all the tasks have finished, the results can be plotted using the plot_options() function. Here we make contour plots of the Asian call and Asian put options as function of the volatility and strike price:\n\n```In : plot_options(sigma_vals, strike_vals, prices['acall'])\n\nIn : plt.figure()\nOut: <matplotlib.figure.Figure object at 0x18c178d0>\n\nIn : plot_options(sigma_vals, strike_vals, prices['aput'])\n```\n\nThese results are shown in the two figures below. On our 15 engines, the entire calculation (15 strike prices, 15 volatilities, 100,000 paths for each) took 37 seconds in parallel, giving a speedup of 14.1x, which is comparable to the speedup observed in our previous example.", null, "", null, "## Conclusion¶\n\nTo conclude these examples, we summarize the key features of IPython’s parallel architecture that have been demonstrated:\n\n• Serial code can be parallelized often with only a few extra lines of code. We have used the DirectView and LoadBalancedView classes for this purpose.\n• The resulting parallel code can be run without ever leaving the IPython’s interactive shell.\n• Any data computed in parallel can be explored interactively through visualization or further numerical calculations.\n• We have run these examples on a cluster running RHEL 5 and Sun GridEngine. IPython’s built in support for SGE (and other batch systems) makes it easy to get started with IPython’s parallel capabilities." ]

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[ "## You are looking at 1-20 of 22 entries  for:\n\n• All: 20/20\n• Mathematics and Computer Science\n• Science and technology\n• Results with images only x\nclear all", null, "", null, "View:\n\nOverview\n\n## 20/20\n\nSubject: Music\n\nThis US group was formed in Los Angeles, California, in 1978 by Steve Allen (guitar, vocals) and Ron Flynt (bass, vocals), two expatriate musicians from Tulsa, Oklahoma. Drummer Mike Gallo ...", null, "## multidimensional scaling Quick reference\n\n### A Dictionary of Statistics (3 ed.)\n\nReference type:\nSubject Reference\nCurrent Version:\n2014\nSubject:\nScience and technology, Mathematics and Computer Science\nLength:\n256 words\nIllustration(s):\n1\n\n... is called the stress (also called Kruskal stress ) of the solution—a low value suggests a good solution. Great Britain Ireland Nether- Lands Belgium France Italy Spain Great Britain 20 11 11 6 7 2 6 Ireland 11 20 11 10 11 10 10 Netherlands 11 11 20 12 11 6 6 Belgium 6 10 12 20 12 11 9 France 7 11 11 12 20 10 12 Italy 2 10 6 11 10 20 13 Spain 6 10 6 9 12 13 20 The proximity matrix shows, for seven European countries, the numbers of food products (out of twenty) that were found to similar extents (i.e. common in both countries or scarce in both...", null, "## tetrahedral number Quick reference\n\n### The Concise Oxford Dictionary of Mathematics (5 ed.)\n\nReference type:\nSubject Reference\nCurrent Version:\n2014\nSubject:\nScience and technology, Mathematics and Computer Science\nLength:\n48 words\nIllustration(s):\n1\n\n...number An integer of the form , where n is a positive integer. This number equals the sum of the first n triangular numbers . The first few tetrahedral numbers are 1, 4, 10 and 20, and the reason for the name can be seen from the...", null, "## equiangular spiral Quick reference\n\n### The Concise Oxford Dictionary of Mathematics (5 ed.)\n\nReference type:\nSubject Reference\nCurrent Version:\n2014\nSubject:\nScience and technology, Mathematics and Computer Science\nLength:\n93 words\nIllustration(s):\n1\n\n...property that the angle α between OP and the tangent at P is constant. In fact, k =cot α . The equation can be written r = kθ +b, and the curve is also called the logarithmic spiral. http://jwilson.coe.uga.edu/EMT668/EMAT6680.F99/Erbas/KURSATgeometrypro/related%20curves/related%20curves.html Animations of related curves, and an interactive manipulation of the spiral in Geometer’s...", null, "## box plot Quick reference\n\n### The Concise Oxford Dictionary of Mathematics (5 ed.)\n\nReference type:\nSubject Reference\nCurrent Version:\n2014\nSubject:\nScience and technology, Mathematics and Computer Science\nLength:\n80 words\nIllustration(s):\n1\n\n...together with lines, sometimes called ‘whiskers’, showing the maximum and minimum observations in the sample. The median is marked on the box by a line. Box plots are particularly useful for comparing several samples. The figure shows box plots for three samples, each of size 20, drawn uniformly from the set of integers from 1 to...", null, "## Anderson–Darling test Quick reference\n\n### A Dictionary of Statistics (3 ed.)\n\nReference type:\nSubject Reference\nCurrent Version:\n2014\nSubject:\nScience and technology, Mathematics and Computer Science\nLength:\n184 words\nIllustration(s):\n1\n\n...sample equivalents. In some cases, as shown in the following table, an adjusted test statistic is required. test statistic upper tail probability 0.10 0.05 0.025 0.01 Specified distribution A 2 1.933 2.492 3.070 3.857 Normal, estimated mean ( n >20) A 2 0.894 1.087 1.285 1.551 Normal, estimated variance ( n >20) A 2 1.743 2.308 2.898 3.702 Normal, estimated mean and variance 0.631 0.752 0.873 1.035 Exponential, estimated mean 1.062 1.321 1.591...", null, "## decision tree Quick reference\n\n### A Dictionary of Statistics (3 ed.)\n\nReference type:\nSubject Reference\nCurrent Version:\n2014\nSubject:\nScience and technology, Mathematics and Computer Science\nLength:\n185 words\nIllustration(s):\n1\n\n...tree A graphical representation of the alternatives in a decision-making problem. As an example , suppose that we are choosing between two machines. One costs 100 units and has a 20% probability of breaking down within a year. The other costs 120 units and has a 5% breakdown probability. A breakdown costs 60 units. Ignoring the possibility of multiple breakdowns, which machine should we buy? Reading the decision tree from right to left, we see that it is cheaper to buy the 100-unit machine. In machine learning the ‘decisions’ are, in effect, the...", null, "## hypothesis test Quick reference\n\n### A Dictionary of Statistics (3 ed.)\n\nReference type:\nSubject Reference\nCurrent Version:\n2014\nSubject:\nScience and technology, Mathematics and Computer Science\nLength:\n991 words\nIllustration(s):\n2\n\n... N( μ , 9) and it is desired to test H 0 : μ  = 20 against H 1 : μ >20, using a sample of size 25. An appropriate statistic is the sample mean X̄ , which has distribution N(20, 9/25) under H 0 , or its standardized value which has distribution N(0, 1). If the desired significance level is 5%, the critical region, from tables of critical values for the standard normal distribution ( see appendix iv ) is Z >1.645. In terms of X̄ , the critical region is X̄ >20 + 1.645 or, equivalently, X̄ >20.99. The probability of making a Type II error is...", null, "## rank correlation coefficient Quick reference\n\n### A Dictionary of Statistics (3 ed.)\n\nReference type:\nSubject Reference\nCurrent Version:\n2014\nSubject:\nScience and technology, Mathematics and Computer Science\nLength:\n121 words\nIllustration(s):\n1\n\n... variables are related. The original data may be ranks , or measurements that are converted to ranks. For example, in a flower show, six sunflower exhibits (A–F) might be given the ranks 1, 3, 2, 5, 6, and 4. If the heights of these exhibits are 2.4, 1.9, 2.2, 1.8, 1.85, and 2.0 m then we might suppose that the heights affected the judge's ranking, since on converting the heights to ranks we get a similar pattern to the judge's ranks: An attraction of using a rank correlation coefficient is that the calculations are simple. Commonly used coefficients are ...", null, "## L-moments Quick reference\n\n### A Dictionary of Statistics (3 ed.)\n\nReference type:\nSubject Reference\nCurrent Version:\n2014\nSubject:\nScience and technology, Mathematics and Computer Science\nLength:\n118 words\nIllustration(s):\n1\n\n... that are computed from linear combinations of the ordered data values x (1) ≤ x (2) ≤⋯≤ x ( n ) (with mean x̄ , which is the first L -moment). Defining the quantities b 1 , b 2 ,…, b n− 1 by the next three L -moments are 2 b 1 − x̄ , 6 b 2 − 6 b 1 + x̄ , and 20 b 3 −30 b 2 +12 b 1 − x̄ . The second L -moment is related to the Gini statistic . An advantage of L -moments is that they can be calculated even when distributions have infinite...", null, "## Pascal's triangle Quick reference\n\n### The Concise Oxford Dictionary of Mathematics (5 ed.)\n\nReference type:\nSubject Reference\nCurrent Version:\n2014\nSubject:\nScience and technology, Mathematics and Computer Science\nLength:\n101 words\nIllustration(s):\n1\n\n...r , with r =0, 1,…, n . With the numbers set out in this fashion, it can be seen how the number is equal to the sum of the two numbers and , which are situated above it to the left and right. For example, equals 35, and is the sum of , which equals 15, and , which equals 20. http://www.mathsisfun.com/pascals‐triangle.html An illustration of some of the patterns in Pascal's...", null, "## stem‐and‐leaf plot Quick reference\n\n### The Concise Oxford Dictionary of Mathematics (5 ed.)\n\nReference type:\nSubject Reference\nCurrent Version:\n2014\nSubject:\nScience and technology, Mathematics and Computer Science\nLength:\n92 words\nIllustration(s):\n1\n\n...plot A method of displaying grouped data , by listing the observations in each group, resulting in something like a histogram on its side. For the observations 45, 25, 67, 49, 12, 9, 45, 34, 37, 61, 23, grouped using class intervals 0–9, 10–19, 20–29, 30–39, 40–49, 50–59 and 60–69, a stem‐and‐leaf plot is shown on the left. If, as in this case, the class intervals are defined by the digit occurring in the ‘tens’ position, the diagram is more commonly written as shown on the right. See figure. stem‐and‐leaf...", null, "## Cochran-Armitage test Quick reference\n\n### A Dictionary of Statistics (3 ed.)\n\nReference type:\nSubject Reference\nCurrent Version:\n2014\nSubject:\nScience and technology, Mathematics and Computer Science\nLength:\n184 words\nIllustration(s):\n1\n\n...after Cochran and Armitage ) for trend in a 2 × k contingency table where the columns correspond to categories of an ordinal variable . Denote the frequency in row i and column j by f ij with i = 1 or 2, and j = 1,…, J , denote the row totals by f 10 and f 20 , the column totals by f 01 ,…, f 0 J , and the grand total by f 00 . The test examines the null hypothesis that, as j increases, the population proportion for a given row consistently increases (or decreases). The test statistic is T given by where t 1 , …, t k are...", null, "## birthday problem Quick reference\n\n### A Dictionary of Statistics (3 ed.)\n\nReference type:\nSubject Reference\nCurrent Version:\n2014\nSubject:\nScience and technology, Mathematics and Computer Science\nLength:\n121 words\nIllustration(s):\n1\n\n...is not 183, but 23. The complementary event is that all n people have different birthdays. The probability of this, p n , is which reduces surprisingly quickly as n increases: n 3 5 9 13 16 19 22 23 26 30 34 40 46 p n 0.99 0.97 0.91 0.81 0.72 0.62 0.52 0.49 0.40 0.29 0.20 0.11...", null, "## Platonic solid Quick reference\n\n### The Concise Oxford Dictionary of Mathematics (5 ed.)\n\nReference type:\nSubject Reference\nCurrent Version:\n2014\nSubject:\nScience and technology, Mathematics and Computer Science\nLength:\n153 words\nIllustration(s):\n1\n\n..., with 4 triangular faces ( p =3, q =3), (ii) the cube , with 6 square faces ( p =4, q =3), (iii) the regular octahedron , with 8 triangular faces ( p =3, q =4), (iv) the regular dodecahedron , with 12 pentagonal faces ( p =5, q =3), (v) the regular icosahedron , with 20 triangular faces ( p =3, q...", null, "## binary prefixes Quick reference\n\n### A Dictionary of Computer Science (7 ed.)\n\nReference type:\nSubject Reference\nCurrent Version:\n2016\nSubject:\nScience and technology, Mathematics and Computer Science\nLength:\n300 words\nIllustration(s):\n1\n\n...In computing, it became common to use the prefix ‘kilo-’ to mean 2 10 , so one kilobit was 1024 bits (not 1000 bits). This was extended to larger prefixes, so ‘mega-’ in computing is taken to be 2 20 (1 048 576) rather than 10 6 (1 000 000). However, there is a variation in usage depending on the context. In discussing memory capacities megabyte generally means 2 20 bytes, but in disk storage (and data transmission) megabyte is often taken to mean 10 6 bytes. (In some contexts, as in the capacity of a floppy disk, it has even been quoted as 1 024 000 bytes,...", null, "## reliability Quick reference\n\n### A Dictionary of Statistics (3 ed.)\n\nReference type:\nSubject Reference\nCurrent Version:\n2014\nSubject:\nScience and technology, Mathematics and Computer Science\nLength:\n276 words\nIllustration(s):\n4\n\n.... Alternatively, if a variety of scores are possible on test j , let s 2 j be the variance of those obtained. An approximation to the reliability coefficient is provided by Cronbach’s alpha , given by and other approximations are provided by the Kuder–Richardson formulae KR 20 and KR 21 (named after the equation numbers in Kuder and Richardson's 1937 paper): An alternative approach is the split-half method , in which each test provides two scores (for example, the score on the even questions and the score on the corresponding odd questions). Let ...", null, "## runs test Quick reference\n\n### A Dictionary of Statistics (3 ed.)\n\nReference type:\nSubject Reference\nCurrent Version:\n2014\nSubject:\nScience and technology, Mathematics and Computer Science\nLength:\n290 words\nIllustration(s):\n3\n\n...null hypothesis, for reasonably large m and n , this has an approximate normal distribution , with mean and variance equal to respectively. Since the count is an integer, a continuity correction of 0.5 will be needed. As an example, suppose that the reaction times, in ms, of 20 girls and 25 boys were: Girls  428, 444, 446, 479, 492, 513, 522, 533, 544, 545, 560, 566, 581, 582, 590, 595, 599, 612, 634, 655 Boys  415, 439, 442, 477, 500, 512, 523, 532, 577, 580, 613, 614, 622, 633, 670, 671, 680, 688, 701, 703, 722, 730, 744, 750, 777 The null hypothesis is...", null, "## Mantel–Haenszel test Quick reference\n\n### A Dictionary of Statistics (3 ed.)\n\nReference type:\nSubject Reference\nCurrent Version:\n2014\nSubject:\nScience and technology, Mathematics and Computer Science\nLength:\n278 words\nIllustration(s):\n2\n\n...The test statistic, M , is computed as follows. Denote by f jkl the number of patients in class l (=1, 2,…, L ) who experience outcome j (=1 or 2) when given treatment k (=1 or 2). Write f 0 kl = f 1 kl + f 2 kl , f j 0l = f j1l + f j2l , and f 00 l = f 10 l + f 20 l . Then Under the null hypothesis, the distribution of M is approximated by a chi-squared distribution with one degree of freedom . When L =1 the test is equivalent to the Yates-corrected chi-squared test ( see two-by-two table ). The ½ term is a continuity correction ....", null, "## wavelet Quick reference\n\n### A Dictionary of Statistics (3 ed.)\n\nReference type:\nSubject Reference\nCurrent Version:\n2014\nSubject:\nScience and technology, Mathematics and Computer Science\nLength:\n394 words\nIllustration(s):\n6\n\n...most used were introduced by Daubechies in 1988 . These wavelets have fractal properties. Other families of wavelets include symmlets and coiflets . Mother wavelets. The D 2 wavelet is the Haar wavelet. The wavelets illustrated here are the D 4 , D 12 , and D 20 wavelets. All are members of the Daubechies...", null, "## chi-squared test Quick reference\n\n### A Dictionary of Statistics (3 ed.)\n\nReference type:\nSubject Reference\nCurrent Version:\n2014\nSubject:\nScience and technology, Mathematics and Computer Science\nLength:\n490 words\nIllustration(s):\n7\n\n...if there are too many small expected frequencies. As an example, suppose it is hypothesized that a type of sweet pea occurs in shades of white, red, pink, and blue, with proportions ¼, p , (¾−3 p ), and 2 p , respectively. A random sample of 120 seeds is sown. All germinate with 20 having white flowers, 10 having red flowers, 40 pink, and 50 blue. The question is whether these results are consistent with the hypothesis. In this case the maximum likelihood estimate of p is 0.15, so the expected frequencies are 30, 18, 36, and 36. Thus There are 4−1−1=2...", null, "", null, "View:" ]

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[ "# Two Progressions\n\nCalculus Level 3\n\n5 distinct positive reals form an arithmetic progression. The 1st, 2nd and 5th term form a geometric progression. If the product of these 5 numbers is $124 \\frac {4} {9}$, what is the product of the 3 terms of the geometric progression?\n\nDetails and assumptions\n\nThe phase \"form an arithmetic progression\" means that the values are consecutive terms of an arithmetic progression. Similarly, \"form a geometric progression\" means that the values are consecutive terms of a geometric progression.\n\n×" ]

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[ "Finite Element Solution of a Stream Function-Vorticity System and Its Application to the Navier Stokes Equations\nAbstract: The finite element solution of a generalized Stokes system in terms of the flow variables stream function and vorticity is studied. This system results from a time discretization of the time-dependent Stokes system in stream function-vorticity formulation, or yet by the application of the characteristics method to solve the Navier-Stokes equations in the same representation. Numerical results presented for both cases illustrate the good behaviour of the adopted approach.\nCite this paper: F. Ghadi, V. Ruas and M. Wakrim, \"Finite Element Solution of a Stream Function-Vorticity System and Its Application to the Navier Stokes Equations,\" Applied Mathematics, Vol. 4 No. 1, 2013, pp. 257-262. doi: 10.4236/am.2013.41A039.\nReferences\n\n   F. A. Ghadi, V. Ruas and M. Wakrim, “A Mixed Finite Element to Solve the Stokes Problem in the Stream Function and Vorticity Formulation,” Hiroshima Mathematical Journal, Vol. 28, No. 3, 1998, pp. 381-398.\n\n   R. Glowinski and O. Pironneau, “Numerical Methods for the First Bi-Harmonic Equation and for the Two-Dimensional Stokes Problem,” SIAM Review, Vol. 21, No. 2, 1979, pp. 167-212. doi:10.1137/1021028\n\n   F. A. Ghadi, V. Ruas and M. Wakrim, “A Mixed Method to Solve the Evolutionary Stokes Problem in the Stream Function and Vorticity Formulation,” Proceedings of LUXFEM, Luxemburg, 13-14 November 2003.\n\n   F. A. Ghadi, V. Ruas and M. Wakrim, “Numerical Solution to the Time—Dependent Incompressible Navier-Stokes Equations by Piecewise Linear Finite Elements,” Journal of Computational and Applied Mathematics, Vol. 215, No. 2, 2008, pp. 429-437. doi:10.1016/j.cam.2006.03.047\n\nTop" ]

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[ "# Frequency Modulation\n\nFREQUENCY MODULATION\n\nFrequency modulation uses the instantaneous frequency of a modulating signal (voice, music, data, etc.) to directly vary the frequency of a carrier signal. Modulation index, b, is used to describe the ratio of maximum frequency deviation of the carrier to the maximum frequency deviation of the modulating signal.", null, "Depending on the modulation index chosen, the carrier and certain sideband frequencies may actually be suppressed. Zero crossings of the Bessel functions, Jn(b), occur where the corresponding sideband, n, disappears for a given modulation index, b. The composite spectrum for a single tone consists of lines at the carrier and upper and lower sidebands (of opposite phase), with amplitudes determined by the Bessel function values at those frequencies.\n\nFM General Equation\nLet the carrier be xc(t) = Xc·cos (wct), and the modulating signal be\nxm(t) = b·sin (wmt)\nThen x(t) = Xc·cos [wct + b·sin (wmt)]\n\nModulation Index\n b = Dw  wm = maximum carrier frequency deviation modulation frequency\n\nNarrowband FM (NBFM)\n\nNarrowband FM is defined as the condition where b is small enough to make all terms after the first two in the series expansion of the FM equation negligible.\n\nNarrowband Approximation: b = Dw/wm < 0.2 (could be as high as 0.5, though)\n\nBW ~ 2wm\n\nWideband FM (WBFM)\n\nWideband FM is defined as when a significant number of sidebands have significant amplitudes.\n\nBW ~ 2Dw\n\nCarson's Rule\n\nJ.R. Carson showed in the 1920's that a good approximation that for both very small and very large b,\n\nBW ~ 2 (Dw + wm) = 2*wm (1 + b)\n\nIn the following examples, the carrier frequency is eleven time the modulation frequency. Red (dashed) lines represent the modulation envelope. Blue (solid) lines represent the modulated carrier.\nModulation Index (b) = 1", null, "Here, the maximum frequency (fmax) causes a maximum deviation of 1*fmax in the carrier. From the modulation index formula:\n\n b = 1 1 = 1\n\nModulation Index (b) = 5", null, "Here, the maximum frequency (fmax) causes a maximum deviation of 5*fmax in the carrier. From the modulation index formula:\n\n b = 5 1 = 5\n\nModulation Index (b) = 25", null, "Here, the maximum frequency (fmax) causes a maximum deviation of 25*fmax in the carrier. From the modulation index formula:\n\n b = 25 1 = 25\nNote: FM waveforms created with MathCAD 4.0 software." ]

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[ "Forecasting: how to detect outliers?\n\nis — of course — left to you to experiment…Do It YourselfPython If you have a pandas DataFrame with one column as the forecast and another one as the demand (the typical output from our exponential smoothing models), we can use this code:df[“Error”] = df[“Forecast”] — df[“Demand”]m = df[“Error”].\n\nmean()s = df[“Error”].\n\nstd()from scipy.\n\nstats import normlimit_high = norm.\n\nppf(0.\n\n99,m,s)+df[“Forecast”]limit_low = norm.\n\nppf(0.\n\n01,m,s)+df[“Forecast”]df[“Updated”] = df[“Demand”].\n\nclip(lower=limit_low,upper=limit_high)print(df)Go the extra mile!If you think back about our idea to analyze the forecast error and make a threshold of acceptable errors, we actually still have a minor issue.\n\nThe threshold we compute is based on the dataset including the outliers.\n\nThis outlier drives the error variation upward so that the acceptable threshold is biased and overestimated.\n\nTo correct this, one could actually shrink the outlier not to the threshold calculated based on the original demand dataset but to a limit calculated on a dataset without this specific outlier.\n\nHere’s the recipe:Populate a first forecast against the historical demand.\n\nCompute the error, the error mean and the error standard deviationCompute the lower & upper acceptable thresholds (based on the error mean and standard deviation).\n\nIdentify outliers just as explained previously.\n\nRe-compute the error mean and standard deviation but excluding the outliers.\n\nUpdate the lower & upper acceptable thresholds based on these new values.\n\nUpdate the outlier values based on the new threshold.\n\nIf we take back our seasonal example from above, we initially had a forecast error mean of 0.\n\n4 and a standard deviation of 3.\n\n22.\n\nIf we remove the point Y2 M11, we obtain an error mean of -0.\n\n1 and a standard deviation of 2.\n\n3.\n\nThat means that now the thresholds are -5.\n\n3,5.\n\n2 around the forecast.\n\nOur outlier in Y2 M11 would then be updated to 10 (instead of 12 with our previous technique).\n\nDo It YourselfWe’ll take back our code from our previous idea and add a new step to update the error mean and standard deviation values.\n\ndf[“Error”] = df[“Forecast”] — df[“Demand”]m = df[“Error”].\n\nmean()s = df[“Error”].\n\nstd()from scipy.\n\nstats import normprob = norm.\n\ncdf(df[“Error”], m, s)outliers = (prob > 0.\n\n99) | (prob < 0.\n\n01)m2 = df[“Error”][~outliers].\n\nmean()s2 = df[“Error”][~outliers].\n\nstd()limit_high = norm.\n\nppf(0.\n\n99,m2,s2)+df[“Forecast”]limit_low = norm.\n\nppf(0.\n\n01,m2,s2)+df[“Forecast”]df[“Updated”] = df[“Demand”].\n\nclip(lower=limit_low,upper=limit_high)print(df)About the authorNicolas Vandeput is a supply chain data scientist specialized in demand forecasting & inventory optimization.\n\nIn 2016, he founded SupChains (www.\n\nsupchains.\n\ncom), his consultancy company; two years later, he co-founded SKU Science (www.\n\nskuscience.\n\ncom), a smart online platform for supply chain management.\n\nIf you are interested in forecast and machine learning, you can buy his book Data Science for Supply Chain Forecast.. More details" ]

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[ "# Speaker impedance & proper speaker loads.\n\nOne of the questions that I often get asked about is the proper use of the extension cabinet speaker output, so this seems like a good time to explain it and I’ll start with the speaker impedance.\n\nFirst you need to verify the impedance of your speakers, both the internal speaker if you have a combo as well as the extension cabinet that you plan to use.  If it’s not marked on the speaker or you’re not sure that the listed impedance is correct, all you need is a multi-meter to measure it.  You will need to keep in mind that a static speaker, or one that’s not being powered by an amp, would only show about 75% of the speaker’s actual impedance.  For example, an 8 ohm speaker may show something like 6.2 ohms while a 16 ohm speaker may only measure at 12.6 ohms.\n\nFiguring out your total speaker load is fairly easy to do and here’s a couple of examples which will help explain the math to you.  For example, when you wire speakers in parallel you’ll find that two 8 ohm speakers in parallel would be a 4 ohm total load (8 ohms + 8 ohms = 4 ohms in parallel).  In the same manner two 16 ohm speakers in parallel would be a 8 ohm total load (16 ohms + 16 ohms = 8 ohms in parallel).\n\nWhen you use series wiring on the speakers it would look like this.  When you wire speakers in series you’ll find that two 4 ohm speakers in series would be a 8 ohm total load (4 ohms + 4 ohms = 8 ohms in series).  In the same manner two 8 ohm speakers in series would be a 16 ohm total load (8 ohms + 8 ohms = 16 ohms in series).\n\nThen you’ll often find series/parallel wiring in multi-speaker cabinets like a 4×12, and in that configuration each pair may be wired in parallel, and then each pair wired in series to arrive at the necessary speaker load." ]

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[ "## What is the formula for moisture content?\n\nThe amount of water is determined by subtracting the dry weight from the initial weight, and the moisture content is then calculated as the amount of water divided by the dry weight or total weight, depending on the reporting method.\n\n## What is percentage moisture content?\n\nWet basis moisture content (designated MW in the text) is described by the percentage equivalent of the ratio of the weight of water (WW) to the total weight of the material (Wt). When the term “moisture content” is used in the food industry it almost always refers to wet basis moisture content.\n\n## What do you mean by moisture content?\n\nWater content or moisture content is the quantity of water contained in a material, such as soil (called soil moisture), rock, ceramics, crops, or wood.\n\n## What is the formula for percentage of moisture content in timber?\n\nThe moisture (water) content in the wood is defined as the weight of the water in damp material divided by the weight of the material in a dry state. To obtain a percentage, the result is multiplied by 100.\n\n## What level of moisture is acceptable?\n\nA range of 5 to 12 percent is considered optimal. A reading of up to 17 percent is generally considered moderate moisture and acceptable. Any reading over 17 percent is considered to be an indication of saturation and the need to replace the drywall and to take preventative measures against future moisture buildup.\n\n## What is optimum moisture content of soil?\n\nThe water content at which a soil can be compacted to the maximum dry unit weight by a given compactive effort. Also called optimum water content.\n\n## What is the principle of moisture content determination?\n\nThe principle of the thermogravimetric method of moisture content determination is defined as the weight loss of mass that occurs as the material is heated. The sample weight is taken prior to heating and again after reaching a steady-state mass subsequent to drying.\n\n## What is dry basis moisture content?\n\nDry basis moisture content (designated Md in the text) is described by the percentage equivalent of the ratio of the weight of water (WW) to the weight of the dry matter (Wd). Dry basis moisture is most commonly used for describing moisture changes during drying.\n\n## Why do we measure moisture content?\n\nMoisture determination is one of the most important and most widely used measurements in the processing and testing of foods. Since the amount of dry matter in a food is inversely related to the amount of moisture it contains, moisture content is of direct economic importance to the processor and the consumer.\n\n## What is difference between moisture content and water content?\n\nMoisture content and water activity are measured for two different purposes. Moisture content— AKA water content— is a measurement of the total amount of water contained in a food, usually expressed as a percentage of the total weight (see calculation).\n\n## What is critical moisture content?\n\nThe critical moisture content (average material moisture content at the end of the constant-rate-drying period) is a function of material properties, the constant-rate of drying, and particle size. A change in material handling method or any operating variable, such as heating rate, may effect mass transfer.\n\n## How is dry matter content calculated?\n\n1) Figure the dry matter percentage. Subtract the moisture content from 100%. In this example 100% – 75.9% = 24.1% dry matter. 2) Convert individual nutrients from “as fed” to “dry matter”.\n\n## What is wood moisture content?\n\nWood is a hygroscopic, meaning it is a material that absorbs water. The moisture content of newly sawn wood is usually 40-200%. In normal use the moisture content of wood varies between 8% and 25% by weight, depending on the relative humidity of the air.\n\n## What is the moisture content of fresh cut wood?\n\n40-200%\n\n### Releated\n\n#### Characteristic equation complex roots\n\nWhat are roots of characteristic equations? discussed in more detail at Linear difference equation#Solution of homogeneous case. The characteristic roots (roots of the characteristic equation) also provide qualitative information about the behavior of the variable whose evolution is described by the dynamic equation. How do I know if my roots are complex? When graphing, if […]\n\n#### Free fall time equation\n\nWhat is the formula for time in free fall? Free fall means that an object is falling freely with no forces acting upon it except gravity, a defined constant, g = -9.8 m/s2. The distance the object falls, or height, h, is 1/2 gravity x the square of the time falling. Velocity is defined as […]" ]

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[ "# Analysis of Nonlinear Partial Differential Equations\n\nLead Research Organisation: University of Oxford\nDepartment Name: Mathematical Institute\n\n### Abstract\n\nPartial differential equations (PDEs) are equations that relate the partial derivatives, usually with respect to space and time coordinates, of unknown quantities. They are ubiquitous in almost all applications of mathematics, where they provide a natural mathematical description of phenomena in the physical, natural and social sciences, often arising from fundamental conservation laws such as for mass, momentum and energy. Significant application areas include geophysics, the bio-sciences, engineering, materials science, physics and chemistry, economics and finance. Length-scales of natural phenomena modelled by PDEs range from sub-atomic to astronomical, and time-scales may range from nanoseconds to millennia. The behaviour of every material object can be modelled either by PDEs, usually at various different length- or time-scales, or by other equations for which similar techniques of analysis and computation apply. A striking example of such an object is Planet Earth itself.Linear PDEs are ones for which linear combinations of solutions are also solutions. For example, the linear wave equation models electromagnetic waves, which can be decomposed into sums of elementary waves of different frequencies, each of these elementary waves also being solutions. However, most of the PDEs that accurately model nature are nonlinear and, in general, there is no way of writing their solutions explicitly. Indeed, whether the equations have solutions, what their properties are, and how they may be computed numerically are difficult questions that can be approached only by methods of mathematical analysis. These involve, among other things, precisely specifying what is meant by a solution and the classes of functions in which solutions are sought, and establishing ways in which approximate solutions can be constructed which can be rigorously shown to converge to actual solutions. The analysis of nonlinear PDEs is thus a crucial ingredient in the understanding of the world about us.As recognized by the recent International Review of Mathematics, the analysis of nonlinear PDEs is an area of mathematics in which the UK, despite some notable experts, lags significantly behind our scientific competitors, both in quantity and overall quality. This has a serious detrimental effect on mathematics as a whole, on the scientific and other disciplines which depend on an understanding of PDEs, and on the knowledge-based economy, which in particular makes increasing use of simulations of PDEs instead of more costly or impractical alternatives such as laboratory testing.The proposal responds to the national need in this crucial research area through the formation of a forward-looking world-class research centre in Oxford, in order to provide a sharper focus for fundamental research in the field in the UK and raise the potential of its successful and durable impact within and outside mathematics. The centre will involve the whole UK research community having interests in nonlinear PDEs, for example through the formation of a national steering committee that will organize nationwide activities such as conferences and workshops.Oxford is an ideal location for such a research centre on account of an existing nucleus of high quality researchers in the field, and very strong research groups both in related areas of mathematics and across the range of disciplines that depend on the understanding of nonlinear PDEs. In addition, two-way knowledge transfer with industry will be achieved using the expertise and facilities of the internationally renowned mathematical modelling group based in OCIAM which, through successful Study Groups with Industry, has a track-record of forging strong links to numerous branches of science, industry, engineering and commerce. The university is committed to the formation of the centre and will provide a significant financial contribution, in particular upgrading one of the EPSRC-funded lectureships to a Chair\n\n### Publications\n\n10 25 50\nKnezevic D (2009) A heterogeneous alternating-direction method for a micro-macro dilute polymeric fluid model in ESAIM: Mathematical Modelling and Numerical Analysis\n\nChrusciel P (2011) A lower bound for the mass of axisymmetric connected black hole data sets in Classical and Quantum Gravity\n\nBall J (2015) A probabilistic model for martensitic avalanches in MATEC Web of Conferences\n\nNiethammer B (2010) A rigorous derivation of mean-field models for diblock copolymer melts in Calculus of Variations and Partial Differential Equations\n\nChrusciel P (2010) A Uniqueness Theorem for Degenerate Kerr-Newman Black Holes in Annales Henri Poincaré\n\nHerrmann M (2010) Action Minimising Fronts in General FPU-type Chains in Journal of Nonlinear Science\n\nBURKE S (2013) AN ADAPTIVE FINITE ELEMENT APPROXIMATION OF A GENERALIZED AMBROSIO-TORTORELLI FUNCTIONAL in Mathematical Models and Methods in Applied Sciences\n\nLuskin M (2009) An Analysis of Node-Based Cluster Summation Rules in the Quasicontinuum Method in SIAM Journal on Numerical Analysis\n\nMielke A (2014) An Approach to Nonlinear Viscoelasticity via Metric Gradient Flows in SIAM Journal on Mathematical Analysis\n\nBall J (2014) An investigation of non-planar austenite-martensite interfaces in Mathematical Models and Methods in Applied Sciences\n\nBulícek M (2014) Analysis and approximation of a strain-limiting nonlinear elastic model in Mathematics and Mechanics of Solids\n\nOrtner C (2008) Analysis of a quasicontinuum method in one dimension in ESAIM: Mathematical Modelling and Numerical Analysis\n\nHudson T (2015) Analysis of Stable Screw Dislocation Configurations in an Antiplane Lattice Model in SIAM Journal on Mathematical Analysis\n\nKirchheim B (2011) Automatic convexity of rank-1 convex functions in Comptes Rendus Mathematique\n\nDescription This was a broad grant designed to help consolidate research on nonlinear partial differential equations in the UK. In particular the Oxford Centre for Nonlinear PDE was founded as a result of the grant and is now a leading international centre. As regards specific research advances, these were in various areas of applications of PDE, for example to fluid and solid mechnaics, liquid crystals, electromagnetism, and relativity.\nExploitation Route Through publications and consultation with current and former members of OxPDE.\nSectors Aerospace, Defence and Marine,Chemicals,Construction,Electronics,Energy,Environment\n\nURL https://www0.maths.ox.ac.uk/groups/oxpde" ]

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[ "# Loan Amortization Excel Template\n\nThe Loan Amortization Excel Template will help you in understanding the process of amortization and will help you calculate the amortization of your loan. It is a very useful guide for all of you. It will help you understand the concept of amortization.", null, "Car Payment Amortization Schedule Spreadsheet Unique Car Loan from loan amortization excel template , source:oregondaysofculture.org\n\nOnce you understand the concept of amortization then you can understand the whole process of loan amortization. You have to pay some amount on the loan every month, which is interest on the amount you borrowed. In the repayment process, the loan amount is reduced by the repayment of the capital amount in cash.\n\nThere is some amount of things which are charged by the lender as the total loan amount. These are Interest, taxes, and charges. The payment of these charges is not always made along with the loan amount. Sometimes they are deducted from the repayment of the loan amount.", null, "Amortization Formula Excel Template amortization schedule template from loan amortization excel template , source:300blackbroth.com\n\nFor the process of amortization, you can use the amortization calculator which will help you in calculating the repayment schedule. The idea behind the calculator is that it will calculate the total payment schedule based on your loan amounts. You have to enter the annual amount of loans. You also have to enter the term of the loan, the number of payments to be made, the number of installments and the monthly payment.\n\nThen the amortization calculator will calculate the amortization of your loan. The repayment of your loan will be calculated based on the repayment schedule. After that, you have to enter the payment amount. Finally, you have to calculate the final balance of your loan.", null, "Payment Schedule Template Free Loan Payment Plan Template Loan from loan amortization excel template , source:thalmus.co\n\nNow that you have understood the concept of an amortization calculator, you can easily calculate your loan payments with the help of the amortization calculator. You have to enter the same information on the calculator. The amortization calculator will calculate the payment schedule. You can calculate the repayment schedule on the mortgage calculator.\n\nThe amortization calculator will also help you calculate the capital repayment and it will also calculate the capital interest on the loan. Also, you can calculate the interest on the loan using the amortization calculator. You have to make the payment on the capital amount on the interest calculated on the calculator.", null, "Car Loan Amortization Schedule Excel Awesome Car Loan Excel Template from loan amortization excel template , source:chicharronesesmeralda.com\n\nRemember that a loan amortization can take a long time. You have to calculate the monthly payments and capital payments. With the help of the amortization calculator, you can calculate all the payments and repayments." ]

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[ "# How to avoid computationally expensive fluid networks in Dymola\n\nWhen building fluid networks in Dymola, it is of paramount importance to place the components in a specific order to achieve a staggered grid arrangement. This will reduce the number of non-linear equations, making the convergence faster and easier.\n\nIn this article, the staggered grid approach in the fluid networks will be explained and the importance of building networks following this rule will be demonstrated through a simple example.\n\nFigure 1 shows a staggered grid, where mass-related fields such as pressure and temperature are defined on one set of grid points while velocity fields are defined on another set of grid points.\n\nIn Dymola, flow resistance models are components where the momentum balance is performed. Inside this models (that can be replaced with any model extended from BaseClasses.FlowModels.PartialStaggeredFlowModel), two different types of functions are used:\n\n1. dp_MFLOW: calculates the mass flow rate through the component out of the pressure drop dp. Generally, this function is numerically best used for the compressible case, where the pressure loss (dp) is known from the pressures defined as state variables in the medium model.\n2. dp_DP: calculates the pressure loss from the mass flow rate. It is numerically best used for the incompressible case, where the mass flow rate is known as state variable in the medium model.\n\nOn the other hand, volume models are the models where mass and energy balances are performed according to the finite volume approach. In the volumes, the pressure is defined as a state.\n\nWhen two flow resistance models are connected in series, the pressure at the connection is unknown and the solver has to iterate at every step to calculate the pressure, leading to large nonlinear systems.\n\nIf two volumes are directly connected together, this means that (due to the stream concept of the connectors) the pressures of the two connected volumes are identical, the temperatures, however, are not equal. This corresponds to volumes that are connected together with a very short distance between them and the solver will need to iterate for some time until the different volume temperatures are equalised (leading to high-index DAE).", null, "Figure 1: Staggered grid.", null, "Figure 2: Example of volume models vs flow resistance models.\n\nMSL pipes can have different structures which can be configured with the parameter ModelStructure on the Advanced tab:", null, "Figure 3: MSL Pipe Advanced tab showing all the possible model structure configurations.\n\nThe model structures included in MSL pipes are:\n\n• av_vb: port_a – volume – flow model – volume – port_b (nNodes-1 momentum balances between nNodes flow segments)\n• a_v_b: port_a – flow model – volume – flow model – port_b (nNodes+1 momentum balances across nNodes flow segments)\n• av_b: port_a – volume – flow model – port_b\n• a_vb: port_a – flow model – volume – port_b\n\nThe default is:\n\n• nNodes=2\n• modelStructure=Types.ModelStructure.av_vb\n\nTo practically demonstrate what has been discussed so far, we can build a simple fluid network to test the MSL pipe model.\n\nThe default settings for the MSL pipe are nNodes=2 and modelStructure av_vb (volume – flow model – volume). This means that to build the staggered grid correctly, we need two orifices (which are flow models), one before and one after the pipe.\n\nDoing so, there will not be nonlinear systems, as shown in the translation tab – Statistics:", null, "Figure 4: Translation tab – Statistics for a staggered grid (nNodes=2).\n\nIn case one wants to discretize the pipe with just one volume, the structure of the pipe could be either a_vb (flow model – volume) or av_b (volume – flow model). In this case, the fluid layout with the pipe between two orifices is not staggered anymore and leads to a nonlinear system of equations, as shown in the translation tab – Statistics.", null, "Figure 5: Translation tab – Statistics for a non staggered grid (nNodes=1).\n\nHowever, if a volume is introduced before the pipe (with structure a_vb), the nonlinear system of equations disappears:", null, "Figure 6: Translation tab – Statistics for a staggered grid (nNodes=1).\n\nThis simple example has shown the effect of a non-staggered arrangement on the number of non-linear systems. When building more complex systems, it is essential to build fluid networks using the structure flow model – volume, to avoid models that are uselessly computationally expensive.\n\nReferences:\n\nModelica Users Guide version 3.2.2\n\nWritten by: Maura Gallarotti – Project Engineer" ]

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[ "# Transistor help!\n\nStatus\nNot open for further replies.\n\n#### hantto\n\n##### Member\nPlease help me. I get so confused when i try to read the BC547 and BD140 datasheets! I want to know what's the max current and voltage that one can feed into transistors BASE whitout them braking!\n\n#### kinjalgp\n\n##### Active Member\nFor BD140 Max. base emitter volatge is 5V and for BC547 it is 6V.\nHfe min. for BD140 is 40 at bias current of 150mA and\nHfe min. for BC547 is 110 at bias current of 2mA.\n\n#### hantto\n\n##### Member\nOK so the resistor for the BC547 will be (let's say i use 11,3V suply) 11,3-2=9,3 (i want to give 2V to the base) and then the 2mA current 9,3/0,002=4650\n\nSo the resistor before the base will be 4150ohm?\n\n(this is just an example to help ME understad a bit more)\n\nAnd thx kinjal!\n\n#### kinjalgp\n\n##### Active Member\nIf you want to give 2V to base, a potential divider should be designed such that it will produce 2V at transistors base as well as it is capable of supplying at least 2mA of current to the base. Lets take an example where Vcc=12V (please use '.' instead of ',') Assuming Emitter is directly grounded,\nApplying KVL to base loop\nVBB - IB*RBB - Vbe = 0\nHere,\nVBB = 2V\nIB = 2mA\nVbe = 0.7V (Silicon)\nSo, 2 - 2m*RBB - 0.7 = 0\nor, RBB = 650 Ohms\n\nRBB is the parallel combination of R1 and R2 in potential divider. Thus to supply 2mA to base, equivalent resistance of R1 and R2 in parallel should be 650 Ohms.\n\nNow let us assume R2 = 820 Ohms.\nUsing potential divider formula,\nVBB = (R2 * Vcc)/(R1 + R2)\n2 = (820 * 12)/(R1 + 820)\nSolving above equation, R1 = 4.1k ~= 3.9k (Standard value)\n\nSo now we have R1 = 3.9K, R2=820 Ohms\nLets see if parallel combination of both these match 650 Ohms which is our requirement.\nRBB = (R1 * R2) / (R1 + R2) = (3.9k * 820) / (3.9k + 820) = 677.54 Ohms which is nearly equal to 650 Ohms.\n\nThus we have designed potential divider bias circuit for BC547 for Vbe = 2V and Ib = 2mA.\n\nHope you got it.\n\n#### hantto\n\n##### Member\nYes, I got it!!!!", null, "", null, "", null, "", null, "THX KINJAL!!!!", null, "Status\nNot open for further replies.\n\nReplies\n7\nViews\n1K\nReplies\n15\nViews\n2K\nReplies\n10\nViews\n2K\nReplies\n8\nViews\n2K\nReplies\n120\nViews\n7K", null, "" ]

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[ "# #061307 Color Information\n\nIn a RGB color space, hex #061307 is composed of 2.4% red, 7.5% green and 2.7% blue. Whereas in a CMYK color space, it is composed of 68.4% cyan, 0% magenta, 63.2% yellow and 92.5% black. It has a hue angle of 124.6 degrees, a saturation of 52% and a lightness of 4.9%. #061307 color hex could be obtained by blending #0c260e with #000000. Closest websafe color is: #000000.\n\n• R 2\n• G 7\n• B 3\nRGB color chart\n• C 68\n• M 0\n• Y 63\n• K 93\nCMYK color chart\n\n#061307 color description : Very dark (mostly black) lime green.\n\n# #061307 Color Conversion\n\nThe hexadecimal color #061307 has RGB values of R:6, G:19, B:7 and CMYK values of C:0.68, M:0, Y:0.63, K:0.93. Its decimal value is 398087.\n\nHex triplet RGB Decimal 061307 `#061307` 6, 19, 7 `rgb(6,19,7)` 2.4, 7.5, 2.7 `rgb(2.4%,7.5%,2.7%)` 68, 0, 63, 93 124.6°, 52, 4.9 `hsl(124.6,52%,4.9%)` 124.6°, 68.4, 7.5 000000 `#000000`\nCIE-LAB 4.695, -6.051, 4.046 0.346, 0.52, 0.283 0.301, 0.452, 0.52 4.695, 7.28, 146.233 4.695, -2.673, 3.167 7.21, -4.043, 2.719 00000110, 00010011, 00000111\n\n# Color Schemes with #061307\n\n• #061307\n``#061307` `rgb(6,19,7)``\n• #130612\n``#130612` `rgb(19,6,18)``\nComplementary Color\n• #0c1306\n``#0c1306` `rgb(12,19,6)``\n• #061307\n``#061307` `rgb(6,19,7)``\n• #06130e\n``#06130e` `rgb(6,19,14)``\nAnalogous Color\n• #13060c\n``#13060c` `rgb(19,6,12)``\n• #061307\n``#061307` `rgb(6,19,7)``\n• #0e0613\n``#0e0613` `rgb(14,6,19)``\nSplit Complementary Color\n• #130706\n``#130706` `rgb(19,7,6)``\n• #061307\n``#061307` `rgb(6,19,7)``\n• #070613\n``#070613` `rgb(7,6,19)``\nTriadic Color\n• #121306\n``#121306` `rgb(18,19,6)``\n• #061307\n``#061307` `rgb(6,19,7)``\n• #070613\n``#070613` `rgb(7,6,19)``\n• #130612\n``#130612` `rgb(19,6,18)``\nTetradic Color\n• #000000\n``#000000` `rgb(0,0,0)``\n• #000000\n``#000000` `rgb(0,0,0)``\n• #000000\n``#000000` `rgb(0,0,0)``\n• #061307\n``#061307` `rgb(6,19,7)``\n• #0c260e\n``#0c260e` `rgb(12,38,14)``\n• #123a15\n``#123a15` `rgb(18,58,21)``\n• #184d1c\n``#184d1c` `rgb(24,77,28)``\nMonochromatic Color\n\n# Alternatives to #061307\n\nBelow, you can see some colors close to #061307. Having a set of related colors can be useful if you need an inspirational alternative to your original color choice.\n\n• #081306\n``#081306` `rgb(8,19,6)``\n• #071306\n``#071306` `rgb(7,19,6)``\n• #061306\n``#061306` `rgb(6,19,6)``\n• #061307\n``#061307` `rgb(6,19,7)``\n• #061308\n``#061308` `rgb(6,19,8)``\n• #061309\n``#061309` `rgb(6,19,9)``\n• #06130a\n``#06130a` `rgb(6,19,10)``\nSimilar Colors\n\n# #061307 Preview\n\nText with hexadecimal color #061307\n\nThis text has a font color of #061307.\n\n``<span style=\"color:#061307;\">Text here</span>``\n#061307 background color\n\nThis paragraph has a background color of #061307.\n\n``<p style=\"background-color:#061307;\">Content here</p>``\n#061307 border color\n\nThis element has a border color of #061307.\n\n``<div style=\"border:1px solid #061307;\">Content here</div>``\nCSS codes\n``.text {color:#061307;}``\n``.background {background-color:#061307;}``\n``.border {border:1px solid #061307;}``\n\n# Shades and Tints of #061307\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, #010402 is the darkest color, while #f4fcf5 is the lightest one.\n\n• #010402\n``#010402` `rgb(1,4,2)``\n• #061307\n``#061307` `rgb(6,19,7)``\n• #0b220c\n``#0b220c` `rgb(11,34,12)``\n• #0f3112\n``#0f3112` `rgb(15,49,18)``\n• #144017\n``#144017` `rgb(20,64,23)``\n• #194f1d\n``#194f1d` `rgb(25,79,29)``\n• #1e5e22\n``#1e5e22` `rgb(30,94,34)``\n• #226c28\n``#226c28` `rgb(34,108,40)``\n• #277b2d\n``#277b2d` `rgb(39,123,45)``\n• #2c8a33\n``#2c8a33` `rgb(44,138,51)``\n• #309938\n``#309938` `rgb(48,153,56)``\n• #35a83e\n``#35a83e` `rgb(53,168,62)``\n• #3ab743\n``#3ab743` `rgb(58,183,67)``\nShade Color Variation\n• #41c34b\n``#41c34b` `rgb(65,195,75)``\n• #50c859\n``#50c859` `rgb(80,200,89)``\n• #5fcd68\n``#5fcd68` `rgb(95,205,104)``\n• #6ed176\n``#6ed176` `rgb(110,209,118)``\n• #7dd684\n``#7dd684` `rgb(125,214,132)``\n• #8cdb92\n``#8cdb92` `rgb(140,219,146)``\n• #9bdfa0\n``#9bdfa0` `rgb(155,223,160)``\n• #aae4ae\n``#aae4ae` `rgb(170,228,174)``\n• #b9e9bc\n``#b9e9bc` `rgb(185,233,188)``\n• #c7edca\n``#c7edca` `rgb(199,237,202)``\n• #d6f2d9\n``#d6f2d9` `rgb(214,242,217)``\n• #e5f7e7\n``#e5f7e7` `rgb(229,247,231)``\n• #f4fcf5\n``#f4fcf5` `rgb(244,252,245)``\nTint Color Variation\n\n# Tones of #061307\n\nA tone is produced by adding gray to any pure hue. In this case, #0c0d0c is the less saturated color, while #001902 is the most saturated one.\n\n• #0c0d0c\n``#0c0d0c` `rgb(12,13,12)``\n• #0b0e0b\n``#0b0e0b` `rgb(11,14,11)``\n• #0a0f0a\n``#0a0f0a` `rgb(10,15,10)``\n• #091009\n``#091009` `rgb(9,16,9)``\n• #081109\n``#081109` `rgb(8,17,9)``\n• #071208\n``#071208` `rgb(7,18,8)``\n• #061307\n``#061307` `rgb(6,19,7)``\n• #051406\n``#051406` `rgb(5,20,6)``\n• #041505\n``#041505` `rgb(4,21,5)``\n• #031605\n``#031605` `rgb(3,22,5)``\n• #021704\n``#021704` `rgb(2,23,4)``\n• #011803\n``#011803` `rgb(1,24,3)``\n• #001902\n``#001902` `rgb(0,25,2)``\nTone Color Variation\n\n# Color Blindness Simulator\n\nBelow, you can see how #061307 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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[ "## DBMS MCQs\n\nIn this section, we are going to cover some of the basic questions based on DBMS, which are frequently asked during an interview. Here we will try to cover almost all the topics related to DBMS.\n\n### 1) Relational database is a collection of\n\n1. Table\n2. Field\n3. Record or Row\n4. Colum\n\nAnswer: a\n\nExplanation: The answer is A. In Relational Database, Field is the column of the table that keeps data vertically, and the row is used to store it horizontally.\n\n### 2) What associates with all information about every entry in the table?\n\n1. Column\n2. Row\n3. Entry\n4. Key\n\nAnswer: b\n\nExplanation: Column is used to store a piece of unique information associated with entries that can be anything such as id, name, number, etc. Key is just a constraint to find anything in relation, and a row shows all the attributes associated with entries.\n\n### 3) In relation to a table, what term is used to refer single row?\n\n1. Tuple\n2. Relation Instance\n3. Attributes\n4. Field\n\nAnswer: a\n\nExplanation: Its answer is A because a tuple is a single entry that shows all attributes associated with the entry in a table.\n\n### 4) The attributes of the table is referred to as:\n\n1. Tuple\n2. Column\n3. Row\n4. Field\n\nAnswer: b\n\nExplanation: The column is referred to as an attribute of a table that contains the unique information associated with an entry.\n\n### 5) Relational algebra is a ______ which takes instances of relations as input and yields instances of relations as output.\n\n1. Relational query language\n2. Structural query language\n3. Procedural query language\n4. Fundamental query language\n\nAnswer: c\n\n### 6) Choose the correct option:\n\n1. Set intersection is the fundamental operation in relational algebra.\n2. An assignment is a fundamental operation in relational algebra.\n3. Select is the fundamental operation in relational algebra.\n4. Rename operation is the fundamental operation in relational algebra.\n\nAnswer: c and d\n\nExplanation: The correct answer is C and D. Because Select, Project, Union, Set different, Cartesian product, and Rename is considered as fundamental operations in relational algebra while Set intersection, Assignment, Natural join is an additional operation in relational algebra.\n\n### 7) Which language is used to define the database?\n\n1. DML\n2. DDL\n3. DCL\n4. TCL\n\nAnswer: a\n\nExplanation: DML(Data manipulating language) is used for accessing and manipulating the data, DCL(Data control language) is used for retrieving the stored data, and TCL(Transactional control language) is used for changes made by the DML statement.\n\n### 8) Which of the following can be considered as the primary key?\n\n1. Name\n2. Roll no.\n3. Marks\n4. Street\n\nAnswer: b\n\nExplanation: The attribute associated with the primary key should be unique. In any class, more than one student can have a common name, more than one student can get equal marks in any subject, and their street can also be common. So here, the roll number can be considered as the primary key.\n\n### 9) The overall description of the database is known as:\n\n1. Data manipulation\n2. Database schema\n3. Data definition\n4. Data integrity\n\nAnswer: b\n\n### 10) A boy name X has rupees 1000 in their bank account. His friend Y needs rupees 200, so he tries to make a transaction. He asks his bank to make the transaction to his friend’s account. After filling all the details and successful transaction, 200 is deducted from X’s account. But his friend Y told him that he doesn’t receive the money. What property of transaction has not been maintained here?\n\n1. Atomicity\n2. Consistency\n3. Isolation\n4. Durability\n\nAnswer: a\n\nExplanation: The correct answer is A because atomicity means data should remain atomic. It means that whenever an operation starts, it should execute fully. The operation should not break in the middle or execute partially.\n\n### 11) Which of the following SQL command mentioned below is used to create a table, delete the table, and alter the table?\n\n1. DML(Data manipulating language)\n2. DDL(Data definition language)\n3. DQL(Data query language)\n4. Relational Schema\n\nAnswer: b\n\nExplanation: DDL command is the right choice because the DML command is used to make any changes in the database, DQL is used to retrieve any information from the database, and relational schema is the structure of relation.\n\n### 12) If we want to remove some information from the database, then which of the following SQL command should we perform?\n\n1. DML\n2. TCL\n3. DCL\n4. DDL\n\nAnswer: a\n\nExplanation: Whenever we need to make any changes like INSERT, DELETE, UPDATE in the database, then we need to perform the DML command.\n\n### 13) SELECT * FROM student;\n\nWhich of the following option is suitable for the given statement?\n\n1. DML\n2. DQL\n3. DDL\n4. DDL\n\nAnswer: b\n\nExplanation: SELECT is used to fetch values from the database\n\n### 14) CREATE TABLE student (name VARCHAR (10), id INTEGER)\n\nChoose the correct option for the given statement.\n\n1. DDL\n2. DML\n3. DQL\n4. TCL\n\nAnswer: a\n\nExplanation: CREATE command is used to build a table in the database.\n\n### 15) To delete all the rows from a table, we need to apply the…… command.\n\n1. Truncate\n2. Remove\n3. Drop table\n4. Delete\n\nAnswer: a\n\nExplanation: Truncate is used to delete complete data without removing the table structure.\n\n### 16) Which of the following command is used to delete a table from the database?\n\n1. DROP\n2. DELETE\n3. TRUNCATE\n4. REMOVE\n\nAnswer: a\n\nExplanation: The DROP command is used to remove the table from the database, DELETE is used to remove one or more rows from a table in the database.\n\n### 17) DELETE FROM X\n\nWhat does the above statement mean?\n\n1. Delete table\n2. Delete row\n3. Delete fields\n4. Remove entry x\n\nAnswer: d\n\nExplanation: The DELETE command removes the table entries.\n\n### 18) The basic data type of SQL varchar(n) has ….. length Unicode characters, and char(n) has …… length Unicode characters.\n\n1. Fixed, Variable\n2. Fixed, Fixed\n3. Variable, Variable\n4. Variable, Fixed\n\nAnswer: d\n\nExplanation: The storage required for the char and varchar variable is different. Char() has a static allocation of space, whereas varchar() has a dynamic allocation of space. So the length of varchar() will vary according to the specified string.\n\n### 19) Which of the following command is used to display any values from the database?\n\n1. SELECT\n2. INSERT\n3. DELETE\n4. TRUNCATE\n\nAnswer: a\n\nExplanation: SELECT command is used to fetch values from the table. The syntax for the SELECT command is SELECT * FROM table_name;\n\n### 20) _________statement returns only distinct values from the table.\n\n1. Different\n2. Distinct\n3. None\n4. All\n\nAnswer: b\n\nExplanation: The table column may contain duplicate values. If we need the unique value from any column, we need to apply a distinct command. The syntax for the distinct command is:\n\n### 21) SELECT * FROM class WHERE marks>90 AND roll_no = 20;\n\nWhat should be the output for the above query?\n\n1. Marks and roll_no\n2. class\n3. Marks\n4. All the columns from the class table.\n\nAnswer: d\n\nExplanation: The “*” character in SQL is used to show all columns of the table.\n\n### 22) Choose the wrong option in this question.\n\n1. SELECT * FROM table_name WHERE column_name;\n2. SELECT column_name FROM table_name;\n3. SELECT column_name FROM table_name WHERE condidtion;\n4. SELECT column_name WHERE condition;\n\nAnswer: d\n\nExplanation: The correct answer is option D because the SELECT command does not produce output without specifying the table name.\n\n1. SELECT\n2. WHERE\n3. FROM\n4. ALL\n\nAnswer: b\n\n### 24) INSERT INTO employee ……(101,xyz,5000);\n\nWhat keyword is missing in the given statement?\n\n1. table\n2. values\n3. column\n4. field\n\nAnswer: b\n\nExplanation: To insert value in the table, we have to use the INSERT keyword with the VALUES clause.\n\n### 25) Which of the following condition allow to join relations?\n\n1. Set\n2. Where\n3. Using\n4. On\n\nAnswer: d\n\nExplanation: On condition is used for join expressions.\n\n### 26) Which of the join operations do not preserve non matched tuples?\n\n1. Left join\n2. Inner join\n3. Right join\n4. Full join\n\nAnswer: b\n\nExplanation: Inner join returns all rows from both tables when there is at least one match is found.\n\n### 27) What type of join is needed when you want to include rows that do not have matching values?\n\n1. Equi-join\n2. Natural join\n3. Outer join\n4. None of the above\n\nAnswer: d\n\nExplanation: An outer join does not require each record in the two joined tables to have a matching record.\n\n### 28) Which joins refer to joining records from the right table with no matching values in the left table?\n\n1. Left join\n2. Right join\n3. Full join\n4. Half join\n\nAnswer: b\n\nExplanation: Right join returns all values from the right table and the matched values from the left table.\n\n### 29) How many tables can be combined with a join?\n\n1. One\n2. Two\n3. Three\n4. Many\n\nAnswer: d\n\nExplanation: Join operation can combine many tables at a time.\n\n### 30) Which join is used to return all tuples from both tables even condition doesn’t satisfy?\n\n1. Inner join\n2. Right join\n3. Natural Join\n4. Full join\n\nAnswer: d\n\nExplanation: In a full outer join, all tuples from both relations are included in the result, irrespective of the matching condition.\n\n### 31) SELECT * FROM employee JOIN takes USING (id);\n\nThe above query is equivalent to:\n\n1. SELECT * FROM employee LEFT JOIN takes USING (id);\n2. SELECT * FROM employee RIGHT JOIN takes USING (id);\n3. SELECT * FROM employee INNER JOIN takes USING (id);\n4. All of the mentioned above\n\nAnswer: c\n\nExplanation: We can also write JOIN instead of INNER JOIN. JOIN is the same as INNER JOIN.\n\n### 32) Which set operation is used to return the common row from relation?\n\n1. Union\n2. Union all\n3. Intersect\n4. Minus\n\nAnswer: c\n\nExplanation: The intersect operation combines two SELECT statements and returns the common rows from both the SELECT statement.\n\n### 33) Which set operation should be used with the SELECT statements to eliminate duplicate rows?\n\n1. Union\n2. Union all\n3. Intersect\n4. Minus\n\nAnswer: a\n\nExplanation: The union operation eliminates the duplicate rows from its result set.\n\n### 34) If we want to retain all duplicates, we must write_______in place of union.\n\n1. Union all\n2. Intersect\n3. Minus\n4. All of the above\n\nAnswer: a\n\nExplanation: Union all returns the result without removing duplicate rows.\n\n### 35) Which of the following makes the transaction permanent in the database?\n\n1. Commit\n2. Rollback\n3. Flashback\n4. View\n\nAnswer: a\n\nExplanation: Commit operation is used to save the work permanently in the database.\n\n### 36) In which state, a transaction executes all its operations successfully?\n\n1. Active\n2. Committed\n3. Partially committed\n4. Aborted\n\nAnswer: b\n\nExplanation: A complete transaction is always committed.\n\n### 37) Which of the following is used to preserve the order of the operation?\n\n1. Schedule\n2. Atomicity\n3. Active state\n4. None of the above\n\nAnswer: a\n\nExplanation: A series of operations from one transaction to another transaction is known as a schedule, which is used to preserve the order of the operation in each of the individual transactions.\n\n### 38) _______will undo all statements up to commit?\n\n1. Rollback\n2. Commit\n3. Partially committed\n4. Abort\n\nAnswer: a\n\nExplanation: The rollback operation is used to undo the work done.\n\n### 39) OLAP stands for\n\n1. Online analytical processing\n2. Online analysis processing\n3. Online transaction processing\n4. Online aggregate processing\n\nAnswer: a\n\nExplanation: OLAP is the manipulation of information to support decision making.\n\n### 40) In which normal form multi-valued attribute, composite attribute and their combination are not allowed?\n\n1. First Normal Form (1NF)\n2. Second Normal Form (2NF)\n3. Third Normal Form (3NF)\n4. Fourth Normal Form (4NF)\n\nAnswer: a\n\nExplanation: The first normal form is used to eliminate duplicate values, and an attribute of a table cannot hold multiple values.\n\n### 41) The characteristics of the table in second normal form (2NF) is:\n\n1. Eliminates any hidden dependency\n2. Have a composite key\n3. Eliminate the possibility of insertion anomalies\n4. None of the above\n\nAnswer: a\n\nExplanation: The relation in the second normal form should be in the first normal form and doesn’t have any partial dependency.\n\n### 42) Which of the following statement is false about the third normal form (3NF)?\n\n1. It is used to remove data duplication.\n2. It contains transitive partial dependency.\n3. It is used to achieve data integrity.\n4. All of the above\n\nAnswer: b\n\nExplanation: A relation will be in 3NF if it is in 2NF, and there should not be a transitive dependency for non-prime attributes.\n\n### 43) Which forms are based on the concept of functional dependency?\n\n1. 1NF\n2. 2NF\n3. 3NF\n4. 4NF\n\nAnswer: c\n\nExplanation: A relation is in third normal form if it holds at least one of the following conditions for every non-trivial functional dependency X ? Y.\n\n• X is a super key.\n• Y is a prime attribute, i.e., each element of Y is part of some candidate key.\n\n### 44) Which of the following is a bottom-up approach to design a database by examining the relationship between attributes?\n\n1. Functional dependency\n2. Database modeling\n3. Normalization\n4. Decomposition\n\nAnswer: c\n\nExplanation: Normalisation is the process of removing redundancy and unwanted data.\n\n### 45) Which of the following header must be included in the java program to establish database connectivity using JDBC?\n\n1. Import java.sql.*;\n2. Import java.sql.odbc.jdbc.*;\n3. Import java.jdbc.*;\n4. Import java.sql.jdbc.*;\n\nAnswer: a\n\nExplanation: The Java program must import java.sql.*, which contains the interface definitions for the functionality provided by JDBC.\n\n### 46) Which of the following is used to access large objects from a database?\n\n1. setBlob()\n2. getBlob()\n3. getClob()\n4. All of the above\n\nAnswer: d\n\nExplanation: These are the predefined functions in SQL to access the large object from a database.\n\n### 47) DriverManager.getConnection(……. , ……. , ……..)\n\nFill in the blanks with valid options.\n\n1. URL or machine name where the server runs, Password, User ID\n2. Password, URL or machine name where the server runs, User ID\n3. URL or machine name where the server runs, User ID, Password\n4. User ID, Password, URL or machine name where the server runs\n\nAnswer: c\n\nExplanation: This connection method is used to open the database because a database must be opened first before performing any action.\n\n### 48) Which of the following statement is used to invoke the function in SQL?\n\n1. Connection statement\n2. Callable statements\n3. Prepared Statements\n4. All of the above\n\nAnswer: b\n\nExplanation: This interface is provided by JDBC that allows invocation of SQL stored procedures and functions.\n\n### 49) Which of the following is a bottom-up approach that combines two or more low-level entities into a high-level entity?\n\n1. Generalization\n2. Specialization\n3. Aggregation\n4. All of the above\n\nAnswer: a\n\nExplanation: Specialization is the opposite of generalization that break high-level entities into low-level entities.\n\n### 50) Which keyword is used to access records in file organization?\n\n1. Primary key\n2. Foreign key\n3. Candidate key\n4. Super key\n\nAnswer: a\n\nExplanation: The collection of records is known as a file.\n\n### 51) ____a condition where two or more transactions are waiting indefinitely for one another to give up locks.\n\n1. Deadlock\n2. Waiting\n3. Idle\n4. Ready\n\nAnswer: a\n\nExplanation: When one data item is waiting for another data item in a transaction, then the system is in a deadlock state.\n\n### 52) Which of the following are the ways of avoiding deadlock?\n\n1. Deadlock detection\n2. Deadlock avoidance\n3. Deadlock prevention\n4. All of the above\n\nAnswer: d\n\n### 53)Which of the following is the transaction failure?\n\n1. Logical errors\n2. Boot errors\n3. Read error\n4. All of the above\n\nAnswer: a\n\nExplanation: Logical errors and syntax errors are the types of transaction failure.\n\n### 54) Which kind of failure loses its data in head crash or failure during a transfer operation?\n\n1. Disk failure\n2. System crash\n3. Transaction failure\n4. All of the above\n\nAnswer: a\n\nExplanation: This kind of error occurs due to the formation of bad sectors, head crashes, and unreachability to the disk.\n\n### 55) The log is a sequence of _____ recording all the update activities in the database.\n\n1. Records\n2. Log records\n3. Entries\n4. Redo\n\nAnswer: b\n\nExplanation: Log records are the most commonly used recovery method.\n\nShare This" ]

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[ "Data\nHoney_bee_Seasonal_mortality\n\n# Honey_bee_Seasonal_mortality\n\nactive ARFF Attribution-NoDerivs (CC BY-ND) Visibility: public Uploaded 30-06-2017 by Carsten Behring\nIssue #Downvotes for this reason By\n\nData from https://doi.org/10.5281/zenodo.269636\n\n### 39 features\n\n ID_api nominal 4758 unique values 0 missing Seasonal_mortality numeric 46 unique values 0 missing Age nominal 4 unique values 0 missing Activity nominal 3 unique values 0 missing Beekeep_for nominal 3 unique values 0 missing Qualif nominal 2 unique values 0 missing Training nominal 2 unique values 0 missing Coop_treat nominal 2 unique values 0 missing Bee_population_size nominal 6 unique values 0 missing Country nominal 16 unique values 0 missing Apiary_Size nominal 5 unique values 0 missing Production nominal 7 unique values 0 missing Apiarist_book nominal 2 unique values 0 missing Org_member nominal 2 unique values 0 missing Continue nominal 2 unique values 0 missing Breed nominal 8 unique values 0 missing Chronic_Depop nominal 2 unique values 0 missing ClinSign_Brood nominal 2 unique values 0 missing ClinSign_Honeybees nominal 2 unique values 0 missing H_Rate_ColMortality nominal 2 unique values 0 missing H_Rate_HoneyMortality nominal 2 unique values 0 missing OtherEvent nominal 2 unique values 0 missing VarroaMites nominal 2 unique values 0 missing QueenProblems nominal 2 unique values 0 missing Management nominal 8 unique values 0 missing Swarm_bought nominal 5 unique values 0 missing Swarm_produced nominal 5 unique values 0 missing Queen_bought nominal 5 unique values 0 missing Queen_produced nominal 5 unique values 0 missing Environment nominal 7 unique values 0 missing VarroosisV2 nominal 2 unique values 0 missing ChronicParalysisV2 nominal 2 unique values 0 missing AmericanFoulbroodV2 nominal 2 unique values 0 missing NosemosisV2 nominal 2 unique values 0 missing EuropeanFoulbroodV2 nominal 2 unique values 0 missing Migration nominal 2 unique values 0 missing Merger nominal 2 unique values 0 missing Winter_Mortality_Class nominal 6 unique values 0 missing Program nominal 2 unique values 0 missing\n\n### 62 properties\n\n4758\nNumber of instances (rows) of the dataset.\n39\nNumber of attributes (columns) of the dataset.\nNumber of distinct values of the target attribute (if it is nominal).\n0\nNumber of missing values in the dataset.\n0\nNumber of instances with at least one value missing.\n1\nNumber of numeric attributes.\n38\nNumber of nominal attributes.\n5.12\nFirst quartile of skewness among attributes of the numeric type.\n3.65\nMean of means among attributes of the numeric type.\n12.18\nFirst quartile of standard deviation of attributes of the numeric type.\nAverage class difference between consecutive instances.\nAverage mutual information between the nominal attributes and the target attribute.\nSecond quartile (Median) of entropy among attributes.\nEntropy of the target attribute values.\nAn estimate of the amount of irrelevant information in the attributes regarding the class. Equals (MeanAttributeEntropy - MeanMutualInformation) divided by MeanMutualInformation.\n31.78\nSecond quartile (Median) of kurtosis among attributes of the numeric type.\n0.01\nNumber of attributes divided by the number of instances.\n128.82\nAverage number of distinct values among the attributes of the nominal type.\n3.65\nSecond quartile (Median) of means among attributes of the numeric type.\nNumber of attributes needed to optimally describe the class (under the assumption of independence among attributes). Equals ClassEntropy divided by MeanMutualInformation.\n5.12\nMean skewness among attributes of the numeric type.\nSecond quartile (Median) of mutual information between the nominal attributes and the target attribute.\nPercentage of instances belonging to the most frequent class.\n12.18\nMean standard deviation of attributes of the numeric type.\n5.12\nSecond quartile (Median) of skewness among attributes of the numeric type.\nNumber of instances belonging to the most frequent class.\nMinimal entropy among attributes.\n56.41\nPercentage of binary attributes.\n12.18\nSecond quartile (Median) of standard deviation of attributes of the numeric type.\nMaximum entropy among attributes.\n31.78\nMinimum kurtosis among attributes of the numeric type.\n0\nPercentage of instances having missing values.\nThird quartile of entropy among attributes.\n31.78\nMaximum kurtosis among attributes of the numeric type.\n3.65\nMinimum of means among attributes of the numeric type.\n0\nPercentage of missing values.\n31.78\nThird quartile of kurtosis among attributes of the numeric type.\n3.65\nMaximum of means among attributes of the numeric type.\nMinimal mutual information between the nominal attributes and the target attribute.\n2.56\nPercentage of numeric attributes.\n3.65\nThird quartile of means among attributes of the numeric type.\nMaximum mutual information between the nominal attributes and the target attribute.\n2\nThe minimal number of distinct values among attributes of the nominal type.\n97.44\nPercentage of nominal attributes.\nThird quartile of mutual information between the nominal attributes and the target attribute.\n4758\nThe maximum number of distinct values among attributes of the nominal type.\n5.12\nMinimum skewness among attributes of the numeric type.\nFirst quartile of entropy among attributes.\n5.12\nThird quartile of skewness among attributes of the numeric type.\n5.12\nMaximum skewness among attributes of the numeric type.\n12.18\nMinimum standard deviation of attributes of the numeric type.\n31.78\nFirst quartile of kurtosis among attributes of the numeric type.\n12.18\nThird quartile of standard deviation of attributes of the numeric type.\n12.18\nMaximum standard deviation of attributes of the numeric type.\nPercentage of instances belonging to the least frequent class.\nNumber of instances belonging to the least frequent class.\n3.65\nFirst quartile of means among attributes of the numeric type.\n771.25\nStandard deviation of the number of distinct values among attributes of the nominal type.\nAverage entropy of the attributes.\n22\nNumber of binary attributes.\nFirst quartile of mutual information between the nominal attributes and the target attribute.\n31.78\nMean kurtosis among attributes of the numeric type." ]

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[ "# Fuzzy Inference System Modeling\n\nBuild fuzzy inference systems and fuzzy trees\n\nFuzzy inference is the process of formulating input/output mappings using fuzzy logic. Fuzzy Logic Toolbox™ software provides tools for creating:\n\n• Type-1 or interval type-2 Mamdani fuzzy inference systems\n\n• Type-1 or interval type-2 Sugeno fuzzy inference systems\n\n• Trees of interconnected fuzzy inference systems\n\n## Apps\n\n Fuzzy Logic Designer Design and test fuzzy inference systems\n\n## Functions\n\nexpand all\n\n#### Type-1 Systems\n\n `mamfis` Mamdani fuzzy inference system `sugfis` Sugeno fuzzy inference system `genfis` Generate fuzzy inference system object from data `genfisOptions` Option set for `genfis` command\n\n#### Type-2 Systems\n\n `mamfistype2` Interval type-2 Mamdani fuzzy inference system `sugfistype2` Interval type-2 Sugeno fuzzy inference system\n\n#### FIS Trees\n\n `fistree` Network of connected fuzzy inference systems\n\n#### FIS Conversion\n\n `convertToSugeno` Convert Mamdani fuzzy inference system into Sugeno fuzzy inference system `convertToType1` Convert type-2 fuzzy inference system into type-1 fuzzy inference system `convertToType2` Convert type-1 fuzzy inference system into type-2 fuzzy inference system `convertToStruct` Convert fuzzy inference system object into a structure `convertfis` Convert previous versions of fuzzy inference data in current format\n `addInput` Add input variable to fuzzy inference system `addOutput` Add output variable to fuzzy inference system `removeInput` Remove input variable from fuzzy inference system `removeOutput` Remove output variable from fuzzy inference system `fisvar` Fuzzy variable\n `mfedit` Open Membership Function Editor `addMF` Add membership function to fuzzy variable `removeMF` Remove membership function from fuzzy variable `fismf` Fuzzy membership function `fismftype2` Interval type-2 fuzzy membership function\n `ruleedit` Open Rule Editor `ruleview` Open Rule Viewer `addRule` Add rule to fuzzy inference system `showrule` Display fuzzy inference system rules `fisrule` Fuzzy rule `update` Update fuzzy rule using fuzzy inference system\n `evalfis` Evaluate fuzzy inference system `evalfisOptions` Option set for `evalfis` function `plotfis` Display fuzzy inference system `plotmf` Plot membership functions for input or output variable `surfview` Open Surface Viewer `gensurf` Generate fuzzy inference system output surface `gensurfOptions` Option set for `gensurf` function\n `readfis` Load fuzzy inference system from file `writeFIS` Save fuzzy inference system to file\n `evalmf` Evaluate fuzzy membership function `gaussmf` Gaussian membership function `gbellmf` Generalized bell-shaped membership function `trimf` Triangular membership function `dsigmf` Difference between two sigmoidal membership functions `gauss2mf` Gaussian combination membership function `pimf` Pi-shaped membership function `psigmf` Product of two sigmoidal membership functions `sigmf` Sigmoidal membership function `smf` S-shaped membership function `trapmf` Trapezoidal membership function `zmf` Z-shaped membership function `defuzz` Defuzzify membership function `probor` Probabilistic OR `fuzarith` Perform fuzzy arithmetic\n\n## Topics\n\n### Fuzzy Logic Basics\n\nWhat Is Fuzzy Logic?\n\nFuzzy logic uses linguistic variables, defined as fuzzy sets, to approximate human reasoning.\n\nFoundations of Fuzzy Logic\n\nA fuzzy logic system is a collection of fuzzy if-then rules that perform logical operations on fuzzy sets.\n\nFuzzy Inference Process\n\nFuzzy inference maps an input space to an output space using a series of fuzzy if-then rules.\n\n### Fuzzy Inference Systems\n\nMamdani and Sugeno Fuzzy Inference Systems\n\nYou can implement either Mamdani or Sugeno fuzzy inference systems using Fuzzy Logic Toolbox software.\n\nType-2 Fuzzy Inference Systems\n\nYou can create and evaluate interval type-2 fuzzy inference systems with additional membership function uncertainty.\n\nFuzzy Trees\n\nYou can implement complex fuzzy inference systems as a collection of smaller interconnected fuzzy systems.\n\n### Build Fuzzy Inference Systems\n\nBuild Fuzzy Systems Using Fuzzy Logic Designer\n\nInteractively construct a fuzzy inference system using the Fuzzy Logic Designer app.\n\nBuild Fuzzy Systems at the Command Line\n\nConstruct a fuzzy inference system at the MATLAB® command line.\n\nBuild Fuzzy Systems Using Custom Functions\n\nYou can replace the built-in membership functions and fuzzy inference functions with your own custom functions." ]

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[ "", null, "# معادلات دیفرانسیل کاربردی با مسائل مقدار مرزی", null, "دسته: ریاضی", null, "", null, "سال انتشار: 2018  |  700 صفحه  |  حجم فایل: 15 مگابایت  |  زبان: انگلیسی\n\nApplied Differential Equations with Boundary Value Problems (Textbooks in Mathematics)\nنویسنده:\nناشر:\nChapman and Hall – CRC\nISBN10:\n1498733654\nISBN13:\n9781498733656\n\nقیمت: 12000 تومان\n\nخرید کتاب توسط کلیه کارت های شتاب امکان پذیر است و بلافاصله پس از خرید، لینک دانلود فایل کتاب در اختیار شما قرار خواهد گرفت.\n\nبرچسب‌ها:  مسائل مقدار مرزی\n\n#### عناوین مرتبط:\n\nApplied Differential Equations with Boundary Value Problems presents a contemporary treatment of ordinary differential equations (ODEs) and an introduction to partial differential equations (PDEs), including their applications in engineering and the sciences. This new edition of the author’s popular textbook adds coverage of boundary value problems. The text covers traditional material, along with novel approaches to mathematical modeling that harness the capabilities of numerical algorithms and popular computer software packages. It contains practical techniques for solving the equations as well as corresponding codes for numerical solvers. Many examples and exercises help students master effective solution techniques, including reliable numerical approximations. This book describes differential equations in the context of applications and presents the main techniques needed for modeling and systems analysis. It teaches students how to formulate a mathematical model, solve differential equations analytically and numerically, analyze them qualitatively, and interpret the results.", null, "", null, "", null, "" ]

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[ "", null, "## Simple Solver 5.5.2\n\ncomputer logic design, simulation, Boolean\n\nGraphical hierarchical logic schematic drawing, Boolean equation and truth table minimization, Computer logic simulation of all circuit types: combinational, sequential, synchronous and asynchronous, Automatic design and simulation of small digital logic circuits from truth table or waveform inputs, Generation of Permutations and Random numbers, Freeware, Shareware\n\n#### Simple Solver5.5.2 details\n\n Author: SimpleSolver Logic License: Freeware Price: FREE Released: Nov 19, 2020 File size: 1.50 MB Downloads: 171 Keywords: computer, logic, digital, design, simulation, Boolean, equation, algebra, truth table, minimize, minimizer, minimization, software, freeware, permutation, random number, combinational, sequential, synchronous, asynchronous Author URL: https://www.simplesolverlogic.com\n\n## Simple Solver for Windows 10 - Full description\n\nSimple Solver is a free Windows application that can simplify computer logic systems, Boolean equations, and truth tables. The application includes six different tools:Logic Design Draw, Logic Simulation, Logic Design Auto, Boolean, Permutation and Random Number. These tools are built on years of engineering design experience and are intended for both educational and industrial usage. The software is available as freeware for download. Logic Design Draw - A hierarchical WYSIWYG tool that enables a user to interactively create a logic schematic diagram and to run circuit simulation. Logic circuits can be very simple, such as and-or logic, or can consist of hundreds of parts. Both basic parts (logic gates, flip-flops) and MSI (Medium Scale Integration) building blocks are provided. Using blocks, large hierarchical designs - such as small computers - can be built. Computer Logic Simulation - Evaluates both logic circuit functionality and timing problems such as flip-flop setup and hold times, race conditions and glitches/spikes. All circuit types and configurations are supported: Combinational, Sequential, Synchronous and Asynchronous. Logic Design Auto - Automatically design small digital logic circuits and state machines from timing diagrams or truth tables. Boolean - Generates minimized Boolean equations from Boolean equation or Truth table inputs. Boolean operator formats are supported for a variety of languages including: ABEL, C, C++, PALASM, VB, Verilog and VHDL. The software uses both Quine-McCluskey and Espresso (UC Berkeley) algorithms to optimize minimization. Permutation - Generates permutations of numbers from a specified base number and a specified number of digits. Can be used for a variety of applications such as generating binary, octal or decimal number tables. Random Number - Generates from 1-99,999 random numbers in a specified range between -99,999 to 99,999.\n\n## Simple Solver for Windows 10 - Post your review", null, "" ]

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