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http://www.gradesaver.com/textbooks/math/prealgebra/prealgebra-7th-edition/chapter-6-review-page-461/57
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## Prealgebra (7th Edition)
$\frac{1}{9}$
2 teachers for 18 students is $\frac{2}{18}$ $\frac{2}{18}=\frac{2\div2}{18\div2}=\frac{1}{9}$
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https://artofproblemsolving.com/wiki/index.php/2017_AMC_8_Problems/Problem_2
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# 2017 AMC 8 Problems/Problem 2
## Problem
Alicia, Brenda, and Colby were the candidates in a recent election for student president. The pie chart below shows how the votes were distributed among the three candidates. If Brenda received $36$ votes, then how many votes were cast all together?
$[asy] draw((-1,0)--(0,0)--(0,1)); draw((0,0)--(0.309, -0.951)); filldraw(arc((0,0), (0,1), (-1,0))--(0,0)--cycle, lightgray); filldraw(arc((0,0), (0.309, -0.951), (0,1))--(0,0)--cycle, gray); draw(arc((0,0), (-1,0), (0.309, -0.951))); label("Colby", (-0.5, 0.5)); label("25\%", (-0.5, 0.3)); label("Alicia", (0.7, 0.2)); label("45\%", (0.7, 0)); label("Brenda", (-0.5, -0.4)); label("30\%", (-0.5, -0.6)); [/asy]$
$\textbf{(A) }70 \qquad \textbf{(B) }84 \qquad \textbf{(C) }100 \qquad \textbf{(D) }106 \qquad \textbf{(E) }120$
## Solution 1
Let $x$ be the total amount of votes casted. From the chart, Brenda received $30\%$ of the votes and had $36$ votes. We can express this relationship as $\frac{30}{100}x=36$. Solving for $x$, we get $x=\boxed{\textbf{(E)}\ 120}.$
## Solution 2
We're being asked for the total number of votes cast -- that represents $100\%$ of the total number of votes. Brenda received $36$ votes, which is $\frac{30}{100} = \frac{3}{10}$ of the total number of votes. Multiplying $36$ by $\frac{10}{3},$ we get the total number of votes, which is $\boxed{\textbf{(E)}\ 120}.$
## Solution 3
If $36$ votes is $\frac{3}{10}$ of all the votes, we can divide that by $3$ to get $12$ as 10%, and then we can multiply the $12$ by $10$ to get to $120$. So, the answer is $\boxed{\textbf{(E)}\ 120}.$
~AllezW
## Video Solution (CREATIVE THINKING!!!)
~Education, the Study of Everything
~savannahsolver
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http://www.cs.cmu.edu/afs/cs.cmu.edu/academic/class/15750-s01/www/notes/lect0118-treaps
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Random Search Trees (Kozen 13) Various schemes have been proposed for using randomization to "balance" search trees. It's based on the fact that a "random" tree is balanced. Two approaches: Skip Lists (Pugh) Treaps (Aragon and Seidel) We'll present treaps. Say we're storing a set of keys 1...n in a binary tree. The keys are stored in symmetric order by their value. So the left subtree of a key k contains only keys less than k, and the right subtree contains only keys greater than k. Say that each key also has a unique priority p(k). (Drawn from a totally ordered set.) Supose we assert that the tree is not only symmetric ordered by key, but also heap ordered by priority, with the highest priority at the root. Lemma: for any assignment of the priorities, there's a unique tree satisfiying the symmetric and heap order constraints. Proof: You can build the required tree simply by inserting the elements in decreasing order of priority into an initially empty tree. We can show it's unique by induction. Say you have two different trees supposedly satisfying these constraints. If they have different nodes at the roots, then the tree with the root with the lower priority must have a violation. If the roots are the same, then the two left subtrees must have the same sets of nodes in them, and similarly for the right subtrees, and they therefore must be identical by induction. QED. We will assume that we can generate a random priority without collisions. Here are the algorithms for the standard operations using Treaps: Search: Normal binary search Insertion: Search down from the root till we come to the null pointer where the new node ought to go. Insert the new node there with a random priority. Now rotate that node up the tree until heap order is restored. Deletion: Say x is the node we want to delete. Rotate x down the tree till it has no children. Then simply remove x. At each step rotate x with its child of highest priority. Given a tree, we'll refer to the _ancestor relation_ among the nodes in the tree. A node a is an ancestor of a node b if the path from b to the root includes a. A tree is heap ordered if and only if for every pair (a,b) where a is an ancestor of b, we have that p(a) >= p(b). Insertion works because as we rotate x (the inserted node) up the tree heap order (among nodes other than x) is preserved. To see this, consider moving x up the tree by doing a left rotation in the following picture: b / \ A x / \ C D becomes: x / \ b D / \ A C The ancestor relation (ignoring node x) is preserved with the exception that node b was an ancestor of all nodes in tree D before, but not after. Therefore if the nodes (excluding x) were heap ordered before the rotation then they are heap ordered after the rotation. Insertion terminates when x has a priority less than its parent. At this point heap order has been restored to all nodes including x. The above picture also helps in understanding deletion. In this case x moves down the tree (the bottom picture is converted by a right rotation into the top picture.) Again the ancestor relation (ignoring x) is preserved, except that b is now an ancestor of all the nodes in subtree D. We chose to rotate b up because its priority is greater than the root of D. Thus heap order is preserved. (Do some examples) Running-time Analysis. Theorem 1: The expected length of a path to a node m in a treap of n nodes is: H[m] + H[n-m+1] - 1 (Where H[k] = 1/1 + 1/2 + 1/3 + ... + 1/k (harmonic number)) To analyze insertions and deletions we'll prove this: Theorem 2: The expected number of rotations in a deletion (or insertion) is less than 2. Note that H[m] <= 1+ln(m) Before we prove these theorems we'll need the following lemma: Ancestor Lemma: Let mm and =m. The ancestor lemma shows that the set of ancestors of m that are m. (Then add 1 for m itself.) So consider a treap built with keys 1...m. Consider the process of building the treap by inserting the nodes in the order they appear in the permutation. Which nodes are ancestors of m? These are the nodes on the right path from the root down to m. A node k appears on that path if and only if: (1) k is inserted before m (2) must be the greatest node inserted so far. So now we can state a property in terms of the permutation inducing the treap. We're interested in each key k in the permutation that has the following properties: (1) k occurs before m in the permutation (2) k is a left-right maximum (greater than all those to its left) So our problem is reduced to computing the expected number of keys in a random permutation that have these properties. Given a random permutation, we can write a check mark next to each element that has these properties. Let X[m] be the expected number of check marks. We know that X[1] = 0. We can write a recurrence for X[] as follows. Generate a random permutation of 1...m by first generating a random permutation of 2...m then inserting 1 randomly in it. After generating the permutation of 2...m we label the keys with check marks. Then we insert 1. This operation does not change any of the already existing check marks because 1 is less than all the other elements. The number of check marks before inserting 1 is just X[m-1]. By linearity of expectation, X[m] is just X[m-1] plus the expected number of check marks on 1. This is just the probability that 1 is marked. This probability is just 1/m because 1 is only marked if it's the first element of the permutation. This gives: X[1] = 0. X[m] = X[m-1] + 1/m. The solution of this recurrence is just X[m] = H[m]-1. So the expected number of ancestors of m, counting m itself is just X[m] + X[n-m+1] + 1 Which is H[m] + H[n-m+1] - 1. QED. Now we can prove Theorem 2. Say we're deleting m. Consider the sum of the number of nodes on the right path from the left child of m and the number of nodes on the left path from the right child of m. This quantity decreases by exactly one on each rotation. (Draw pictures). So our goal is to evaluate the expected value of this sum, which is the sum of the expected values of each part. Let the two subtrees of m be A and B. Consider the treap that would be built by using only elements 1...m. Let A' be the left subtree of m in this treap. By the ancestor lemma, the set of nodes in A is the same as the set of nodes in A'. In fact A and A' are equal, because the same nodes are being inserted in the same order. Therefore we can do just what we did in the proof of theorem 1. We simply restrict ourselves to what happens to the stuff in 1...m, compute the expected length of the path, and add it to the expectation for the other side. So the question is, which nodes end up being on the right path of A? For a node k to end up there, it has to be inserted after m. And it has to be greater than any node inserted so far (except m). Here's the picture. x \ x \ x \ m / x \ x \ k If k was less than any of the xs, it would not end up there. If k is greater than any x, it does end up there. And once on the path of interest, it stays there. So suppose we take a random permutation of 1...m, and we check any item k that occurs after m but is greater than any element to its left (except m). What is the expected number of checks? Let Y[m] be this quantity. Y[1] = 0. One way of generating these marks is to first generate a random permutation of 2...m, install the marks, then insert 1 in a random position. The insertion of 1 does not change any of the marks installed before. The expected number of marks on 1 is the probably that 1 is marked. This only happens if the permutation starts out [m1...]. The probability of this happening is 1/m(m-1). Thus we get the following recurrence. Y[1] = 0 Y[m] = Y[m-1] + 1/m(m-1) The solution of this is just Y[m] = (m-1)/m. QED.
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# Mass, Force and Gravity Mass – Amount of stuff (atoms/molecules) in an object. Measured in kg. Force – Measures the gravitational pull on something. Measured.
## Presentation on theme: "Mass, Force and Gravity Mass – Amount of stuff (atoms/molecules) in an object. Measured in kg. Force – Measures the gravitational pull on something. Measured."— Presentation transcript:
Mass, Force and Gravity Mass – Amount of stuff (atoms/molecules) in an object. Measured in kg. Force – Measures the gravitational pull on something. Measured in N (newtons) Earths Gravitational Field Strength – 10N/kg. (The gravitational field strength at the surface of the Earth produces a force of approximately 10 N on every mass of 1 kg)
Gravitational field Earth’s gravitational force makes the ball move down towards the centre of the Earth. Sun’s gravitational force makes the Earth move in an orbit around the Sun.
Putting it all together The apple has mass (kg) It gets pulled to the ground at a force (this can be measured in Newtons) The earth’s gravitational field strength pulls the apple towards it
Now… Weight – Is a measure of force pulling something to the ground due to gravity Hang on a minute…that sounds very familiar…
Weight = Force Weight is measured in N (newtons)
Putting it all together The apple has mass (kg) It gets pulled to the ground at a force (this can be measured in Newtons) = WEIGHT The earth’s gravitational field strength pulls the apple towards it
How can we calculate weight? We can use an equation we already know to help us: g ( gravitational field strength ) = force / mass If Force = Weight Weight = Mass x g ( gravitational field strength ) Memorise: W = m X g
This powerpoint was kindly donated to www.worldofteaching.com www.worldofteaching.com http://www.worldofteaching.comhttp://www.worldofteaching.com is home to over a thousand powerpoints submitted by teachers. This is a completely free site and requires no registration. Please visit and I hope it will help in your teaching.
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# NCERT Solutions For Class 11 Maths Chapter 1 Sets Ex 1.1
Here, Below you all know about NCERT Solutions for Class 11 Maths Chapter 1 Sets Ex 1.1 Question Answer. I know many of you confuse about finding Chapter 1 Sets Ex 1.1 Of Class 11 NCERT Solutions. So, Read the full post below and get your solutions.
## NCERT Solutions for Class 11 Maths Chapter 1 Sets Ex 1.1
NCERT TEXTBOOK EXERCISES
Ex 1.1 Class 11 Maths Question 1. Which of the following are sets? Justify your answer.
(i) The collection of all the months of a year beginning with the letter J.
(ii) The collection of ten most talented writers of India.
(iii) A team of eleven best-cricket batsmen of the world.
(iv) The collection of all boys in your class.
(v) The collection of all natural numbers less than 100.
(vi) A collection of novels written by the writer Munshi Prem Chand.
(vii) The collection of all even integers.
(viii) The collection of questions in this Chapter,
(ix) A collection of most dangerous animals of world.
Solution.
(i) The collection of all months of a year beginning with J is {J anuary, June, July}, which is well defined and hence it forms a set.
(ii) The collection of most talented writers of India is not well defined because opinions about ‘most talented writers’ vary from person to person and hence it does not form a set.
(iii) A team of eleven best-cricket batsmen of the world is not well defined because opinions about ‘best-cricket batsmen’ vary from person to person and hence it does not form a set.
(iv) The collection of all boys in your class is well defined and hence it forms a set.
(v) The collection of all natural numbers less than 100 is {1, 2, 3, 4,…………, 99}, which is well
defined and hence it forms a set.
(vi) A collection of novels written by the writer Munshi Prem Chand is well defined and hence it forms a set.
(vii) The collection of all even integers is {…………..,-4, -2, 0, 2, 4,……….. } which is well defined and hence it forms a set.
(viii) The collection of questions in this chapter is well defined and hence it forms a set.
(ix) A collection of most dangerous animals of the world is not well defined because opinions about ‘most dangerous animals’ vary from person to person and hence it does not form a set.
Ex 1.1 Class 11 Maths Question 2. Let A = {1, 2, 3, 4, 5, 6). Insert the appropriate symbol ∈ or ∉ in the blank spaces:
(i) 5…A
(ii) 8…A
(iii) 0…A
(iv) 4…A
(v) 2…A
(vi) 10…A
Solution.
(i) Since 5 is the element of A. ∴ 5 ∉ A.
(ii) As 8 is not the element of A. ∴ 8 ∉ A
(iii) As 0 is not the element of A ∴ 0 ∈ A.
(iv) 4 is the element of A ∴ 4 ∈ A.
(v) 2 is the element of A ∴ 2 ∈ A.
(vi) 10 is not the element of A ∴ 10 ∉ A.
Ex 1.1 Class 11 Maths Question 3. Write the following sets in roster form:
(i) A = {x: x is an integer and -3 < x < 7}
(ii) B = {x: x is a natural number less than 6}
(iii) C = {x: x is a two-digit natural number such that the sum of its digits is 8}
(v) E = The set of all letters in the word TRIGONOMETRY
(vi) F = The set of all letters in the word
Solution.
(i) Integers lying between -3 and 7 are -2, -1, 0, 1, 2, ……….. , 6
∴ A = {-2,-1, 6}.
(ii) Natural numbers less than 6 are 1, 2, 3, 4, 5.
∴ B = 11, 2, 3, 4, 5}
(iii) Two digit natural numbers such that the sum of its digits is 8 are 17, 26, 35, 44, 53, 62, 71, 80.
∴ C= (17, 26, 35, 44, 53, 62, 71,80}
(iv) Prime number divisors of 60 are 2, 3, 5.
∴ D = (2, 3, 5}
(v) Word TRIGONOMETRY is formed by using the letters T, R, I, G, O, N, M, E, Y.
∴ E = (T, R, I, G, N, O, M, E, Y}
(vi) Word BETTER is formed by using the letters B, E, T, R
∴ F = (B, E, T, R}
Ex 1.1 Class 11 Maths Question 4. Write the following sets in the set-builder form:
(i) {3, 6, 9, 12}
(ii) {2, 4, 8, 16, 32}
(iii) {5, 25, 125, 625}
(iv) {2,4,6,…}
(v) {1,4,9,……..,100}
Solution.
(i) Let A = (3, 6, 9, 12}
All elements of the set are natural numbers that are multiples of 3.
∴ A = (x : x = 3n, n∈N and 1 ≤ n ≤4}
(ii) Let B = (2, 4, 8, 16, 32} = (21, 22, 23, 24, 25}
∴ B = {x : x = 2n, n ∈ N and 1 ≤ n ≤ 5}
(iii) Let C = (5, 25, 125, 625} = (51, 52, 53, 54}
∴ C = {x : x = 5n, n ∈ N and 1 ≤ n ≤ 4}
(iv) Let D = (2, 4, 6,……………..}
All elements of the set are even natural numbers.
∴ D = (x: x is an even natural number)
(v) Let E = {1,4,9,……….,100}
All elements of the set are perfect squares.
∴ E = {x: x = n2, n ∈ N and 1 ≤ n ≤ 10}
Ex 1.1 Class 11 Maths Question 5. List all the elements of the following sets:
(i) A = {x: x is an odd natural number}
(ii) B = {x: x is an integer, −1/2 < x < 9/2}
(iii) C = {x: x is an integer, x2 ≤ 4}
(iv) D = {x: x is a letter in the word “LOYAL”}
(v) E = {x: x is a month of a year not having 31 days}
(vi) F = {x : x is a consonant in the English alphabet which precedes k}.
Solution.
(i) A = {x: x is an odd natural number}
∴ A = {1, 3, 5, 7,……………}
(ii) B = {x: x is an integer, −1/2 < x < 9/2}
∴ B = { 0, 1, 2, 3, 4}
(iii) C = {x: x is an integer, x2 ≤ 4}
x2 ≤ 4⇒ -2 ≤ x ≤ 2
∴ C = {-2, -1, 0, 1, 2}
(iv) D = {x: x is a letter in the word “LOYAL”}
∴ D = {L, O, Y, A}
(v) E = {x: x is a month of a year not having 31 days}
∴ E = {February, April, June, September, November}
(vi) F = {x: x is a consonant in the English alphabet which precedes k}
∴ F = {b, c, d, f, g, h, j}
Ex 1.1 Class 11 Maths Question 6. Match each of the set on the left in the roster form with the same set on the right described in set-builder form:
i. {1,2, 3,6} (a) {x: x is a prime number and a divisor of 6}
ii. {2,3} (b) {x : x is an odd natural number less than 10}
iii. {M, A, T, H, E,I,C, S} (c) {x: x is natural number and divisor of 6} iv. {1, 3, 5, 7, 9} (d) {x : x is a letter of the word MATHEMATICS}.
Solution.
(i) → (c),
(ii) → (a),
(iii) → (d),
(iv) → (b).
The sets which are in set-builder form can be written in roster form as follows:
(a)
{x : x is a prime number and a divisor of 6} = {2, 3}
(b) {x: x is an odd natural number less than 10} = {1, 3, 5, 7, 9}
(c) {x : x is natural number and divisor of 6} = {1, 2, 3, 6}
(d) {x: x is a letter of the word MATHEMATICS} = {M, A, T, H, E, I, C, S}
## NCERT Solutions for Class 11 Maths Chapter 1 Sets Exercise 1.1 PDF
For NCERT Solutions for Class 11 Maths Chapter 1 Sets Ex 1.1, you may click on the link below and get your NCERT Solutions for Class 11 Maths Chapter 1 Sets Exercise pdf file.
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# heat generation due to viscosity in a 3D fluid flow
Consider an arbitrary 3D fluid flow:
$$\vec{\nu}=\vec{V}\left( \vec{x} ,t \right) \tag{1}$$
where velocity at each point $\vec{\nu}$ is a function $\vec{V}$ of position $\vec{x}$ and time $t$ (non-steady). Due to the viscosity $\mu$ there is heat being generated at each point in space. I know this heat should be a function of velocity, viscosity and div/curl of velocity:
$$\dot{q}=F\left(\vec{\nabla},\vec{\nu},\mu \right) \tag{2}$$
But I can't just find it out.
For a 2D problem with unidirectional horizontal velocity of:
$$\nu_x=V_x\left(x,y,t\right) \tag{3}$$
I think I can write
$$\dot{q}=\mu \frac{\partial \nu_x}{\partial y}\nu_x \tag{4}$$
Where $\tau_x=\mu \frac{\partial \nu_x}{\partial y}$ is the shear force due to viscosity. But I'm not quite sure if that's correct.
I would appreciate if you could help me know what is the correct form for equation 2.
P.S. I'm not quite sure but I think it should be something like:
$$\dot{q}=\mu \left( \vec{\nu} \times \vec{\nabla}\right).\vec{\nu} \tag{5}$$
It's not what you have written. The rate of viscous heat generation per unit volume is given by $$2\mu\mathbf{E}:\mathbf{E}$$ where $\mathbf{E}$ is the rate of deformation tensor. For the shear flow you mentioned, this reduces to $\mu (dv/dy)^2$. See Transport Phenomena by Bird, et al for the full development.
• I don't understand this notation. how should I write $E$? and what type of operator $:$ is?
• $\mathbf{E}=(\nabla \mathbf{v}+(\nabla \mathbf{v})^T)/2$ (the velocity gradient tensor plus its transpose divided by 2), and $\mathbf{E}:\mathbf{E}$ is the double contraction of the rate of deformation tensor with itself. See the reference I provided, if you are not familiar working with tensors. Commented Feb 19, 2018 at 14:30
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Truncated Platonics
There are several Archimedean solids that are formed by the truncation (cutting off) of each corner of a Platonic solid.
These can be shown in successive truncations from one shape to its dual.
Original Truncation Rectification Bitruncation Birectification (dual)
Truncated Tetrahedron
Tetratetrahedron
aka Octahedron
Truncated Tetrahedron
Truncated Cube
Cuboctahedron
Truncated Octahedron
Truncated Dodecahedron
Icosidodecahedron
Above images from Wikipedia.com
Initial truncations remove a pyramid from each corner the original object, the base of which is a regular polygon. If the original face is a triangle, the resultant edges will be $$\frac13$$ the original edge; if square, $$\sqrt\frac12$$; and if a pentagon, $$\sqrt\frac15$$.
Note: Each truncation makes the object smaller, but for the math below, the presumption is an object with edge length of 1.
Complete truncation (also called “rectified”) completely removes original edges, new faces meet a the midpoint of original edges.
Note that the left objects are “sitting” on a face, and the right objects are “standing” on a vertex. This is the result of matching the vertices of one to the midpoints of the dual.
In quasi-regular polyhedra, truncation is not necessarily resultant in regular faces. Adjustments are made to make the faces become regular, these are called rhombi-truncations.
The dihedral angles of the “main” faces (red/red or yellow/yellow) of the truncated object are the same as the original object’s dihedral angles. If there are only three faces at a vertex, the vertex angles are easily found with the formula: $$\Large{\cos\theta=\frac{\cos\epsilon}{\cos(\frac{\mu}{2})}}$$, where θ is the vertex angle of the newly created face to the opposing edge, ε is the interior angle of the new main face, and μ is the interior angle of the newly created face.
The dihedral angles of the new faces are a bit harder. We have to do each separately.
Since each truncation removes a pyramid with a regular base, we can calculate the dihedral angle of one side to the base, giving us the complimentary angle to that of the dihedral angle of the new/main faces of the polyhedra. Remember that if A+B=180, then cos A = —cos B, so we can find the cosine of the pyramid’s dihedral angle, then negate it.
I’ve already done the math for the tetrahedron, the cosine of the dihedral is $$\frac13$$, or $$70.528779\ldots^\circ$$, so the dihedral cosine for the hexagon-triangle faces of the truncated tetrahedron would be $$-\frac13$$, or $$109.47122\ldots^\circ$$. The two do add up to 180° and the latter is the same as the rectified tetrahedron (aka octahedron), therefore we are on the right track.
For the trunc. cube, a triangular based pyramid with right angles at the apex, would have a base dihedral of $$cos^{-1}(\frac{1}{\sqrt3})$$, so the octagon-triangle dihedral would be $$cos^{-1}(\frac{-1}{\sqrt3})$$, or 125.2643897…°.
The trunc. octahedron has a square pyramid (Johnson solid J1) removed that equates to a half octahedron. The dihedral for J1 is $$\cos^{-1}(\frac{1}{\sqrt3})$$, so the trunc. octahedron hex-square dihedral would be $$\cos^{-1}(\frac{-1}{\sqrt3})$$, or 125.2643897…°.
Yes, these are the same two angles, which can be seen in the cuboctahedron, all edges are square-triangles, so must be the same. It is also evident when you realize that corresponding faces are parallel from one shape to the next during truncation.
For a trunc. dodecahedron, the decagon-decagon faces have the same dihedral as the dodecahedron. The decagon-triangle edges have the same dihedral as the pent-hex edges of the trunc. icosahedron, 142.622632°, but we will check this later below.
The vertex angle θ is found by
\begin{align}\cos\theta &=\frac{\cos108}{\cos\frac{108}{2}}=\frac{-\cos72}{\cos54} \\&=-\frac{\frac{\sqrt5-1}{4}}{\sqrt{\frac{5-\sqrt5}{8}}} \\&=-\frac{\sqrt5-1}{4}\cdot\sqrt{\frac{8}{5-\sqrt5}} \\&=-\frac{\sqrt8 \cdot (\sqrt5-1)}{4\sqrt{5-\sqrt5}} \\&=-\frac{\sqrt8 \cdot \sqrt{(\sqrt5-1)^2}}{4\sqrt{5-\sqrt5}} \\&=-\frac{\sqrt8 \cdot \sqrt{6-2\sqrt5} }{4\sqrt{5-\sqrt5}} \\&=-\frac{\sqrt8 \cdot \sqrt2\cdot\sqrt{3-\sqrt5} }{4\sqrt{5-\sqrt5}} \\&=-\frac{\sqrt{16} \cdot\sqrt{3-\sqrt5} }{4\sqrt{5-\sqrt5}} \\&=-\frac{4 \cdot\sqrt{3-\sqrt5} }{4\sqrt{5-\sqrt5}} \\&=-\frac{\sqrt{3-\sqrt5} }{\sqrt{5-\sqrt5}} \\&=-\sqrt{\frac{3-\sqrt5 }{5-\sqrt5} } \\&=-\sqrt{\frac{3-\sqrt5 }{5-\sqrt5} \cdot \frac{5+\sqrt5}{5+\sqrt5} } \\&=-\sqrt{\frac{10-2\sqrt5 }{20} } \\&=-\sqrt{\frac{5-\sqrt5 }{10} } \\ \end{align}
So the vertex angle is 121.7174744…°
The trunc. icosahedron we already did the work for, the vertex angles are 148.2825255…°.
We already know the dihedrals of the cuboctahedron and icosidodecahedron are the same as their truncated counterparts, but what if we wanted to verify that?
Figure 1: Icosidodecahedron vertex
As Fig. 1 shows, we can draw a rectangle onto the surface of a icosidodecahedron, with points ABCD, which creates a pyramid with apex E. As lines AD and BC cross the width of the pentagons, their length is then $$Phi\ (\phi)\ or\ \frac{1+\sqrt5}{2}$$. The green line is also Φ and crosses AD at point F. Point G is the midpoint of line DE, making DG=EG=1/2.
Notice that the blue and green lines are parallel to the upper and left sides of the pentagon, making a parallelogram, making AF=1, so $$DF=\phi-1 = \frac{\sqrt5-1}{2}$$. ∠DGF is 90°, we can find FG,
\begin{align}DG^2+FG^2 &=DF^2 \\FG^2&=DF^2-DG^2 \\&=\left(\frac{\sqrt5-1}{2}\right)^2-\frac1{2^2} \\&=\frac{6-2\sqrt5}{4} – \frac14 \\&=\frac{5-2\sqrt5}{4} \\FG&=\frac{\sqrt{5-2\sqrt5}} {2} \end{align}
The distance between F and C is
\begin{align}CF^2&=CD^2+DF^2 \\&=1+(\frac{\sqrt5-1}{2})^2 \\&=\frac44 + \frac{6-2\sqrt5}{4} \\&=\frac{10-2\sqrt5}{4} \\CF&=\frac{\sqrt{10-2\sqrt5}} {2} \end{align}
Line CG is the same as the height of the triangle, $$\frac{\sqrt3}{2}$$. Now we can find the dihedral (∠FGC)
\begin{align}\cos\angle FGC&=\frac{FG^2+CG^2-CF^2}{2\cdot FG\cdot CG} \\&=\frac{ \frac{5-2\sqrt5}{4}+\frac34-\frac{10-2\sqrt5}{4}} {2\cdot\frac{\sqrt{5-2\sqrt5}} {2}\cdot\frac{\sqrt3}{2}} \\&=\frac{ \frac{5-2\sqrt5+3-10+2\sqrt5}{4}} {(\sqrt{5-2\sqrt5})\cdot\frac{\sqrt3}{2}} \\&=\frac{-\frac24}{\sqrt{5-2\sqrt5}\cdot\frac{\sqrt3}{2}} = -\frac24\div \left(\frac{\sqrt3\sqrt{5-2\sqrt5}} {2}\right) \\&=-\frac12 \cdot \left(\frac{2}{\sqrt3\sqrt{5-2\sqrt5}}\right) \\&=\frac{-1}{\sqrt3 \sqrt{5-2\sqrt5}} \\&=\frac{-1}{\sqrt3 \sqrt{5-2\sqrt5}}\cdot \frac{\sqrt{5+2\sqrt5}}{\sqrt{5+2\sqrt5}} \\&=\frac{-\sqrt{5+2\sqrt5}}{\sqrt3 \sqrt{25-4\sqrt5-20-4\sqrt5}} \\&=\frac{-\sqrt{5+2\sqrt5}}{\sqrt3 \sqrt5} \\&=-\sqrt{\frac{5+2\sqrt5}{15} } \\\angle FGC&=142.62263\ldots^\circ \end{align}
Figure 2: Cuboctahedron vertex
Just like the icosidodecahedron, we can draw a rectangle onto the surface of a cuboctahedron (Fig. 2), with points WXYZ, which creates a pyramid with apex V. As lines WZ and XY cross the diagonals of the squares, their length is then $$\sqrt2$$. The green line is also $$\sqrt2$$ and crosses WZ at point T. Point S is the midpoint of line WV, making WS=VS=1/2.
Line ST is half the width of the square, so its length is ½. Lines TV and TW are $$\frac{\sqrt2}{2}\ or \frac1{\sqrt2}$$.
The distance between T and X is
\begin{align}TX^2&=WX^2+TW^2 \\&=1+(\frac1{\sqrt2})^2 \\&=\frac22 + \frac12 \\&=\frac32 \\TX&=\sqrt{\frac32} \end{align}
Line SX is the same as the height of the triangle, $$\frac{\sqrt3}{2}$$. Now we can find the dihedral (∠TSX)
\begin{align}\cos\angle TSX&=\frac{ST^2+SX^2-TX^2}{2\cdot ST\cdot SX}=\frac{ \left(\frac12\right)^2+\left(\frac{\sqrt3}{2}\right)^2-\left(\sqrt\frac32\right)^2} {2\cdot\frac12\cdot\frac{\sqrt3}{2}} \\&=\frac{\frac14+\frac34-\frac64} {\frac{\sqrt3}{2}} \\&=\frac{-\frac24} {\frac{\sqrt3}{2}} = -\frac24 \cdot \frac2{\sqrt3} \\&=\frac{-1}{\sqrt3} \\\angle TSX&=125.2643897\ldots^\circ \end{align}
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# Python中的运算符函数用法详细介绍|S1
2021年4月6日19:25:00 发表评论 616 次浏览
Python在模块"operator"下具有用于许多数学, 逻辑, 关系, 按位等操作的预定义函数。本文介绍了一些基本功能。
sub(a, b):-这个函数返回给定参数的差值。
3.mul(a, b):-这个函数返回给定参数的乘积。
``````# Python code to demonstrate working of
# importing operator module
import operator
# Initializing variables
a = 4
b = 3
print ( "The addition of numbers is :" , end = "");
# using sub() to subtract two numbers
print ( "The difference of numbers is :" , end = "");
print (operator.sub(a, b))
# using mul() to multiply two numbers
print ( "The product of numbers is :" , end = "");
print (operator.mul(a, b))``````
``````The addition of numbers is :7
The difference of numbers is :1
The product of numbers is :12``````
4. truediv(a,b):-这个函数返回给定参数的分割。
5. floordiv(a,b):-这个函数还返回给定参数的分割。但该值是最小值，即返回最大的小整数。
6. pow(a,b):-这个函数返回给定参数的取幂。
7. mod(a,b):-这个函数返回给定参数的模数。
``````# Python code to demonstrate working of
# truediv(), floordiv(), pow(), mod()
# importing operator module
import operator
# Initializing variables
a = 5
b = 2
# using truediv() to divide two numbers
print ( "The true division of numbers is : " , end = "");
print (operator.truediv(a, b))
# using floordiv() to divide two numbers
print ( "The floor division of numbers is : " , end = "");
print (operator.floordiv(a, b))
# using pow() to exponentiate two numbers
print ( "The exponentiation of numbers is : " , end = "");
print (operator. pow (a, b))
# using mod() to take modulus of two numbers
print ( "The modulus of numbers is : " , end = "");
print (operator.mod(a, b))``````
``````The true division of numbers is : 2.5
The floor division of numbers is : 2
The exponentiation of numbers is : 25
The modulus of numbers is : 1``````
8. lt(a, b):—这个函数用于检查a是否小于b。如果a小于b则返回true，否则返回false。
9. le(a, b):-这个函数用于检查a是否小于或等于b。如果a小于或等于b则返回true，否则返回false。
10. eq(a, b):-这个函数用于检查a是否等于b。如果a等于b则返回true，否则返回false。
``````# Python code to demonstrate working of
# lt(), le() and eq()
# importing operator module
import operator
# Initializing variables
a = 3
b = 3
# using lt() to check if a is less than b
if (operator.lt(a, b)):
print ( "3 is less than 3" )
else : print ( "3 is not less than 3" )
# using le() to check if a is less than or equal to b
if (operator.le(a, b)):
print ( "3 is less than or equal to 3" )
else : print ( "3 is not less than or equal to 3" )
# using eq() to check if a is equal to b
if (operator.eq(a, b)):
print ( "3 is equal to 3" )
else : print ( "3 is not equal to 3" )``````
``````3 is not less than 3
3 is less than or equal to 3
3 is equal to 3``````
11. gt(a,b):-这个函数用来检查a是否大于b。如果a大于b则返回true，否则返回false。
12. ge(a,b):—该函数用于检查a是否大于等于b。如果a大于或等于b则返回true，否则返回false。
13. ne(a,b):—该函数用于检查a是否等于b或是否相等。如果a不等于b则返回true，否则返回false。
``````# Python code to demonstrate working of
# gt(), ge() and ne()
# importing operator module
import operator
# Initializing variables
a = 4
b = 3
# using gt() to check if a is greater than b
if (operator.gt(a, b)):
print ( "4 is greater than 3" )
else : print ( "4 is not greater than 3" )
# using ge() to check if a is greater than or equal to b
if (operator.ge(a, b)):
print ( "4 is greater than or equal to 3" )
else : print ( "4 is not greater than or equal to 3" )
# using ne() to check if a is not equal to b
if (operator.ne(a, b)):
print ( "4 is not equal to 3" )
else : print ( "4 is equal to 3" )``````
``````4 is greater than 3
4 is greater than or equal to 3
4 is not equal to 3``````
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EASEL BY TPT
# Equivalent Fraction Games - Comparing Fractions - Differentiated 14 Levels!
3rd - 5th, Homeschool
Subjects
Standards
Resource Type
Formats Included
• PDF
Pages
14 games, 4 pages of detailed Teacher Notes & Tips, 23 Total Pages
### Description
Would you love to have Fraction practice that wasn't difficult and confusing? Wouldn't it be great if your 4th grade students actually asked for fraction activities? Try these fraction games!
Differentiate for all the many levels of understanding about fractions in your classroom with these fun and engaging Mystery Themed fraction games! They are even NO PREP!
There are 14 different levels from recognizing fractions and equivalent fractions to converting mixed to improper fractions.
You'll have all the review you need for test prep with these!
Motivate your students to practice and continue learning about fractions with these games that will give them confidence while also enriching and deepening their understandings!
▶️Here's What's Included:
★ Fraction Models #1 - Match the fraction to the shaded whole –Circles only
★ Fraction Models #2 - Match the fraction to the shaded whole – Mixed shapes
★ Fraction Models #3 - Match the fraction of the set that is hats
★ Fraction Models #4 - Match the fraction to the model – a whole and a fractional part
★ Fractions on a Number Line #1 - Match the fraction to its point on the number line
★ Fractions on a Number Line #2 - Match the mixed number to its point on the number line
★ Equivalent Fractions #1 - Match the fraction models that are equivalent
★ Equivalent Fractions #2 - Match the equivalent fractions – denominators to 16
★ Comparing Fractions #1 - Match a fraction model to the fraction model that is greater or less than – like denominators to 10
★ Comparing Fractions #2 - Match a fraction to the fraction that is greater or less than – like denominators to 10
★ Comparing Fractions #3 - Match a fraction to the fraction that is greater or less than – unlike denominators to 12
★ Simplifying Fractions #1 - Match a fraction to the same fraction in simplest terms – denominators to 14
★ Simplifying Fractions #2 - Match a fraction to the same fraction in simplest terms – denominators to 56
★ Converting from Mixed to Improper Fractions #1 - Match a mixed number to its improper fraction – denominators to 8
❤️Why you’ll L-O-V-E these:
• Just print. Grab some dice and counters. Place in an envelope or plastic bag. Done!
• Games increase in difficulty to meet all the needs in your classroom.
• They’re fun!
• They’re challenging!
• They’re Common Core aligned.
• They’re perfect for small groups, math centers, homework, or even as informal assessment tools.
Great for homeschooling families too!
Just print, laminate, or use sheet protectors, and use them year after year!
Looking for more fun activities for reviewing math skills? Check out these popular resources:
Early Finishers and Gifted: Math Challenges Bundled!
Let’s connect!
Blog: Sum Math Fun
Instagram: @summathfun
Pinterest: SumMathFun
❤️ If you are not following me, please join me! I LOVE to give away freebies!!! Plus, followers can take advantage of Early Bird Specials! All new products are 50% off for the first two days of posting. Click the green star under my name on my store home page or on the right side of any product page to become a follower. As a follower, you will receive an email notification when I post new products. Then you can ⭐snap them up at the Early Bird price!⭐
Thanks for taking your time to peek inside this resource!
Have fun Mathing!
-Leah
Total Pages
14 games, 4 pages of detailed Teacher Notes & Tips, 23 Total Pages
N/A
Teaching Duration
Lifelong tool
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### Standards
to see state-specific standards (only available in the US).
Understand a fraction 1/𝘣 as the quantity formed by 1 part when a whole is partitioned into 𝘣 equal parts; understand a fraction 𝘢/𝑏 as the quantity formed by 𝘢 parts of size 1/𝘣.
Understand a fraction as a number on the number line; represent fractions on a number line diagram.
Explain equivalence of fractions in special cases, and compare fractions by reasoning about their size.
Explain why a fraction 𝘢/𝘣 is equivalent to a fraction (𝘯 × 𝘢)/(𝘯 × 𝘣) by using visual fraction models, with attention to how the number and size of the parts differ even though the two fractions themselves are the same size. Use this principle to recognize and generate equivalent fractions.
Compare two fractions with different numerators and different denominators, e.g., by creating common denominators or numerators, or by comparing to a benchmark fraction such as 1/2. Recognize that comparisons are valid only when the two fractions refer to the same whole. Record the results of comparisons with symbols >, =, or <, and justify the conclusions, e.g., by using a visual fraction model.
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Home
# Cross rates bid ask calculation
Currency Cross Rate Calculation Given two exchange rates B/A and A/C, the cross rate are: B/C = B/A multiplied by A/C and C/B = 1 divided by (B/A multiplied by A/C) And you need to use the same side when calculating bid/ask, e.g. B.C Bid = A.C.bid * A.B.bid B.C Ask = A.C.ask * A.B.as Cross Rate Calculation: A. Required: The Rate for S\$. Hence S\$ will be Denominator:. It is assumed that the bid and ask rates are equal. B. Required: The Rate for Hence € will be the Denominator on the lhs:. It is assumed that the bid rate and ask rate are... C. The Result of 2.25 S\$ Per € is.
### ObjectLab-Kit - Currency Cross Rate Calculatio
1. You just have to multiply the two bid prices with your cross rate calculator to get the cross rate. For example: In the case of the GBP/CHF. The bid prices are as follows: GBP/USD=1.5700, USD/CHF=0.9300. Thus the cross rate (GBP/CHF) will be 1.5700*0.9300=1.4601. At times, the USD might be the base or quote currency of both pairings. When this is the case, reciprocal paring is done where one of the currencies is flipped
2. If we have exchange rates in the form of bid-ask quote, we can derive the bid-ask cross rate by multiplying the bid rate of one currency with the ask of the other such that the common currencies cancel out and vice versa. This point is illustrated in Example 2 below
3. The current quote in the market is €1 = \$1.3300 / 1.3302. The bid-ask spread, in this case, is 2 pips —or the smallest price move a given exchange rate makes based on market convention. The spread..
4. Calculate the Denom in one of four ways, depending on the DirectTermIndicator of the cross rate exchanges: Both currencies are direct (have a DirectTermIndicator of T): Denom = Target Currency Offer Price / Originating Currency Bid Price . Example of Cross Rate Currency Calculation: AUD/EUR (Australian Dollar, Euro) with USD as the CrossRateCurrency, both in Direct term
5. Because we are ignoring the bid/ask spread and transaction costs to simplify the math in this example, there is no reason to believe that it would be exact. It is also true that arbitrage is not a perfect equalizer because the market is not perfectly efficient. But what if these cross rates didn't equalize. Suppose, for instance, that 1 GBP was exactly equal to 2 USD, with all other cross rates remaining the same. Obviously, the above equation would not hold, but it would present an.
6. The bid equals the division of the bid of the cross rate terms currency (top), by the offer of the base (bottom). To find the offer, divide the offer of the terms currency by the bid of the base
7. The idea of cross rates implies two exchange rates with a common currency, which enables you to calculate the exchange rate between the remaining two currencies. Financial media provide information only about the most frequently used exchange rates. Therefore, you may not have all the exchange rate information you need
In order to understand how to calculate cross rates, it is important to first be exposed to common cross rate pairings in order to develop comfortability with the concept. When two currencies are being valued against each other, they become a cross-rate pairing. The pairing is then compared to a base currency (e.g., U.S. dollar), creating a cross rate. Some of the more popular cross rates not. Bid Ask EURUSD: 1.3798: 1.3858: USDAUD: 1.0432: 1.0502: The EURAUD Bid rate = Multiply the term currency bid by the base currency ask = 1.3798 x 1.0432 = 1.4394: this is the rate at which the market buys EUR and sellsAUD: The EURAUD Ask rate = Multiply the term currency ask by the base currency bid = 1.3858 x 1.0502 = 1.455 Currency Exchange Rates Cross Table. This page displays the latest currency exchange rates for the top worldwide currencies and also the top european currencies. Click for British Pound exchange. Forex: Bid and Offer Rates. Exchange rates are commonly expressed as two rates, the bid price and the offer price, for example: Rates shown in the financial press are the average (mid-point) of the bid and offer rates. The bid price is the rate at which the bank quoting the price, the market marker will buy the base currency from a customer.
### How to Calculate Cross and Forward Rate? Exchange Rate
• imum JPY/CHF bid and the maximum ask cross rate that the bank would quote? A6. First calculate the JPY/CHF bid rate, the rate at which the bank buys CHF for JPY. Doing the calculations in two parts, we have: 1. The bank sells JPY, and it buys USD at JPY/USD 245. 2. The bank sells USD, and it buys CHF at CHF/USD 2.5140. Thus the rate is: JPY/USD 245 CHF/USD 2.5140 = JPY/CHFbid 97.
• Foreign exchange rates of major world currencies. Compare key cross rates and currency exchange rates of U.S. Dollars, Euros, British Pounds, and others
### How to Easily Calculate Cross Currency Rates Market
13. A term used to describe the average rate agreedupon when conducting foreign exchange. The middle rate is calculated using the medianaverage of the bid and offer rates. The middle rate is the average of bid and askrates. For direct quotes,◦ S (bid) = M - c◦ S (ask) = M + c◦ Where, M= Mid rate and c = one side average spread orcost of transaction14Mrs The Ask (or Buy) rate is the higher amount. It will cost you more dollars when buying euros. Many currency information sites provide the midpoint rate, which is the average of the Bid and Ask rates for a currency pair. At OANDA, we default to the Bid price for our applications, as it more accurately mimics the rate that you would be charged if.
### Cross Rate Definition Formula Exampl
e Estimate the cross exchange rates bid and ask rates separately as at 8 th. E estimate the cross exchange rates bid and ask rates. School University of Melbourne; Course Title FINANCE MISC; Uploaded By 1026248450A. Pages 12 Ratings 100% (1) 1 out of 1 people found this document helpful; This preview shows page 6 - 9 out of 12 pages.. What must the S cross rate be Cross Bid rates with Bid Ask Spreads How many. What must the s cross rate be cross bid rates with. School Pennsylvania State University; Course Title FIN 407; Type. Test Prep. Uploaded By erkirsch; Pages 43 Ratings 75% (4) 3 out of 4 people found this document helpful; This preview shows page 27 - 32 out of 43 pages..
1. Calculate the new SEKJPY cross rate 3 A Bid 87280 Ask 87317 B Bid 85240 Ask from BLB 201 at University of Johannesbur
2. Cross-Rate Trading. Find the currency against currency bid, ask and bid-ask spread for the currencies listed below. Show your work! (It may be helpful to use equations 5.12 and 5.13 on page 125 and 126. The methods are demonstrated on page 125 - 126 in Example 5.2 and Exhibit 5.7 parts a & b for finding euros-to-pounds spread
3. imum JPY/CHF bid and the maximum ask cross rate that the bank would quote? A6. First calculate the JPY/CHF bid rate, the rate at which the bank buys CHF for JPY. Doing the calculations in two parts, we have: 1. The bank sells JPY, and it buys USD at JPY/USD 245. 2. The bank sells USD, and it buys CHF at CHF/USD 2.5140. Thus the rate is: JPY/USD 245 CHF/USD 2.5140 = JPY/CHFbid 97.
4. ate currency risk over the 30-day period. b. Calculate the following: • The CHF/ZAR cross-currency rate Omni would use in valuing the.
### Cross Rate Currency Calculation - Oracl
1. ute and then performs an 8-Hour Time-Weighted-Average-Price (TWAP) over the series of
2. In the bid-ask formula, we find out the difference between the price the sellers ask and the price the buyer's bid. source: NSE India. We can see from the bid-ask example of Reliance Industries. For a buy quantity of 47, the bid price is 925.25, whereas the asking price is 925.30. Bid-Ask = 925.30 - 925.25 = 0.05
3. The bid price for the trader is always lower than the ask price, because that's how forex dealers make money. If you want to buy currency, you have to pay the higher ask price, but if you want to sell currency, you have to sell it at the lower bid price. So if you were to buy currency, then immediately sell it back to the same dealer, the dealer would make money, and you would lose money. Thus.
4. Currency converter uses cross rates to deliver other currencies values, which means you can calculate the values of the EUR (Euro) and USD (United States Dollar) to any other currency. Currency codes are standardized by ISO 4217:2015 and represented by three-letter alphabetic code followed by the full name of the currency. 1 JPY = 0.0075 EUR: Japanese yen: ↔: Euro: 1 JPY = 0.0075 EUR: 1 EUR.
5. Know your base Bid ask spread Implied cross rates Triangular arbitrage Forward. Know your base bid ask spread implied cross rates. School Texas Christian University; Course Title ECON MISC; Uploaded By xuandinhbank; Pages 22 This preview shows page 7 - 16 out of 22 pages..
6. show how \$ 10,000 available with you can be used to generate risk-less profits. Calculation of Cross Rate - BID [DEM/£] = Bid [\$/£] x Bid[S Fr/\$] x Bid[DEM/S Fr] = Bid[\$/£] x 1/Ask[\$/SFr] x 1/Ask[SFr/DEM] = 1.3670 x 1/0.8803 x 1/1.0078 = 1.54086 ASK [DEM/£] = Ask[\$/£] x Ask[DEM/SFr] = Ask[\$/£] x 1/Bid[\$/SFr] x 1/Bid[SFr/DEM] = 1.3708 x 1/0.8790 x 1/1.0030 = 1.55483 Cross Rate - DEM.
CME Group will calculate and disseminate the midpoint of the CME Globex bid-ask spread during the 60-second interval ending at 10:00 a.m. New York time on a real time basis - sampling at least once per second (i.e., a minimum of 60 observations). The CME Group FX fixing price is the average of the midpoints. If no bid-ask spreads on CME Globex are available during the 60-second interval. Triangular arbitrage cross rate To be more specific, suppose you're looking for a triangular arbitrage opportunity by spotting 3 different currencies: USD, EUR and GBP. Suppose that 1 EUR is worth 1,0910 USD, 1 EUR is worth 0,7413 GBP and 1 USD is worth 0,6794 GBP as shown in the provided Excel spreadsheet below exchange rate is the benchmark price the market uses to express the underlying value of the currency. Rates for dates other than the spot are always calculated relative to the spot rate. Listed below are the various value dates available in the market-they are all determined relative to the deal date. Assume the deal date is Monday, December 12
### Currency Cross Rates and Triangular Arbitrage in the FX
• Bid. Sell 1 USD . Ask. Buy 1 USD. MIN : AVG : MAX : These values represent the daily average of the Bid and Ask rates OANDA receives from many data sources. Recent Trends. USD/EUR average daily bid prices. 30 days: 60 days: 90 days: INTERACTIVE GRAPH. OANDA's currency calculator tools use OANDA Rates ™, the touchstone foreign exchange rates compiled from leading market data contributors. Our.
• MarketWatch.com is an award-winning web site, featuring market overview and tools for foreign exchange rates/currencies, key cross rates and currency convertor
• The following spot rates are quoted: USD/EUR 0.7824 - 89 TTD/EUR 8.2965 - 5745 Using a cross rate, calculate the bid-ask spread for the TTD/USD exchange rate. This question hasn't been solved yet Ask an expert Ask an expert Ask an expert done loading. 1. The following spot rates are quoted: USD/EUR 0.7824 - 89.
### How to Calculate Cross Rates - dummie
You can also calculate your unrealized profits and losses on open positions. Just substitute the current bid or ask rate for the action you will take when closing out the position. For example, if you bought 100,000 Euros at 1.2178 and the current bid rate is 1.2173, you have an unrealized loss of \$50. (\$1.2173 - \$1.2178) X 100,000 = -\$5 Doing the calculation. In order to calculate the yield, start with the quoted ask price, which is typically stated in terms that assume a face value of \$100. Subtract \$100 minus the ask price, and. Another example: For an EU resident the direct bid-ask quote for USD is 1.3391 − 1.3392. To get the indirect quote, we just need to find the inverse of the prices and then switch their places. So, for an EU resident, the indirect quotation rate for the euro and USD would be (0.7467 − 0.7468) Returns are calculated as follows: For time t (a holding period), let. t' = time of last available price < t; r(t) = return on purchase at t', sale at t; p(t) = last sale price or closing bid/ask average at time t; d(t) = dividend amount for t; f(t) = factor to adjust price in period t; p(t') = last sale price or closing bid/ask average at time of last available price < t; t' is.
### Cross Rates - Overview, How Pairings Work, How to Calculat
• R Risk-free interest rates to expiration Q(K i) The average of the bid quote and ask quote for each option with strike K i. Time to Expiration for Constituent Options The VIX Index measures the 30-day expected volatility of the S&P 500 Index. The components of the VIX Index are at- and out-of-the-money put and call options with more than 23 days and less than 37 days to a Friday SPX expiration.
• ated NBP and Euro-deno
• fact, reflect bid or ask prices which fluctuate during the day. As quite a few pieces of academic literature after Black Monday documented, the key to understanding second-by-second price changes is the investigation of a relationship between trading activity and price changes. In other words this is the price impact of trading activity, which can be labeled liquidation or liquidity risk.
• Forex glossary: ask-bid, Fibonacci calculator Fibonacci calculator for Forex; Currency rates Current currency rates and quotes charts; Economics news Latest financial news from the Forex world; Deposit/Withdrawal. Credit Card Transfers We accept Visa, MasterCard; Deposit by bank Deposit and withdrawal to bank account; Electronic transfers A wide range of payment systems; Cryptocurrencies.
• SPY is the most highly liquid stock or ETF in the market. The bid price at the time of writing is 357.98 and the ask price is 357.99. That's a \$0.01 spread or basically no spread at all, especially when taken in percentage terms. MSFT is another highly liquid stock and the spreads there are very good also at only \$0.21 or about 0.09%
• The Bid Ask Spread is the separation between buyers and sellers. If someone is willing to Bid in a stock at \$10.50 but a seller is only willing to post an Ask price of \$10.55, then the Bid Ask Spread is \$0.05. In order for a transaction to occur, someone must either sell to the buyer at the lower (Bid) price, or someone must buy from the sell at the higher (Ask) price. Alternatively another.
The 7.91 under the word Ask in the table comes from the discount rate calculated on the bill. The discount rate is defined as: 360 / x F F P discount rate where X = days to maturity In our case (using the ask price): .0791 7.91% 360 / 91 100 100 98 ( ) ask discount rate 4) The 8.11 in the table under the word bid uses the same discount rate calculation as above except it uses the bid price. PathOptimizer calculates the optimal arbitrage path (maximizing rate of return) with cplex algorithm given bid-ask prices of each crypto trading pair fetched from ccxt. It takes in exchanges as a required parameter and some other optional parameters that affects the path constraint. It can be used to monitor multiple exchanges as long as it's supported by ccxt and you put it in exhcanges.py So, what do you think the bid is out of the two numbers above? Well if you guessed it right, the number in red is the bid number. The bid is the price you are willing to buy the security. That leaves one other number which is in green - the ask price. The simple way of thinking about the ask is the price you are willing to sell the security
In total, there are 5 different Euribor rates (until November 1st 2013 there were 15 Euribor rates). See current Euribor rates for an overview of all rates. Next to that there is also a 1-day European interbank interest rate called Eonia. On this site you will find lots of information about Euribor and the different Euribor rates. We do offer. Bid-Ask spreads are far less than on corporate bonds, and even governments in most cases. Swap spreads are around 5 bps, the lowest in any market. 2. Swap spreads (the difference between the fixed and floating leg) do not display the volatile cyclical behavior of corporate bond spreads. 3. The quoted swap rates do not reflect credit rating differences between counterparties. We call these. Calculating Currency Appreciation or Depreciation. Given 2 exchange rates in terms of a Base Currency and a Quote Currency we can calculate appreciation and depreciation between them using the percentage change calculation. Letting V 1 be the starting rate and V 2 the final rate To calculate a cross-rate swap can be even more complicated. For example, to calculate a CHF/JPY swap, a forward trader must calculate each leg of the swap by triangulating USD/CHF and USD/JPY outright rates. The CHF/JPY spot rate is then subtracted from the resultant CHF/JPY outright rates to give CHF/JPY forward points. In practice, traders use tools and spreadsheets to speed up this process. Capitalization Rate for property A = Net Operating Income / Current Market Value of property. Capitalization Rate for property A = \$50000 / \$1500000. Capitalization Rate for property A = 3.33%. Capitalization Rate for property B is calculated using below formula
### What is a Cross Rate & How To Derive One Western Union
On Forex, online exchange rates are presented in the form of pairs so that the rate of one currency is expressed in the number of units of the other. The most popular trading pair is undoubtedly EUR/USD; its trading volume covers more than half of the market. You can see the current exchange rate of the dollar against the euro on Forex online below in the quotes chart. Monitor the current. Rate Calculator. Transitioning from being an employee to being a contractor can take some thought. For example, how does an annual salary as an employee translate to an hourly rate as a contractor? Let us help you figure that out! Just enter in a few details below to find out what hourly rate you should target. Salary (annual): \$ Annual salary you would like to target. Time off (days): Number.
### Video: Currency Exchange Rates Cross Tabl
bid and ask - The average (mean) bid and ask quotes over the date range. The date range average is calculated using the average daily bid and ask quotes over the entire date range, including the start and end dates; spot - For Central Bank exchange rates only. The average (mean) spot rate over the date range. This is calculated just as bid. In April 2010, the Mayor introduced a two pence (2p) business rates supplement on larger non-domestic properties in London. Since April 2017, this has only applied to business premises with a rateable value of over £70,000. This is helping to pay for Crossrail, a vital new east-west train link that will provide a major boost to London's economy interest rate differential between the two currencies concerned. Using the example of the U.S. Dollar and the Ethiopian Birr with a spot exchange rate of USD-ETB=9.8600 and one-year interest rates of 3.23% and 6.50% respectively for the U.S. and Ethiopia, we can calculate the one year forward rate as follows Salvage Value =INR 100,000 - (INR 10,000 * 7) Salvage Value =INR 100,000 - 70,000; Salvage Value = INR 30,000 Therefore, the salvage value of the machinery after its effective life of usage is INR 30,000.. Salvage Value Formula - Example #
The indicator plots two lines: Bid and Ask prices. The drawing of each of them can be turned off in the options of the indicator. The indicator saves the prices of the current symbol, received from the broker in a text file with the following format: Server time, Bid price and Ask price: 2010.03.26 19:43:02 1.33955 1.3396 Once the burden rate is calculated, it can be used to determine pricing in a number of different pricing structures, but don't forget to add in profit before you present the bid to the customer! Customers Are Key The final aspect of bidding in snow and ice, and perhaps the most important, can have a profound affect on the snow and ice business. Forming a consistent bidding process is null.
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https://www.teachoo.com/2964/654/Misc-7---Write-the-statement-in-five-different-ways/category/Miscellaneous/
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Miscellaneous
Mathematical Reasoning
Serial order wise
### Transcript
Misc 7 Write the following statement in five different ways, conveying the same meaning. p: If triangle is equiangular, then it is an obtuse angled triangle. The given statement can be written in five different ways as follows. A triangle is equiangular implies that it is an obtuse-angled triangle. (ii) A triangle is equiangular only if it is an obtuse-angled triangle. (iii) For a triangle to be equiangular, it is necessary that the triangle is an obtuse-angled triangle. (iv) For a triangle to be an obtuse-angled triangle, it is sufficient that the triangle is equiangular. (v) If a triangle is not an obtuse-angled triangle, then the triangle is not equiangular.
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https://physics.stackexchange.com/questions/681382/binding-energy-for-a-uniform-sphere-bound-by-a-yukawa-potential-rather-than-an-i
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# Binding energy for a uniform sphere bound by a Yukawa potential rather than an inverse-square-law potential?
The binding energy of a uniform sphere that interacts through a force that follows the inverse square law is proportional to $$a\frac{C^2}{R}$$, where $$C$$ is whatever charge determines the strength of the force on the particle, such as mass or electric charge, $$R$$ is the radius, and $$a$$. For gravity, this looks like $$\frac{3GM^2}{5R}$$. What would the equation be for a Yukawa potential though? I thought the equation might be $$a\frac{C^2e^{-bR}}R$$, where $$a$$ and $$b$$ are constants, but when I predicted the binding energy by using an integral to calculate potential for a given radius and plotting the results, I couldn't get $$a\frac{C^2e^{-bR}}R$$ to fit the data.
The equation I used to predict the binding energy of the sphere for a given radius was this: $$\frac{\int_0^R4\pi r^2\left(\int_{-R}^R\left(\int_0^\sqrt{1-x^2}\left(2\pi y \frac{e^{-\sqrt{(x-r)^2+y^2}}}{\sqrt{(x-r)^2+y^2}}\right)dy\right)dx\right)dr}{\left(\frac43\pi R^3\right)^2}$$ The outermost integral variable, $$r$$, represents a radius from the center of the sphere that is less than the radius $$R$$. The $$4\pi r^2$$ bit is because each radius $$r$$ also represents an equipotential shell. The rest of the equation treats $$r$$ like a distance to a certain point, and determines the potential at that point. $$\int_{-R}^R\int_0^\sqrt{1-x^2}2\pi y...dydx$$ maps each point in the sphere and calculates each point's "contribution" to the potential at point $$r$$. I divide the whole thing by the volume squared because I want the total "charge" to remain constant.
Per this Math.SE post, given a Yukawa potential with range $$1/a$$, a spherical shell of mass $$m$$ with radius $$r$$ behaves like a point mass potential at the origin, except with mass $$m~\sinh(ar)/(ar).$$
If we nest these then the binding energy can be found as follows: bring a small mass from infinity to the origin for free, forming a little ball of density $$\rho$$ and radius $$\delta r$$, bring another small mass from infinity to the radius $$\delta r$$ to grow the sphere a little more, paying some energy, and so on. To keep a constant density the mass we have to move each time is $$\mathrm dm=\rho~4\pi r^2\mathrm dr,$$ while the mass already moved is $$m(r)=\frac43\pi r^3~\rho$$. So we need to know two things:
First, what is the effective mass of a uniform ball, instead of a spherical shell? That's a straightforward integration by parts:$$m_{\text{eff}}(R)=\int_0^R\mathrm dr ~4\pi r^2\rho~\frac{\sinh(ar)}{ar} = \frac{4\pi\rho}{a^3}~\big(aR\cosh(aR)-\sinh(aR)\big)$$ Then assuming your potential looks like $$U(r)=-G\frac{m_1m_2}{r}~e^{-ar}$$ the binding energy is some integral $$E= G\int_0^R\mathrm dr~4\pi r^2\rho~m_{\text{eff}}(r)\frac{e^{-ar}}{r}$$ Calling $$4\pi\rho/a^3=\mu$$ this gives $$E= G\mu^2~a\int_0^{aR}\mathrm dx~x~e^{-x}(x \cosh x-\sinh x)$$This is technically not a hard integral, it is just a couple more integrations by parts and the integrand actually simplifies when you substitute in the definition of cosh/sinh... But it is a rather big expression so I trust Wolfram Alpha more than I trust my pen and paper, and Wolfram Alpha says it is $$E= {G\mu^2a \over 12}\left( 2 (aR)^3 - 3 (aR)^2 - 3 e^{-2 aR} (aR + 1)^2 + 3 \right)$$ Wolfram Alpha is also kind enough to give us a check on this expression via the Taylor series, which it says is$$E=G\mu^2a\left(\frac{(aR)^5}{15} -\frac{(aR)^6}{18}+\frac{(aR)^7}{35}-\dots\right)$$ which indeed limits in the infinite range $$a=0$$ case to $$(3/5) GM^2/R$$ (remember that $$\mu\sim1/a^3$$).
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In any ΔABC, prove that
Question:
In any ΔABC, prove that
$a^{2} \sin (B-C)=\left(b^{2}-c^{2}\right) \sin A$
Solution:
Need to prove: $a^{2} \sin (B-C)=\left(b^{2}-c^{2}\right) \sin A$
We know that, $\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=2 R$ where R is the circumradius.
Therefore,
$a=2 R \sin A \ldots(a)$
Similarly, b = 2R sinB and c = 2R sinC
From Right hand side,
$=\left(b^{2}-c^{2}\right) \sin A$
$=\left\{(2 R \sin B)^{2}-(2 R \sin C)^{2}\right\} \sin A$
$=4 R^{2}\left(\sin ^{2} B-\sin ^{2} C\right) \sin A$
We know, $\sin ^{2} B-\sin ^{2} C=\sin (B+C) \sin (B-C)$
So,
$=4 \mathrm{R}^{2}(\sin (\mathrm{B}+\mathrm{C}) \sin (\mathrm{B}-\mathrm{C})) \sin \mathrm{A}$
$=4 \mathrm{R}^{2}(\sin (\pi-\mathrm{A}) \sin (\mathrm{B}-\mathrm{C})) \sin \mathrm{A}[\mathrm{As}, \mathrm{A}+\mathrm{B}+\mathrm{C}=\pi]$
$=4 \mathrm{R}^{2}(\sin \mathrm{A} \sin (\mathrm{B}-\mathrm{C})) \sin \mathrm{A}[\mathrm{As}, \sin (\pi-\theta)=\sin \theta]$
$=4 \mathrm{R}^{2} \sin ^{2} \mathrm{~A} \sin (\mathrm{B}-\mathrm{C})$
$=\mathrm{a}^{2} \sin (\mathrm{B}-\mathrm{C})[$ From $(\mathrm{a})]$
$=$ Left hand side. [Proved]
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Physics
The Physics Philes, lesson 110: Ring My Decibel
OK, friends. Let’s talk about wave intensity.
Wave intensity (denoted with an I) is the average rate of time energy is transported per unit area across a surface that is perpendicular to the direction of propagation. So, if we’re talking about a transverse wave on a string, the wave might be propagating right to left. That means that the relevant surface for calculating wave propagation is the surface that goes up and down. We can express the wave intensity by using concepts we already understand: displacement amplitude and pressure amplitude.
Let’s start with displacement amplitude. We have a sound wave moving in the positive x-direction. We know that the power of a sound wave is the product of the force and velocity. That means that the power per unit area is the product of the force per unit area and the particle velocity. In math, it looks something like this:
Remember that particle velocity is the velocity at a particular time t of a portion of the wave medium at a particular coordinate x. It’s important to keep this concept distinct from wave velocity. A wave continues to move through a medium, but the individual particles just slosh up and down or side to side a bit. The particles don’t move along with the wave. Also, the max wave speed and the max particle speed can be very different. So don’t mix them up!
The wave intensity is the time average of the above equation. So, for any x value the average of the sin^2 function over one period T = 2π/ω is a half. If we weave that into the product of the force per unit area and the particle velocity, we get:
We can now use a couple of relationships – ω = vk and v^2 = B/ρ – to further transform the equation:
This is the equation for intensity of a sinusoidal sound wave in terms of displacement amplitude!
But there is another, more useful way to express wave intensity. This time, we’ll use pressure amplitude. Finding this equation is comparatively easy. All we need is the equation for pressure amplitude and ω = vk.
If you’ll remember, the pressure amplitude is
Using this equation and ω = vk, we can rewrite the wave intensity equation:
Now we can use the wave speed relationship v^2 = B/ρ to further rewrite the equation:
This is the intensity of a sinusoidal sound wave in terms of pressure amplitude!
If sound waves are emitted in all directions equally we need to deal with something called the inverse square law. The intensity of the sound wave is proportional to 1/r^2, where r is the distance from the source of the sound wave. However, if the sound goes mostly in one direction, the inverse square law doesn’t apply. In that case, the intensity of the sound wave will decrease more slowly. In addition, the inverse square law doesn’t apply indoors because the sound will bounce off the walls and ceiling.
Finally, let’s talk briefly about the decibel scale. The decibel scale is a logarithmic intensity scale. The sound intensity level β is defined as
Sound intensity levels are expressed in decibels, which is one-tenth of a bel. We can use sound level meters to measure the sound intensity level, because the human ear isn’t equally sensitive to all frequencies, some meters weight the frequencies unequally. This is called the dBA scale, and it does not emphasize very low and very high frequencies because we aren’t as sensitive to those wavelengths.
Welp, we made it through sound intensity! I’m proud of us. Next week we’ll take what we know about normal modes and apply it sound waves. It’ll be fun!
Featured image credit: flattop341 via Flickr
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# Why is one algorithm faster than the other if both have performance of O(1)
If I have a very large number X, `(978678567445456455878909099775523213255436445691200897623461676543789287 948754875435250392049320487584759329 454754875487589457244076477592458249)`
And I have two algorithms that can compute whether or not the the input number is divisible by 4 (i.e. n % 4 == 0). Since the modulo operation takes O(1), why would one algorithm be better than the other if they are both O(1). How could I prove that the one which only compares the last two digits(2) is in fact better than the other?
algorithm 1:
``````n:= Input
if n divisible by 4, let output :=0
else output :=1
``````
algorithm 2:
``````m:=last two digits of input n
if m divisible by 4, let output:=0
else output:=1
``````
• You don't seem to understand what O(1) means. If you want to compare two algorithms, you should err.. test them:) – Martin James May 11 '16 at 0:39
• O(1) describes a function that relates input size (number of bits) to run time. In particular, it says that the input size doesn't affect the run time. It says nothing about the run time itself. So one O(1) algorithm may produce an answer in one millisecond and another in 10,000 years. Which one do you think is "better?" Moreover, as others have pointed out, the modulo operation is not O(1) for any realistic machine model. I.e., the size of the input matters very much indeed. – Gene May 11 '16 at 5:20
## 2 Answers
1. Algorithm 1 is not O(1). It's running time is proportional to the number of bits of the input number, so it is O(log `n`).
2. It makes no sense to talk about the running time of a function with constant input. Running time is about asymptotics, not about how long it takes for a specific single instance.
3. Big O notation does not distinguish between constants. Instead, two running times are the "same" only if they are within a constant multiple of each other. In other words, even though two algorithms may have the "same" big O notation (actually big theta notation if we want to be more precise), one might actually be a constant times faster than the other in the asymptotic case.
• Thankyou, but why is it O(log n) and not simply O(n) if proportional to number of bits? Is this because the modulo function is a involves division? – Als May 11 '16 at 11:55
• @Als because n has O( log n ) bits (more precisely, it has log base 2 of n bits) – TheGreatContini May 11 '16 at 12:14
• Since the number in the OP is denoted by `X` wouldn't it be reasonable to replace `O(log n)` by `O(log X)`? In fact, since `n` usually denotes the input size, I would assume `n = log X`. – Dmitri Chubarov May 12 '16 at 7:14
Depending on the key type a C++ Map is considerably faster at finding things for small datasets than it's Hash. The Map is O(log n) and the Hash is O(1). The reason is Big O is dealing with the asymptotic performance of the algorithm ie if we plot the performance of the algorithm with some ever increasing value of `N` what curve does it approach.
When we hash something like a string we likely compute each byte in turn. That could 10 or 100 bytes where we need to perform some form of computation to find the bucket where the item is to be stored. Note however that this computation is a fixed size. As the number of strings `N` is increased it in no way affects the computation of the hash ie `N` is not impacting the time to find the bucket and insert the item. If you plot it as `N` increases the time to hash the function and locate a bucket is constant ie O(1)
When we use a Map we need to compare each string against the next one in the tree. This time as we add strings to the tree we have more strings we need to compare against ie as `N` increases we're in some way affecting the amount of comparisons that we need to make in order to find or insert a string. When there are a few strings in the Map this is actually faster than a Hash (Note, It was for me a few years ago using C++ and Urls, not sure if it's still the same now). This is because for a few strings it needs to do less work than the Hash does. This time because it's a binary tree we have a logarithmic affect ie the amount of computation required is increasing and if you plotted it you'd find it's following a logarithmic curve ie it's O(log n).
The reason is that Big O is not about fixed sized datasets it's about describing the behavior of an algorithm as `N` increases.
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Similar Triangles Robert Leverett Feb 07, 2007 08:10 PST
ENTS, The geometrical concept of similar triangles provides a powerful tool for tree measurers. However, as with the percent slope method, it is often misapplied to tree height. But, used correctly, the method gives us one more tool to work with. Let's look at the concept of similar triangles. Similar triangles are triangles that have the same overall shape. One is just a blown-up or reduced version of another. Because similar triangles have the same shape, corresponding angles of similar triangles are equal and corresponding sides are proportional in length. It is the condition of proportionality of side length that allows us to compute tree height. But how does the process work? Assume we have two triangles that are similar. Triangle ABC is the larger and abc the smaller where the lengths of the sides are designated by A, B, and C for the larger and a, b, and c for the smaller. The condition of proportionality leads to the following relationships: A/B = a/b, A/C = a/c, B/C = b/c. These are all algebraic expressions and can be manipulated algebraically to get equivalent forms such as B/A = b/a, B/b = A/a, and b/B = a/A, etc. Now suppose that we form a big triangle ABC such that A is the tree's height and B is the baseline from the measurer to the trunk. Now, if we can form a smaller similar triangle abc where we can measure sides a, b, and c, we can measure the baseline to the tree B and can compute the tree's height by using the A/B = a/b proportionality relationship. Algebraically rearranging, we get A = B(a/b). This last formula is what often accompanies diagrams showing how to measure tree height using similar triangles. If side A is vertical and the two triangles are truly similar, then the process works. However, what happens when side A is not vertical (the line from the crown-point to the base is not vertical)? Then the process does not work and that will be the case when the crown-point is not directly over the base of the tree. Sound familiar? Similar triangles can also be used very productively for determining crown-point offset. However, that determination requires a mult-step process that will be explained with diagrams at the April ENTS event at Cook Forest SP. In fact, all the material in these e-mails will be brought together in what I hope will be our first crack in producing an ENTS Guide to Dendromorphometry. BTW, Dendromorphometry was a term created by Gary Beluzo, partly in jest and partly in seriousness to describe the energy with we Ents expend pursuing our passion. Gary, is of course, a first class dendromorphometrist. Gary once made up certificates for John Knuerr and myself. However, the idea of formalizing Dendromorphometry as an official ENTS discipline has merit. ENTS could issue certificates to bonafide dendromorphometrists - both in fun and in all seriousness. Bob Robert T. Leverett Cofounder, Eastern Native Tree Society
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# Domain and Range
38 teachers like this lesson
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## Objective
SWBAT identify the domain and range of a function. SWBAT analyze the domain and range of continuous and discrete functions.
#### Big Idea
Students will use real world examples to solidify their understanding of continuous and discrete inputs and outputs.
## Do-Now
10 minutes
Students will complete today's Do Now. These problems review operations with integers, giving my students the opportunity to maintain or continue to develop fluency (MP6). While students are working, I will circulate around the room passing back the graded Exit Cards from our last class.
After about 4 minutes, I will call the problems from the Do-Now aloud, and students will echo back the answer. Then, we will quickly discuss the responses to the yesterday's Exit Cards.
Next a student volunteer will read today's lesson objective, "SWBAT identify the domain and range of a function. SWBAT analyze the domain and range of continuous and discrete functions."
## Whole Group Discussion + Practice
40 minutes
For the majority of my students domain and range are new concepts. To introduce this topic to them I will use the chart on Slide 3 of Domain and Range. I begin by asking five or six volunteers to tell me their name and their birth date. I continue asking for volunteers until I get two students with the same birth month. To reinforce what we learned yesterday about Functions, I will then ask students to decide if the relation, name and birth month, is a function.
Next, I will ask a student to read Slide 4 aloud. Then, I will allow students to come up with their own definition of domain. I will list the domain of the Name/Birthday example as:
{Jan, Feb, Mar, Apr, May, Jun, Jul, Aug, Sep, Oct, Nov, Dec}
I stress the word "possible" in the definition of domain. I will ask students to briefly brainstorm with a partner why this word is important in the definition.
On Slide 8 I will call on additional student volunteers to supply their birthday and their names. Again, I will ask the class to decide if this is a function. A student will read Slide 9 aloud, and we will discuss as a group the definition of the range of a function.
Domain: {1, 2, 3, ... 29, 30, 31}
Range: {Students names in the class}
To continue the discussion of "possible input values" from the definition, I will ask students to state the domain of our example on Slide 8, and justify if the number 32 is in our domain.
Slide 11: Next we will talk about what it means for a function to be continuous or discrete. I will ask students to examine the word "continuous" to find the hidden word (continue). I will tell students that a continuous function is a function where the input values include all possible values within a parameter. I will guide students to examine the domain values first to determine whether a function is continuous or discrete.
Teaching Note:
The concept of "infinity" and what it means for a line to to extend endlessly in a direction is often brand new for my students. This idea requires them to think abstractly, and view a coordinate plane as an infinite plane that has no ending, not just the 10x10 box that is commonly seen on a paper. This misconception is typically evident when I ask students to identify the domain and range of the line y=x, and they describe the domain using the numerical window constraints.
To help students visualize that a linear function will go on forever, I will project this site on my board. (To make the graph easier to view as a whole group, I use the settings to place the graph in "projector mode" as well as zooming in the screen using my browser settings). I plan to graph a variety of functions and zoom out many times to show students that the line never "stops" making the domain and range all real numbers. I will frequently use this resource throughout the year to reinforce this idea with multiple standards.
2015 Update: This lesson felt like something was missing. I have now created two sets of guided notes to be used over two days to support students as they learn this concept: Day OneDay Two.
## Partner Practice: Versatiles
20 minutes
Students will work in pairs to practice finding the domain and range of a function using the Domain and Range Versatiles handout. Students will match each question to the box on page two that corresponds to its correct response.
To assist those who are still struggling with the domain and range of a continuous function, I will pass out colored pencils and tell students to shade the x and y axis with a different color below the function on the graph.
## Closing
10 minutes
To bring this lesson to closure, we will have a brief summary discussion. I will ask students to explain the definition of domain and range in their own words, and, to give an example of a situation that would have constraints or a situation where all of the input values are not possible. I will also ask the class to elaborate on their understanding of continuous and discrete functions.
Students will then complete an Exit Card for today's lesson.
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# Functional Determinant of a matrix of operators
How would we calculate the functional determinant of a matrix with both continuous and discrete indices; such as
$$$$O =\begin{pmatrix} a(t) & \frac{d}{dt} +b(t) & c(t) \\ \frac{d}{dt}+ b(t) & 0 & d(t) \\ c(t) & d(t) & 1 \end{pmatrix}$$$$
subject to some boundary conditions e.g. $$\vec{\phi}(0)=\vec{\phi}(1)=0$$. Where $$\vec{\phi}$$ is a vector of functions?
I know that for a simpler problem e.g. $$O =\frac{d}{dt} + b(t)$$ the solution is just to find the product of the eigenvalues of $$O$$ (again with some boundary conditions) using the eigenvalue equation $$O \psi = \lambda \psi$$ and then divide by the products of the eigenvalues of $$O'=\frac{d}{dt}$$ with matching boundary conditions. My guess is that I'll eventually need to do something like this
$$$$\text{Det}[O]=\frac{\text{det} \begin{pmatrix} a(t) & \frac{d}{dt} +b(t) & c(t) \\ \frac{d}{dt}+ b(t) & 0 & d(t) \\ c(t) & d(t) & 1 \end{pmatrix}}{\text{det}\begin{pmatrix} 0 & \frac{d}{dt} & 0 \\ \frac{d}{dt} & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix}}$$$$
Which means I will need to calculate the eigenvalues for both infinite-dimensional matrices.
I have tried to calculate $$$$\text{det}\begin{pmatrix} 0 & \frac{d}{dt} & 0 \\ \frac{d}{dt} & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix}$$$$ with some boundry condition of the form $$\vec{\phi}(0) =0, \vec{\phi}(1)=1$$, by using the eigenvalue equation, getting a pair of differential equations and solving for them, but I find that this doesn't give me a discrete set of eigenvalues to multiply together? Should I be dividing by a different determinant, or just using a different method completely?
Any help or resource recommendations would be appreciated (I have so far not been able to find any online resources dealing with this type of, presumably quite common, problem?)
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# not so good with word problems
G
#### Guest
##### Guest
it says to translate to an equation and solve,
it says,the product of 2 consecutive even positive intergers is 120. find the integers.....so.....what do i do?
#### sgrabbit
##### New member
2 consecutive numbers mean that they are followed by each other like 2,3,and they have to be even,and if you multiply them together they make 120.
#### Denis
##### Senior Member
ccastillo said:
it says to translate to an equation and solve,
it says,the product of 2 consecutive even positive intergers is 120. find the integers.....so.....what do i do?
n = lower even number; then n+2 is the next one; so:
n(n+2) = 120
n^2 + 2n - 120 = 0
Now solve that.
#### JOSE JAIME-RODRIGUEZ
##### New member
all you need to do is multiply consecutive even numbers for example 8 with 10 until you get it!!!!!!!!!!
It works for me and it should work for you!!!
HINT:multiply 10 and another consecutive even integer
hope this helps!!!!!!!!
____________
-Jose
#### TchrWill
##### Full Member
ccastillo said:
it says to translate to an equation and solve,
it says,the product of 2 consecutive even positive intergers is 120. find the integers.....so.....what do i do?
The square root of 120 is 10.95 making the two numbers 10 and 12.
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# SOLUTION: DETERMINE WHETHER (8,-9) IS A SOLUTION OF 8X+3Y=-1
Algebra -> Algebra -> Systems-of-equations -> SOLUTION: DETERMINE WHETHER (8,-9) IS A SOLUTION OF 8X+3Y=-1 Log On
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Algebra: Systems of equations that are not linear Solvers Lessons Answers archive Quiz In Depth
Click here to see ALL problems on Systems-of-equations Question 263071: DETERMINE WHETHER (8,-9) IS A SOLUTION OF 8X+3Y=-1Answer by richwmiller(9143) (Show Source): You can put this solution on YOUR website!Why are you yelling? Use both upper and lower case. Just plug 8 in for x and -9 for y 8*8+3*-9=-1 64-27=-1 64=26 Obviously not equal so the point (8,-9) is not on the line 8x+3y=-1
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http://perplexus.info/show.php?pid=6821&cid=43945
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Drop into the Bucket (Posted on 2010-02-03)
Consider a bucket in the shape of a cube 1 foot on a side and filled with water.
A smaller cube shaped container, open at the top, is pushed straight down into the bucket without rotating it. At first it displaces some water which spills out of the bucket but when this container is pushed down far enough the extra water will pour into it.
If this container is very small it will be completely filled and sink to the bottom. If it is very big it will not end up with much water in it. What dimensions of this cubic container will maximize the volume that ends up inside of it.
See The Solution Submitted by Jer Rating: 4.0000 (3 votes)
Comments: ( Back to comment list | You must be logged in to post comments.)
solution | Comment 2 of 9 |
Let the side of the smaller cube be s.
The "very small cube" will hold s^3 cubic units of water.
But the water level in the larger cube must not go below the top of the smaller cube in order to reach this limit. That happens when s^3 is indeed the amount of water at a higher level than the top of the smaller cube. But s^3 of water has also been lost outside the larger cube.
So at that point, 2*s^3 + s = 1.
DEFDBL A-Z
s = .6
DO
s = (1 - s) ^ (1 / 3) / 2
PRINT s, s ^ 3
LOOP
quickly finds s = .4175611742406833.
At that point the cube is completely filled with the water that is at a higher level than its top. Any larger, and there'd be less water at a higher level, and so the contents would go down. The cube of this s has volume .0728048532199146.
So each dimension of the smaller cube is the solution to 2*s^3 + s = 1, and this is approximately .4175611742406833.
Posted by Charlie on 2010-02-03 14:21:56
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# Demystifying Calculus: An Introduction and Derivative Problems
Comprehensive Definition, Description, Examples & RulesÂ
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## Introduction to Calculus
Calculus is called infinitesimal or the Calculus of infinitesimal is a part of mathematics that focuses on change. Issac Newton and Gottfried Wilhelm Leibniz from Germany developed and define Calculus independently in the 17th century. Those who want to study physics, biology, economics, chemistry, or actuarial science need Calculus. Let’s check in detail what is Calculus:
## What Is Calculus?
Calculus is used in mathematics to get optimal math solutions. Calculus checks the rate/cost of change or changes of values that are related to a function. Calculus handles the properties of derivatives and integrals of quantities such as volume, velocity, area, acceleration, etc.Â
Calculus is classified into two main concepts-Â
• Differential CalculusÂ
• Integral CalculusÂ
calculus meaning concepts are focused on the idea of limit and function. Differential Calculus calculates the change in the quality while integral calculus calculates the quality when the rate of change is known.Â
## Basic Calculus Concepts
Limits and Continuity
Limit and continuity in calculus is one of the important calculus concepts. The limit and continuity of calculus focus on determining a value that function approaches a specific point. For example f(x)=3×+1 finds the limit as x nearing 2, it is the same as finding a number that f(x)=3x+1 approaches when nearing 2.Â
Differentiation and Integration
Differentiation refers to the study of the rate of change in Quantity Y with respect to quantity X. Derivative or differentiation f(x) given byÂ
F'(a)=lim h->0 f(a+h)-f(a)/hÂ
Integration refers to a process to find the definite integrals and indefinite integrals. It is the average rate/cost of change of a quantity. Integration is represented as:Â
F(x) dx=f(x) +cÂ
Fundamental Theorem of Calculus
The Theorem of Calculus links the differentiation with Integration. It states that Let A(x) be area function and f be a continuous function with closed interval [a,b]. Then A'(x)= f(x), for all x E[a,b].Â
## Intro to Calculus for Beginners
Let’s comprehend basic Calculus for Beginners:Â
Building Blocks of Calculus: Calculus has 4 main building blocks- discrete iteration, discrete limit, continuous limit, and continuous iteration. The basic calculus runs around these factors in mathematics that you will work on.Â
Preparing for Calculus Studies
When studying calculus studies, you need to work on concepts to understand the core of calculus. To prepare for calculus studies, you need to understand the basics & intro to calculus, practice, and work on your mistakes.Â
### Calculus Explained
Calculus in mathematics revolves around limits, continuity, derivatives, and integration. Calculus is used by engineers in various fields to determine quantities as the rate of change. Calculus is important in mathematics calculations where variables change over time.Â
Understanding the Core Principles
Calculus follows some principles that you need to know. These are some of the core principles of calculus:
• DepositionÂ
• TransformationÂ
• RigidityÂ
• SymmetryÂ
## Calculus Derivative Problems
Derivative measure or calculate the instantaneous rate of change of function.Â
Let’s check step by step calculation of calculus derivative problems:Â
Step 1: Write function as f(x)Â
Step 2: Compute equation f(x+h) – f (x). Combine all the like terms that is if the common factor of the terms is h, factor the expression by eliminating the common factor of the term i.e. h.Â
Step 3: Simplify f(x+h) -f(x)/h. H->0 in the last, so we have to cancel 0-factor h.
Step 4: Compute lim h->0 f(x+h)-f(x)/h. By letting h->0.Â
For instance, Let f(x) = ax2 +bx+c. Compute f 0 (x)
f(x+h) = a(x+h) 2+b(x+h)+c = a(x 2+2xh+h 2 )+bx+bh+c = ax2+2 axh+ah2+bx+bh+c.Â
Step 2: Use algebra to single out the factor h. f(x+h)−f(x) = (ax2+2axh+ah2+bx+bh+c)−(ax2+bx+c) = 2axh+ah2+bh = h(2ax+ah+b).Â
Step 3: Cancel the zero factor h.
f(x + h) − f(x) h = h(2ax + ah + b) h = 2ax + ah + b.Â
Step 4: Let h → 0 in the resulting expression
f 0 (x) = lim h→0 f(x + h) − f(x) h = lim h→0 2ax + ah + b = 2ax + 0 + b = 2ax + b.
## Differentiation in Calculus
Differentiation in Calculus follows some rules and techniques. Here are the list of rules to follow:Â
1. Power Rule
2. Sum and Difference Rule
3. Product Rule
4. Quotient Rule
5. Chain Rule
Derivatives of Common FunctionsÂ
Here are some derivatives of common functions:Â
1.Derivative Of f(x) = ax^n
2.Derivative Of f(x) = sin x And f(x) = cos x
3.Derivative of f(x) = e^x
4.Derivative Of f(x) = ln x
## Common Mistakes in Calculus
Identifying and Avoiding Errors in Calculus Problems
There are a few mistakes in calculus that students are most likely to make, such as:Â
• Dropping the limit notationsÂ
• Wrong use of Integration formula
• Dropping absolute value during Integration.Â
• Wrong derivative notion.Â
• Wrong integration notation.Â
• In general loss of notation.Â
• Dropping constant of Integration.Â
• Misconceptions of 1\0 and 1\infinity.Â
• Using or treating infinity as a number.Â
Tips for Successful Calculus Studies
To avoid mistakes in calculus make sure you are working on various:Â
• Make sure you take extra precautions when working with minus signs.Â
• Practice.Â
• Mark important information.Â
• Work on your problems again and again.Â
## Challenges and Advanced Calculus Topics
Calculus can be tough, students might face challenges when working on calculus such as:Â
• The equations are generally more complex than usual.Â
• Calculus is a complicated concept compared to others in mathematics.Â
• Students fail to fix their basics which leads to problems later.Â
There are some advanced concepts and topics in calculus that students can check as follows:Â
• Vector and vector calculus
• Linear approximation of vector-valued functionsÂ
• Derivative matrix
• Valued functions
• Line integralsÂ
• Multiple integralsÂ
• Theorems of Green, and more.Â
Preparing for Further Studies in Calculus
Students who want to pursue their education in the fields of physics, mathematics, and chemistry at a higher level will have to opt for further studies in calculus. Those who want to study calculus at a higher level can opt for a master’s and Ph.D. in the field.Â
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## Key Takeaways
1. Calculus is part of mathematics to calculates variables that change after a certain time.Â
2. Calculus has main basics or building blocks which include limits & continuity, derivatives & integration.Â
3. Calculus has advance topics like vectors that students will learn to use after understanding the basics.Â
4. Calculus can be challenging when it comes to mathematical equations.Â
5. There are some mistakes like the wrong use of formulas in calculus that students need to avoid with effective tips in mind.Â
6. Students who want to study in fields like physics, chemistry, and mathematics have to study calculus at a higher level.
## Quiz
Check your score in the end
Quiz
Check your score in the end
Question of
#### Question comes here
Differentiation refers to the rate of change instantaneously that breaks down function for the instant with respect to quantity. Integration on the other hand is known as the average rate of change.
Derivative in calculus is the rate of change in Y quantity with respect to X quantity.
When performing calculus, make sure you do not miss minus signs. Treat minus signs in the equation carefully, and regularly identify your mistakes to avoid them later.Â
Yes! There is detailed information online and tutorials given on calculus. Moreover, you can practice and work on the concept using resources such as the Edulyte worksheet.
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x Turn on thread page Beta
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# Don't understand this likelihood ratio question (Statistics) watch
1. Hey guys. I have a question about a Beta(a,a) distribution. It's to do with the abilities of students and estimating the true value. We have observed the ability of m students, theta(w) = (theta1(w), theta2(w).....thetam(w)) are our observations.
The question asks me to find the likelihood function and then asks a question about whether a given statistic is sufficient or not. However I am stuck on the third part which says:
c) Construct the likelihood ratio W for Θ0 = {1}, Θ1 = { 0.5 , 2}.
I have no idea what to do here, any ideas?
2. (Original post by pineapplechemist)
Hey guys. I have a question about a Beta(a,a) distribution. It's to do with the abilities of students and estimating the true value. We have observed the ability of m students, theta(w) = (theta1(w), theta2(w).....thetam(w)) are our observations.
The question asks me to find the likelihood function and then asks a question about whether a given statistic is sufficient or not. However I am stuck on the third part which says:
c) Construct the likelihood ratio W for Θ0 = {1}, Θ1 = { 0.5 , 2}.
I have no idea what to do here, any ideas?
So presumably you are able to write down the likelihood function as a function of ? Now simply write down the definition of the likelihood ratio (that is, the formula that involves taking suprema on top and bottom over the relevant parameter values). Take the required suprema and you are home.
3. (Original post by Gregorius)
So presumably you are able to write down the likelihood function as a function of ? Now simply write down the definition of the likelihood ratio (that is, the formula that involves taking suprema on top and bottom over the relevant parameter values). Take the required suprema and you are home.
My likelihood is (a-1)(sum from i=1 to M)log(thetai) + (a-1)(sum from i=1 to M)log(1-thetai) - M log B(a,a). B is the beta function: is this correct?
I know the formula but I'm confused specifically by the parameter spaces given. What does it mean to have a paramter space of 0.5 and 2? Am I taking the supremum of all possible values given by having a Beta(0.5,0.5) distribution or a Beta(2,2) distribution?
4. (Original post by pineapplechemist)
My likelihood is (a-1)(sum from i=1 to M)log(thetai) + (a-1)(sum from i=1 to M)log(1-thetai) - M log B(a,a). B is the beta function: is this correct?
I know the formula but I'm confused specifically by the parameter spaces given. What does it mean to have a paramter space of 0.5 and 2? Am I taking the supremum of all possible values given by having a Beta(0.5,0.5) distribution or a Beta(2,2) distribution?
You've got the log likelihood there - if you continue with that, you'll need to subtract rather than divide.
The supremum is a supremum over in the parameter space. One of the parameter spaces has a single element, so that is easy, just set . For the other one, you have two possibilities for . Which one maximizes the likelihood?
5. (Original post by Gregorius)
You've got the log likelihood there - if you continue with that, you'll need to subtract rather than divide.
The supremum is a supremum over in the parameter space. One of the parameter spaces has a single element, so that is easy, just set . For the other one, you have two possibilities for . Which one maximizes the likelihood?
For the Null parameter space do I literally sub 1 into my p.d.f? This gives me a value of 1. What about for the other parameter space with 1/2 and 2? Do I need to find the MLE or do I simply sub both in to my p.d.f and choose the bigger one? Sorry, I really don't understand this very well.
6. (Original post by pineapplechemist)
For the Null parameter space do I literally sub 1 into my p.d.f? This gives me a value of 1. What about for the other parameter space with 1/2 and 2? Do I need to find the MLE or do I simply sub both in to my p.d.f and choose the bigger one? Sorry, I really don't understand this very well.
You are going along the right lines - it is a matter of substituting the parameter values into the likelihood function. The likelihood is
If you plug you will get a value of one for the likelihood. What do you get if you set then in turn?
7. (Original post by Gregorius)
You are going along the right lines - it is a matter of substituting the parameter values into the likelihood function. The likelihood is
If you plug you will get a value of one for the likelihood. What do you get if you set then in turn?
But the value of the likelihood function for alpha = 2 and alpha = 1/2 is dependent on the values of theta_i ?
For example when theta_i are all close to 0 (or 1) then the supremum of the likelihood is when alpha=1/2. When the theta_1 are near 1/2 then we should take alpha=2
So how do we take supremum
8. I get this
Attached Images
9. (Original post by Namch)
I get this
That looks the right sort of thing. If you set
then
and
The decision as to which is bigger simply then depends upon whether is bigger or smaller than
10. Woah when 1/2 is the argument of the gamma function we have sqrt pi. I didnt know
11. (Original post by Namch)
Woah when 1/2 is the argument of the gamma function we have sqrt pi. I didnt know
Yes,
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# News from Mathnasium of Waterloo
### Number Sense Part 1
May 3, 2018
Ever hear your child or another student talk about math class and say, “That stuff doesn’t make sense to me!”? That’s pretty typical, and it’s because children don’t get enough context when they learn material. Providing context helps students integrate new ideas and information in a way they understand. Unfortunately, children often experience “fragmented learning” (learning without a sense of context) rather than “integrated learning.” To fill in the gap between fragmented and integrated learning, students need to establish Number Sense.
So, Number Sense … what is it?
Number Sense is the ability to appreciate the size and scale of numbers in the context of the question at hand.
What does Number Sense involve? We’ll explain.
There are three major elements that fall under number sense:
countingwholes and parts, and proportional thinking.
The basis of Number Sense begins with counting, so we’ll focus on that today. (We’ll do wholes and parts and proportional thinking in future articles.)
Counting, simply put, is the ability to count from any number, to any number, by any number, both forward and backward.
When asked to explain what counting is, many students will respond by counting from 1 (1, 2, 3, …), though counting should technically start at 0 instead of 1.
After learning to count by 1s, kids need to learn to count starting from other numbers, for instance, beginning at 28 (28, 29, 30, …). How about counting by 2s? Starting from 2 (0, 2, 4, 6, …) is easy to understand. But can our kids count by 2s when starting from the number 3 (3, 5, 7, …)?
After a good deal of practice, an experienced counter will be able to count to 250 by 1s forward and backward; to 300 by 2s, 5s, and 10s; and to 3,000 by 100s—starting at any point.
As children become experienced counters, they should also learn how to group the numbers they count. Ask your child questions such as:
• How much altogether is a group of 9 balls and a group of 7 balls?
• If you have a group of 10 and take away a group of 7, that leaves how many?
• If I give you 5 groups of 10 pennies, how many pennies will you have?
• How many groups of 4 does it take to make 12?
The importance of thinking about counting this way is its connection with the basic math operations: addition (counting how much altogether), subtraction (counting how much is left or how far apart), multiplication (counting in equal groups), and division (counting how many of these are inside of that).
As children learn to think in larger and larger groups, several good things happen.
• Being able to see one group of 24 as 1 group of 24 … 2 groups of 12 … 3 groups of 8 … 4 groups of 6 … 8 groups of 3 … 12 groups of 2 … or 24 groups of 1 allows students to almost effortlessly transition to knowing their times tables.
• Thinking in groups of 10 is the foundation of understanding place value in the decimal number system, as well as our monetary system (10 pennies make 1 dime …) and the metric system (10 centimeters make 1 decimeter …).
• Seeing the relationship between the size of two groups is the foundation of proportional thinking (literally “reasoning in groups, according to amount”—seeing in groups).
Once children learn the basis for counting, they can progress with confidence to wholes and parts and proportional thinking. At Mathnasium™, we’ve developed a method that teaches these fundamentals so that children can finally understand math in a way that makes sense to them.
Remember, children don't hate math; they hate being confused, frustrated and embarassed by math. Once they understand math, a passion will follow naturally.
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## raden_zaikaria 2 years ago what is the domain and range mean?
1. Slahrichi101
Hold on-- ill get it for you
2. Slahrichi101
ok-- if you google it up its just going to confuse you even more, so my genius brain and i put it in a simple simple definition Domain- the x place Range- the y place example: (2,5) the domain is 2 and the range is 5
3. swathi540
The domain is the set of all possible x-values which will make the function "work", and will output real y-values The range is the resulting y-values we get after substituting all the possible x-values.
4. Slahrichi101
Thank you ;)
is that domain ---> Object... Range ---> image
6. Slahrichi101
In a way, but mathematically its just domain is x and range is y
7. hewsmike
A function is the totality of associations/links between members of one set and another. You define one set as the domain, requiring that all it's members must have an association with only one member of the second set, which we call the range. We say that the function is 'single valued'. |dw:1360014364411:dw| Not all members of the the range need have an association with the members of the domain. That subset of the range which do have an association with at least one domain member is called the image. |dw:1360014481078:dw| If there are members of the range which don't have an association with a member of the domain then the function is deemed as an 'into'. So the image is a proper subset of the range. |dw:1360014618709:dw| If all range members have an association with some domain member then the function is deemed as 'onto'. So the image and the range are the same set. |dw:1360014744956:dw|
8. Slahrichi101
9. hewsmike
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Permutation Indexby Isai Damier, Android Engineer @ Google
```#=======================================================================
# Author: Isai Damier
# Title: Permutation Index
# Project: geekviewpoint
# Package: algorithms
#
# Statement:
# Given a permutation of a set, return the index of the permutation.
#
# Sample Input: [3, 1, 2]
# Sample Output: 4
#
# Time Complexity of Solution:
# Best = Average = Worst = O(n^2)
# Space Complexity of Solution:
# Best = Average = Worst = O(1)
#
# Details:
# A permutation is an ordering of the elements of a set. So for a set
# constituted of the elements 1,2,3; then [1,2,3] and [2,1,3] are two
# different permutations of the set. A set of length n has n!
# permutations. For example, if a set contains 3 elements, it has
# 3! = 3*2*1 = 6 permutations. The following algorithm uses the
# relation between permutation and factorial to find the index of a
# given permutation of a set.
#
# Illustrating by manually getting the index of [2, 4, 3, 1].
# Since this is a 4-element set, we know there are 4! permutations
# (4! = 4*3*2*1). If the set only had 3 elements, we would have 3*2*1
# permutations. If the set only had 2 elements, we would have 2!=2*1
# permutations; and so on.
#
# ASIDE: The decimal system of counting is a positional system.
# A 3-element decimal number, for instance, has the following three
# positional weights: hundred, ten, unit. Hence, we know the value of
# the number 472 because we understand: 4*hundred + 7*ten + 2*unit.
#
# If we treat our 4-element set as a positional system, then we get
# the following positional weights: 3!, 2!, 1!, 0. So that the index
# of [2, 4, 3, 1] is: x*3!+y*2!+z*1!+w*0. Presently it suffices to
# find the values of x,y,z to calculate the index (we ignore w because
# it is paired with 0). x,y,z are counters: the number of succeeding
# elements less than the element being considered. For example, in
# [2, 4, 3, 1], there are two succeeding elements less than 4
# (namely 3 and 1). For 2 it's 1 (1); for 4 it's 2 (3 and 1); for 3
# it's 1 (1); for 1 it's 0.
#
# Now we can calculate the index of [2, 4, 3, 1] as: x=1, y=2, z=1:
# x*3!+y*2!+z*1!+w*0 = 1*3! + 2*2! + 1*1! = 6 + 4 + 1 = 11.
#
# Now that we have our algorithm, the trick is implementing it. As you
# may imagine there are a number of possible implementations. The
# presented implementation focuses on using constant auxiliary memory:
# memory = O(1) and time = O(n^2). The code is written for readability.
#=======================================================================
def permutationIndex( permutation ):
index = 0
position = 2 # position 1 is paired with factor 0 and so is skipped
factor = 1
for p in range( len( permutation ) - 2, -1, -1 ):
successors = 0
for q in range( p + 1, len( permutation ) ):
if permutation[p] > permutation[q]:
successors += 1
index += ( successors * factor )
factor *= position
position += 1
return index```
```import unittest
from algorithms import numbers as algorithm
class Test( unittest.TestCase ):
def testPermutationIndex( self ):
three = [[1, 2, 3], [1, 3, 2], [2, 1, 3], [2, 3, 1],
[3, 1, 2], [3, 2, 1]]
for i in range( len( three ) ) :
self.assertEquals( i, algorithm.permutationIndex( three[i] ) )
four = [[1, 2, 3, 4], [1, 2, 4, 3], [1, 3, 2, 4], [1, 3, 4, 2],
[1, 4, 2, 3], [1, 4, 3, 2], [2, 1, 3, 4], [2, 1, 4, 3],
[2, 3, 1, 4], [2, 3, 4, 1], [2, 4, 1, 3], [2, 4, 3, 1],
[3, 1, 2, 4], [3, 1, 4, 2], [3, 2, 1, 4], [3, 2, 4, 1],
[3, 4, 1, 2], [3, 4, 2, 1], [4, 1, 2, 3], [4, 1, 3, 2],
[4, 2, 1, 3], [4, 2, 3, 1], [4, 3, 1, 2], [4, 3, 2, 1]]
for i in range( len( four ) ) :
self.assertEquals( i, algorithm.permutationIndex( four[i] ) )
```
| 1,352 | 3,747 |
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| 3.859375 | 4 |
CC-MAIN-2023-40
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latest
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en
| 0.862988 |
https://boards.straightdope.com/t/cecils-tube-through-the-earth-column-a-2nd-question/25604
| 1,708,794,765,000,000,000 |
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crawl-data/CC-MAIN-2024-10/segments/1707947474541.96/warc/CC-MAIN-20240224144416-20240224174416-00892.warc.gz
| 148,164,337 | 8,683 |
# Cecil's "Tube through the earth" column, a 2nd question
In regards to: “What if you fell through a tube in the Earth?” column…
What would happen if this tube were opened so that both ends opened into the oceans letting the seas drain in and fill the tube?
clarification: I am inquiring about the effects this would have on a) the oceans themselves b)the center of the earth and c) the earth overall
I’m going to ignore any possible question of the change in sea level, since that would require knowing the shape and size of the tube, as well as math. I’m just not up to it.
But it’s interesting to think of a bottomless sea channel. The pressure would increase as you went down, but only to a certain point. At some point, the attraction from the mass of earth and water above you would cancel out some of the gravity acting on you and the water, and the pressure would start to decrease. In the center, the pressure would be zero, I believe.
What would happen then if you blew a big bubble? I think it would stay in the center of the earth, offering a dry place for travellers through the tube.
Caveat: This is all off the top of my speculating head. I fully expect it to be clarified, if not totally demolished, by those with the stamina to check the math.
Interesting, but I think there would be pressure at the center. It’s true that the gravity at the center would pull you equally in all directions, and thus cancel out. But all the water extending up the tube (the tube would be going up in opposite directions) is attracted to the center.
The pressure at the center would be the integral of the mass of the water times the gravity at the given radius, from 0 to the radius of the Earth. That gravity function would be a little weird. It would increase with r^2, but decrease with r^3 (the amount of mass still attracting you to the center, all of the Earth’s mass outside your current radius perfectly cancels).
Think of it this way. Starting at the surface and sinking toward the center, the pressure will increase. For a given depth that you sink, the pressure will increase a certain amount. That amount will shrink as you get closer to the center. The pressure increase from r=2000 meters to r=1000 meters will be tiny, but there is never some opposing force that will cause the pressure to decrease as you sink farther.
[ At some point, the attraction from the mass of earth and water above you would cancel out some of the gravity acting on you and the water, and the pressure would start to decrease. In the center, the pressure would be zero, I believe. ]
you are ignoring the pressure caused by gravity of all the water on top. the pressure would be great. Think of a all water planet. Drop a piece of lead in it, it would sink to the bottom and be suspended by the force of gravity in the center (all G forces cancel) but the pressure is still there, a buble would eventually work its way to a side by curents and start upward. Also in the case of the earth, I think the water would boil for probally tens of thousands of years and act as a underwater gyser.
Everything I state is IMHO, I have never been to a planet made entirely of water, nor have I drilled a hole through the earth ocean to ocean.
Can a water/air planet exist? (no land) or would it deform under gravity of star/moons enough to rip it apart?
If the tube was full of air and the water entered from each end at the same time and under the same pressure, the air trapped in the tunnel would eventually be compressed into it’s components’ liquid state. Which would be at the middle of the earth and would be lighter than the salt water pushing against it. Would the various factors (rotation of the earth at it’s center)(pressure of the salt water), (whatever), cause the unsalted liquid to remain in the center, separating the columns of salt water?
I think gas giants like Jupiter and Saturn are kind of like that - at the very least, they have no land. I’m not a planetary scientist, but from what I remember their interiors deep down become at least vaguely liquid. The pressures and temperatures are incredible, so it’s not quite like anything you or I have actually laid eyes on and handled.
This is a pretty half-baked answer, and somebody can probably expand on and/or correct it.
Saltire: here’s a simple rule for determining the pressure at any point. Imagine moving everything a differential amount in one direction, leaving a vacuum. The amount of work need to do this is equal to the pressure in that direction times the volume of the vacuum left. So the center of the Earth would ideed have a very large pressure, since moving water away from the center of the Earth would take a lot of work.
tcburnett: you would have to fill the tunnel very quickly to have the air liquify before it bubble out. Mere gravity would not be enough; you’d need some powerful explosives.
jupiter is a gas planet. toward the center of jupiter hydrogen exists as a liquid metal, not unlike mercury. chunks of rocky material are at the center of this core of molten metalic hydrogen. scientists have no way of knowing how large this rocky mass is, but some have speculated that it might be the size of the earth. It stands to reason that a similar rocky mass would exist at the center of other gas giants and possibly the sun as well.
I always thought that the sun was so hot that it would vapourise and even plasmorise (is that a word, and if it is, is it spelt right?) absolutely anything that even approached the centre. Arthur C. Clarke in 2061: Oddesy three had the pressures of Jupiter compress carbon at the centre of the planet to form the largest diamond in the solar system? This seems pretty far out, I admit, but hey, this guy proposed satelites and geostationary orbit. So I wouldn’t be inclined to dismiss it entirely out of hand. Does diamond count as rock?
I also don’t think that it’s possible for an all-water planet to exist, especially one with an aptmosphere - maybe a solid ball of ice, or a world entirely covered by water, but not 100% liquid water. I doubt that it would be dense enough to attract an aptmosphere, or even to hold itself together. We all know how the moon affects water on earth (tides), so it would be likely that any large mass nearby would tear it apart (including the sun - which would be necessary to keep it liquid, so its a kind of catch 22)
Just an interesting sidetrack - what would happen if you drilled a hole through the earth with one end in the ocean and one end on land? would the water fill only to around the centre? would it go halfway back out the other side? or all the way out the other side (assuming that there is enough water to do so).
I kinda suck at science, but I’m fond of astronomy and SF is a minor passion of mine, so I hope that there is some validity to what I say…
thanks
[QUOTE]
*Originally posted by big_yellow_kingswood *
**
The affect of the moon on water on the earth is extremely small. The tidal effect only distorts the shape of the earth by a couple meters–and the radius of the earth is about 6378136 meters. In order to tear each other apart through tidal effects, two planets, and/or a sun, would have to be very close together.
[QUOTE]
*Originally posted by big_yellow_kingswood *
**
All the way out the other side. The surface of the water will move to conform to an equipotential surface. Ignoring, of course, the obvious problems of actually creating such a hole. That’d be a hell of a waterfall.
6376000 or so of which is not water. (given a estimated sea depth of about 2000 metres, which is excessive for most parts of the world, I believe) The question posed by k2dave (and I’m going off the OP, I know) was
and I’m saying that while the moons gravitational effect on earths water is minimal, it is easily measurable. If there was only water, the effect would be a lot bigger as the whole planet would be dynamic enough to be affected, and not just the top 0.1%. Feel free to work out the maths to prove this right or wrong, cause I can’t be bothered.
[QUOTE]
*Originally posted by big_yellow_kingswood *
**
I have worked it out and it is wrong, but I’m not going to include the maths here, cause you’ll not bother to read them.
The earth under your feet deforms with the tide as well. It just doesn’t deform as quite as readily as the water–otherwise you wouldn’t even notice the movement of the water, as they would rise and fall together. They both try to fit the equipotential surface. The additional effect, due to the deformation, is much smaller, unless the second planet is very close.
| 1,915 | 8,595 |
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| 3.578125 | 4 |
CC-MAIN-2024-10
|
latest
|
en
| 0.96839 |
https://kat1055.com/force-and-fan-carts-gizmo-answer-key-activity-c/
| 1,701,604,190,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2023-50/segments/1700679100499.43/warc/CC-MAIN-20231203094028-20231203124028-00039.warc.gz
| 380,317,748 | 12,418 |
# Exploring Force And Fan Carts Gizmo Answer Key Activity C
## Introduction
In the year 2023, the Force and Fan Carts Gizmo Answer Key Activity C has become a popular topic in the field of science education. This activity is designed to help students learn about force and motion, and how these concepts apply to real-world scenarios.
### What is the Force and Fan Carts Gizmo?
The Force and Fan Carts Gizmo is an interactive simulation that allows students to explore the relationship between force, mass, and acceleration. The Gizmo consists of a virtual cart that can be pushed or pulled by a variety of forces, including fans, springs, and gravity.
### Activity C: Fan Cart Physics
Activity C of the Force and Fan Carts Gizmo focuses on the physics of fan carts. Students are asked to investigate how the force of a fan affects the motion of a cart, and how changes in mass and friction can impact the results.
## Key Concepts
### Force
Force can be defined as any push or pull that causes an object to change its motion. In the context of the Force and Fan Carts Gizmo, force is represented by the fan that propels the cart forward.
### Mass
Mass is the amount of matter in an object. In the Gizmo, mass is represented by the weight of the cart and any additional objects that are placed on it.
### Acceleration
Acceleration is the rate at which an object\’s velocity changes over time. In the Gizmo, acceleration is affected by the force of the fan and the mass of the cart.
## How to Use the Gizmo
To use the Force and Fan Carts Gizmo, students first need to log in to their account on the website. Once logged in, they can access the Gizmo and begin exploring the physics of fan carts.
### Setting Up the Experiment
To set up the experiment, students need to choose the mass of the cart and the force of the fan. They can also adjust the friction by changing the surface on which the cart is placed.
### Collecting Data
As the fan propels the cart forward, students can collect data on the cart\’s acceleration and velocity. They can also measure the distance the cart travels over a set period of time.
## Conclusion
The Force and Fan Carts Gizmo Answer Key Activity C is a valuable tool for teaching students about force and motion. By using the Gizmo, students can explore the physics of fan carts in a safe and interactive way, and gain a better understanding of how these concepts apply to the real world.
| 529 | 2,439 |
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| 3.953125 | 4 |
CC-MAIN-2023-50
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latest
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en
| 0.925664 |
https://oeis.org/A185186
| 1,576,391,809,000,000,000 |
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crawl-data/CC-MAIN-2019-51/segments/1575541301598.62/warc/CC-MAIN-20191215042926-20191215070926-00121.warc.gz
| 482,348,952 | 5,060 |
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A185186 Numbers divisible by at least one of their digits other than 1. 2
2, 3, 4, 5, 6, 7, 8, 9, 12, 15, 20, 22, 24, 25, 26, 28, 30, 32, 33, 35, 36, 39, 40, 42, 44, 45, 48, 50, 52, 55, 60, 62, 63, 64, 65, 66, 70, 72, 75, 77, 80, 82, 84, 85, 88, 90, 92, 93, 95, 96, 99, 102, 104, 105, 112, 115, 120, 122, 123, 124, 125, 126, 128, 132 (list; graph; refs; listen; history; text; internal format)
OFFSET 1,1 COMMENTS The only primes in the sequence are 2, 3, 5, 7. No repunits are eligible. Also, an interesting class of non-eligible integers consists of some powers of 2, 3 and 7: "2, 4, 8-less" powers of 2, 2^m = 1, 16, 65536 with m = 0, 4, 16 (a subsequence of A034293); "3, 9-less" powers of 3, 3^m = {1, 27, 81, 177147, 1162261467}, with m = {0, 3, 4, 11, 19} (a subsequence of A131629); "seven-less" powers of 7, 7^m, with m = 0, 2, 3, 4, 7, 16, 22, 24, 39 (see 6th row of A136291 Array read by rows: each row is a sequence of numbers k such that n^k does not contain the digit n). Asymptotic density 27/35 = 0.771... - Charles R Greathouse IV, Mar 11, 2011 The asymptotic density of numbers having a prime digit is 1 for each prime digit. The asymptotic density of numbers being divisible by 2, 3, 5 or 7 is -Sum_{d|210, d>1}((-1)^omega(d) / d) = 27/35. Also, the asymptotic density of numbers divisible by the first n primes is r(n) where r(1) = 1/2 and r(n) = r(n - 1) + (1 - r(n - 1)) / prime(n). - David A. Corneth, May 28 2017 LINKS Giovanni Resta, Table of n, a(n) for n = 1..10000 MATHEMATICA digDivQ[n_] := AnyTrue[IntegerDigits[n], # > 1 && Mod[n, #] == 0 &]; Select[Range[200], digDivQ] (* Giovanni Resta, May 27 2017 *) PROG (PARI) is(n) = my(d = vecsort(digits(n), , 8), t = 1); while(t<=#d&&d[t] < 2, t++); sum(i=t, #d, n%d[i]==0) > 0 \\ David A. Corneth, May 27 2017 CROSSREFS Cf. A187398, A187516, A187238, A187533, A187534, A187551. Sequence in context: A008816 A002271 A048381 * A115569 A064653 A130588 Adjacent sequences: A185183 A185184 A185185 * A185187 A185188 A185189 KEYWORD nonn,base AUTHOR Zak Seidov, Mar 11 2011 EXTENSIONS Name edited by Alonso del Arte, May 16 2017 STATUS approved
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Last modified December 15 00:30 EST 2019. Contains 329988 sequences. (Running on oeis4.)
| 1,089 | 2,936 |
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| 3.609375 | 4 |
CC-MAIN-2019-51
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longest
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en
| 0.863571 |
http://mathforum.org/library/drmath/view/60810.html
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crawl-data/CC-MAIN-2017-22/segments/1495463608058.57/warc/CC-MAIN-20170525102240-20170525122240-00235.warc.gz
| 251,880,889 | 3,614 |
Associated Topics || Dr. Math Home || Search Dr. Math
### Real Life Uses of Quadratic Equations
```Date: 06/13/2002 at 14:08:52
From: Eric Veneto
Subject: Practical uses for quadratic equations
Can you give me a real life problem that would require you to solve a
quadratic equation? In High School, we only saw equations, not word
problems. If we learn it, it should have practical use, right?
```
```
Date: 06/13/2002 at 17:26:45
From: Doctor Ian
Subject: Re: Practical uses for quadratic equations
Hi Eric,
Let's say I'm shooting my pistol at a watermelon 50 yards away.
I know that the bullet comes out traveling 750 feet per second,
and I know that the target is 150 feet away, so I can estimate
the time of flight to be about 150 feet divided by 750 feet per
second, or about 1/5 of a second.
Now, I know that the bullet will drop according to
h = (1/2)gt^2
= (1/2)(-32 ft/sec^2)(1/5 sec)^2
= -16/25 feet
or a little more than half a foot. Which means that I have to
aim about a half a foot above the melon in order to hit it.
But that's not really solving a quadratic equation, is it? It's
just taking a square root. So let's change the problem a
little. I see an old abandoned refrigerator 800 yards away, and
I want to know whether I can hit it or not.
If I aim at a 45 degree angle, the vertical component of velocity
will be (750 ft/sec) * sin(45 degrees), or about 530 feet per
second. Again, we have a parabolic arc, so the height above the
ground after t seconds will be
h = (530 ft/sec)t + (1/2)(-32 ft/sec^2)t^2
Obviously the height will be zero when t=0; I want to know when
it will be zero again at the other end! So I set the height
equal to zero, and I get
0 = (530 ft/sec)t + (1/2)(-32 ft/sec^2)t^2
Now, this is still too easy, because I can just divide both
sides by t, and then I don't have a quadratic equation anymore.
So let's say that the refrigerator is at the top of a cliff that
looks to be about 20 feet high. Now I want to know when the
height will be 20 feet:
20 = (530 ft/sec)t + (1/2)(-32 ft/sec^2)t^2
So let's say I solve this, and find the values of t. (There will
be two: one for when the bullet reaches a height of 20 feet on
the way up, and the other when it lands at the top of the cliff.)
I can multiply that by the _horizontal_ component of the velocity
(also 530 feet per second), and if I get a distance of more than
800 yards, then the refrigerator is in range. If not, it's not,
and I shouldn't waste expensive ammunition shooting at it.
What could be more practical than that?
- Doctor Ian, The Math Forum
http://mathforum.org/dr.math/
```
Associated Topics:
| 750 | 2,638 |
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| 4.28125 | 4 |
CC-MAIN-2017-22
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longest
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en
| 0.936394 |
https://www.jiskha.com/display.cgi?id=1289000435
| 1,516,138,356,000,000,000 |
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| 929,091,651 | 3,809 |
# physics
posted by .
After fixing a flat tire on a bicycle you give the wheel a spin.
(a) If its initial angular speed was 5.41 rad/s and it rotated 13.7 revolutions before coming to rest, what was its average angular acceleration?
• physics -
wf^2=wi^2+2*alpha*displacement
solve for angular acceleration, in rad/s^2
• physics -
13.7 * 2 pi = 86.08 radians total
so time to stop = 86.08/2.705 = 32.8 seconds
d = Vi t +(1/2) a t^2
86.08 = 5.41 (32.8) +.5 a (32.8)^2
86.08 = 172 + 538 a
-86 = 538 a
## Similar Questions
After fixing a flat tire on a bicycle you give the wheel a spin. (a) If its initial angular speed was 5.17 rad/s and it rotated 10.6 revolutions before coming to rest, what was its average angular acceleration?
2. ### college physics (inertia)
As a result of friction, the angular speed of a wheel changes with time according to dθ/dt = ω0 *e^-(σ*t) where ω0 and σ are constants. The angular speed changes from an initial angular speed of 3.08 rad/s …
3. ### college physics (inertia)
As a result of friction, the angular speed of a wheel changes with time according to dθ/dt = ω0 *e^-(σ*t) where ω0 and σ are constants. The angular speed changes from an initial angular speed of 3.08 rad/s …
4. ### college physics (inertia)
As a result of friction, the angular speed of a wheel changes with time according to dθ/dt = ω0 *e^-(σ*t) where ω0 and σ are constants. The angular speed changes from an initial angular speed of 3.08 rad/s …
5. ### physics
A person is riding a bicycle, the wheels of a bicycle have an angular velocity of +20.5 rad/s. Then, the brakes are applied. In coming to rest, each wheel makes an angular displacement of +12.0 revolutions. (a) How much time does it …
6. ### physics, bike
After fixing a flat tire on a bicycle you give the wheel a spin. Its initial angular speed was 6.50 and it rotated 15.2 revolutions before coming to rest.What was its average angular acceleration?
7. ### Physics
A person is riding a bicycle, the wheels of a bicycle have an angular velocity of +18.5 rad/s. Then, the brakes are applied. In coming to rest, each wheel makes an angular displacement of +14.5 revolutions. (a) How much time does it …
8. ### Physics
After fixing a flat tire on a bicycle you give the wheel a spin. Its initial angular speed was 7.35 rad/s and it rotated 15.0 revolutions before coming to rest. What was the average angular acceleration?
9. ### physics
A person is riding a bicycle, the wheels of a bicycle have an angular velocity of +24.0 rad/s. Then, the brakes are applied. In coming to rest, each wheel makes an angular displacement of +16.0 revolutions. (a) How much time does it …
10. ### physics
As a result of friction, the angular speed of a wheel changes with time according to d(theta)/dt = w0 e^-(aplpha*t) where w0 and alpha are constants. The angular speed changes from an initial angular speed of 4.87 rad/s to 4.27 rad/s …
More Similar Questions
| 789 | 2,939 |
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| 3.65625 | 4 |
CC-MAIN-2018-05
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latest
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| 0.894796 |
https://calculator.academy/conduit-shrink-calculator/
| 1,713,459,537,000,000,000 |
text/html
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crawl-data/CC-MAIN-2024-18/segments/1712296817222.1/warc/CC-MAIN-20240418160034-20240418190034-00695.warc.gz
| 135,708,712 | 56,633 |
Enter the offset distance (in) and the shrink per inch (in/in) into the Calculator. The calculator will evaluate the Conduit Shrink.
## Conduit Shrink Formula
CS = OD * SPI
Variables:
• CS is the Conduit Shrink (in)
• OD is the offset distance (in)
• SPI is the shrink per inch (in/in)
To calculate Conduit Shrink, multiply the offset distance by the shrink per inch.
## How to Calculate Conduit Shrink?
The following steps outline how to calculate the Conduit Shrink.
1. First, determine the offset distance (in).
2. Next, determine the shrink per inch (in/in).
3. Next, gather the formula from above = CS = OD * SPI.
4. Finally, calculate the Conduit Shrink.
5. After inserting the variables and calculating the result, check your answer with the calculator above.
Example Problem :
Use the following variables as an example problem to test your knowledge.
offset distance (in) = 30
shrink per inch (in/in) = 0.67
## FAQs
What is conduit shrink in electrical installations?
Conduit shrink refers to the reduction in length of electrical conduit due to bends or curves made during installation. It’s calculated by multiplying the offset distance by the shrink per inch to accommodate the change in length and ensure accurate conduit runs.
Why is calculating conduit shrink important?
Calculating conduit shrink is crucial for ensuring that electrical conduit installations are precise and efficient. It helps in avoiding material wastage, ensuring that conduits fit perfectly within the designated paths, and maintaining the integrity of electrical systems.
Can conduit shrink calculation be used for all types of conduits?
Yes, the basic principle of calculating conduit shrink can be applied to different types of conduits, including PVC, steel, and aluminum. However, the shrink per inch value may vary based on the conduit material and bending method used.
Are there tools available to help calculate conduit shrink?
Yes, there are various calculators and software tools available that can help in accurately calculating conduit shrink. These tools often require inputs like the offset distance and shrink per inch, similar to the manual calculation method described.
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## Elementary Algebra
$\frac{(x+1)}{3x}$
To factor the numerator, we use the rule $a^{2}-b^{2}=(a+b)(a-b)$. Then, we factor the expression in the denominator in order to cancel out common factors in the numerator and the denominator: $\frac{x^{2}-1}{3x^{2}-3x}$ =$\frac{x^{2}-1^{2}}{3x(x-1)}$ =$\frac{(x+1)(x-1)}{3x(x-1)}$ =$\frac{(x+1)}{3x}$
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# Calculating a Taylor Polynomial of a mystery function
I need to calculate a taylor polynomial for a function $f:\mathbb{R} \to \mathbb{R}$ where we know the following $$f\text{ }''(x)+f(x)=e^{-x} \text{ } \forall x$$ $$f(0)=0$$ $$f\text{ }'(0)=2$$
How would I even start?
-
You want to find the $n$-th derivative of $f(x)$ at $0$. Once you know all of those, you will know the series. Start from $f''(x)+f(x)=e^{-x}$. Put $x=0$. So $f''(0)+f(0)=e^{-0}$, and therefore $f''(0)=1$. Next, differentiate both sides of given equation. We get $f'''(x)+f'(x)=-e^{-x}$. Put $x=0$. We get $f'''(0)=-3$. Differentiate again. We get $f^(4)(x)+f''(x)=e^{-x}$. Put $x=0$. We get $f^{(4)}(0)=?$. Differentiate again. A pattern will become apparent, maybe. – André Nicolas Nov 30 '11 at 6:44
@Andre This was the most straightforward approach. Thanks! – pigishpig Dec 1 '11 at 22:50
Think about the procedure described by J.M. and Didier Piau. It is what you need to do in general. (The solve the DE then expand approach is ordinarily not feasible, so that answer is less relevant.) – André Nicolas Dec 2 '11 at 0:13
The Frobenius method for solving differential equations is easily done: assume an ansatz
$$f(x)=c_0+\sum_{k=1}^\infty c_k x^k$$
and you have the derivatives
\begin{align*} f^\prime(x)&=\sum_{k=1}^\infty k c_k x^{k-1}=c_1+\sum_{k=1}^\infty (k+1)c_{k+1} x^k\\ f^{\prime\prime}(x)&=\sum_{k=1}^\infty k(k+1)c_{k+1} x^{k-1} \end{align*}
From these, you have $c_0=0$ and $c_1=2$ (why?); a relation for the other $c_k$ can be derived by comparing the series coefficients of $f(x)+f^{\prime\prime}(x)$ with the coefficients of $\exp(-x)$.
-
We have the following
$$f''(x) + f(x) = e^{-x}$$ and $f(0) = 0$, $f'(0) = 2$.
And thus we need to find $f^{(n)}(0)$ to construct the Taylor series.
Note that we already have two values and can find $f''(0)$ since
$$f''(0) + f(0) = e^{-0}$$
$$f''(0) +0 = 1$$
$$f''(0) = 1$$
So now we differentiate the original equation and get:
$$f'''(x) + f'(x) = -e^{-x}$$
But since we know $f'(0) = 2$, then
$$f'''(0) + f'(0) = -e^{-0}$$
$$f'''(0) + 2 = -1$$
$$f'''(0) = -3$$
And we have our third value. Differentiating one more time gives
$$f^{IV}(x) + f''(x) = e^{-x}$$
So again we have
$$f^{IV}(0) + f''(0) =1$$
$$f^{IV}(0) + 1 =1$$
$$f^{IV}(0) =0$$
Using this twice more you'll get
$$f^{V}(0) =2$$ $$f^{VI}(0) =1$$ $$f^{VII}(0) =-3$$
In general the equation is saying that
$$f^{(2n+2)}(0) + f^{(2n)}(0) = 1$$
$$f^{(2n+1)}(0) + f^{(2n-1)}(0) = -1$$
which will allow you to get all values.
A little summary of the already known values:
$f(0) = 0$
$f'(0) = 2$
$f''(0) = 1$
$f'''(0) = -3$
$f^{IV}(0) = 0$
$f^{V}(0) = 2$
$f^{VI}(0) = 1$
$f^{VII}(0) = -3$
Do you see a pattern?
-
Hint: Assume that $$f(x)=\sum\limits_{n\geqslant0}a_n\frac{x^n}{n!},\qquad g(x)=\sum\limits_{n\geqslant0}b_n\frac{x^n}{n!},$$ and that $g=f''+f$. Then $$f''(x)=\sum\limits_{n\geqslant0}a_{n+2}\frac{x^n}{n!},$$ hence, for every $n\geqslant0$, $b_n$ is...
To do: Identify $b_n$ for every $n\geqslant0$ and translate the initial conditions on $f$ in terms of $a_0$ and $a_1$.
-
This is a 2nd order linear non-homogeneous ODE with solution $f(x)=\frac{-1}{2}\cos x + \frac{5}{2} \sin x + \frac{1}{2} e^{-x}$. With the Taylor series for $\sin(x),\cos(x),$ and $e^{-x}$ in hand, you should be able to compute the series for $f$ straightforwardly.
-
Great. Now, how did you obtain your explicit solution? – J. M. Dec 1 '11 at 5:40
@Adam Consider expanding a little if you don't want downvotes. – Pedro Tamaroff Feb 17 '12 at 3:52
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# If x, y, and n are positive integers, is (x/y)^n greater
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If x, y, and n are positive integers, is (x/y)^n greater than 1,000 ?
(1) x = y^3 and n > y.
(2) x > 5y and n > x.
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Re: is (x/y)^n greater than 1,000 ? [#permalink]
### Show Tags
17 Nov 2009, 08:35
SMAbbas wrote:
If x, y, and n are positive integers, is (x/y)^n greater than 1,000 ?
(1) x = y^3 and n > y.
(2) x > 5y and n > x.
B
1. (x/y)^n = (y^3/y)^n= y^2n. we can get values both greater or less than 1000 hence insuff
2. x>5y so if y =1 then least value of x is 6 and n>x>6. ie minimum value we will get is 6^7. hence suff
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Re: is (x/y)^n greater than 1,000 ? [#permalink]
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17 Nov 2009, 08:42
kp1811 wrote:
SMAbbas wrote:
If x, y, and n are positive integers, is (x/y)^n greater than 1,000 ?
(1) x = y^3 and n > y.
(2) x > 5y and n > x.
B
1. (x/y)^n = (y^3/y)^n= y^2n. we can get values both greater or less than 1000 hence insuff
2. x>5y so if y =1 then least value of x is 6 and n>x>6. ie minimum value we will get is 6^7. hence suff
Also getting B
1) let's assume y = 1 then x = 1 and doesn't matter what n is it will be less than 1000
y = 2 x = 8 n can be anything that makes it >1000
insufficient
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Re: is (x/y)^n greater than 1,000 ? [#permalink]
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17 Nov 2009, 09:01
Getting B as well. Similar reasoning as above. Whats the OA?
_________________
Click below to check out some great tips and tricks to help you deal with problems on Remainders!
http://gmatclub.com/forum/compilation-of-tips-and-tricks-to-deal-with-remainders-86714.html#p651942
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1) Translating the English to Math : http://gmatclub.com/forum/word-problems-made-easy-87346.html
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3) 'Distance/Speed/Time' Word Problems Made Easy : http://gmatclub.com/forum/distance-speed-time-word-problems-made-easy-87481.html
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Re: is (x/y)^n greater than 1,000 ? [#permalink]
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17 Nov 2009, 11:28
OA is B
Thanks a lot !
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Re: is (x/y)^n greater than 1,000 ? [#permalink] 17 Nov 2009, 11:28
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# If x, y, and n are positive integers, is (x/y)^n greater
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https://seeyouranswer.com/solved-consider-the-following-reaction-2h2o2aq%E2%86%922h2olo2g-use-the-graph-to-calculate-the-following/
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# [SOLVED] Consider the following reaction: 2H2O2(aq)→2H2O(l)+O2(g) Use the graph to calculate the following.?
Consider the following reaction: 2H2O2(aq)→2H2O(l)+O2(g) Use the graph to calculate the following.?
The instantaneous rate of formation of O2 at 50s.
2. If the initial volume of the H2O2 solution is 1.6 L , what total amount of O2 (in moles) is formed in the first 50 s of reaction?
This is the best answer for Consider the following reaction: 2H2O2(aq)→2H2O(l)+O2(g) Use the graph to calculate the following.?
for the general reaction
.. . A –> B + C
rate can be defined as
.. rate = – d[A] / dt
*********
for this reaction..
.. 2 H2O2 —> 2 H2O + 1 O2
rate of reaction = -d[H2O2] / dt = -slope of the tangent to any point on the curve [H2O2] vs t
so.. do this
.. (1) draw a tangent to the curve at t=50,
.. (2) pick any 2 data points on that tangent line and calc the slope
.. .. . (Y2 – Y1) / (X2 – X1)
.. (3) the rate of the rxn… with respect to H2O2 consumption = – slope of that tangent line
.. (4) from the balanced equation, 2 H2O2 –> 1 O2
.. …..so the rate of O2 production = 1/2 the rate of H2O2 consumption.
rather than guess at the tangent line, I actually, read the [H2O2] concentrations for times 0, 10, 20, 30, etc and plotted time vs ln[H2O2]…. which turned out to be linear with
.. y = -0.029424x – 0.008231…. ..r^2 = 0.999726
so that I could calculate the [H2O2] value at say t = 48sec and t = 52 sec and use those to generate the slope of the tangent line… looked something like this
.. . .t.. .. . ..ln[H2O2].. .. [H2O2]
.. . .48.. .. .-1.42538.. .0.240416
.. . .52.. .. .-1.54348.. .0.213637
and slope = (0.240416 – 0.213637) / (48 – 52) = -0.006695 M H2O2 / sec
so that rate of reaction (in terms of H2O2 consumption = +0.006695 M / sec
and rate of generation of O2 = 0.5x+0.006695 M (H2O2) / sec = 3.35×10^-3 M/sec
i.e… 3.35×10^-3 moles O2 / (L x sec)
********
question #2…
At t = 50 sec
.. [H2O2] = 0.225M
so from t = 0 sec to t = 50sec, the concentration of H2O2 has decreased from 1.0M to 0.225M
agreed?..
If we stated with 1.6L solution, we started with
.. 1.6L x (1mol H2O2 / 1L) = 1.6mol H2O2
now IF…. IF.. all that H2O2 decomposed, we would have generated
.. 1.6mol H2O
and that would add this much water to our solution
.. 1.6mol x (18g / mol) x (0.001L / g) = 0.03L of H2O
in other words.. IF all the H2O2 was consumed, our amount of water present would add to the volume of water we started with but NOT significantly.. So… this is where I going with this.
.. “we can ignore the volume of water produced by the rxn”
that makes things MUCH easier
.. initial moles H2O2 = 1.6mol H2O2
.. final moles H2O2 = 1.6L x (0.225mol / L) = 0.36mol H2O2
and
.. moles H2O2 consumed = 1.6 – 0.36 = 1.24mol H2O2
and since 2 H2O2 –> 1 O2
.. moles O2 produced = 0.5 x 1.24 = 0.62 mol O2 produced… (2 sig fig)
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# LeetCode – Populating Next Right Pointers in Each Node Solution
### 题目：
You are given a perfect binary tree where all leaves are on the same level, and every parent has two children. The binary tree has the following definition:
```struct Node {
int val;
Node *left;
Node *right;
Node *next;
}
```
Populate each next pointer to point to its next right node. If there is no next right node, the next pointer should be set to `NULL`.
Initially, all next pointers are set to `NULL`.
Follow up:
• You may only use constant extra space.
• Recursive approach is fine, you may assume implicit stack space does not count as extra space for this problem.
Example 1:
Input:
``` root = [1,2,3,4,5,6,7]
```
Output:
``` [1,#,2,3,#,4,5,6,7,#]
```
Explanation:
```Given the above perfect binary tree (Figure A), your function should populate each next pointer to point to its next right node, just like in Figure B. The serialized output is in level order as connected by the next pointers, with '#' signifying the end of each level.
```
Constraints:
• The number of nodes in the given tree is less than `4096`.
• `-1000 <= node.val <= 1000`
### 实现：
由于是完全二叉树，所以有左，右子树同时存在，左子树next节点为对应的右子树； 右子树的next节点为其父节点的next节点的左子树 。 父节点next为空，则为空。
• 递归方式
``````
// Definition for a Node.
class Node {
public:
int val;
Node* left;
Node* right;
Node* next;
Node() : val(0), left(NULL), right(NULL), next(NULL) {}
Node(int _val) : val(_val), left(NULL), right(NULL), next(NULL) {}
Node(int _val, Node* _left, Node* _right, Node* _next)
: val(_val), left(_left), right(_right), next(_next) {}
};
*/
/* Recursive
class Solution {
public:
Node* connect(Node* root) {
//check and set next
if (root==NULL) {
return NULL;
}
if (root->left) {
root ->left -> next = root->right;
}
if (root->right) {
root ->right -> next = root->next ? root->next->left : NULL;
}
//left
connect(root->left);
//right
connect(root->right);
return root;
}
};
``````
• 水平遍历, 由于是完全二叉树， 水平遍历取出的节点的next节点为队列下一个节点， 对于最后一个节点 size-1进行特殊处理即可。
``````class Solution {
public:
Node* connect(Node* root) {
if (root == NULL) {
return NULL;
}
queue<Node *> q;
q.push(root);
while (!q.empty()) {
size_t size = q.size();
for (size_t i=0; i<size; i++){
Node* node = q.front();
q.pop();
if (i != size-1)
node->next = q.front();
else
node->next = NULL;
if (node->left) {
q.push(node->left);
}
if (node->right) {
q.push(node->right);
}
}
}
return root;
}
};``````
• 其他解法： 思路是一样的是通过两层循环实现，使用两个指针，一个指针pre指向每层的第一个节点， 通过cur节点进行遍历， 其左子树next=cur->right, 其右子树next=cur->next->left。 完成本次遍历，重新设置每层第一个节点left，进行下一层遍历。
``````
class Solution {
public:
Node* connect(Node* root) {
if (root == NULL){
return root;
}
Node* pre = root ;
Node* cur = NULL;
while (pre->left){
cur = pre;
while (cur) {
cur->left->next = cur->right;
if (cur->next) {
cur->right->next = cur->next->left;
}
cur = cur->next;
}
pre = pre->left;
}
return root;
}
};``````
Comments are closed.
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# Week 5 Maths
Hi Year 6,
I’ve been watching your ‘Tests of Divisibility’ videos on BHTV and can see we are getting much better at talking about Maths; well done!
BRONZE:
‘The number 840 is divisible by 2, 3, 5, 6, 7, 9 and 10’
Is this statement true? Use the information in the videos to help you prove it. Write a few sentences in your homework book for each of these possible factors of 840.
SILVER:
Now that you are fantastic at talking about Maths, I would like you to write about it. Write a paragraph, including examples, for the Tests of Divisibility for either 6, 7 or 9. Use the BHTV videos to help you.
GOLD:
We still need to add some videos for the Tests of Divisibility for 4, 8, 11 and 12. Please choose two of these Tests of Divisibility and write a paragraph for each with examples. Remember to use excellent Maths vocabulary including the words ‘factor’ and ‘multiple’.
## 44 thoughts on “Week 5 Maths”
1. Nasim Sherif
silver:
Test of divisibility for 6.
rule of 6
the test of divisibility for 6 is that you think of a digit and add with other number for example 3+3=6 or you can do 4+2=6, but if you make a mistake on adding your hole divisibility will be wrong.
by Nasim Sherif :)
2. Rayan Mohamed
SILVER: Test of divisibility for 6.
If a number is divisible by 2 and 3 then it will also be divisible by 6. For example 114, I know that this is divisible by 2 because 114 is an even number and 4 is in the 2 times table because 1+1+4=6 and 6 is a multiple of therefore it is divisible b
3. Abena Daffour
The test of divisible for 10:)
The test of divisibility for 10 is the units all ways have to be 0.it does not matter if it is 3 tens so for example,500 is a multiple because of its units(which is 0).632 is not a multiple because of its units.So I hope you like my explanation :).
4. Junior Kom Yawli
The test of divisibility for 8.The last three digits must be divisible by 8 for an example 1412 is not because 412 is not a multiple of 8.On the other hand 12816 is divisible by the 8s and 816 is a multiple of 8.
The test of divisibility for 4.The last two digits must be a multiple of 4 for example 940 is divisible by 4 and 40 is a multiple of 4.However 543 is not divisible by 4 and 43 is not a multiple of 4
5. Musa Ahmed
test of divasablity for 4 times table
the test of divasability for 4 times table is that it has to be in the for times table and deviasable by 4.Also it has to end in 2,4,6,8,0 because they are even and are at the the end of each number in the four times table.for example 24 is in the four times table because it ends in 4.
6. Junior Kom Yawli
GOLD
The test of divisibility for 8.The last three digits must be divisible by 8 for an example 1412 is not because 412 is not a multiple of 8.On the other hand 12816 is divisible by the 8s and 816 is a multiple of 8.
The test of divisibility for 4.The last two digits must be a multiple of 4 for example 940 is divisible by 4 and 40 is a multiple of 4.However 543 is not divisible by 4 and 43 is not a multiple of 4.
7. Humayun Jamshed
tests of divisibility for the twelve times table are that the intagets have to be divisible by 3 and 4 also. For,example 848. 6 + 4=10 +8 = 18
18 divided. the last two digits are 48 so you do 48 divided by 4 = 12.
test of divisibility for 6:
the number is divisible by 2 or 3 e.g.
114(it is even,1+1+4=6 and 6 divided by 3=2) yes
308(it is even, but 3+0+8=11 and 11 divided by 3= 3 and 2 thirds)NO ;)
9. Janani Balamurali
Gold
The test of divisibility for the 8 times tables is when you look at the last three digits. The last three digits in the number has to be a multiple of the eights REMEMBER: you can only use this rule for numbers, which have more than 3 digits.
For example 1432 the last digits in the number is 1432{underlined numbers}.The methods you can use chunking and bus stop to work this out.
432 /8 =54
400/8=50 32/8=4
50+4=54
10. Ahsan Khan
Silver:
Divisibility Test for 9:
In order for a number to be divisible by 9 The digits must add up to 9.
An example is 45 it has to add up to 9 so 4+5=9 the numer 45 is divisible by 9 this ha s completed the divisibility for 9.
11. Patrycja Pawlowska
It was meant to say 6 is a multiple of 3. :)
12. Patrycja Pawlowska
SILVER: Test of divisibility for 6.
If a number is divisible by 2 and 3 then it will also be divisible by 6. For example 114, I know that this is divisible by 2 because 114 is an even number and 4 is in the 2 times table ; 1+1+4=6. 6 is a multiple of therefore it is divisible by 6. :)
By Patrycja :)
13. Aishah Hussain
Tests of divisibility for the 8 times table
To work this out you must look at the last three digits of the number. If this three digit number is a multiple of 8, the number is divisible by 8. For example the number 56,214. The last three digits in this number are 214. 214 divided by 8 is 26 remainder 6. This means that 56,214 is not divisible by 8. Another example is 912,624. The last three digits are 624. 624 divided by 8 is 78 which means 912,624 is divisible by 8.
Test of divisibility for the 12 times table
To work this out you have to know whether the number is divisible by 2,3 AND 4. If this is true, the number is divisible by 12. For example the number 1224. The number is divisible by 2 because its units are 4, it is a multiple of three because the sum of the digits is 9 (1+2+2+4=9) and it is divisible by 4 because the last two digits are 24 which is a multiple of 4. Therefore, 1224 is divisible by 12.
Aishah Hussain :)
14. Ismaeel Dean
silver
test of divisibility for 6
if a number is divisible by 6 then the numbers need to be in the 3 or 2 times table because they are all divisible by eachother. Also the end number has to end in the 3 or 2 times table.
by Ismaeel
you have to add 2 because the 2 is in the times tabele it is 2,4,6,8,10,12,14,16,18,20 is in the times table.
16. Sarah Kauser
Rule of divisibility for 6
Silver challenge
The rules of divisibility for 6 are that the number always has to be divisible by 2 or 3,or in the 2 or 3 times table,this will prove it is divisible by 6.For example if you have 114,it is even and we already know it is divisible by 2 but not 3.However 1 add 1 add 4 equals 6 so if you do 6 divided by 3 it will equal 2 this means it is in the 3 times table and it is divisible by 3.On the other hand,308 is even,but 3 add 0 add 8 equals 11 divided by 3 equals 3 so no.That is the rule of divisibility for 6.
BY SARAH ;)
17. Zakariya Hasan
Test of Divisibility:
Gold
4
the test of divisibility for the 4 times tables,the last two digits must be divisable by 4,for example:752 is divisable by 4 because 4 X 13 = 52.
12
The test of divisibility for the 12 times tables,the number must be divisable by 3 and 4 for example: 144 is divisable by 12 because 1+4+4 = 9 and 9 is in the 3 times tables and the last two digits are divisable by 4 because 4 X 11 = 44.
18. Keeleigh Hoban
egzample:
is 4829 divisible by 3 ?
4+8+2+9=23
is 23 in the three times table?
no 23 is not in the 3 x table :)
19. Salan Sharif
Welcome to year 6
Bronze challenge rule 2.
16 is divisible by 2 because 2 it is in the 2 time table
20. Aamina Habib
SILVER:
SIX
The prime factors of 6 are 2 and 3,so for a number to be divisible by six it must also be divisible by 2 and 3.
23,908
This nuber is even therefore it is divisible by 2. 2+3+9+0+8=22, 3 does not divide evenly into this number so this number is not divisible by 3.
154,608
This number is even therefore it is divisible by . 1+5+4+6+0+8=24 24 is divisible by 3 because 3×8=24 which means this number is divisible by 6.
21. Keeleigh Hoban
silver ;)
test of divisibility for 6
if a number is divisible by 6 then the numbers need to be in the 3 or 2 times table because they are all divisible by eachother. Also the end number needs to be even because 6 is a even number.
by keeleigh :)
22. Ayub Khan
SILVER
Rule for 9
The test of divisibility for 9 is that all the numbers that are divisible by 9 have to be in the 9 times table. For example 81 is divisible by 9 because 9 times 9 = 81. On the other hand, 76 is not divisible by 9 because if you check in the 9 times table you will not see 76.
By Ayub Khan:)
23. Bilal Mujahid
silver
test of divisibility for 6
if the number is divisible by 3 which is when all the numbers are added together in units equals a number in the 3 time table is divisible by 3.aswell as if it is divisible by 2 which is when the last digit Is even it is divisible by 6
24. Hamzah Khan
Gold:
The test of divisibility for the 11 times tables.
If you sum up every second digit of a number and if the answer is either 11 or 0 then it is divisible by 11. For example 918,082 is divisible by11 because 9-1=8 and add 8 is 16 and 8-2 is 6 and 16 add 6 =22.22 = 2×11.
25. Muryam Khalid
Silver:
How to find the test of divisibility for 7:
To find the test of divisibility for 7 you need to look at the last digit which is the units.You then need to double that number and then subtract it from the rest of the number,however you do not include the unots.For example,if we had 172 we will have to look at the 2 because it is the units {last digit}.Then we will have to subtract it from 17.17-2=15.15 is not in the 7 x’s table and if we try to divide it we will end up with a remainder.A number that does work however is 777.Double 7=14.77-14=63.63 is a multiple of 7.
Thank you for reading:) By Muryam.
• Muryam Khalid
*units
26. Florida Malaj
Bronze challenge>RULE FOR 2.
8+4+0=12.1+2=3>840 is not divisible by 2 because 1+2=3 and 3 is not in the 2 times table or divisible by 2.
RULE FOR 3>is that you have to look at the sum of the digits.840 is not divisible by 3 because 40 is not in the 3 times or divisible by 3.
RULE FOR 5>is that if the number ends in 0 or 5.840 is divisible by 5 because it does end in 0.
RULE FOR 10>is if it ends in 0.840 is divisible by 10 because it ends in 0.
27. Ojasvi Nag
Gold
Test of Divisibility for the number 8:
To work out if a large number is divisible by eight, you should know that the last three digits have to be divisible by 8. If the last three digits are divisible by eight, then the whole number is a multiple of eight! For example, “Is 38,864 divisible by 8?” It sure is! This is because the last three digits of the number (864) are divisible by eight and equal 108. This implies that, the number, 38,864 is divisible by eight! However, if this rule does not work for a number, that means the number is not a multiple of eight. For instance, ” Does 7,966 divide by 8?” Of course not! That’s because 966 (the three last digits) are not divisible by 8 and the answer comes in decimal or comes with a remainder. It implies that the whole number is not a multiple of 8!
28. Ojasvi Nag
Gold
Test of Divisibility for the number 8:
To work out if a large number is divisible by eight, you should know that the last three digits have to be divisible by 8. If the last three digits are divisible by eight, then the whole number is a multiple of eight! For example, “Is 38,864 divisible by 8?” It sure is! This is because the last three digits of the number (864) are divisible by eight and equal 108. This implies that, the number, 38,864 is divisible by eight! However, if this rule does not work for a number, that means the number is not a multiple of eight. For instance, ” Does 7,966 divide by 8?” Of course not! That’s because 966 (the three last digits) are not divisible by 8 and the answer comes in decimal or comes with a remainder. It implies that the whole number is not a multiple of 8!
• Ojasvi Nag
Test of divisibility for the number 12:
If a number is exactly divisible by 3 and 4, then that same number is a multiple of 12! For example, 336 is divisible by 3 (112) and 4 (84), consequently meaning that 336 is a factor of 12! Although, for some numbers (such as 298) this rule does not work. 298 ➗ 4 = 74.5 1nd 298 ➗ 3 = 99.3. This points to the fact that 298 is not divisible by 12!
Bye!!
• Ojasvi Nag
Test of divisibility for the number 12:
If a number is exactly divisible by 3 and 4, then that same number is a multiple of 12! For example, 336 is divisible by 3 (112) and 4 (84), consequently meaning that 336 is a factor of 12! Although, for some numbers (such as 298) this rule does not work. 298 ➗ 4 = 74.5 1nd 298 ➗ 3 = 99.3. This points to the fact that 298 is not divisible by 12!
Bye!!
• Ojasvi Nag
Where it says 1nd, that is supposed to be 1 and.
29. Aaisha Raja
silver:
rule for 9:
to find the test of divisibility for 9 you need to add all the factors together for example:
198 1+9+8=18
after you have done that find out if the number is in the 9 x table 18 is in the 9 x table because 18 divided by 9 =2.But if you had the number 215 it would not be divisible by 9 because 2+1+5=8 6 is not in the 9 x table because the first number of the 9 x table is 9.That is how you find the test of divisibility for the 9 x table. :)
• Omar Hussain
Silver challenge…
The test of the divisibility by 9 is that all the numbers in the 9 times table are divisible by 9. You have to look at the last 2 digits so you know if the number is in the 9 times table. For example 279 is divisible by 9 because 2+7+9=18 and 18 is in the 9 times table.
The test of divisibility by 7 is that all of the numbers in the 7 times table are divisible by 7. You have to look at the tens and unit to decide if the number is or not divisible by 7. For example 4791 is divisible by 7 because 4+7+9+1=21 and 21 is in the 7 times table
By Omar Hussain 6S :) :)
30. Aaisha Raja
silver:
rule for 9:
to find the test of divisibility for 9 you need to add all the factors together for example:
198 1+9+8=18
after you have done that find out if the number is in the 9 x table 18 is in the 9 x table because 18 divided by 9 =2.But if you had the number 215 it would not be divisible by 9 because 2+1+5=8 6 is not in the 9 x table because the first number of the 9 x table is 9.
31. Maryam Hameed
silver
rule for 9
To find out if a number is divisible by nine you need to add all the digits together.45 is divisible by nine because it is in the nine times table and if you add all the digits together like this 4+5=9 it equals nine.725 0n the other hand, is not divisible by nine because when you add all the digits together it equals 7+2+5=14. 14 is not a multiple of nine.
32. Mohamed El Safi
Gold:
The test of divisibility for the 11 times table.If you sum every second digits and if the answer is 11 or 0 it is divisible by 11.For example: 3542 is divisible by 11 because 3(the first number)add 4(the third number)=7 and 5(the second number number)add 2(the forth number)=7 so 7-7=0 so it is divisible by 11 because it ends with either 0 or 11.7271 is divisible by 11 because 7(the first number)add 7(the third number)=14 and 2(the second number number)add 1(the forth number)=3 so 14-3=11 so it is divisible by 11 because it ends with either 0 or 11.6219 is not divisible by 11 because 6(the first number)add 1(the third number)=7 and 2(the second number number)add 9(the forth number)=11 so 11-7=4 because it doesn’t equal 11 or 0 so it is not divisible by 11.
Divisibility for 8 times table.The last three digits must be divisible by 8.For example: 109816 is because 816(the last three digits) dived by 8 equal 102 and it is a whole number.216302 is not because 302(the last three digits) divided by 8 equal 37.75 and it is not a whole number.
:)(heart)(hug)(cool)
:)
SILVER:
RULE FOR SEVEN..
If you want to know the divisibility test for seven then read on and find out. First you look at the last number and double it. then you take that number away from the number that you doubled, and if the answer you got is divisible by seven then the number is also divisible by seven.
For Example: 777, double 7 =14 and 77-14=63, and 63 is divisible by 7. So that means 777 is divisible by 7.
Thank you for reading and I hope you have learnt something
SILVER :):
rule for 6
If you doing a test of divisibility for 6 it has to be in the 3 and 2x time table so all the numbers should be even. For example 34 is even but is it in the 3x times table? If you want to find it out use the test divisibility of 3 so 34 is not a multiple of 3. I know this because I added it so for example;3+4=7. So it not a multiple of 6.
FROM SABIREN…. :) :)
37. Bilal Hussain
SILVER:
RULE FOR NINE
For a number to be divisible by 9 you have to add the digits and if they equal 9 then it is divisible by 9 if not then they are not divisible by 9.FOR example 18 is divisible by 9 because 1+8=9 but 73 ain’t because 7+3=10. That is how you find out if a number is divisible by 9. ;)
• J McCabe
Thanks Bilal. You’re the first one! :)
Remember, “ain’t” isn’t a word though! We want to sound like grown up Mathematicians when we write about Maths.
• J McCabe
Also, what did you think about the videos?
Were they clear and easy to follow?
Do you think someone in a younger group or someone’s parents could follow them?
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## Andre sometimes mows Lawns on the weekend to make extra money. Two weeks ago he mowed a neighbor’s lawn for half an hour and earned \$10. Las
Question
Andre sometimes mows Lawns on the weekend to make extra money. Two weeks ago he mowed a neighbor’s lawn for half an hour and earned \$10. Last week he mowed at his uncle’s lawn for 3 1/2 hours and earned \$30. This week he mowed the lawn of a community center for 2 hours and \$30 which job paid better than others
in progress 0
2 months 2021-10-11T11:27:39+00:00 1 Answer 0 views 0
1. Answer: he had a better pay when he mowed his neighbor’s lawn.
Step-by-step explanation:
Two weeks ago he mowed a neighbor’s lawn for half an hour and earned \$10. This means that the amount that he would earn per hour is
10/0.5 = \$20
Last week he mowed at his uncle’s lawn for 3 1/2 hours and earned \$30. This means that the amount he would earn per hour is
30/3.5 = \$8.6
This week he mowed the lawn of a community center for 2 hours and \$30. This means that the amount he would earn per hour is
30/2 = \$15
Therefore, he had a better pay when he mowed his neighbor’s lawn.
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# 1955 AHSME Problems/Problem 1
## Problem
Which one of the following is not equivalent to $0.000000375$?
$\textbf{(A)}\ 3.75\times 10^{-7}\qquad\textbf{(B)}\ 3\frac{3}{4}\times 10^{-7}\qquad\textbf{(C)}\ 375\times 10^{-9}\\ \textbf{(D)}\ \frac{3}{8}\times 10^{-7}\qquad\textbf{(E)}\ \frac{3}{80000000}$
## Solution
First of all, $0.000000375 = 3.75 \times 10^{-7}$ in scientific notation. This eliminates $\textbf{(A)}\ 3.75\times 10^{-7}$, $\textbf{(B)}\ 3\frac{3}{4}\times 10^{-7}$, and $\textbf{(C)}\ 375\times 10^{-9}$ immediately. $\textbf{(E)}\ \frac{3}{80000000}$ is a bit harder, but it can be rewritten as $\frac{3}{8} \cdot 10^{-7}$, which is the same as $3.75 \times 10^{-7}$, so the answer is \boxed{(A) 3.75\times 10^{-7}}
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## Subtracting a number is the same as adding its opposite. Example: -13 – 7 = -13 + (- 7) = -20 Example: 17- 8 = 17 + (-8) = 12 Example: -13 – (- 7) = -13 + 7 = -6 Example: 17 – (-8) = 17 + 8 = 25 Example: 67 – (- 45) = 67 + 45 = 112 Example: -17 – (-30) = -17 + 8 = -13
Mavis Marble uses her new Mitra Card ( the card that does the math for you ) to buy material and yarn. She has been using it for 4 months. She just got a letter in the mail saying her account is now at -355. "This can't be. My card says I have a credit of \$240." Mavis will need to compare her Mitra card charges with her payments. Find the correct balance for her Mitra account.
Mitra Charges Payments Balance -80 sewing material -(-60) 60 -120 sewing material -(-60) 120 -60 sewing material -(-60) 180 -95 sewing material -(-60) 240
Find the difference and input the correct balance.
Alpha | Go on to the Interactive Activity at the Red Box Division
writer: Kevin Gallagher/mC
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https://17calculus.com/precalculus/complex-numbers/rationalizing/
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17Calculus Precalculus - Rationalizing Complex Numbers
17Calculus
Rationalizing Complex Numbers
Remember with square roots, you would perform an operation called rationalizing. This usually occurs when you want to remove a square root from the denominator of a fraction. The idea is to multiply by something called the conjugate. This is what it looks like. If you have $$1+3\sqrt{2}$$, the conjugate is $$1-3\sqrt{2}$$, i.e. you change the sign in the middle. The beauty of this that when you you multiply these together, the square root disappears.
Example: $$(1+3\sqrt{2})(1-3\sqrt{2}) =$$ $$1(1-3\sqrt{2}) +3\sqrt{2}(1-3\sqrt{2}) =$$ $$1-3\sqrt{2} + 3\sqrt{2} -9(2) = -17$$
Notice in the last example, that the square root of two is gone. That is the point of rationalizing. We can do the same thing with complex numbers.
Example: $$(1+3i)(1-3i) =$$ $$1(1-3i) +3i(1-3i) =$$ $$1-3i + 3i -9i^2 = 1-9(-1) = 10$$
When using the material on this site, check with your instructor to see what they require. Their requirements come first, so make sure your notation and work follow their specifications.
DISCLAIMER - 17Calculus owners and contributors are not responsible for how the material, videos, practice problems, exams, links or anything on this site are used or how they affect the grades or projects of any individual or organization. We have worked, to the best of our ability, to ensure accurate and correct information on each page and solutions to practice problems and exams. However, we do not guarantee 100% accuracy. It is each individual's responsibility to verify correctness and to determine what different instructors and organizations expect. How each person chooses to use the material on this site is up to that person as well as the responsibility for how it impacts grades, projects and understanding of calculus, math or any other subject. In short, use this site wisely by questioning and verifying everything. If you see something that is incorrect, contact us right away so that we can correct it.
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https://dsp.stackexchange.com/questions/84176/number-of-periods-of-signal-required-when-doing-an-fft
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# Number of periods of signal required when doing an FFT
I'm using numpy.fft in python to compute Fast Fourier Transforms. In particular, I'm using rfft as I have a real signal and don't need negative frequencies.
My question is: my signal has 184320 points currently and encompasses multiple periods. Should I be truncating the signal after a certain number of periods for best results? I know the sample rate in which the data was collected. What's sparking this question is that applying various windows is changing the amplitudes at each frequency of interest quite significantly, and I'm wondering if perhaps I should be truncating my signal.
• what's the sampling rate?
– Jdip
Aug 16, 2022 at 19:10
• @Jdip, my sampling rate is 15360 Hz
– Bern
Aug 16, 2022 at 19:26
• Read my edited answer. If you still have trouble, you can post the frequencies you’re trying to get an amplitude for and I’ll help you out further ;)
– Jdip
Aug 16, 2022 at 19:31
• Looking to compare my results so the frequencies I'm interested in are from 60 - 2940 (in increment steps of 60)
– Bern
Aug 16, 2022 at 19:55
• Ok so you need frequency bins $k=f_n \cdot nfft/15360$ at $f_n = n \cdot 60$ so $nfft = 256$ is the smallest chunk of your data you can use (assuming your signal is stationary, I.e the frequencies don’t change over time). You can also just use 15360 samples chunk, I.e 1s of your data, giving you exactly 1hz per bin
– Jdip
Aug 16, 2022 at 20:10
Choose an appropriate FFT length (for your signal, such as 2^18), and scale your FFT result by $$1/sum(\texttt{window})$$ (for no window, that would be $$1/N$$ with $$N$$ your input signal length).
EDIT That's if you are set on using the FFT as an amplitude estimation tool. As pointed out by @OverLordGoldDragon, there are better tools for that.
Each frequency bin $$k$$ has center frequency $$f_{k_{(\texttt{Hz})}} = kdF_{(\texttt{Hz})}$$, with $$dF_{(\texttt{Hz})} = f_s/\texttt{Nfft}$$ the spectral precision, $$\texttt{Nfft}$$ the length of the DFT, i.e the length of your input sequence (padded or not), and $$f_s$$ the sampling frequency.
Here's an example: let $$x(t)$$ be a $$0.5s$$ long ($$\texttt{Nfft} = 22050$$) signal sampled at $$44100 \texttt{Hz}$$. Its DFT, $$X[k]$$, has precision $$dF = 2 \texttt{Hz}$$ with no padding. That means that component at $$2000 \texttt{Hz}$$ would exactly fall at bin $$k = f / dF = f \cdot \texttt{Nfft}/f_s = 2000 \cdot 22050 / 44100 = 1000$$
But a component at $$2001 \texttt{Hz}$$ would fall between bins 1000 and 1001 (because $$f/dF = 1000.5$$ so the fft output will spread between bins $$1000$$ and $$1001$$).
By padding your signal to twice its original length, you can double the signal length $$\texttt{Nfft}$$, effectively doubling the frequency precision $$dF$$ to $$1\texttt{Hz}$$, and a $$2000 \texttt{Hz}$$ component now falls at the center of bin $$k = f \cdot \texttt{Nfft} /f_s = 2000 \cdot 44100 / 44100 = 2000$$ and a component at $$2001\texttt{Hz}$$ falls at the center of bin $$2001$$ by the same calculations.
All this to say, if you know the frequencies of interest, you can deduce the exact frequency precision $$dF$$ you need, from there you can deduce the input length you need by doing $$\texttt{Nfft} = f_s/dF$$ and use that to guide how much padding/truncating you need to apply.
• Thanks for the insight. I'm working on a project with the requirement that it be done using FFT. In the future I'll definitely check out the tools proposed. The signal that I'm dealing with can be considered as a bunch of sine waves added together (with a slight phase) so that the resulting waveform isn't a perfect sine wave. I read online that numpy fft doesn't require it to be an integer power of 2. Would you recommend padding with zeros to make it 2^18 length?
– Bern
Aug 16, 2022 at 17:55
• You don’t need to pad, no. You might want to pad however to get better frequency precision, which in your case might be useful to get exact amplitude measurements. The power of 2 length requirement is only for efficiency, so unless you’re worried about that, no need.
– Jdip
Aug 16, 2022 at 17:57
• good to know. That isn't much of a concern but I'll keep it in the back of my mind for the future. Going back to the idea of truncating the signal, do you have any more guidance on how to properly do that? I'm currently unsure on what point to do so, which of course is changing quite significantly the output amplitudes.
– Bern
Aug 16, 2022 at 18:03
• Why do you want to truncate your signal? And it shouldn’t change the signal amplitude if you make sure to scale your result like I mentioned…
– Jdip
Aug 16, 2022 at 18:04
• I'm attempting to window the my spectrum to get "cleaner" results. One of the things I'm noticing is that the application of windows changes the output amplitudes at frequencies of interest, quite significantly. It was recommended to me to try looking at only a chunk of my signal to see if that helps at all. I'm pretty new to this stuff so figured I'd give it a shot, but if you have any suggestions for windowing, I'd be grateful for those too!
– Bern
Aug 16, 2022 at 18:07
Title: no such thing. The DFT (which is what FFT computes) doesn't require anything beyond the input being finite in length and values.
Body: keep truncating until the FFT is perfect impulses (just peaks, no leaky stuff). To the DFT, the input is a perfect sine if there's an integer number of cycles of it. Note however, truncation will change the peak's location, and its value, which need accounting for: frequency normalization for former, amplitude for latter via *= N_orig / N_trunc (for pure sines only). If the signal is measured rather than synthetically generated, then you'll most likely never get a perfect impulse.
If the goal is reliable amplitude estimation, FFT is a suboptimal tool; I recommend time-frequency analysis (CWT, STFT), which also don't suffer from the 'measured' problem. Details here under "Modulation Model vs Fourier Transform".
• //"The DFT (which is what FFT computes) doesn't require anything beyond the input being finite in length and values.// --- That's true, but because of the nature of the basis functions of the DFT, certain properties are inherent. Aug 16, 2022 at 22:43
Number of periods of signal required when doing an FFT: $$1$$
• Of course Robert's answer is as right and short as it could ever be. To add to my previous comments, in your case, your lowest frequency of interest being 60Hz, at a sampling rate $f_s = 15360$, a full period would be 1/60 = 0.0167s, which is exactly 256 samples.
– Jdip
Aug 16, 2022 at 22:26
• yup. never said anything about sample rate. just something about the inherent nature of the DFT. Aug 16, 2022 at 22:36
• "Lengths of sides required to make a rectangle must sum to its perimeter". Aug 16, 2022 at 22:57
• I wish there was a +🤣 option for voting.
– Peter K.
Aug 17, 2022 at 0:02
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Implicant
In Boolean logic, an implicant is a "covering" (sum term or product term) of one or more minterms in a sum of products (or maxterms in product of sums) of a Boolean function. Formally, a product term P in a sum of products is an implicant of the Boolean function F if P implies F. More precisely:
P implies F (and thus is an implicant of F) if F also takes the value 1 whenever P equals 1.
where
This means that P ${\displaystyle \leq }$ F with respect to the natural ordering of the Boolean space. For instance, the function
${\displaystyle f(x,y,z,w)=xy+yz+w}$
is implied by ${\displaystyle xy}$, by ${\displaystyle xyz}$, by ${\displaystyle xyzw}$, by ${\displaystyle w}$ and many others; these are the implicants of ${\displaystyle f}$.
Prime implicant
A prime implicant of a function is an implicant that cannot be covered by a more general, (more reduced - meaning with fewer literals) implicant. W. V. Quine defined a prime implicant of F to be an implicant that is minimal - that is, the removal of any literal from P results in a non-implicant for F. Essential prime implicants (aka core prime implicants) are prime implicants that cover an output of the function that no combination of other prime implicants is able to cover.
Using the example above, one can easily see that while ${\displaystyle xy}$ (and others) is a prime implicant, ${\displaystyle xyz}$ and ${\displaystyle xyzw}$ are not. From the latter, multiple literals can be removed to make it prime:
• ${\displaystyle x}$, ${\displaystyle y}$ and ${\displaystyle z}$ can be removed, yielding ${\displaystyle w}$.
• Alternatively, ${\displaystyle z}$ and ${\displaystyle w}$ can be removed, yielding ${\displaystyle xy}$.
• Finally, ${\displaystyle x}$ and ${\displaystyle w}$ can be removed, yielding ${\displaystyle yz}$.
The process of removing literals from a Boolean term is called expanding the term. Expanding by one literal doubles the number of input combinations for which the term is true (in binary Boolean algebra). Using the example function above, we may expand ${\displaystyle xyz}$ to ${\displaystyle xy}$ or to ${\displaystyle yz}$ without changing the cover of ${\displaystyle f}$.[1]
The sum of all prime implicants of a Boolean function is called its complete sum, minimal covering sum, or Blake canonical form.
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## Thomas' Calculus 13th Edition
$\frac{31\pi}{6}$
$V=\int^{2\pi}_0 \int^{\pi/2}_0 \int^2_ {cos\phi} p^2 sin \phi dp d\phi d\theta$ =$\frac{1}{3} \int^{2\pi}_0 \int^{\pi/2}_0 (8-cos^3\phi) sin\phi$ $d\phi$ $d\theta$ =$\frac{1}{3}\int^{2\pi}_0 [-8cos\phi+\frac{cos^4\phi}{4}]^{\pi/2}_0 d\theta$ =$\frac{1}{3}\int^{2\pi}_0 (8-\frac{1}{4})d\theta$ =$(\frac{31}{12}(2\pi))$ =$\frac{31\pi}{6}$
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# Why does a piece of iron sink while an iron boat doesn't?
2020-08-21 09:35:47
Buoyancy Law: the object immersed in the liquid is subject to upward buoyancy, and the buoyancy is equal to the gravity of the object to displace the liquid.
That is, f float = g row = m row g = ρ liquid GV row g row: drain the gravity of liquid
M row: drain the mass of liquid
ρ liquid: density of liquid
V row: drain the volume of liquid
(i.e. volume immersed in liquid)
If it is a solid iron block, it must sink. Because the ship's interior is empty, the ship's displacement in the water will be very large. When the gravity of the discharged water is equal to the weight of the ship, the ship will float
It's a matter of buoyancy, because steel is denser than water, and it's solid, so it sinks. However, the ship is not solid, and the volume of water discharged by it is much larger than that of steel used, so row g will be larger than material g, so the ship will not sink
This has something to do with the volume of the object
The same amount of steel
When the volume of an object is large and its density is small, it will float up,
When the volume is small and the density is high, it will sink
Ships are made of steel. The density of steel is higher than that of water. If it is solid, it will sink to the bottom of the water. But if it is hollow, its average density will be less than that of water, so that it will float on the water surface.
Source: Jiangmen shipyard http://www.zhongyucy.com/
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+0
# geometry
0
88
1
What is the total area in blue knowing that the area of the large triangle is 1?
May 19, 2020
#1
+21953
+1
I believe that the first (largest) blue areas of the triangle form 3/4ths of the complete triangle.
The next blue area forms 3/4ths of 1/4th of 1/4th of the triangle.
The next blue area forms 3/4ths of 1/4th of 1/4th of 1/4th of 1/4th of the triangle.
Etc.
(3/4) + (3/4)(1/4)2 + (3/4)(1/4)4 + (3/4)(1/4)6 + ...
This forms a geometric sequence with first term = 3/4 and with common ratio = 1/16:
Sum = (3/4) / [ 1 - 1/16 ] = (3/4) / (15/16) = 4/5.
May 19, 2020
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You must complete Depreciation of Fixed Asset I to unlock this Lesson.
METHODS OF DEPRECIATION
THE DIMINISHING OR REDUCING BALANCE METHOD OF DEPRECIATION
In this method, a fixed percentage is written off the diminishing balance of the asset yearly. The total depreciation is spread over the anticipated useful life of the asset by annual installments of diminishing amount. This method of depreciation is advantageous because depreciation is more scientifically provided for.
Depreciation rate is computed using this formula: $$\sqrt[1 -N]{\frac{\text{residual value}}{cost}}$$
Where N = the number of years (useful life of the asset)
Example: A motor van cost N10,000 in 2005. It has an expected life span of four years and the estimated residual value of N256. What is the depreciation rate?
SOLUTION
Rate of Depreciation, $$R = \sqrt[1 -N]{\frac{\text{residual value}}{cost}}$$
$$R = \sqrt[1 -4]{\frac{₦250}{₦10,000}}$$
The depreciation calculation applied to each of four years of use would be:
DEPRECIATION SCHEDULE
Accounting entries (with modern method)
The following accounts will be prepared:
1. Motor van account
2. Provision for depreciation account
3. Profit and loss account
4. Balance sheet extract
MOTOR VAN ACCOUNT
PROVISION FOR DEPRECIATION – MOTOR VAN
PROFIT AND LOSS ACCOUNT
BALANCE SHEET
EVALUATION:
1. Define diminishing method of depreciation
2. Mention four accounts to be prepared in the above method
DISPOSAL OF ASSETS
Fixed assets can be sold in the course of the business.
Lesson tags: Financial Accounting Lesson Notes, Financial Accounting Objective Questions, SS1 Financial Accounting, SS1 Financial Accounting Evaluation Questions, SS1 Financial Accounting Evaluation Questions Third Term, SS1 Financial Accounting Objective Questions, SS1 Financial Accounting Objective Questions Third Term, SS1 Financial Accounting Third Term
Back to: Financial Accounting – SS1 > Third Term
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Topic: layer logic: a new dimension to logic?
Replies: 12 Last Post: May 26, 2013 10:56 AM
Messages: [ Previous | Next ]
Oskar Trestone Posts: 22 From: Germany Registered: 9/17/11
Re: layer logic: a new dimension to logic (incl. Cantor)
Posted: Jan 8, 2012 1:11 PM
Hello,
elementary everyday logic should be enough to understand my layer theory.
I myself have studied mathematics and philosophy - but some 20 years ago -
and books and long articles about logic are too boring for me (as would be my thread ...)
To understand my layer theory, I think it is important to understand my motivation, to look for a new kind of logic:
The liar´s paradox always has fascinated me - as a sign that to logic "the jury is still out"
and that in spite of a thousand year old rather stable tradition
we still have not reached full understanding of our ways of thinking
My new logic, layer logic, shows a way around this paradox by using (meta) layers (or meta levels).
The key idea of layer logic is, that there is a kind of additional new dimension in logic, the layers.
These layers are hierarchically arranged and have discrete values 0,1,2,3, ?
A statement has not a truth value "true" or "false" any more, but a truth value in every layer
and some statements (like the liar statement) have different truth values in different layers.
This is near to the idea and solution of Alfred Tarski (but I do not know details of his proposals and layers/(meta)levels)).
The following two ideas are my own invention (at least I think so):
- Layer 0 (Zero), and the idea that every statement has truth value "u" (=undefined) at layer 0.
- Meta statements, especially the idea that every meta statement has an identical truth value in all layers > 0
(that means it is almost a classic statement).
The liar paradox is connected with some other problems:
Russell´s set; Cantor´s paradox of the power set / diagonalization;
Gödel´s incompleteness theorem, the halting Problem of informatics, EPR and Bell´s theorem in physics
(You can look up this (and other references below) all in Wikipedia for example,
but especially the last three or four problems and their proofs are hard technical stuff ?
and it is not necessary to study them for the understanding of layer theory)
If we have a look at the connected proofs to this problems,
we always find a proof by contradiction (connected with the law of the excluded middle (tertium non datur)).
One try to come around the problems with the liar (or the proofs by contradiction)
was multivalued logic, but the extended liar ( "this statement has not the truth value "true" " ) still is a problem there.
Even as I am using a third truth value "u" at layer 0 (It helps to start symmetrically),
I think that this is not the most important part of layer theory.
Another try to change logic and mathematics was intuitionistic or constructive logic/mathematics.
http://en.wikipedia....ki/Intuitionism
Here proofs by contradiction are not valid anymore.
But as this complicated the life of mathematicians a lot, most did not follow this line,
although Gödel´s incompleteness theorem had shaken classical mathematics (like ZFS).
In level theory we would have a third way:
It is (at least in some points) not so radical as constructive mathematics and proofs by contradiction are still possible,
but only within one layer.
As almost all classical proofs to the problems above use multiple layers, they are no longer valid.
The reconstruction of most parts of classical mathematics is possible in layer theory, with some exceptions:
I could not show, that there is only one prime factorisation for every natural numbers over all levels.
The square root of 2 might be rational (with different fractions in different layers).
The nicest part of layer theory in my opinion is the set theory, here we have the set of all sets as a set and only one kind of infinity.
But as I am more a philosopher than a mathematician,
there is still a lot to be controlled and proofed
and a better and complete formalisation is yet to be done (help welcome!).
And if the theory is valid, my next question is,
where and how we can use it to solve problems
(even up to now unsolvable ones and not only in mathematics ?)
Yours
Trestone
Date Subject Author
9/17/11 Oskar Trestone
9/18/11 Oskar Trestone
9/19/11 Oskar Trestone
9/21/11 Oskar Trestone
9/21/11 frogfoot
10/1/11 Oskar Trestone
10/8/11 Oskar Trestone
10/15/11 Oskar Trestone
12/26/11 Oskar Trestone
12/28/11 Oskar Trestone
1/8/12 Oskar Trestone
12/28/12 Oskar Trestone
5/26/13 Oskar Trestone
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Mar11 awarded Popular Question Nov10 awarded Popular Question Sep24 awarded Autobiographer Jul2 awarded Curious Nov28 accepted Is it possible to do a Hasse Diagram for divisibility on the following set Nov27 revised Is it possible to do a Hasse Diagram for divisibility on the following set edited body Nov27 asked Is it possible to do a Hasse Diagram for divisibility on the following set Nov17 accepted Is it possible to simplify the following combination? Nov4 asked Is it possible to simplify the following combination? Oct2 accepted Find a counterexample for the following statement (explanation please, I already have the solution) Oct2 asked Find a counterexample for the following statement (explanation please, I already have the solution) Oct2 comment Use mathematical induction to prove that 9 divides $n^3 + (n + 1)^3 + (n + 2)^3$; Looking for explanation, I already have the solution. @labbhattacharjee the problem asks for that procedure Oct2 accepted Use mathematical induction to prove that 9 divides $n^3 + (n + 1)^3 + (n + 2)^3$; Looking for explanation, I already have the solution. Oct2 asked Use mathematical induction to prove that 9 divides $n^3 + (n + 1)^3 + (n + 2)^3$; Looking for explanation, I already have the solution. Sep18 accepted Is $\{1, 2, 3\}\times \Bbb Z$ uncountable? Sep18 comment Is $\{1, 2, 3\}\times \Bbb Z$ uncountable? That was perfect, thank you! Sep18 revised Is $\{1, 2, 3\}\times \Bbb Z$ uncountable? edited title Sep18 comment Is $\{1, 2, 3\}\times \Bbb Z$ uncountable? I'm really new to Discrete Mathematics (first week) so I'm having a bit of trouble understanding what you mean by (a, b) -> 3b + a - 1, could you possibly give me a brief explanation? Sep18 comment Is $\{1, 2, 3\}\times \Bbb Z$ uncountable? no, unfortunately it was as I wrote it. Sep18 asked Is $\{1, 2, 3\}\times \Bbb Z$ uncountable?
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# Partitioning a Rectangle into Congruent Isosceles Triangles
Is it possible to partition any rectangle into congruent isosceles triangles?
No. Note that the acute angle of your triangle must divide $\pi/2$ (look at a corner), so there are countably many such triangles (up to similarity), and hence you get only a countable set of possible ratios of sides.
• Brilliant! $\mbox{}$ – Joseph O'Rourke Nov 6 '10 at 21:03
• You seem to be saying that there is a one-to-one correspondence between classes of similar triangles and classes of rectangles of sides of a given ratio. Perhaps I'm misunderstanding your argument, but if I'm not, I don't see why that has to be true. Can you elaborate a bit? Many thanks! – John Iskra Nov 8 '10 at 20:43
• Not one-to-one, but if we fix triangle (with sides a and b), then both sides of rectangle are linear combinations of a,b with integer coefficients. So, there are at most countably many of them for fixed a,b. – Fedor Petrov Nov 8 '10 at 21:56
• Got it. Thank you! That is a really nice proof. – John Iskra Nov 9 '10 at 2:31
If the length divided by the width is rational, then yes. Just partition the rectangle into congruent squares and cut each square along a diagonal.
• Is this a necessary condition as well? – Jeremy West Nov 6 '10 at 0:22
• The argument for necessity would run something like this: 1. By looking at a corner, convince yourself that the triangles must be right (i.e. a half-square). (Some angle on the triangle must divide $90^\circ$; then analyze by cases which angle it can be.) 2. Using irrationally of $\sqrt2$, argue that for any rectangle tiled with half-squares, all short (say) sides must be parallel/perpendicular to each other (as opposed to at $45^\circ$). 3. Whether all short sides are parallel to the sides of the rectangle or perpendicular, you still win. – Theo Johnson-Freyd Nov 6 '10 at 1:04
• Yeah. Thank you to those who have posted so far: I did know, though, that the answer is 'yes' in the cases listed above. In fact, relaxing any of the hypotheses seems to give a positive answer. That is, if the triangles need not be isosceles, if they need not be congruent etc. I was asked this question without the congruency condition by a sophomore student I have. It seems like, potentially, a good research project, but I didn't know whether it was 1. known 2. trivial or 3. too hard. This is outside my area of specialization, so I thought I'd post here to get some sense of things. – John Iskra Nov 6 '10 at 1:37
• If you knew all that, you could have saved people some work by mentioning it in your original statement. – Gerry Myerson Nov 6 '10 at 1:53
• @Gerry: some was equal to about epsilon here, so no worries as far as I am concerned. @unknown: There is nothing wrong with extremely long questions on MO. One reason why I like MO is that I can learn a lot from just reading questions, and in this sense longer questions are better than short ones. Concerning your points {1,2,3}: I am not sure whether the answer is known, but combining Theo's comment with my answer gives a complete characterization. If one can thrash out the details of Theo's comment, I think it would be neither trivial (2) or too hard (3). Probably 2.4. – Tony Huynh Nov 6 '10 at 12:01
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Guided Lessons
# Ordering Fish
In this fun fishy lesson plan, students will love getting their creative juices flowing as they make fish pictures all the while practicing how to order numbers. Can be used as a stand alone or support lesson for the **It's Raining Numbers!** lesson plan.
This lesson can be used as a pre-lesson for the It’s Raining Numbers! lesson plan.
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This lesson can be used as a pre-lesson for the It’s Raining Numbers! lesson plan.
Students will be able to count, label, and order numbers from 0-10.
##### Language
Students will be able to explain how to count, label, and order numbers using number labels.
(5 minutes)
• Gather the class together for a read aloud.
• Introduce the book, Click, Click, Splash, Splash by Doreen Cronin and read it aloud.
(3 minutes)
• Say, "Now we are going to practice counting and putting numbers in order. Order is another way to say organize or to put things in a way that makes sense."
• Model counting from 1-10 in order using the precut numbered fish you created prior to the start of the lesson (can also use number cards if preferred). Hold up each fish 0-10 in order, say the number aloud, and then place on the board for students to see.
• Repeat this process, this time having the students echo count after you.
(5 minutes)
• Tell the class that now they will get a chance to put the fish in order.
• Place the second set of numbered fish on the board (under the first set) in mixed up order.
• Ask the students to look at the numbers and turn to a partner to share the number that should go first. Then ask 1-2 pairs to share their thinking aloud. Have the rest of the class give a thumbs up if they agree or a thumbs down if they disagree.
• Repeat with the remaining numbered fish until they are correctly ordered from 1-10.
(15 minutes)
• Explain that now students will get a chance to make their own number fish, just like in the story.
• Model how to draw ten fish on the blue paper, then cut out the numbers from the Counting to Ten worksheet and glue the numbers onto your fish in order. When finished, show the students how to draw a line from 1-10 in order. Practice counting your fish to check that they were ordered correctly.
• Ask students to turn and talk to share the directions with a partner in their own words.
• Review instructions:
• Draw ten fish
• Cut out numbers
• Glue numbers onto fish
• Draw a line from 1-10 in order
• Pass out materials and have students work independently.
(3 minutes)
• Say, "Now we are going to practice counting and putting numbers in order. Order is another way to say organize or to put things in a way that makes sense."
Beginning
• Work with a small group of students to order additional sets of number cards. As a group discuss strategies to order the numbers (number line, hundreds chart, song, etc).
• Have students mix up the number cards on their paper (without gluing them) and then trade with a partner to put back in order. Ask the pairs to share with each other how they organized the numbers.
• Provide number cards 1-20 and have students work together to put in order.
(5 minutes)
• Have students trade work with a partner to check that they ordered their fish from 1-10.
• Collect work samples to assess if students were able to label, order, and count their numbers correctly.
• Listen to students count aloud to assess ability to count to ten in order.
(2 minutes)
• Gather the class back together and invite 2-3 students to share their picture with the class. Have the student point to the numbers 1-10 and as a group count chorally.
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##### Need help with Pure statistics with integers, include every step!
label Statistics
account_circle Unassigned
schedule 1 Day
account_balance_wallet \$5
Q1> a,b, and c are integers and a<b<c. S is the set of all integers from a to b, inclusive. Q is the set of all integers from b to c, inclusive. The median of Set S is (3/4)b. The median of set Q is (7/8)c. If R is the set of all integers from a to c, inclusive, what fraction of c is the median of set R?
Dec 15th, 2015
Median of S =(a+b)/2
3/4b = (a+ b)/2
3/2b = a+ b
a= 3/2b-b
a=b/2
Median of Q=(b+c)/2
7/8c = (b+c)/2
b+c= 7/4c
c= 7/4c-c
b= 3/4c
Median of R = (a+c)/2
but a= b/2
a= 3/4c/2
a= 3/8c
median of R = (3/8c +c)/2
= (11/16)c
Dec 15th, 2015
...
Dec 15th, 2015
...
Dec 15th, 2015
Oct 22nd, 2017
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Effect of Magnet on Current carrying Wire - Study24x7
# Effect of Magnet on Current carrying Wire
Updated on 21 January 2022
Class12NonMedical
Updated on 21 January 2022
Magnetic Force is defined as the force or the influence of a magnet on electric currents. The effect of a magnetic field is seen in permanent magnets like iron which attracts or repels the object. Magnetic field changes with the change in location. The electric currents are created through the magnetic fields with the help of electromagnets and wires.
The magnetic force on a current carrying wire problem is calculated by F=I→l×→B.
A magnetic force acts on a current carrying conductor because the current or the magnetic force is transmitted to the conductor. This is based on the right-hand rule.
There is also a formula for calculating the magnetic field due to current carrying wire. This is calculated by F=Bilsinθ F = Bi l sin θ. In this formula, ‘l’ is the length of the wire, ‘i’ is the current and θ is the angle between the current and the magnetic field. In the case of the magnetic field due to the current carrying conductor, the current is reversed. It in turn encircles the conductor and is perpendicular to it.
Now students often wonder, what is the force of the current carrying conductor in a magnetic field experiment? Well, the force of the current carrying conductor in a magnetic field experiment is calculated by F = IlB sin θ. A conductor experiences a force in a magnetic field experiment. This is determined by Fleming's left-hand rule. The left-hand rule is the way to determine the motion of a conductor when the magnetic field direction and the current direction are known.
## What happens when a current carrying conductor is placed in a magnetic field?
This is a valid question and is often asked by the students. If the current carrying conductor is placed in a magnetic field, then a mechanical and a magnetic force or pressure is exerted which results in the conductor moving a little from its initial position. The force is exerted in the right direction as per Fleming’s left-hand rule as stated above (basically in a reverse direction).
Students are often in confusion regarding the effect of the magnetic field produced by a current carrying conductor. Well, the answer to this question will be that the magnetic field is produced by the electric current. The magnetic field is represented as a circular diagram (concentric circles). The direction of the magnetic field can be measured by a compass. In the pictorial representation, it is represented as a straight line intersecting the circles in the middle. It is shown in the diagram below.
Now, let’s talk about the force on a current carrying conductor in a magnetic field Class 12th. Students are often confused as to which force is exerted mechanical, magnetic or some other force. The formula has also been stated above to calculate the force.
The formula is F = IlB sin θ. The force exerted is a magnetic force. There is one more way to calculate the force with the help of Fleming’s right-hand rule. The right-hand rule is undertaken to find the direction of the force exerted on a positive moving charge. The formula to calculate this force is F = (qvdB sin θ) (nAl). In this formula, A is the area covered by the wire in the cross-section.
The effect of a magnet on a current carrying wire is one of the important topics in Physics. Students must know the detailed explanation of the magnetism topic. There are various topics covered in this chapter like the magnetic force on a current carrying wire problem and many more. You can take help from reference books or online articles if you have any doubts. Thus, magnetism is one of the important chapters and students must be through with it to attempt this topic easily in the exam.
Related Article:
Difference between Distance and Displacement
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11 January, 04:02
# On a track for remote controlled racing cars racing car A completes the track in 28 seconds while racing car B completes it in 24 seconds. if they both start at the same time after how many seconds will they be side by side again?
+3
1. 11 January, 04:15
0
We'll take out LCM of 24 seconds and 28 seconds.
that is, 168 seconds.
That means,
Car A and Car B will be side by side to each other after 2 mins 48 seconds after start.
2. 11 January, 04:31
0
After 168 seconds, both the cars will be side by side again.
Step-by-step explanation:
On a track for remote controlled racing cars racing car A completes the track in 28 seconds while racing car B completes it in 24 seconds.
To know after how many seconds will they be side by side again, we will find the LCM of 24 and 28.
24 = 2 x 2 x 2 x 3
28 = 2 x 2 x 7
LCM is = 2 x 2 x 2 x 3 x 7 = 168
Hence, after 168 seconds, both the cars will be side by side again.
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# Ch 1.3 – Order of Operations
## Presentation on theme: "Ch 1.3 – Order of Operations"— Presentation transcript:
Ch 1.3 – Order of Operations
Objective: To simplify expressions using the order of operations.
Order of Operations (PEMDAS)
The Order of Operations are rules that explain how to evaluate mathematical expressions. Parenthesis Exponents Multiplication & Division Addition & Subtraction
Example = 2 + 3 * 4 = 5 * 4 = 20 = 2 + 12 = 14 This is WRONG!
This is CORRECT! 1 * * 4 = * 4 = * 4 = 20 Multiply – Add – Multiply NO! 1 * * 4 = = Multiply – Multiply - Add YES!
Order of Operations Parenthesis Exponents Multiply / Divide
1) 6 + 5(8 - 2) 6 + 5(6) = 36 Exponents 2) = 77 Multiply / Divide 3) = 32 Add / Subtract 4) 7 + 8 = 15
Try These Simplify each expression below. 1) 6 + 5(8 - 2) 2) 3) 4)
Simplify the following.
1) 3) (4 + 18)2 - 10 (22)2 - 10 -1 + 5 = 34 4 2) 4) 24 - 2[ ] 8 + 20 24 - 2[3] 4 = 18 28 = 7 4
Evaluate the expression if a = 2, b = 4, and c = 8
1) ab - c 3) 2 (2)(4) - 8 4 + 3(2) 8 - 8 4 + 3(4) 2 = 10 2) 3a(c - b) 4) 3(2)(8 - 4) 2(4 + 8) 3(2)(4) 2 2(12) 24 24 = = 12 2 2
Fact: Scientific calculators and computers follow the order of operations. Basic calculators do not.
3 + 4 x 2 = 3 + 4 x 2 = = * 7 x 2 = 14 3 + 8 = 11 3 + 8 = 11
Symbolic Differences in Calculators and Computers
Scientific Calculator Expression Computer + + + - - - x x * / 4 ^ 2 4 2 4 x 2 y
Key Sequences and Computer Input
Calculator Key Sequence Expression Computer 3(8 - 2) 4 3 6 - 2 6 + 8 2 3 - 1
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# Order of Operation Games Online
Page content
The following section contains some of the best websites, but if you need some background prep before hitting these resources, feel free to learn about how to solve equations using order of operation first.
## Top Websites For Order of Operation
Understanding order of operation can be very hard for some children. The internet provides so many fantastic tools and resources that can be of great help. Let’s take a look at some of these great order of operations games online that can help in understanding.
**
Funbrain.com’s Operations Order is a simple online practice game. The game gives you three numbers, then there is a problem in which there are two operations, for example ___ + ____ x _____ = 30. The goal is to place the numbers in the correct spot using order of operations to get the correct answer. The website also provides immediate feedback and error correction, which is very helpful.
AAA Math provides a great deal of information on order of operation. There is an instruction section, a practice section, and a play section. The practice section provides you with a problem which you must solve. This site also provides immediate feedback. In the play section on this website, the same types of problems are used in the practice section, but there are goals like determining how many problems can you do in 60 seconds.
Dositey.com has a great game to play for order of operation practice with adding, subtracting, and multiplying only. This is great for students that are around grades three or four. This is such a fun game. There is a square with 16 smaller squares, each of which holds a number. Around the squares are problems, you need to match the problem to the answer in order to reveal a picture. This is great for practicing with the younger children because of the entertainment value of it.
Calculator Chaos uses some higher level thinking skills. Some of the keys have fallen off the calculator and you need to figure out how to get a given answer using only the certain keys that are remaining on the calculator. Yet another fun game using order of operation.
Homeschoolmath.net is another great website. You are given a problem with two operations signs missing in which you must fill in to make the problem true. You can set the minimum and maximum number values depending on the grade level of the student so the problems can be easy with whole numbers, to harder problems with integers. It uses only the four basic operations.
As you can see there are many great order of operations games online. If you need to help your child on homework, or if you are a teacher looking for some fun games for your students to play, check some of these out. It will definitely be worth their time.
## References
Calculator Chaos https://www.mathplayground.com/calculator_chaos.html
Wikimedia commons / Math
## This post is part of the series: Math Games
Math games are a great way to teach concepts like order of operation, Algebra, division, adding, and so on.
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melalyvanealy
19
# which value of x makes this equation true (-2)(3)(×)=0
(2) Answers
pamelacoyer
When you're multiplying anything together to get zero, one of the factors must be 0. That means that x has to be 0 because none of the other factors are 0.
JadeVolmink627
(-2)(3)(x) = 0 Simplifies to 0. Anything equal to zero is zero. Particularly equations like this one.
Add answer
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# How do you count 5 working days?
As we can see from the above calculation, there are 52 Mondays in the year 2021.
## How long is 4 days from now?
What does 4 days ago mean? 1. “Yesterday I met Bob. Read also : How many Friday the 13th are there in a year? He said he bought an expensive smartphone four days ago.” (four days before today*) 2. “
Which date is 240 days away? Do you want to calculate the date that is exactly two hundred and forty days from January 22, 2022 without counting? Your start date is January 22, 2022, so that means 240 days later is September 19, 2022.
What date is it in 4 weeks? In 4 weeks, the date will be March 18, 2022, which is also 672 hours.
## How do I calculate the number of days excluding Saturday and Sunday in Excel?
=NETWORKDAYS(A2,B2) Then type the Enter key and you will count the number of working days excluding Sunday and Saturday between the two dates. Read also : Est-ce que Le Grinch est sur Netflix ? See screenshot: Note: In the above formula, A2 indicates the start date and B2 the end date.
How to calculate days excluding weekends in Excel? To add days excluding weekends, you can do as follows: Select a blank cell and type this formula =WORKDAY(A2,B2), and press Enter key to get the result. Tip: In the formula, A2 is the start date, B2 is the days you want to add. Now the end date is shown which adds 45 working days excluding weekends.
How to calculate days excluding Sundays in Excel? Count days excluding Sundays with formula Select a blank cell, here is C2, and type this formula =B2-A2-INT((B2-A2-WEEKDAY(B2) 1)/7) into it, and then press Enter, a date will be displayed.
Also to discover
## What are 2/5 working days?
In the US, business days are typically Monday through Friday (weekdays), excluding weekends and federal (national) holidays. See the article : How much is Qt in C? 5 working days are basically Monday to Friday, while Saturday and Sunday are for rest.
What does 5 working days mean? Monday to Friday are working days. So 5 working days means five days, not including Saturday and Sunday.
What does 2/5 working days mean? A business day is defined as a day commencing Monday through Friday (generally from 9 a.m. to 5 p.m. for any business day). … So to calculate 2 business days from now, count two consecutive days that fall between Monday and Friday.
What does 2 to 3 working days mean? In the United States, this means three consecutive business days, with business days being Monday, Tuesday, Wednesday, Thursday, and Friday. However, it is a tricky term, as many businesses are now also open on Saturdays and Sundays.
## What are working days?
A working day is any day you work. Usually workday refers to one of the days in the work week, the period of (often five) days that do not fall on a weekend and on which many people work. To see also : Is 200ml the same as 20cl? The word working day means the same as working day, and working week means the same as working week.
Is Friday and Saturday a working day? A business day is an official business day. Monday to Friday are considered working days, but not public holidays* and weekends.
What are considered working days? A business day is normally Monday to Friday, 9:00 AM to 5:00 PM, excluding holidays. Consumers often have business days to settle or settle financial transactions or to deliver goods or services.
## How many days is 7 working days?
Working days are Monday to Friday and exclude weekends. Seven working days are 7 days excluding weekends. This may interest you : 36000 a year is how much an hour. In other words, one week plus two days would be 7 business days.
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5 Replies Latest reply on Feb 17, 2020 2:56 AM by lui_gough
# Understanding decibels a bit more
I get most of this - I expect it'll be clearer and faster with practice, but I get the principal that it is logarithmic to powers of 10. I also understand the underlying principal of decibel gain/loss related to a ratio:
db = 10log10(Pout/Pin) for Power
db = 20log10(Vout/Vin) for Volts RMS (or, generically, a ratio of amplitude)
So for power, a +3db gain is a doubling of power; for volts, a +6db gain is a doubling of volts (or amplitude.)
I then come across a statement like this:
"dBm: This is the value of a signal as referenced to 1 milliwatt. It is an absolute value. A 0 dBm signal equals 1 milliwatt. A 1 watt signal is equal to +30 dBm. If you had a 100-watt transmitter, that would equate to a +50.0 dBm level. If you had a 2 microvolt signal, that would equate to a -100 dBm signal. If you have a 0.5 microvolt (uV) signal, that would equate to -112 dBm."
I can follow the first part, no problem. If 0dBm is 0.001W, then for 1W, dBm = 10log10(1/0.001) = 30; 100W, dBm = 10log10(100/0.001) = 50. Generically 1W is 10^3 from 1mW and log10(1000) is 3 so 10x3 = 30; ditto 100W ~ 10^5 so log10(100000) is 5 so 10x5 = 50.
The second part of the statement though feels like it's making an assumption, one that maybe is so common to those in the know that it doesn't need stating?? That doesn't include me.
To relate a voltage value to a power value I need to know the load in Ohms or Amps don't I? Take as an example, because it works with that logic (nearly), an assumed 50Ohm load, then from the formula P=V^2/R:
2uV @ 50Ohms: P = (2x10^-6)^2 / 50 = (2^-6)^2 / 50 = (4^-12) / 50 = 8^-14.
dBm = 10log10(8^-14 / 0.001) = -100 (actually, pretty much -101)
0.5uV @ 50Ohms (not showing working out!): P = 5^-15
dBm = 10log10(5^-15 / 0.001) = 113 (although the statement above has 112)
I thought working with decibels was intended to be quick so I think I'm missing a trick here. I couldn't work out in my head that 0.5uV is -112dBm (or -113dBm) so I feel like I'm doing something wrong.
• ###### Re: Understanding decibels a bit more
Hi Andrew,
andrewj wrote:
The second part of the statement though feels like it's making an assumption, one that maybe is so common to those in the know that it doesn't need stating?? That doesn't include me.
You're totally correct, it is making an assumption. It ought to be saying "a 50 ohm load with 2 microvolt rms across it would be consuming -100dBm of power" or something similar, to be more accurately worded. Usually it's assumed though that the load is 50 ohm (but doesn't need to be. Although it's unusual for it not to be 50 ohm when there's a discussion with dBm featured).
Also, you're correct on the next part too, 0.5 uV would be -113 dBm, and the -112 dBm statement was incorrect.
It's annoying, messed up calculations on the Internet sadly, making it harder to learn : (
EDIT: Some of it could be down to people using paper charts or slide rules maybe! : ) I don't know if people had tricks to convert some popular values pre-calculator, but some of it maybe was based off charts etc. Quite a lot of electronic stuff used to be nomograph based too.
5 of 5 people found this helpful
• ###### Re: Understanding decibels a bit more
The way it is written makes it seem like there is a quick trick to knowing the answer - is that the case? I've also seen "...range of -110dBm to +16.5dBm (0.7uV to 1.5V RMS)" written like it's obvious. For me I can only do it by following through the formula above.
At least for one of the very few times in my life I got the maths right
• ###### Re: Understanding decibels a bit more
Hi Andrew,
There's no quick trick, or at least I'm not aware of it! : )
I only remember some dBm approximations just out of familiarity, e.g. 0 dBm being 1mW, or 20 dBm being quite high (WiFi power output level aka 100 mW), -90 to -110 dBm being typical received radio power strengths (and knowing this is of the order of a microvolt or so), -145 dBm or lower being GPS received strength, and so on. Converting between that and volts rms is impossible mentally for me at least.
The reason people maybe write it along with the microvolt value, is because some radio receivers specify their sensitivity in microvolts. I can't (directly) measure that anyway : )
What is worth doing is putting your formula into an Excel file maybe (if you're not already doing it), and slowly keep adding sheets, building it up with all the stuff you come across. I've done that for years, my file is not understandable by anyone, but it works for me! I've even got wire gauge tables in there, and (just looking at it now) for some reason I cannot recall, there's a sheet with hexagon area, and diameter of a circle which would enclose the hexagon.. I have no idea why I ever needed that.
1 of 1 people found this helpful
• ###### Re: Understanding decibels a bit more
Next time I take the car for a service I’ll let them know I think the brake pads are -3dB worn, see what they say!
Building up a spreadsheet is a great idea - I do loads of calculations in them so should keep them. I reckon your hex/circle areas are from when you used to be the bad boy of the local quilting scene judging by the questions my wife asks me.
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Blog
# The Quest for 700: Weekly GMAT Challenge (Answer)
Yesterday, Manhattan GMAT posted a GMAT question on our blog. Today, they have followed up with the answer:
As in so many of these problems, the first task is careful reading. Note that x is not itself the integer (except in the case of 1/1); x is the reciprocal of an integer. You might even make a quick table corresponding to the values in the answer choices:
Integer x 1 1/1 = 1 2 1/2 3 1/3 4 1/4 5 1/5
Now, just keep adding columns. Add one for y, defined as –x2:
Integer x y = –x2 1 1/1 = 1 –1 2 1/2 –1/4 3 1/3 –1/9 4 1/4 –1/16 5 1/5 –1/25
Notice that according to PEMDAS, you apply the squaring operation (Exponent) before the negative sign (Subtraction).
Add one more column for xy. For clarity, we’ll use the caret symbol (^) to indicate exponents. Notice that the negative sign undoes the reciprocal in x, leaving you with an integer base.
Integer x y = –x2 xy 1 1/1 = 1 –1 1-1 = 1 2 1/2 –1/4 (1/2)^(–1/4) = 2^(1/4) 3 1/3 –1/9 (1/3)^(–1/9) = 3^(1/9) 4 1/4 –1/16 (1/4)^(–1/16) = 4^(1/16) 5 1/5 –1/25 (1/5)^(–1/25) = 5^(1/25)
Finally, you are left with the task: which of those numbers in the last column is largest? Remember, fractional exponents are roots, so the second number is the fourth root of 2, and so on.
You can quickly eliminate 1, since the positive roots (to any degree) of any integer greater than 1 are always greater than 1. For instance, what is the 25th root of 5? It must be a number bigger than 1, even if only slightly, because that number times itself 25 times in all produces 5. So you can eliminate A.
Let’s compare the second and fourth numbers, because the base of 4 can be rewritten as 22. The fourth answer becomes then (2^2)^(1/16) = 2^(1/8), which is less than 2^(1/4). The number that solves z8 = 2 is smaller than the one that solves z4 = 2. So the answer can’t be D.
What about B versus C? Raise both numbers to the 36th power, so that we eliminate fractional exponents.
2^(1/4)^36 = 2^9
3^(1/9)^36 = 3^4
Which of the results is bigger? 2^9 = 512, while 3^4 = only 81. So the original numbers must be in that same order of size; B is bigger than C. C is out.
A similar argument takes out E (raise both B and E to the 100th power):
2^(1/4)^100 = 2^25
5^(1/25)^100 = 5^4 = 125 = much smaller than 2^25.
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## Algebra and Trigonometry 10th Edition
$θ=30°$
$sin~θ=\frac{opp}{hyp}=\frac{1250}{2500}=\frac{1}{2}$ Given that $sin~30°=\frac{1}{2}$, we can conclude that $θ=30°$
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If the trial function satisfying certain boundary conditions is assumed in advance, and the approximate solution is carried out on this basis, the difficulty of solution will be reduced..
Such a calculator can store a lot of exam related content, but of course it can't be brought into the examination room. Then, are the other non storage calculators all right? Theoretically, this is the case, but some calculators have too simple calculation functions. They only have the calculation functions of addition, subtraction, multiplication and division, and do not have high-end functions such as root opening. Such calculators are also not recommended for candidates. They should still use scientific calculators without storage functions.
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# Pre-Algebra
This course is designed for the middle school student preparing to enter a high school level algebra course. Students taking this course should have strong skills in arithmetic to include fraction operations.
The first part of Pre-Algebra will explore areas of Real Numbers (positive/negative numbers), solving equations, graphing and writing lines and inequalities.
Part 2 of the course deals with topics such as functions and relations, systems, quadratic equations, polynomials, powers and exponents and rational/radical expressions. There are also sections on basic geometry to include circles, area, volume, Distance Formula and Pythagorean Theorem.
### Chapter 1: Introduction to Algebra Terms and Concepts
This chapter introduces students to basic terms and concepts used in algebra. Time is taken to ensure the student understands basic number operations, variables and their applications. Additionally, the student gains a fundamental sense of equations, inequalities and their solutions.
### Sections:
• 1.1 Number Operations
• 1.2 Variables
• 1.3 Order of Operations
### Chapter 2: Real Numbers and Simplifying Variable Expressions
This chapter focuses on getting the student to master working with the Real Numbers. Students learn the rules of integers and practice through many examples. Also, students will learn to apply the Distributive Property and simplify variable expressions by combining like terms.
### Sections:
• 2.1 Real Number System
### Chapter 3: Fractions and Decimals
This chapter reviews how to work with fractions. Although it is assumed students posses these skills, time is taken to ensure the student has mastered the procedures to perform operations involving fractions.
### Chapter 4: Solving Equations
The chapter breaks down the steps to solve multi-step linear equations. Students will build up their skills as they progress from one and two-step equations to more advance equations. Core concepts involved will be reviewed to include the Distributive Property and combining like terms.
### Sections:
• 4.1 One Step Equations
### Chapter 5: Inequalities
In this chapter students will apply their equation solving skills to solve linear inequalities. Basic concepts and terms are introduced first, along with how to graph inequalities.
### Sections:
• 5.1 Linear Inequalities
### Chapter 6: Graphing Linear Equations
This very important chapter walks the student step-by-step to master how to graph linear equations. Concepts involving the coordinate plane, slope and methods to graph lines are thoroughly reviewed and introduced. Upon completion of the chapter students will gain the necessary knowledge and skills needed to learn more advance topics involving linear equations.
### Sections:
• 6.3 The Slope of a Line
• 6.5 XY Intercept Method
### Chapter 7: Writing the Equations of Lines
This chapter builds on the student’s prior knowledge and skill of linear equations. Various methods to find and write the equation of a line are introduced and practiced. The chapter focuses on the proper way to set-up and use formulas to write linear equations. Additional related topics are explored to include linear models, linear regression and word problems.
### Chapter 8: Introduction to Systems
Understanding systems and the methods to solve them are vital in algebra. This chapter introduces the student to systems and what their solutions represent. Techniques to solve systems will build from the student’s prior knowledge of solving linear equations. Upon completion of the chapter a student will have a solid skill set in systems that prepares them for Algebra 1 and 2.
### Chapter 9: Absolute Value
Absolute value problems can be challenging for some students to grasp. Time is taken to teach students core concepts and build understanding. Students will learn how to graph absolute value functions and apply the steps to solve absolute value equations/inequalities.
### Chapter 10: Powers and Exponents
This chapter covers the rules of powers and exponents a student needs to learn in algebra. Also, important applications of these rules are covered to include scientific notation, compound interest and exponential growth and decay.
### Sections:
• 10.4 Scientific Notation
• 10.5 Compound Interest
### Chapter 11: Polynomials
Polynomials are the key building blocks of algebra. The chapter starts by covering the parts of a polynomial and related terminology. Students then learn how to perform various polynomial operations, and a special focus is placed on avoiding common mistakes. Lastly, there is a section dedicated to introduce quadratic equations which are extremely important in more advance topics in algebra.
### Sections:
• 11.3 Multiplying Polynomials
### Chapter 12: Rational Expressions
This chapter takes the student through fundamental rational expressions to include ratios, proportions, percent and variation. Students will learn different methods to solve rational expression problems. The section on simplifying rational algebraic expressions starts by reviewing basic examples using numbers before introducing variable examples.
### Sections:
• 12.1 Ratios and Proportions
• 12.2 Percent
### Chapter 13: Introduction to Functions and Relations
Functions and relations transcend all through mathematics. This chapter explains core concepts at the Pre-Algebra level and prepares the student for more advance study of the topic. Time is taken to explain the difference between a function and relation; and introduce the student to the language of functions to include the domain, range and linear/nonlinear functions.
### Sections:
• 13.2 Graphing Functions
### Chapter 14: Area and Volume
This chapter explores the link between algebra and geometry. Basic formulas for area and volume are covered. Also, a section is dedicated to explain concepts related to circles to include area, volume, diameter, radius, and pi.
### Sections:
• 14.1 Area of Basic Figures
• 14.4 Volume of Basic Figures
### Chapter 15: Right Triangle Theorems and Formulas
Properties of right triangles form the basis of many more advance topics in algebra, geometry and trigonometry. Sections in the chapter introduce fundamental theorems and formulas to include the Distance and Mid-Point formula and the Pythagorean Theorem.
### Sections:
• 15.2 The Pythagorean Theorem
### Course Completion:
TabletClass makes course management for your student easy and secure! Although we do not issue grades or transcripts, we do provide all the materials you need so that you can generate them on your own, including:
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Final Exam
Final Exam Key
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https://www.wyzant.com/resources/answers/741652/how-to-apply-the-empirical-rule-68-95-99-7-rule
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Robert J.
# How to apply the empirical rule (68–95–99.7 rule)
How to apply the empirical rule (68–95–99.7 rule) to the following scenario:
I have spent a significant amount of time trying to do the impossible (i.e., mathematically beat a game with a negative expectation). Yes, I know, mathematicians for hundreds of years have failed at this (but that is only partially true). I can give examples, if need be, where games conventionally thought to be unbeatable have been beaten. My formulas do not involve the simplistic use of input/variable change the way let’s say card counting or other such formulas have been used to win at certain ‘games’ of chance. My dad taught me calculus by 7th grade and ‘card counting’ of all things by 8th grade, so my love of math and ‘beating games of chance’ has been an ongoing love affair over my short life : - )
I have not as yet been able to program/code the formulas to test large data sets, as coding my formulas is proving to be challenging, so I have been performing longhand calculations (very time consuming and tedious). My starting points are: I may only start with a small sum of money and when wagering, I am never increasing the spread between bets (i.e., I am winning by flat betting, but I also created changes to the formulas that allow me to 1.5x, 2x, 4x, etc my flat bet if I wish and still consistently have a positive outcome (i.e., a profit/net gain).
I have now multiplied my ‘random’ hypothetical starting sum 57x which means the odds of this occurring randomly are low: X/(X + (57X)) = 1 / (1 + 57) = 0.01724 (a 1.724%) chance my success is random (assuming I am using that correct algebra to calculate this. I have been told that when I reach 1% or lower (which will take a fair amount of time testing data manually) this is considered by most mathematicians to be ‘near certainty’ that the result I am achieving is ‘not random’. I am current in some areas of math and not current in other areas (i.e., I still have yet to take the courses or teach myself) but I am wondering what variables to plug into the empirical rule to see if I am within 3 standard deviations.
I typically retain 18% of every dollar I win on paper… meaning I will win $100, then lose$82, then win $100 and lose$82, with the retained amount of $18 (as an average) accumulating over time as I get through larger and larger data sets. In other words, the$18 that I retain per win/loss cycle adds to the sum of $retained (won), and I now have accumulated 57x my random starting sum. E.g. I started with$100 and now have $5700. What numbers can I plug into 68–95–99.7 rule? Using this formula: Pr(μ-2σ ≤ X ≤ μ + 2σ) ≈ 0.9545 And Pr(μ-3σ ≤ X ≤ μ + 3σ) ≈ 0.9973 I know where Χ is “an observation from a normally distributed random variable”, but in my case/scenario what is “X”? The number of ‘Sums” [multiples of my starting sum] won? The amount ($) retained (i.e., the net $retained per win/loss cycle) per$100 of gross winnings? I know it is likely none of these, but what value should I use for X?
μ is the mean of the distribution, and σ is its standard deviation, but in my scenario what numbers do I plug into μ and σ?
My goal in asking this question is to have some method for determining when I am within three standard deviations and therefore my result will be considered "significant and able to qualify as a discovery.
Please do not ask me to share my formulas, as they have significant ‘real world’ value that I plan to apply ‘for public good’. I may be young, but as a kid growing up these days, I am not gullible just because I am young : -) [no insult intended]. The internet has made us aware and savvy in terms of gaining real world skills at a young age. Also, I have nearly lost the use of my writing arm because I have spent so many thousands of hours after school not moving and just writing, that I feel like my left arm is atrophying, so for that reason alone I would not share my formulas. I actually want to use what I have discovered to earn a PhD, but not till I have achieved the level of monetization I seek to apply towards philanthropy, before I share the formulas (if that is required to attain my PhD).
I hope there are some math majors/professors out there who understand statistics as applicable to games of negative expectation and who can help me learn what values to plug into X, μ and σ so I can apply the empirical rule/68–95–99.7 rule to learn when I have attained ‘near certainty’ regarding when my formulas/equations are yielding results that are ‘not random’. I do have a feeling that for the results I am achieving to be considered "significant" I may have to apply the convention of a five-sigma effect (99.99994% confidence) before my results/formulas qualify as a discovery. At the very least, they do currently have real world value. Once I learn how to code my formulas to test data sets in the billions, I can then quickly ascertain where I sit in terms of whether I can achieve the five-sigma effect... (not practical testing/calculating longhand). I am still trying to decide what programming language to learn in order to have the most flexibility in achieving the results I seek. I know there are a lot of similarities in coding languages these days, but I feel one has to be better than the next to assist me in achieving my goals (which in all fairness, I of course have not been able to fully elaborate on in an open forum/environment).
Thank you in advance for any assistance you can provide.
By:
Tutor
4.9 (195)
I taught Statistics I at Suffolk County Community College for 11 years
Robert J.
Hi Terri... thank you for the reply and kind words. It's funny, all the math professionals and others whose work, theories and research I've studied could only be 'read about' 'after the fact' for the obvious reasons. It's lonely having to be stealth, but a requisite of the field, till the goals are achieved and finally thereafter getting to publish the theorems etc... In any event, thank you so much for the reply, and yes it was helpful, thank you!!
Report
02/22/20
Terri M.
Another avenue for you to explore - take a look at Game Theory and Operations Research. I received my Masters in OR but never figured out how to get an entry level job in that field. So I became an Actuary :)
Report
02/23/20
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Clsoccer - Page 2 of 3 - Cl Soccer Blog
## How To Improve Your Soccer Bets Using The Rateform Method
How To Improve Your Soccer Bets is a series of articles that describe some well known and well used statistical approaches which will help the soccer punter make more educated bets. All these techniques have its own benefits and pitfalls and using these in isolation will enhance your odds of winning. However, together they will prove invaluable in your battle with the bookies. In every article, we will describe in detail how a particular method works giving you enough information that you go ahead and make your own forecasts. We’ll also provide you with advice as to where you can already find sites using this technique in incorporating their weekly predictions.
The statistical methods described in this group of articles will allow you to arrive at a better choice about the match, or games, which you are betting on.
In this article, we will be describing the well-known Rateform method. The premise of this Rateform technique is that each of both teams getting involved in a game is given a point score that is based on their existing form. Although this seems very simplistic, and it is, its calculation is a little more involved.
Listed below are the basic principles…
1. Each team is given a point score representing their current form.
2. At the beginning of the year each team is given 1000 points.
3. The home team supplies more points into the kitty than the away team.
4. This is done in order to reflect the home teams advantage of playing at home.
5. The winning team takes the kitty unless the end result is a draw in which case both sides share the points. In the case of a draw the away side benefits points and the house side loses points.
This is how a normal calculation is made…
Home team points total = HTPOINTS
Away team points total = ATPOINTS
Home team contribution to kitty = HTKITTY = 7% of HTPOINTS
Away staff donation to kitty = ATKITTY = 5% of ATPOINTS
The kitty consequently = KITTY = HTKITTY + ATKITTY
In the event, the end result is really a HOME WIN afterward
Away team receives zero points added to ATPOINTS and looses its 5% contribution to the kitty
If outcome is an AWAY WIN then
Home team receives zero points added to HTPOINTS and looses its 7 percent donation into the kitty
In the event, the result is a DRAW afterward
Let us look at an easy example of how to apply this…
To figure the possible outcome of a match based on rateform the away rateform is subtracted in the home rateform and the distinction is contrasted against put values for home, away and draw possibilities.
For example we can make our set values as follows:
AWAY WIN -250 DRAW 150 HOME WIN
So, in case the RATEFORM DIFFERENCE is significantly less than -250 then the match is very likely to be an off win, in case the RATEFORM DIFFERENCE is higher than 150 then the game is likely to be a home win, and if the RATEFORM DIFFERENCE lies involving then the result is the most likely to be a draw.
Of course, you may opt to use unique values to those shown above and by experimentation, you may come up with better values to use. You could also opt not to reset every group points to 1000 in the beginning of the year but rather have them carry over into another season. Another option would be to have a rolling calculation where you calculate the rateform for each team on say the last six months.
If you have the required skills you could go away and construct your own spreadsheet of data or even write a bit of software to take in fixtures and results and put on the Rateform method for your data. Or, if you are lazy like me, then you can grab some free software that already does this for you.
## How To Improve Your Soccer Bets Using The Footyforecast Method
How To Improve Your Soccer Bets is really a string of articles that describe some well known and well used statistical techniques which will help the soccer punter make more informed bets. Each of the techniques has its own benefits and pitfalls and using them in isolation will improve your odds of winning. However, together they’ll prove invaluable in your conflict with the bookies. In every article, we’ll explain in detail how a particular method works giving you enough information that you go ahead and create your own forecasts. We will also provide you with advice regarding where you can already find websites using this technique in incorporating their weekly predictions.
The statistical methods described in this group of posts should enable you to reach a better decision about the match, or games, which you are betting on.
Within this guide, we will be describing the Footyforecast technique. The Footyforecast method was originally designed for the English Football Pools and tries to remove those matches which will not be drawn, leaving you with a shorter list of games from which to choose your 8 from 11. This method was introduced to the world in 1999 on the original Footyforecast site. This method is similar to the Simple Sequence method which is described in another of our posts in this sequence.
Listed below are the basic rules…
For every team work out the following,
• 1. Work out the total number of points obtained for the previous N matches.
• 2. Work out the maximum number of possible points for the previous N games.
• 3. Divide the total amount of points obtained by the most available and multiply by 100.
• 4. Calculate the forecast value.
In (1) and (2) preceding N matches could be all of the home games to the home side and all of the away games for the off side.
HOME POINTS = number of factors for home team from last N games
AWAY POINTS = amount of points for away team from last N games
HOMEVAL = (HOMEPOINTS / (POINTSFORWIN * N)) *100
AWAYVAL = (AWAYPOINTS / (POINTSFORWIN * N)) *100
To figure the possible outcome of a match based on the Footyforecast method the value is contrasted with the following…
A value between 50 and 100 gives an increased chance of a home win the closer to 100.
A value between 50 and 0 gives an increasing likelihood of an away win the closer to 0.
There are a few factors to consider, for example, the amount of matches to utilize and whether to utilize all matches or just home for the home side and just a way for offside to name but two.
By plotting actual resulting draws contrary to the prediction it is likely to create two threshold values, one for away wins and one for home wins, any worth involving these thresholds are likely drawn. All games out these thresholds will be less inclined to be drawn. For example, a value of 40 or less for away wins and a value of 60 or more for home wins. This would mean any games falling between 41 and 59 could be drawn.
This method does, with careful tuning by the consumer is to remove many matches which will not be draws providing you with a brief list to choose from. This procedure is best used where an English Pools Plan is to be utilized.
Here’s a worked example…
The values shown are the points gained by the team for each game in a succession of four recent matches, you obviously could choose more games to base your calculations on.
DC Utd
H3 = 1
H2 = 1
H1 = 0 (most recent match)
Leeds Utd
A4 = 1 (oldest game)
A3 = 3
A2 = 0
A1 = 3 (latest game)
Using only home games for the home side and just away games for off side…
FFPHome = ((3 + 1 + 1 + 0) / 12) * 100 = 42
In case our threshold values are 40 and 60 afterward for this match the prediction lies in the anticipated draw area and at the lower end significance that if it isn’t a draw the most likely another result would be an off win. This may be interpreted as an X2 prediction, i.e. draw or away win, which some bookies will take as a wager.
Of course, you may opt to use different values to those shown above and by experimenting you may develop with greater values to use. You might also opt to utilize all home and away games played by each team in your calculations instead of just home games for the home team and away games for the away team. You might decide to have different thresholds than those shown above. You could also find it beneficial to plot real results from the Footyforecast method predictions to see how many actual draws fall in the off win, draw, and house win forecast zones.
If you have the necessary abilities you can go away and build your personal spreadsheet of data or even write a bit of software to take in results and fixtures and apply the Footyforecast method to your data. Or, if you are lazy like me, you could grab some free software that already does this for you. If this last option is for you then see 1X2Monster where you can download an FREE copy of this Footyforecast 2.0 software which implements all of the statistical methods described in this series of posts. You will also have the ability to download FREE weekly database updates to your applications, how cool is that?
## How To Find A Top Quality Soccer Ball
It’s widely thought to be the most popular game in the world. Played by millions of people from just about any walk of life, and in nearly every country, soccer — soccer or futbol as it’s often known beyond the boundaries of the United States — is a sport which can be played by almost anyone. Aside from the skills needed to maneuver the ball using your entire body, but not the palms — except for its purpose keeper — football owes much of its international appeal to the fact that all you need to play is a open area and a single ball.
The hills of equipment needed to play different sports — such as golf, tennis, baseball, American football, and so many others — are an anathema to the soccer player. It is the ball, and just the ball, which one needs to play the game. However, the soccer ball is not a generic thing; there are a number of sorts of soccer balls on the market, and each has its own attributes that distinguish it from the other balls.
At first glance it’s tough to tell 1 soccer ball from another. Normally assembled in exactly the exact same distinctive style, the signature pentagonal and hexagonal panels create a soccer ball immediately recognizable. But once you are attempting to tell the difference between one ball and yet another, the very first place to start is the cover.
Before, full grain leather has been used to make a high excellent soccer ball, but real leather tends to absorb water readily, plus a wet ball is a hefty ball which performs much differently than it had been meant to play with. Today, the first quality soccer balls are made of synthetic leather. Although there are lots of variations of synthetic leather, they’re generally all a derivative of polyurethane or polyvinyl chloride. The top chunks — these used in contest and by professionals — are nearly always made of their polyurethane construction, while cheap practice balls are more likely to be poly plastic sheeting.
The way the panels of the soccer ball are stitched together is just another indication of the quality of the ball. A premium excellent ball will be hand woven with polyester cord or Kevlar reinforced polyester. Hand stitching allows the panels to be sewn tighter, which makes for a more durable and longer lasting soccer ball. Second-tier soccer balls are often stitched too, but the sewing is done by machine, therefore, it lacks the uncompromising quality that a hand stitched ball will probably have. Cheap balls are usually not stitched in any way, and rather are held together by gluing the panels on the liner of the ball.
Soccer balls come in different sizes as well: Size 3, Size 4, 5, and Size 5. Size 4 balls are the next size up, and gamers between the ages of eight and twelve use this size of the ball. Size 5 balls would be the typical dimensions of adult play and are the typical size balls for all international play.
When shopping for a soccer ball it is important to have a watch because of the dimensions and construction. If you are unsure that the ball you’re thinking about is of good quality, look to find out if the ball is accepted by either FIFA or NFHS. FIFA, Fédération Internationale de Football Association, and NFHS, the National Federation of State High School Association, equally approve balls that satisfy the strict specifications that are outlined by every organization. If you purchase a ball that is approved by either governing body then you are almost assured of a ball that is of top quality in both construction and performance.
## How To Improve Your Soccer Dribbling Skills
Soccer is a great game to be played, but it is also a great game to be watched. Someone that’s not knowledgeable about soccer might wonder just what it is making it the most popular sport on Earth.
The same as basketball has its slam dunks and alley oops, baseball its own home boxing or runs its own knockouts, football has a lot of key components which make it a highly enjoyable game: goals passes, tackles and dribbling.
It is the last one that I value most, since I believe that a good dribble isn’t just spectacular, but it can be particularly effective in a game, so I’m going to dedicate the next article to explaining how football dribbling works, how you are able to train it, concentrate on a couple of special dribbling moves and determine what soccer abilities affect your dribbling skill.
## Soccer Dribbling — Introduction into the World of Ankle Breakers
By definition, soccer dribbling is a method utilized by the ball carrier to pass the ball beyond an immediate competitor, without devoting ownership. This might be a tight dictionary-like statement, but in fact, soccer dribbling is as easy in theory as that: do whatever you can (from the bounds of the football laws) to get past your opponent and maintain possession of the ball. However, “getting beyond your competition” requires a little explaining.
In a first glance, which may seem restricted to downhill exceeding your competitor on the pitch, even when in truth dribbling can be done sideways, or even back towards your own half (when you would like to avoid a handle like) and in many cases it’s only a way of clearing up the room to find a pass or shot at. This doesn’t necessarily have the ball on the opposing side of your competitor, but it will avoid him, letting you continue the drama to your teammates.
There are lots of sorts of dribbles which have different purposes from the game and by forms of dribbles, I do not mean specific moves, or specific tricks, but rather game mechanisms involving dribbles which have a particular end-goal.
Stress Avoiding Dribbles — This is possibly the most common type of dribble from the game and as a central midfielder almost every touch of the ball that you get will have to be followed closely by a pressure avoiding dribble. This entails a quick flick of the ball in a place with a few vacant space when being under pressure from a competition and its intention is to give you a few moments to perform a pass, or sometimes a shot.
An essential facet of pressure preventing dribbles is understanding where you are going to move the ball even before you really get to it. Take into account that you’ll likely only have a couple of moments to pass or shoot the ball after such a dribble because the defender will probably be quick to place himself in front of the ball again, or try to handle you decisively.
Great examples of players using this kind of dribble effectively include Ronaldinho, Clarence Seedorf or even Michael Ballack.
Rate Dribbles — Speed dribbles are mostly popular with wing-backs since the wings are often more open and free and they allow a fast player to virtually throw the ball forward and run to get it again, smoking one or two opponents in the process.
Speed dribbles aren’t very fancy concerning ball control, however, the dribbler should concentrate on pushing the ball forward just right so that he doesn’t lose ownership to an opposing defender or throw the ball out of the playing bonds.
Receiving Dribbles — This is probably one of the most effective tactics to dribble, but it can also be the toughest: moving about your opponent straight from getting the ball. It can throw a whole defense off balance and create havoc for the opposing group, but you really have to be a proficient player to produce great getting dribbles.
First of all, you have to have impressive vision on the pitch, knowing precisely where your competitors are and where you’ll get some open space on the field, even before the ball reaches your foot.
Second, you need to be in excellent control of the ball receiving technique, since shoving it too soft or too difficult will ruin the dribble. Last but not the least, you need to use your body to receive your direct opponent off balance, letting you turn and push the ball into open space unhindered.
You are probably tempted to state that ball management is the only skill involved with dribbling, but in truth, there are a lot more factors that have to be added up to a thriving dribble.
Ball control is the ability to maneuver with the ball, without losing possession, so it essentially requires you to understand how hard to hit the ball when moving forward with it, in addition to organizing your own body movement in this way that lets you follow up and get the ball to your feet again.
Ball control also refers to being able to snare or receive a ball without pushing it too further away from your body, and that, in our instance, is extremely useful in receiving dribbles. Ball control is important for all types of players when dribbling, regardless of their position on the pitch.
Strength — Strength is an important element in receiving dribbles and positional ones, as it allows you to utilize your own body for a wall between the ball and your opponent, repositioning him or her so you create an advantage towards the enemy target.
If you look at some of the players who are considered the best dribbles out there today, such as Ronaldinho or Zinedine Zidane, you’ll see that they very often use their body in dribbles, in order to transcend an opponent or alleviate themselves of stress and locate a pass or a shot.
In what regards soccer dribbling, power is most important for central midfielders, that is under constant pressure from the opposing group and for attackers who will have to use their bodies to shield the ball before they could unleash a dribble from the box.
Speed — Being faster than your competitor is one thing, but knowing how to find the ball past him AND gain back possession is the key to a successful rate dribble. If the rate difference is greatly in the favor of the attacker, he can simply push the ball forwards along the pitch at an empty region and run for it.
If space is limited, or when the speed difference isn’t too great between the ball carrier and the guardian, the instant you initiate the dribble is extremely important.
You will need to start increasing the pace when the guardian is off equilibrium (possibly after you throw him off-balance using a movement in the body) and be sure you have sufficient space to maneuver around him.
Weaker Foot Ability — Dribbles often involve using both feet and assorted parts of the foot (the sole, the instep, the backheel, the front, the exterior or the inside) so as to work properly and your weaker foot can often cause you difficulties.
Being unsure of your weaker foot will probably leave you quite vulnerable in front of a smart guardian because he will know there is just one direction you can dribble towards utilizing your strong foot.
You are able to train dribbling separately, with a teammate or within a group. Be sure to use both feet, so you know how to dribble in each direction and using your weaker foot also.
Pairing up with a teammate is also a wonderful way to practice your soccer dribbling. This technique has two advantages: first of all, it lets you train your dribbles against a real competitor, which affects quite some parameters in the way you train in comparison to averting ground obstacles.
Second, assuming you change functions with your teammate every now and then, permitting him to become the dribbler, you’ll also function as a defender and will find out how defenders consider when needing to deal with a competitor’s dribble.
Understanding your “enemy” will help you read your match opponents such as an open book and you’ll soon understand what to do in almost any game scenario where a dribble is demanded.
Group dribbling training sessions largely concentrate on pressure preventing dribbles rather than separately going around one competitor. Practicing at a group simulates match requirements best and it is a good way to work out your dribbling skill obviously, instead of pressure like you would have with an individual or paired training.
## Soccer Nutrition
Consider an average human anatomy as a small production shop. It does not expect a lot of prime materials, but in addition, it fails to carry out spectacularly. If a typical human’s body is a production shop, an athlete’s body is an advanced mill that can create outstanding performances, but also needs wagons of prime substance.
Your aim as a soccer player is to get that factory of yours working at max efficiency but it will also be your role to provide it with the proper (and the ideal quantity of) prime substances. This is where soccer nutrition comes into play and if you follow along to the content below, I’ll supply you with a few of the most important information regarding what you need to eat and when you need to eat it.
Proteins are the building blocks of life and they’re the ones responsible for muscular growth. As a football player, muscular growth should be of overriding importance, so your football diet might have to include plenty of proteins. Proteins normally come from animal products, like eggs, milk or meat, but they could also be found in soya, lettuce and some other specific veggies.
The carbohydrates are your body’s fuel supply and exactly like a car needs gas to function properly, so does your body need carbohydrates for each and every one of its functions.
Every motion which you just do eats up a bit of the carbohydrate fuel…move your hand to scratch your head, you lost some carbs…knock on the door…you lost some carbohydrates…even when we blink we shed a few carbs; you can imagine the amount of fuel needed in a football match then. But carbohydrates aren’t just energy boosters, they are also a part of a lot of bodily functions, such as the immune system, blood clotting, muscle growth and development.
## Soccer Nutrition — Pre Game and Post Game Soccer Diets
The problem seeing pre-game and post-game meals is quite debated, largely since the Internet is crammed with a lot of misleading advice on the subject. Generally, your operation in a game will be determined by what you eat beginning with 2-3 days before the match day and the final meal you require a few hours before the game begins. It’s a good idea to consume carbohydrate-rich food a couple of days ahead of the game and you can also receive a rich carbohydrate snack 3-4 hours prior to the match starts.
You will also need to look for foods which have a low glycemic index (low GI) prior to the match, since they are shown to expand your endurance and attempt capabilities for a short period of time. Just make sure that the very low GI food you consume pre match aren’t also high fibred, as dense fiber is absorbed harder by the human body and that food will just dangle around on your stomach as you run. Therefore a good pre-match football diet should comprise:
• Low GI foods
• High Carbohydrate foods
After the game is finished, the extensive effort will have you diminished and your muscles will be very vulnerable, because you consumed all the carbs in them. That’s the reason why a rich carbohydrate meal is vital after a game. Studies suggest that the body is better at consuming these fats and other nutrients from food 1 to 2 hours after prolonged work.
High GI foods can also help you regain your muscles faster and you will not feel tired and conquer all day long, as the carbohydrates and glycemin from the post game meal may work towards getting you back
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مثالی از قانون AND
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Examples of the AND Rule
In the last video we’ve learned about the simplified AND rule. AND just means multiply when we’re dealing with independent events. So, in this video we’re going to look at some examples of this. So, first of all, what is the probability of tossing three coins and getting heads, heads, heads? Well, first of all, notice that the three coins are independent, whatever happens to one coin is gonna have no influence on another coin.
So, we have three independent events, probability is one half for getting heads in one case, one half, one half, and so the three one halves just get multiplied for a probability of one eighth. One eighth is the probability of tossing three heads. Another question.
What is the probability of rolling two six-sided dice and getting snake eyes? That’s one on each die. So snake eyes would look something like this. Well, again, the two die don’t influence each other. So, these are independent events also. The probability of getting a one on one six sided die is one sixth, so one sixth on one, one six on the other, they’re independent so that means and means multiply.
Multiply them we get 1 over 36, the probability of rolling snake eyes when you roll 2 dice. Three cards are selected from a full deck, each time with replacement, what is the probability of selecting three spades in a row? So again. Let’s make sure that we are perfectly clear on this idea of with replacement.
What that means is I shuffle the deck, I pull a card out, I look at it. Then I put that card back in the deck, shuffle it again. So that the second card I pick is being picked from a full shuffled deck. There are 52 cards what I make the second pick. Now I’ll look at that second card, I record it, I put it back into the deck, I shuffle again, so now when I pick the third choice I’m also picking from a full deck of 52 cards, a full shuffled deck, so all three cards are picked from exactly the same conditions.
A full shuffled 52 card deck. That’s what’s going on with replacement. Well, of course, in a modern deck there are four suits. There are hearts, diamonds, clubs and spades and those are evenly divided. So the chance of picking one spade is one quarter. And then, on the second pick it’s also one quarter.
Then on the third pick it’s also one quarter, because we’re selecting with replacement because each time the choice is made from the same initial conditions the events are independent, the choices are independent. And that’s always true with replacement. That means if we get to multiply, we get 1 over 64.
The probability of selecting three spades in a row. Finally, we get this algebraic equation. We don’t even know what the events are. But we’re just told that they’re independent. Sometimes problem will just tell you A and B are independent. Probability of A is 0.6, probability of B is 0.8, what does the probability of A or B mean?
Okay, well this is a bit of a curve-ball. We are talking about an And rule and here we have an Or rule, what’s going on here? Well let’s think about this. This is the general Or rule. And incidentally, if the events are independent that does, that means they can not be mutually exclusive.
If the events are mutually exclusive they are definitely having an effect on one another. Because if one happens it prevents the other one from happening. So that is definitely a very different case from independent. If they’re independent, they’re not mutually exclusive, we need to use the generalized and rule.
So, let’s think about this. We know the probability of A, we know the probability of B. The probability of A and B, well we can get that, because the events are independent, we can get this by multiplying. That would be 0.6 times 0.8, the probability of A and B would be the probability of A times the probability of B, 0.6 times 0.8 is 0.48.
So now we know all 3 terms and now we can say we have 0.6 plus 0.8 minus 0.48. So this is 1.4 minus 0.48 and this winds up being 0.92. Which is the probability of A or B.
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# Count Unique Items in a Column Based on Unique Items in Two Other Columns
1. ## Count Unique Items in a Column Based on Unique Items in Two Other Columns
I'm using this array formula to count the number of Unique Items in Column B for each Unique Item in Column A
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I need to expand this formula so that it now counts the number of unique Colours in column C for each unique Animal in column B with a unique letter in column A. So when the entries are changed in cells J14 and J15 it gives the answer in cell J16. Please see sample attached.
Many thanks
2. ## Re: Count Unique Items in a Column Based on Unique Items in Two Other Columns
Seems to me this is far simpler with a Pivot table and avoids all the messy stuff associated with complex formulae - see attached
3. ## Re: Count Unique Items in a Column Based on Unique Items in Two Other Columns
Hi Richard,
It would be, but I can't use a Pivot Table for what I need to achieve which is why I'm tying to find a formulaic solution.
Many thanks
4. ## Re: Count Unique Items in a Column Based on Unique Items in Two Other Columns
But if I could use a Pivot Table to identify the count then the next problem I have is then being able to populate different cells with the results, so for example, based on the pivot table we know there are 3 colours for Cat with the Letter B, what I then need to be able to do is for specific cells to be populated with those colours, e.g. F4 with the first colour, F5 with the second colour and F6 with the third colour and so on, bearing in mind there could be any number of unique colours that meet the criteria.
Many thanks
5. ## Re: Count Unique Items in a Column Based on Unique Items in Two Other Columns
Not sure what results you expect.
See if this does what you want.
Array entered**:
=SUM(IF(FREQUENCY(IF((A2:A27=J14)*(B2:B27=J15),MATCH(C2:C27,C2:C27,0)),ROW(C2:C27)-ROW(C2)+1),1))
** array formulas need to be entered using the key
combination of CTRL,SHIFT,ENTER (not just ENTER).
Hold down both the CTRL key and the SHIFT key
then hit ENTER.
6. ## Re: Count Unique Items in a Column Based on Unique Items in Two Other Columns
Try this array formula.
=SUM(IFERROR(1/COUNTIFS(A2:A27,J14,B2:B27,J15,Colour,Colour),0))
Although I think it only works because all of the colours are unique, if you have the same colour for several animals I think results could be wrong.
7. ## Re: Count Unique Items in a Column Based on Unique Items in Two Other Columns
Hi,
The point about a pivot table is that the view of the data can be simply changed by dragging and dropping field names around. If you need colours reported in the table for a particular letter and animal then move those two fields into the Report Filter area of the PT - see attached for an example.
OK the colours may not be in F4, F5 & F6 but they are certainly in F5, F6 & F7 with the total of 3 in F8. But if you just want the total 3 then just drag the colours field into the Report Filter area.
8. ## Re: Count Unique Items in a Column Based on Unique Items in Two Other Columns
Hi Richard,
I understand what you are saying and I think you are right, I can probably achieve what I need by adding the pivot to a hidden sheet and pulling the values from there.
Many thanks for your help, I think I was trying to over complicte this, though I'm curious to see if it can be achieved with a formula alone but for now I'll adopt the PT route I think.
9. ## Re: Count Unique Items in a Column Based on Unique Items in Two Other Columns
Originally Posted by HangMan
I'm curious to see if it can be achieved with a formula alone
Did you try either of the formulas that were suggested?
10. ## Re: Count Unique Items in a Column Based on Unique Items in Two Other Columns
Hi Tony and Jason,
Apologies, sometimes the forum behaves strangely, I didn't receive any notification of either of your repies, only Richards, so subsequently I didn't see either of your replies until now.
Many thanks, both solutions work perfectly and overcome my issue with not wanting to use a pivot table...
Very much appreciated, thank you...
11. ## Re: Count Unique Items in a Column Based on Unique Items in Two Other Columns
Good deal. Thanks for the feedback!
12. ## Re: Count Unique Items in a Column Based on Unique Items in Two Other Columns
If you're referring to email notifications then you only get notified of the first new post since your last visit.
On the forum, User CP, you only see the latest post at the time you refreshed the page, and it will only be shown in bold if the post was submitted after you last viewed the thread.
It is down to you to check if you have had one reply, or several.
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What Is an Absolute Value?
An absolute value, or modulus, of a number is the non-negative value of the number without regard to its sign.
The absolute value of a number is denoted by || either side of the number.
For example, the aboslute value of 2 is denoted below, and is equal to 2:
What Is the Absolute Value of a Positive Number?
The absolute value of a positive number is just that positive number. For example:
What Is the Absolute Value of a Negative Number?
The absolute value of a negative number is the non-negative value of that number. It is that number with the minus sign ( - ) removed. For example:
Visualizing Absolute Values on a Number Line
On a number line, the absolute value of a number is given by its distance from zero.
For example, it can be seen that 3 is 3 away from zero, and so has an absolute value of 3. It can also be seen that -2 is 2 away from 0, and so has an absolute value of 2:
show
THE ABSOLUTE VALUE OF ZERO
The absolute value of 0 is 0. 0 is neither a positive nor negative number.
As the absolute value of any other number is always positive, the absolute value of any number must be greater or equal to 0.
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# Thread: Simplify expression
1. ## Simplify expression
Hi there
Can anyone tell me how to simplify this expression
d/sqrt(2*A*d/h)*A+sqrt(2*A*d/h)/2*h
to this expression
sqrt(2*A*d*h)
Thank you i advance
2. ## Re: Simplify expression
Would be more easier to read if we had some brackets to make the expression less ambiguous. Otherwise, there's a bit of guess work involved.
3. ## Re: Simplify expression
Sorry. Would this be all right:
4. ## Re: Simplify expression
Hello, mrtn!
$\text{Show that: }\:\left(\frac{d}{\sqrt{\frac{2Ad}{h}}}\times A\right) + \left(\frac{\sqrt{\frac{2Ad}{h}}}{2} \times h\right) \;=\;\sqrt{2Adh}$
The first product is:
. . $\displaystyle\dfrac{d}{\frac{\sqrt{2}\sqrt{A}\sqrt {d}}{\sqrt{h}}} \cdot A \;=\; \frac{Ad\sqrt{h}}{\sqrt{2}\sqrt{A}\sqrt{d}} \;=\; \frac{\sqrt{A}\sqrt{d}\sqrt{h}}{\sqrt{2}} \;=\;\frac{\sqrt{Adh}}{\sqrt{2}}$
The second product is:
. . $\displaystyle\dfrac{\frac{\sqrt{2}\sqrt{A}\sqrt{d} }{\sqrt{h}}}{2} \cdot h \;=\; \dfrac{\sqrt{2}\sqrt{A}\sqrt{d}\,h}{2\sqrt{h}}\;= \; \frac{\sqrt{A}\sqrt{d}\sqrt{h}}{\sqrt{2}} \;=\; \frac{\sqrt{Adh}}{\sqrt{2}}$
Their sum is:
. . $\displaystyle\frac{\sqrt{Adh}}{\sqrt{2}} + \frac{\sqrt{Adh}}{\sqrt{2}} \;=\; \frac{2\sqrt{Adh}}{\sqrt{2}} \;=\;\sqrt{2}\sqrt{Adh} \;=\;\sqrt{2Adh}$
5. ## Re: Simplify expression
i like algibra
6. ## Re: Simplify expression
Thank you Soroban.
My next challenge is to simplify the first product as you just wrote (I know I am on deep water):
Show that
I need to know step by step, which math rules are used etc. to get the final expressions.
Do you know any software, which are able to help me with this?
Or else I can keep you guys busy, like forever.
7. ## Re: Simplify expression
algebra.help gets you half way there I am student and I am sure there is a law to follow for algebra hope the link helps you to help yourself
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Print Version
Solvers Puzzles Basic Strategies Tough Strategies Diabolical Strategies Extreme Strategies Depreciated Strategies Str8ts Other
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Double Line/Box Reduction
This article is part two of the new Double Intersection Removal strategy specifically for Jigsaw Sudoku puzzles. I recommend reading part one on Double Pointing Pairs first. I am indebted to Brain Hobson's observations for this excellent prod in the right direction and I am appreciating Jigsaws in a whole new way.
With Line/Box Reduction we take one box and one line (a row or column) and look for a specific number that only exists in the overlap of that box and line. So by looking along a row you may see that all the 2s are in the same box (or shape). If that is the case we know the solution will be one of those cells - otherwise the row would lose that number altogether. From this idea we can remove the rest of those candidates from the box or shape in question.
With Jigsaw Puzzles we have shapes that overlap rows and columns in a much more chaotic and interesting way. Unlike normal Sudoku, we can ask, what if we have two rows (or columns) where candidates X are present in just two shapes. In that case, we can be certain that there will be two solutions (one for both boxes and both rows/columns) so any other candidate X in those shapes can simply go. (The doubling up theme is not new. Consider how X-Wings are a doubled up version of Hidden Pairs).
Here is a very illustrative puzzle with lots of examples - both of the normal Intersection Removal and the doubled up version. Down the center are two columns where 2 is present in shapes 4 and 5 - and just those two shapes. Shapes 4 and 5 have many other possible 2s - but they can all be removed. The solver returns the result:Double Line/Box ReductionThe two columns 4 and 5 contain 2 in just boxes 1 and 2, so...- 2 can be removed from A1- 2 can be removed from A2- 2 can be removed from B2- 2 can be removed from H6- 2 can be removed from J8 Double Line/Box Reduction on 2: Load Example or : From the Start
In almost the next step we can remove a bunch of 3s as well, this time but looking at the first two rows and how they overlap with boxes 1 and 8.Double Line/Box ReductionThe two rows 1 and 2 contain 3 in just boxes 1 and 8, so...- 3 can be removed from D4- 3 can be removed from D5- 3 can be removed from E5 Double Line/Box Reduction on 3: Load Example or : From the Start
In exactly the same way doubled up Pointing Pairs work - the two rows and columns do not need to be adjacent - they must only overlap the same two shapes at some point. Here is an example with 4s being removed on a "tough" grade puzzle where the rows are not adjacent.Double Line/Box ReductionThe two rows 4 and 6 contain 4 in just boxes 5 and 6, so...- 4 can be removed from G4- 4 can be removed from H4 Double Line/Box Reduction on 4: Load Example or : From the Start
Go back to Double Pointing Pairs Continue to Law of Leftovers
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## Friday 26-Jul-2013
### ... by: ralph maier
in todays puzzle thereis a trivial exampple of 4 line box reduction of4s in column 6 7 89 so the 4s in ABC5 can be eliminated
## Wednesday 10-Jul-2013
### ... by: ralph maier
Dear Mr Stuart
I do Jigsaw seldom mostly Str8ts and Sudoku. Anyhow I saw for the first time DOUBLE box REDUCTION. This can be genmeralized to N lines or boxes.
Andrew Stuart writes:
Correct, this can be extended to triples or further. Depends on the shapes. I mention it on the next page but I've not extended the solver that far.
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Enter the total number of x-rays performed and the total cost of running the x-ray into the calculator to determine the cost per x-ray.
## Cost Per X-Ray Formula
The following equation is used to calculate the Cost Per X-Ray.
CPX = TRC / #XR
• Where CPX is the cost per x-ray
• TRC is the total running costs, including labor and machine costs
• #XR is the number of x-rays performed
## What is Cost Per X-Ray?
Definition:
A cost per x-ray is the total cost a center spends in labor, operational, machine, and overhead, per x-ray performed.
## How to Calculate Cost Per X-Ray?
Example Problem:
The following example outlines the steps and information needed to calculate Cost Per X-Ray.
First, determine the total running costs. In this example, the total running costs are \$5,000.00 for the week.
Next, determine the number of x-rays performed. For this problem, the number of x-rays over the week was 20.
Finally, calculate the cost per x-ray using the formula above:
CPX = TRC / #XR
CPX = \$5,000.00 / 20
CPX = \$250 / x-ray
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## Linear algebra for balancing chemical reaction equations
The balancing of chemical reaction equations is familiar for many, either from high school chemistry or introductory university chemistry. A simple example of such a problem is the reaction equation for the combustion of methane in oxygen:
$aCH_4 + bO_2 \rightarrow cCO_2 + dH_2 O$,
for which we are supposed to find coefficients a,b,c and d that make the reaction possible in terms of conservation of the number of atoms of any element that is present. In this case, it is rather easy to see that the number of $CO_2$ molecules must be equal to the number of $CH_4$ molecules, and that the number of water molecules must be twice that. Therefore the possible equations are
$CH_4 + 2O_2 \rightarrow CO_2 + 2H_2 O$ , or
$2CH_4 + 4O_2 \rightarrow 2CO_2 + 4H_2 O$, or
$3CH_4 + 6O_2 \rightarrow 3CO_2 + 6H_2 O$ ,
or any other “multiple” of the first equation (obtained by multiplying all the coefficients in the equation with the same number).
Usually this kind of balancing is done mainly by intuition and trying to guess the right coefficients, but this can be difficult if the reactants and product contain many different elements, each of which are present in more than one kind of molecule. One example of a more complicated (but still quite simple) reaction is the reaction of chlorine dioxide ($ClO_2$) with alkaline water to form chlorate ($ClO_3^-$), chlorite ($ClO_2^-$) and water:
$aClO_2 + bOH^- \rightarrow cClO_2^- + dClO_3^- + eH_2 O$ .
Now, an easily applicable way to solve a set of acceptable coefficients a,b,c,d and e is to explicitly write the balancing condition equations for each element that appears in the equation. These balancing conditions are:
$a = c + d$ (balancing of chlorine)
$2a + b = 2c + 3d + e$ (balancing of oxygen)
$b = 2e$ (balancing of hydrogen)
$b = c + d$ (balancing of electric charge)
This can be solved with many kinds of mathematical software, either free (there’s lots of free math programs especially for Linux) or commercial (like Wolfram Mathematica).
But there’s one another thing. Because the number of unknowns in the system of equations above is larger than the number of equations by one, we need to decide the value of one of the coefficients $a, \dots , b$ arbitrarily. So let’s say that $a = 1$. Now the set of equations is:
$a = 1$ (arbitrary choice of one coeffiecient)
$a = c + d$ (balancing of chlorine)
$2a + b = 2c + 3d + e$ (balancing of oxygen)
$b = 2e$ (balancing of hydrogen)
$b = c + d$ (balancing of electric charge)
Using the column vector
the system of equations can be written in matrix form
This can be solved in Mathematica by writing the command
Solve[{a == 1, a == c + d, 2a + b == 2*c + 3*d + e, b == 2e, b == c + d}]
and the result is
$a = 2$,
$b = 2$,
$c = 1$,
$d = 1$,
$e = 1$,
which leads to the reaction equation
$2ClO_2 + 2OH^- \rightarrow ClO_2^- + ClO_3^- + H_2 O$
which doesn’t even need to be multiplied with any constant to make all the coefficients integers (usually that is necessary).
IMAGE SOURCE: The feature image of an Erlenmeyer flask on this blog post is a public domain image from Wikipedia, address https://en.wikipedia.org/wiki/Erlenmeyer_flask#/media/File:Duran_erlenmeyer_flask_narrow_neck_50ml.jpg .
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# Inverse Of Function
Inverse Of Function In mathematics, an inverse function is a function that undoes another function: If an input x into the function produces an output y, then putting y into the inverse function g produces the output x, and vice versa. i.e., (x)=y, and g(y)=x. More directly, g((x))=x, meaning g(x) composed with (x) leaves x unchanged. A function that has an inverse is called invertible; the inverse function is then uniquely determined by and is denoted by 1 (read f inverse, not to be confused with exponentiation). Definitions :- Instead of considering the inverses for individual inputs and outputs, one can think of the function as sending the whole set of inputs, the domain, to a set of outputs, the range. Let be a function whose domain is the set X, and whose range is the set Y. Then is invertible if there exists a function g with domain Y and range X, with the property: If is invertible, the function g is unique; in other words, there can be at most one function g satisfying this property. That function g is then called the inverse of , denoted by 1.
## Know More About :- Reducing Fractions Calculator
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Stated otherwise, a function is invertible if and only if its inverse relation is a function on the range Y, in which case the inverse relation is the inverse function. Not all functions have an inverse. For this rule to be applicable, each element y Y must correspond to no more than one x X; a function with this property is called one-to-one, or information-preserving, or an injection. In this if we inverse the function we will get the same result as q. We can find the inverse just by the swapping of value p and q and we get a new relation which represents the inverse of a function. But we have to note that it is not necessary that an inverse is always a function. Let us assume that we are given a function say f(p) and getting the output q then we will denote the inverse function with q input as f-1 (q). For understanding the inverse function in a better manner let us take another example. Assume a function f(p)=2p+6 then what will be its inverse function lets take a look: f(p) = 2p + 6 p ----> 2*p --- > 2*p + 6 This function will be processed in this manner. For calculating its inverse just does the opposite of each task or inverse of each task like replace addition with subtraction and multiplication with division. f-1(q) will be (q - 6)/2 < ----q-6 <---- q
## So the inverse function f-1(q) of function f(p) is:
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f-1(q) = (q - 6)/2 Mathematics approach: step 1: step 2: f(p) = 2p + 6 Put q at the place of f(p) q = 2p + 6 step 3: Take Integer value at left side q - 6 = 2p step 4: Divide left side by the coefficient of p (q - 6) /2 = p Step 5: p = (q - 6) / 2
Step 6: Put f-1(q) at the place of p. f-1(q) = (q - 6) /2 This is the all about inverse of function.
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# calculate the expected return and standard deviation of a portfolio consisting of equal parts
17 / 01 / 2019 Research Papers
This paper circulates around the core theme of calculate the expected return and standard deviation of a portfolio consisting of equal parts together with its essential aspects. It has been reviewed and purchased by the majority of students thus, this paper is rated 4.8 out of 5 points by the students. In addition to this, the price of this paper commences from £ 99. To get this paper written from the scratch, order this assignment now. 100% confidential, 100% plagiarism-free.
The following are monthly percentage price changes for four market indexes. Month DJIA S&P… 1 answer below » The following are monthly percentage price changes for four market indexes. Month DJIA S&P 500 Russell 2000 Nikkei 1 0.03 0.02 0.04 4.00 2 9.07 0.06 0.10 (0.02) 3 (0.02) (0.01) (0.04) 0.07 4 0.01 0.03 0.03 0.02 5 0.05 0.04 0.11 0.02 6 (0.06) (0.04) (0.08) 0.06 Compute the following: a. Average monthly rate of return for each index. b. Standard deviation for each index c. Covariance between the rates of return for the following indexes: DJIA – S&P500 S&P 500 – Russell 2000 S&P 500 – Nikkei Russell 2000 – Nikkei d. The correlation coefficients for the same four combinations e. View complete question » The following are monthly percentage price changes for four market indexes. Month DJIA S&P 500 Russell 2000 Nikkei 1 0.03 0.02 0.04 4.00 2 9.07 0.06 0.10 (0.02) 3 (0.02) (0.01) (0.04) 0.07 4 0.01 0.03 0.03 0.02 5 0.05 0.04 0.11 0.02 6 (0.06) (0.04) (0.08) 0.06 Compute the following: a. Average monthly rate of return for each index. b. Standard deviation for each index c. Covariance between the rates of return for the following indexes: DJIA – S&P500 S&P 500 – Russell 2000 S&P 500 – Nikkei Russell 2000 – Nikkei d. The correlation coefficients for the same four combinations e. Using the answers from parts (a), (b), and (d), calculate the expected return and standard deviation of a portfolio consisting of equal parts of: (1) the S&P and the Russell 2000, and (2) the S&P and the Nikkei. Discuss the two portfolios. please show process step by step thanks Additional Requirements Level of Detail: Show all work View less » Aug 01 2015 06:08 PM
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# Integration of root tanx + root cotx | Find ∫(√tanx+√cotx) dx
The integration of root tanx + root cotx is equal to √2sin-1(sinx−cosx)+C. Here we learn how to integrate √tanx+√cotx.
The integral formula of $\sqrt{\tan x}+\sqrt{\cot x}$ is given by
## Find ∫(√tanx+√cotx) dx
$\int (\sqrt{\tan x}+\sqrt{\cot x}) dx$
= $\int \Big(\dfrac{\sqrt{\sin x}}{\sqrt{\cos x}}+\dfrac{\sqrt{\cos x}}{\sqrt{\sin x}} \Big) dx$
= $\int \dfrac{\sin x + \cos x}{\sqrt{\sin x} \sqrt{\cos x}} dx$
= $\sqrt{2} \int \dfrac{\sin x + \cos x}{\sqrt{2\sin x \cos x}} dx$
= $\sqrt{2} \int \dfrac{\sin x + \cos x}{\sqrt{2\sin x \cos x+1-\sin^2x-\cos^2x}} dx$ as we know sin2x+cos2x=1.
= $\sqrt{2} \int \dfrac{\sin x + \cos x}{\sqrt{1-(\sin x -\cos x)^2}} dx$
[ Put sinx -cosx =t, so that (cosx +sinx)dx =dt ]
= $\sqrt{2} \int \dfrac{dt}{\sqrt{1-t^2}}$
= $\sqrt{2} \sin^{-1}t+C$
= $\sqrt{2} \sin^{-1}(\sin x -\cos x)+C$
So the integral of root(tanx) + root(cotx) is equal to √2sin-1(sinx−cosx)+C, where C is an integral constant.
More Integrals:
Find ∫e3x dx
## FAQs
### Q1: What is the integration of √tanx+√cotx?
Answer: The integration of √tanx+√cotx is equal to √2sin-1(sinx−cosx)+C.
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Probability
Get Probability essential facts below. View Videos or join the Probability discussion. Add Probability to your PopFlock.com topic list for future reference or share this resource on social media.
Probability
The probabilities of rolling several numbers using two dice.
Probability is the branch of mathematics concerning numerical descriptions of how likely an event is to occur, or how likely it is that a proposition is true. The probability of an event is a number between 0 and 1, where, roughly speaking, 0 indicates impossibility of the event and 1 indicates certainty.[note 1][1][2] The higher the probability of an event, the more likely it is that the event will occur. A simple example is the tossing of a fair (unbiased) coin. Since the coin is fair, the two outcomes ("heads" and "tails") are both equally probable; the probability of "heads" equals the probability of "tails"; and since no other outcomes are possible, the probability of either "heads" or "tails" is 1/2 (which could also be written as 0.5 or 50%).
These concepts have been given an axiomatic mathematical formalization in probability theory, which is used widely in areas of study such as statistics, mathematics, science, finance, gambling, artificial intelligence, machine learning, computer science, game theory, and philosophy to, for example, draw inferences about the expected frequency of events. Probability theory is also used to describe the underlying mechanics and regularities of complex systems.[3]
## Interpretations
When dealing with experiments that are random and well-defined in a purely theoretical setting (like tossing a fair coin), probabilities can be numerically described by the number of desired outcomes, divided by the total number of all outcomes. For example, tossing a fair coin twice will yield "head-head", "head-tail", "tail-head", and "tail-tail" outcomes. The probability of getting an outcome of "head-head" is 1 out of 4 outcomes, or, in numerical terms, 1/4, 0.25 or 25%. However, when it comes to practical application, there are two major competing categories of probability interpretations, whose adherents hold different views about the fundamental nature of probability:
• Objectivists assign numbers to describe some objective or physical state of affairs. The most popular version of objective probability is frequentist probability, which claims that the probability of a random event denotes the relative frequency of occurrence of an experiment's outcome when the experiment is repeated indefinitely. This interpretation considers probability to be the relative frequency "in the long run" of outcomes.[4] A modification of this is propensity probability, which interprets probability as the tendency of some experiment to yield a certain outcome, even if it is performed only once.
• Subjectivists assign numbers per subjective probability, that is, as a degree of belief.[5] The degree of belief has been interpreted as "the price at which you would buy or sell a bet that pays 1 unit of utility if E, 0 if not E."[6] The most popular version of subjective probability is Bayesian probability, which includes expert knowledge as well as experimental data to produce probabilities. The expert knowledge is represented by some (subjective) prior probability distribution. These data are incorporated in a likelihood function. The product of the prior and the likelihood, when normalized, results in a posterior probability distribution that incorporates all the information known to date.[7] By Aumann's agreement theorem, Bayesian agents whose prior beliefs are similar will end up with similar posterior beliefs. However, sufficiently different priors can lead to different conclusions, regardless of how much information the agents share.[8]
## Etymology
The word probability derives from the Latin probabilitas, which can also mean "probity", a measure of the authority of a witness in a legal case in Europe, and often correlated with the witness's nobility. In a sense, this differs much from the modern meaning of probability, which in contrast is a measure of the weight of empirical evidence, and is arrived at from inductive reasoning and statistical inference.[9]
## History
The scientific study of probability is a modern development of mathematics. Gambling shows that there has been an interest in quantifying the ideas of probability for millennia, but exact mathematical descriptions arose much later. There are reasons for the slow development of the mathematics of probability. Whereas games of chance provided the impetus for the mathematical study of probability, fundamental issues [note 2] are still obscured by the superstitions of gamblers.[10]
According to Richard Jeffrey, "Before the middle of the seventeenth century, the term 'probable' (Latin probabilis) meant approvable, and was applied in that sense, univocally, to opinion and to action. A probable action or opinion was one such as sensible people would undertake or hold, in the circumstances."[11] However, in legal contexts especially, 'probable' could also apply to propositions for which there was good evidence.[12]
Al-Kindi's Book of Cryptographic Messages contains the earliest known use of statistical inference (9th century)
The early form of statistical inference were developed by Middle Eastern mathematicians studying cryptography between the 8th and 13th centuries. Al-Khalil (717-786) wrote the Book of Cryptographic Messages which contains the first use of permutations and combinations to list all possible Arabic words with and without vowels. Al-Kindi (801-873) made the earliest known use of statistical inference in his work on cryptanalysis and frequency analysis. An important contribution of Ibn Adlan (1187-1268) was on sample size for use of frequency analysis.[13]
Gerolamo Cardano (16th century)
Christiaan Huygens published one of the first books on probability (17th century)
The sixteenth-century Italian polymath Gerolamo Cardano demonstrated the efficacy of defining odds as the ratio of favourable to unfavourable outcomes (which implies that the probability of an event is given by the ratio of favourable outcomes to the total number of possible outcomes[14]). Aside from the elementary work by Cardano, the doctrine of probabilities dates to the correspondence of Pierre de Fermat and Blaise Pascal (1654). Christiaan Huygens (1657) gave the earliest known scientific treatment of the subject.[15] Jakob Bernoulli's Ars Conjectandi (posthumous, 1713) and Abraham de Moivre's Doctrine of Chances (1718) treated the subject as a branch of mathematics.[16] See Ian Hacking's The Emergence of Probability[9] and James Franklin's The Science of Conjecture[17] for histories of the early development of the very concept of mathematical probability.
The theory of errors may be traced back to Roger Cotes's Opera Miscellanea (posthumous, 1722), but a memoir prepared by Thomas Simpson in 1755 (printed 1756) first applied the theory to the discussion of errors of observation.[18] The reprint (1757) of this memoir lays down the axioms that positive and negative errors are equally probable, and that certain assignable limits define the range of all errors. Simpson also discusses continuous errors and describes a probability curve.
The first two laws of error that were proposed both originated with Pierre-Simon Laplace. The first law was published in 1774, and stated that the frequency of an error could be expressed as an exponential function of the numerical magnitude of the error--disregarding sign. The second law of error was proposed in 1778 by Laplace, and stated that the frequency of the error is an exponential function of the square of the error.[19] The second law of error is called the normal distribution or the Gauss law. "It is difficult historically to attribute that law to Gauss, who in spite of his well-known precocity had probably not made this discovery before he was two years old."[19]
Daniel Bernoulli (1778) introduced the principle of the maximum product of the probabilities of a system of concurrent errors.
Carl Friedrich Gauss
Adrien-Marie Legendre (1805) developed the method of least squares, and introduced it in his Nouvelles méthodes pour la détermination des orbites des comètes (New Methods for Determining the Orbits of Comets).[20] In ignorance of Legendre's contribution, an Irish-American writer, Robert Adrain, editor of "The Analyst" (1808), first deduced the law of facility of error,
${\displaystyle \phi (x)=ce^{-h^{2}x^{2}},}$
where ${\displaystyle h}$ is a constant depending on precision of observation, and ${\displaystyle c}$ is a scale factor ensuring that the area under the curve equals 1. He gave two proofs, the second being essentially the same as John Herschel's (1850).[] Gauss gave the first proof that seems to have been known in Europe (the third after Adrain's) in 1809. Further proofs were given by Laplace (1810, 1812), Gauss (1823), James Ivory (1825, 1826), Hagen (1837), Friedrich Bessel (1838), W.F. Donkin (1844, 1856), and Morgan Crofton (1870). Other contributors were Ellis (1844), De Morgan (1864), Glaisher (1872), and Giovanni Schiaparelli (1875). Peters's (1856) formula[clarification needed] for r, the probable error of a single observation, is well known.
In the nineteenth century, authors on the general theory included Laplace, Sylvestre Lacroix (1816), Littrow (1833), Adolphe Quetelet (1853), Richard Dedekind (1860), Helmert (1872), Hermann Laurent (1873), Liagre, Didion and Karl Pearson. Augustus De Morgan and George Boole improved the exposition of the theory.
In 1906, Andrey Markov introduced[21] the notion of Markov chains, which played an important role in stochastic processes theory and its applications. The modern theory of probability based on the measure theory was developed by Andrey Kolmogorov in 1931.[22]
On the geometric side, contributors to The Educational Times were influential (Miller, Crofton, McColl, Wolstenholme, Watson, and Artemas Martin).[23] See integral geometry for more info.
## Theory
Like other theories, the theory of probability is a representation of its concepts in formal terms--that is, in terms that can be considered separately from their meaning. These formal terms are manipulated by the rules of mathematics and logic, and any results are interpreted or translated back into the problem domain.
There have been at least two successful attempts to formalize probability, namely the Kolmogorov formulation and the Cox formulation. In Kolmogorov's formulation (see also probability space), sets are interpreted as events and probability as a measure on a class of sets. In Cox's theorem, probability is taken as a primitive (i.e., not further analyzed), and the emphasis is on constructing a consistent assignment of probability values to propositions. In both cases, the laws of probability are the same, except for technical details.
There are other methods for quantifying uncertainty, such as the Dempster-Shafer theory or possibility theory, but those are essentially different and not compatible with the usually-understood laws of probability.
## Applications
Probability theory is applied in everyday life in risk assessment and modeling. The insurance industry and markets use actuarial science to determine pricing and make trading decisions. Governments apply probabilistic methods in environmental regulation, entitlement analysis, and financial regulation.
An example of the use of probability theory in equity trading is the effect of the perceived probability of any widespread Middle East conflict on oil prices, which have ripple effects in the economy as a whole. An assessment by a commodity trader that a war is more likely can send that commodity's prices up or down, and signals other traders of that opinion. Accordingly, the probabilities are neither assessed independently nor necessarily rationally. The theory of behavioral finance emerged to describe the effect of such groupthink on pricing, on policy, and on peace and conflict.[24]
In addition to financial assessment, probability can be used to analyze trends in biology (e.g., disease spread) as well as ecology (e.g., biological Punnett squares). As with finance, risk assessment can be used as a statistical tool to calculate the likelihood of undesirable events occurring, and can assist with implementing protocols to avoid encountering such circumstances. Probability is used to design games of chance so that casinos can make a guaranteed profit, yet provide payouts to players that are frequent enough to encourage continued play.[25]
The discovery of rigorous methods to assess and combine probability assessments has changed society.[26][]
Another significant application of probability theory in everyday life is reliability. Many consumer products, such as automobiles and consumer electronics, use reliability theory in product design to reduce the probability of failure. Failure probability may influence a manufacturer's decisions on a product's warranty.[27]
The cache language model and other statistical language models that are used in natural language processing are also examples of applications of probability theory.
## Mathematical treatment
Calculation of probability (risk) vs odds
Consider an experiment that can produce a number of results. The collection of all possible results is called the sample space of the experiment, sometimes denoted as ${\displaystyle \Omega }$.[28] The power set of the sample space is formed by considering all different collections of possible results. For example, rolling a die can produce six possible results. One collection of possible results gives an odd number on the die. Thus, the subset {1,3,5} is an element of the power set of the sample space of dice rolls. These collections are called "events". In this case, {1,3,5} is the event that the die falls on some odd number. If the results that actually occur fall in a given event, the event is said to have occurred.
A probability is a way of assigning every event a value between zero and one, with the requirement that the event made up of all possible results (in our example, the event {1,2,3,4,5,6}) is assigned a value of one. To qualify as a probability, the assignment of values must satisfy the requirement that for any collection of mutually exclusive events (events with no common results, such as the events {1,6}, {3}, and {2,4}), the probability that at least one of the events will occur is given by the sum of the probabilities of all the individual events.[29]
The probability of an event A is written as ${\displaystyle P(A)}$,[28][30] ${\displaystyle p(A)}$, or ${\displaystyle {\text{Pr}}(A)}$.[31] This mathematical definition of probability can extend to infinite sample spaces, and even uncountable sample spaces, using the concept of a measure.
The opposite or complement of an event A is the event [not A] (that is, the event of A not occurring), often denoted as ${\displaystyle A',A^{c}}$,[28] ${\displaystyle {\overline {A}},A^{\complement },\neg A}$, or ${\displaystyle {\sim }A}$; its probability is given by .[32] As an example, the chance of not rolling a six on a six-sided die is ${\displaystyle =1-{\tfrac {1}{6}}={\tfrac {5}{6}}}$. For a more comprehensive treatment, see Complementary event.
If two events A and B occur on a single performance of an experiment, this is called the intersection or joint probability of A and B, denoted as ${\displaystyle P(A\cap B)}$.[28]
### Independent events
If two events, A and B are independent then the joint probability is[30]
${\displaystyle P(A{\mbox{ and }}B)=P(A\cap B)=P(A)P(B).}$
For example, if two coins are flipped, then the chance of both being heads is ${\displaystyle {\tfrac {1}{2}}\times {\tfrac {1}{2}}={\tfrac {1}{4}}}$.[33]
### Mutually exclusive events
If either event A or event B can occur but never both simultaneously, then they are called mutually exclusive events.
If two events are mutually exclusive, then the probability of both occurring is denoted as ${\displaystyle P(A\cap B)}$ and
${\displaystyle P(A{\mbox{ and }}B)=P(A\cap B)=0}$
If two events are mutually exclusive, then the probability of either occurring is denoted as ${\displaystyle P(A\cup B)}$ and
${\displaystyle P(A{\mbox{ or }}B)=P(A\cup B)=P(A)+P(B)-P(A\cap B)=P(A)+P(B)-0=P(A)+P(B)}$
For example, the chance of rolling a 1 or 2 on a six-sided die is ${\displaystyle P(1{\mbox{ or }}2)=P(1)+P(2)={\tfrac {1}{6}}+{\tfrac {1}{6}}={\tfrac {1}{3}}.}$
### Not mutually exclusive events
If the events are not mutually exclusive then
${\displaystyle P\left(A{\hbox{ or }}B\right)=P(A\cup B)=P\left(A\right)+P\left(B\right)-P\left(A{\mbox{ and }}B\right).}$
For example, when drawing a single card at random from a regular deck of cards, the chance of getting a heart or a face card (J,Q,K) (or one that is both) is ${\displaystyle {\tfrac {13}{52}}+{\tfrac {12}{52}}-{\tfrac {3}{52}}={\tfrac {11}{26}}}$, since among the 52 cards of a deck, 13 are hearts, 12 are face cards, and 3 are both: here the possibilities included in the "3 that are both" are included in each of the "13 hearts" and the "12 face cards", but should only be counted once.
### Conditional probability
Conditional probability is the probability of some event A, given the occurrence of some other event B. Conditional probability is written ${\displaystyle P(A\mid B)}$,[28] and is read "the probability of A, given B". It is defined by[34]
${\displaystyle P(A\mid B)={\frac {P(A\cap B)}{P(B)}}.\,}$
If ${\displaystyle P(B)=0}$ then ${\displaystyle P(A\mid B)}$ is formally undefined by this expression. However, it is possible to define a conditional probability for some zero-probability events using a ?-algebra of such events (such as those arising from a continuous random variable).[]
For example, in a bag of 2 red balls and 2 blue balls (4 balls in total), the probability of taking a red ball is ${\displaystyle 1/2}$; however, when taking a second ball, the probability of it being either a red ball or a blue ball depends on the ball previously taken. For example, if a red ball was taken, then the probability of picking a red ball again would be ${\displaystyle 1/3}$, since only 1 red and 2 blue balls would have been remaining.
### Inverse probability
In probability theory and applications, Bayes' rule relates the odds of event ${\displaystyle A_{1}}$ to event ${\displaystyle A_{2}}$, before (prior to) and after (posterior to) conditioning on another event ${\displaystyle B}$. The odds on ${\displaystyle A_{1}}$ to event ${\displaystyle A_{2}}$ is simply the ratio of the probabilities of the two events. When arbitrarily many events ${\displaystyle A}$ are of interest, not just two, the rule can be rephrased as posterior is proportional to prior times likelihood, ${\displaystyle P(A|B)\propto P(A)P(B|A)}$ where the proportionality symbol means that the left hand side is proportional to (i.e., equals a constant times) the right hand side as ${\displaystyle A}$ varies, for fixed or given ${\displaystyle B}$ (Lee, 2012; Bertsch McGrayne, 2012). In this form it goes back to Laplace (1774) and to Cournot (1843); see Fienberg (2005). See Inverse probability and Bayes' rule.
### Summary of probabilities
Summary of probabilities
Event Probability
A ${\displaystyle P(A)\in [0,1]\,}$
not A ${\displaystyle P(A^{\complement })=1-P(A)\,}$
A or B {\displaystyle {\begin{aligned}P(A\cup B)&=P(A)+P(B)-P(A\cap B)\\P(A\cup B)&=P(A)+P(B)\qquad {\mbox{if A and B are mutually exclusive}}\\\end{aligned}}}
A and B {\displaystyle {\begin{aligned}P(A\cap B)&=P(A|B)P(B)=P(B|A)P(A)\\P(A\cap B)&=P(A)P(B)\qquad {\mbox{if A and B are independent}}\\\end{aligned}}}
A given B ${\displaystyle P(A\mid B)={\frac {P(A\cap B)}{P(B)}}={\frac {P(B|A)P(A)}{P(B)}}\,}$
## Relation to randomness and probability in quantum mechanics
In a deterministic universe, based on Newtonian concepts, there would be no probability if all conditions were known (Laplace's demon), (but there are situations in which sensitivity to initial conditions exceeds our ability to measure them, i.e. know them). In the case of a roulette wheel, if the force of the hand and the period of that force are known, the number on which the ball will stop would be a certainty (though as a practical matter, this would likely be true only of a roulette wheel that had not been exactly levelled - as Thomas A. Bass' Newtonian Casino revealed). This also assumes knowledge of inertia and friction of the wheel, weight, smoothness and roundness of the ball, variations in hand speed during the turning and so forth. A probabilistic description can thus be more useful than Newtonian mechanics for analyzing the pattern of outcomes of repeated rolls of a roulette wheel. Physicists face the same situation in kinetic theory of gases, where the system, while deterministic in principle, is so complex (with the number of molecules typically the order of magnitude of the Avogadro constant ) that only a statistical description of its properties is feasible.
Probability theory is required to describe quantum phenomena.[35] A revolutionary discovery of early 20th century physics was the random character of all physical processes that occur at sub-atomic scales and are governed by the laws of quantum mechanics. The objective wave function evolves deterministically but, according to the Copenhagen interpretation, it deals with probabilities of observing, the outcome being explained by a wave function collapse when an observation is made. However, the loss of determinism for the sake of instrumentalism did not meet with universal approval. Albert Einstein famously remarked in a letter to Max Born: "I am convinced that God does not play dice".[36] Like Einstein, Erwin Schrödinger, who discovered the wave function, believed quantum mechanics is a statistical approximation of an underlying deterministic reality.[37] In some modern interpretations of the statistical mechanics of measurement, quantum decoherence is invoked to account for the appearance of subjectively probabilistic experimental outcomes.
In law
## Notes
1. ^ Strictly speaking, a probability of 0 indicates that an event almost never takes place, whereas a probability of 1 indicates than an event almost certainly takes place. This is an important distinction when the sample space is infinite. For example, for the continuous uniform distribution on the real interval [5, 10], there are an infinite number of possible outcomes, and the probability of any given outcome being observed -- for instance, exactly 7 -- is 0. This means that when we make an observation, it will almost surely not be exactly 7. However, it does not mean that exactly 7 is impossible. Ultimately some specific outcome (with probability 0) will be observed, and one possibility for that specific outcome is exactly 7.
2. ^ In the context of the book that this is quoted from, it is the theory of probability and the logic behind it that governs the phenomena of such things compared to rash predictions that rely on pure luck or mythological arguments such as gods of luck helping the winner of the game.
## References
1. ^ "Kendall's Advanced Theory of Statistics, Volume 1: Distribution Theory", Alan Stuart and Keith Ord, 6th Ed, (2009), ISBN 978-0-534-24312-8.
2. ^ William Feller, An Introduction to Probability Theory and Its Applications, (Vol 1), 3rd Ed, (1968), Wiley, ISBN 0-471-25708-7.
3. ^ Probability Theory The Britannica website
4. ^ Hacking, Ian (1965). The Logic of Statistical Inference. Cambridge University Press. ISBN 978-0-521-05165-1.[page needed]
5. ^ Finetti, Bruno de (1970). "Logical foundations and measurement of subjective probability". Acta Psychologica. 34: 129-145. doi:10.1016/0001-6918(70)90012-0.
6. ^ Hájek, Alan (21 October 2002). Edward N. Zalta (ed.). "Interpretations of Probability". The Stanford Encyclopedia of Philosophy (Winter 2012 ed.). Retrieved 2013.
7. ^ Hogg, Robert V.; Craig, Allen; McKean, Joseph W. (2004). Introduction to Mathematical Statistics (6th ed.). Upper Saddle River: Pearson. ISBN 978-0-13-008507-8.[page needed]
8. ^ Jaynes, E.T. (2003). "Section 5.3 Converging and diverging views". In Bretthorst, G. Larry (ed.). Probability Theory: The Logic of Science (1 ed.). Cambridge University Press. ISBN 978-0-521-59271-0.
9. ^ a b Hacking, I. (2006) The Emergence of Probability: A Philosophical Study of Early Ideas about Probability, Induction and Statistical Inference, Cambridge University Press, ISBN 978-0-521-68557-3[page needed]
10. ^ Freund, John. (1973) Introduction to Probability. Dickenson ISBN 978-0-8221-0078-2 (p. 1)
11. ^ Jeffrey, R.C., Probability and the Art of Judgment, Cambridge University Press. (1992). pp. 54-55 . ISBN 0-521-39459-7
12. ^ Franklin, J. (2001) The Science of Conjecture: Evidence and Probability Before Pascal, Johns Hopkins University Press. (pp. 22, 113, 127)
13. ^ Broemeling, Lyle D. (1 November 2011). "An Account of Early Statistical Inference in Arab Cryptology". The American Statistician. 65 (4): 255-257. doi:10.1198/tas.2011.10191. S2CID 123537702.
14. ^ Some laws and problems in classical probability and how Cardano anticipated them Gorrochum, P. Chance magazine 2012
15. ^ Abrams, William, A Brief History of Probability, Second Moment, retrieved 2008
16. ^ Ivancevic, Vladimir G.; Ivancevic, Tijana T. (2008). Quantum leap : from Dirac and Feynman, across the universe, to human body and mind. Singapore ; Hackensack, NJ: World Scientific. p. 16. ISBN 978-981-281-927-7.
17. ^ Franklin, James (2001). The Science of Conjecture: Evidence and Probability Before Pascal. Johns Hopkins University Press. ISBN 978-0-8018-6569-5.
18. ^ Shoesmith, Eddie (November 1985). "Thomas Simpson and the arithmetic mean". Historia Mathematica. 12 (4): 352-355. doi:10.1016/0315-0860(85)90044-8.
19. ^ a b Wilson EB (1923) "First and second laws of error". Journal of the American Statistical Association, 18, 143
20. ^ Seneta, Eugene William. ""Adrien-Marie Legendre" (version 9)". StatProb: The Encyclopedia Sponsored by Statistics and Probability Societies. Archived from the original on 3 February 2016. Retrieved 2016.
21. ^ Weber, Richard. "Markov Chains" (PDF). Statistical Laboratory. University of Cambridge.
22. ^ Vitanyi, Paul M.B. (1988). "Andrei Nikolaevich Kolmogorov". CWI Quarterly (1): 3-18. Retrieved 2016.
23. ^ Wilcox, Rand R. (10 May 2016). Understanding and applying basic statistical methods using R. Hoboken, New Jersey. ISBN 978-1-119-06140-3. OCLC 949759319.
24. ^ Singh, Laurie (2010) "Whither Efficient Markets? Efficient Market Theory and Behavioral Finance". The Finance Professionals' Post, 2010.
25. ^ Gao, J.Z.; Fong, D.; Liu, X. (April 2011). "Mathematical analyses of casino rebate systems for VIP gambling". International Gambling Studies. 11 (1): 93-106. doi:10.1080/14459795.2011.552575. S2CID 144540412.
26. ^ "Data: Data Analysis, Probability and Statistics, and Graphing". archon.educ.kent.edu. Retrieved 2017.
27. ^ Gorman, Michael F. (2010). "Management Insights". Management Science. 56: iv-vii. doi:10.1287/mnsc.1090.1132.
28. "List of Probability and Statistics Symbols". Math Vault. 26 April 2020. Retrieved 2020.
29. ^ Ross, Sheldon M. (2010). A First course in Probability (8th ed.). Pearson Prentice Hall. pp. 26-27. ISBN 9780136033134.
30. ^ a b Weisstein, Eric W. "Probability". mathworld.wolfram.com. Retrieved 2020.
31. ^ Olofsson (2005) p. 8.
32. ^ Olofsson (2005), p. 9
33. ^ Olofsson (2005) p. 35.
34. ^ Olofsson (2005) p. 29.
35. ^ Burgin, Mark (2010). "Interpretations of Negative Probabilities". p. 1. arXiv:1008.1287v1 [physics.data-an].
36. ^ Jedenfalls bin ich überzeugt, daß der Alte nicht würfelt. Letter to Max Born, 4 December 1926, in: Einstein/Born Briefwechsel 1916-1955.
37. ^ Moore, W.J. (1992). Schrödinger: Life and Thought. Cambridge University Press. p. 479. ISBN 978-0-521-43767-7.
## Bibliography
• Kallenberg, O. (2005) Probabilistic Symmetries and Invariance Principles. Springer-Verlag, New York. 510 pp. ISBN 0-387-25115-4
• Kallenberg, O. (2002) Foundations of Modern Probability, 2nd ed. Springer Series in Statistics. 650 pp. ISBN 0-387-95313-2
• Olofsson, Peter (2005) Probability, Statistics, and Stochastic Processes, Wiley-Interscience. 504 pp ISBN 0-471-67969-0.
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# Thread: Problem from my test! Help!
1. ## Problem from my test! Help!
Hey have a look at these
You have 2 standard 52 card decks of cards that have been shuffled randomly. You turn over the first card from the first deck, and the first card from the second deck,
(i) What is the probability they match?
(ii)You keep turning over the top card on each deck, until you have gone through the whole deck. How many matches do you expect to find?
(iii) You have a thousand letters and envelopes addressed to people, the envelopes get mixed up and the letters get randomly put into envelopes. How many do you expect to end up in the right envelope?
I thought there was a 1/52 chance for the first part but the other parts were confusing. Can anyone help? I'd really appreciate it
2. Hello, konix!
I hope this wasn't a homework problem.
The theory required is quite advanced.
. . In fact, I won't into the details here . . .
You have 2 standard 52 card decks of cards that have been shuffled randomly.
You turn over the first card from the first deck, and the first card from the second deck.
(i) What is the probability they match?
(ii)You keep turning over the top card on each deck, until you have gone through
the whole deck. How many matches do you expect to find?
(iii) You have a thousand letters and envelopes addressed to people,
the envelopes get mixed up and the letters get randomly put into envelopes.
How many do you expect to end up in the right envelope?
(i) The first card can be any card: .$\displaystyle \frac{52}{52} \:=\:1$
. . .The other card must match: .$\displaystyle \frac{1}{52}$
The probability that the two top cards match is: .$\displaystyle (1)\left(\frac{1}{52}\right) \:=\:\frac{1}{52}$ . You were right!
{ii) and (iii) This is a classic (very old) problem.
It has been called "The Mis-addressed Envelopes" or "The Careless Hatcheck Girl".
A permutation of objects in which no object is in its proper place
. . is called a "complete permuatation" or, more commonly, a disarrangment.
You might try to discern a pattern, but it's very elusive . . .
. . . $\displaystyle \begin{array}{ccc}\text{objects} & \text{disarrangements} \\ 1 & 0 \\ 2 & 1 \\ 3 & 2 \\ 4 & 9 \\ 5 & 44 \\ 6 & 265 \\ \vdots & \vdots \end{array}$
With a deck of 6 cards, there are 6! = 720 possible orders for the first deck.
There are 265 orders for the second deck to have no matches.
Hence, there are: .$\displaystyle 720 - 265 \:=\:455$ orders with some match.
Therefore: .$\displaystyle P(\text{at least one match}) \:=\:\frac{455}{720} \:=\:0.63194 \:\approx\:\frac{5}{8}$
With 52 cards, we need the number of possible disarrangements, $\displaystyle d(52)$
. . and divide by 52!, the number of possible orderings of the cards.
This is the probability of no matches.
. . Subtract from one (1), to get the probability of some matches.
Skipping all the underlying theory, as the number of objects gets larger,
. . the probability of getting no matches approaches $\displaystyle :\frac{1}{e} \:=\:0.367879441 ...$
Maybe you can apply all this to your problem . . .
Edit: I was reminded that the proper term is derangement
. . . . That's true . . . *sigh*
3. thanks a million, should be able to do it now
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# Probability
• Dec 31st 2009, 05:54 AM
gustavodecastro
Probability
Hello,
I'm needing some help here, please. Suppose that
$
P_n \rightarrow P
$
weakly, where both measures are defined in $R^2$, and that $P_1$ is the marginal of $P_n$.
By Portmanteau's theorem, we have
$
P(A X R) \leq \liminf P_n(A X R) = P_1(A)
$
for all A open. So far, ok. The problem arises now. I've been told that all the above implies
$
P(A X R) = P_1(A)
$
Can someone help me to prove that statement?
*******************
The real problem is the following:
If
$P_n \rightarrow P$
weakly, and if
$P_1$ and $P_2$ are the marginals of $P_n$,
the limit $P$ must have the same marginals.
***************
Thank you very much and happy new year!
• Dec 31st 2009, 08:35 AM
Moo
Hello,
What about that ? (I'm not sure at all)
We know that $P_n$ converges weakly to $P$ iff for any bounded function f, $E[f(X_{n,1},X_{n,2})] \to E[f(X_1,X_2)]$, where [tex](X_{n,1},X_{n,2}) \sim P_n[tex] and $(X_1,X_2) \sim P$
Now let's define the bounded function f : $f=g\circ \pi_1$, where g is any bounded function, and $\pi_1 ~:~ (x_1,x_2)\mapsto x_1$
Then we have (for any bouned function g) $E[g(X_{n,1})]=E[f(X_{n,1},X_{n,2})] \to E[f(X_1,X_2)]=E[g(X_1)]$
which proves that the marginal of $P_n$ converges to the marginal of $P$, so since the marginal of $P_n$ doesn't depend on n, the marginal of P is $P_1$ ?
(I hope you don't get confused with the index of P, because one designs the sequence, the other designs the marginal, but I guess you'll be able to make the difference).
I didn't use the porte-manteau theorem, because the version you have don't seem to get you anywhere...
There are many points that are available on the internet (apart from the liminf stuff).
I have something that says that if, for an open O set of R (more generally, a borelian of R), P( frontier of O )=0, then P_n(O) converges to P(O)
It deals with R, but it should work with Rē.
Then I'm not too sure... You may be able to apply the porte-manteau theorem in its "closed" version : for any closed set C, limsup P_n(C) < P(C)
and consider (R\A)\D, where D is the frontier of A (which is a closed set contained in R\A, by definition)
But I don't know if it's a valuable idea...
again, I'm not sure at all whether my method is correct or not. So just get inspired by it, and if you find something false, tell me (Worried)
• Jan 2nd 2010, 09:47 AM
gustavodecastro
Thank you, Moo.
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# Search by Topic
#### Resources tagged with Properties of numbers similar to Indivisible:
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### There are 65 results
Broad Topics > Numbers and the Number System > Properties of numbers
### Oh! Hidden Inside?
##### Stage: 3 Challenge Level:
Find the number which has 8 divisors, such that the product of the divisors is 331776.
### Can You Find a Perfect Number?
##### Stage: 2 and 3
Can you find any perfect numbers? Read this article to find out more...
### Helen's Conjecture
##### Stage: 3 Challenge Level:
Helen made the conjecture that "every multiple of six has more factors than the two numbers either side of it". Is this conjecture true?
### Two Much
##### Stage: 3 Challenge Level:
Explain why the arithmetic sequence 1, 14, 27, 40, ... contains many terms of the form 222...2 where only the digit 2 appears.
### One to Eight
##### Stage: 3 Challenge Level:
Complete the following expressions so that each one gives a four digit number as the product of two two digit numbers and uses the digits 1 to 8 once and only once.
### Counting Factors
##### Stage: 3 Challenge Level:
Is there an efficient way to work out how many factors a large number has?
### Satisfying Statements
##### Stage: 2 and 3 Challenge Level:
Can you find any two-digit numbers that satisfy all of these statements?
### Thirty Six Exactly
##### Stage: 3 Challenge Level:
The number 12 = 2^2 × 3 has 6 factors. What is the smallest natural number with exactly 36 factors?
### Times Right
##### Stage: 3 and 4 Challenge Level:
Using the digits 1, 2, 3, 4, 5, 6, 7 and 8, mulitply a two two digit numbers are multiplied to give a four digit number, so that the expression is correct. How many different solutions can you find?
### Three Neighbours
##### Stage: 2 Challenge Level:
Look at three 'next door neighbours' amongst the counting numbers. Add them together. What do you notice?
### X Marks the Spot
##### Stage: 3 Challenge Level:
When the number x 1 x x x is multiplied by 417 this gives the answer 9 x x x 0 5 7. Find the missing digits, each of which is represented by an "x" .
### Even So
##### Stage: 3 Challenge Level:
Find some triples of whole numbers a, b and c such that a^2 + b^2 + c^2 is a multiple of 4. Is it necessarily the case that a, b and c must all be even? If so, can you explain why?
### Elevenses
##### Stage: 3 Challenge Level:
How many pairs of numbers can you find that add up to a multiple of 11? Do you notice anything interesting about your results?
### Like Powers
##### Stage: 3 Challenge Level:
Investigate $1^n + 19^n + 20^n + 51^n + 57^n + 80^n + 82^n$ and $2^n + 12^n + 31^n + 40^n + 69^n + 71^n + 85^n$ for different values of n.
### Arrange the Digits
##### Stage: 3 Challenge Level:
Can you arrange the digits 1,2,3,4,5,6,7,8,9 into three 3-digit numbers such that their total is close to 1500?
### Six Times Five
##### Stage: 3 Challenge Level:
How many six digit numbers are there which DO NOT contain a 5?
### Cogs
##### Stage: 3 Challenge Level:
A and B are two interlocking cogwheels having p teeth and q teeth respectively. One tooth on B is painted red. Find the values of p and q for which the red tooth on B contacts every gap on the. . . .
##### Stage: 3 Challenge Level:
Visitors to Earth from the distant planet of Zub-Zorna were amazed when they found out that when the digits in this multiplication were reversed, the answer was the same! Find a way to explain. . . .
### Table Patterns Go Wild!
##### Stage: 2 Challenge Level:
Nearly all of us have made table patterns on hundred squares, that is 10 by 10 grids. This problem looks at the patterns on differently sized square grids.
### Multiply the Addition Square
##### Stage: 3 Challenge Level:
If you take a three by three square on a 1-10 addition square and multiply the diagonally opposite numbers together, what is the difference between these products. Why?
### Chameleons
##### Stage: 3 Challenge Level:
Whenever two chameleons of different colours meet they change colour to the third colour. Describe the shortest sequence of meetings in which all the chameleons change to green if you start with 12. . . .
### The Patent Solution
##### Stage: 3 Challenge Level:
A combination mechanism for a safe comprises thirty-two tumblers numbered from one to thirty-two in such a way that the numbers in each wheel total 132... Could you open the safe?
### Always, Sometimes or Never? Number
##### Stage: 2 Challenge Level:
Are these statements always true, sometimes true or never true?
### Happy Octopus
##### Stage: 3 Challenge Level:
This investigation is about happy numbers in the World of the Octopus where all numbers are written in base 8 ... Find all the fixed points and cycles for the happy number sequences in base 8.
### Four Coloured Lights
##### Stage: 3 Challenge Level:
Imagine a machine with four coloured lights which respond to different rules. Can you find the smallest possible number which will make all four colours light up?
### Slippy Numbers
##### Stage: 3 Challenge Level:
The number 10112359550561797752808988764044943820224719 is called a 'slippy number' because, when the last digit 9 is moved to the front, the new number produced is the slippy number multiplied by 9.
### Which Numbers? (2)
##### Stage: 2 Challenge Level:
I am thinking of three sets of numbers less than 101. Can you find all the numbers in each set from these clues?
### Which Numbers? (1)
##### Stage: 2 Challenge Level:
I am thinking of three sets of numbers less than 101. They are the red set, the green set and the blue set. Can you find all the numbers in the sets from these clues?
### Small Change
##### Stage: 3 Challenge Level:
In how many ways can a pound (value 100 pence) be changed into some combination of 1, 2, 5, 10, 20 and 50 pence coins?
### Water Lilies
##### Stage: 3 Challenge Level:
There are some water lilies in a lake. The area that they cover doubles in size every day. After 17 days the whole lake is covered. How long did it take them to cover half the lake?
### Snail One Hundred
##### Stage: 1 and 2 Challenge Level:
This is a game in which your counters move in a spiral round the snail's shell. It is about understanding tens and units.
### Escape from the Castle
##### Stage: 2 Challenge Level:
Skippy and Anna are locked in a room in a large castle. The key to that room, and all the other rooms, is a number. The numbers are locked away in a problem. Can you help them to get out?
### Cinema Problem
##### Stage: 3 Challenge Level:
A cinema has 100 seats. Show how it is possible to sell exactly 100 tickets and take exactly £100 if the prices are £10 for adults, 50p for pensioners and 10p for children.
### Lesser Digits
##### Stage: 3 Challenge Level:
How many positive integers less than or equal to 4000 can be written down without using the digits 7, 8 or 9?
### Numbers Numbers Everywhere!
##### Stage: 1 and 2
Bernard Bagnall recommends some primary school problems which use numbers from the environment around us, from clocks to house numbers.
### Lastly - Well
##### Stage: 3 Challenge Level:
What are the last two digits of 2^(2^2003)?
### Alphabet Soup
##### Stage: 3 Challenge Level:
This challenge is to make up YOUR OWN alphanumeric. Each letter represents a digit and where the same letter appears more than once it must represent the same digit each time.
### Triangular Triples
##### Stage: 3 Challenge Level:
Show that 8778, 10296 and 13530 are three triangular numbers and that they form a Pythagorean triple.
### Special Numbers
##### Stage: 3 Challenge Level:
My two digit number is special because adding the sum of its digits to the product of its digits gives me my original number. What could my number be?
### Not a Polite Question
##### Stage: 3 Challenge Level:
When asked how old she was, the teacher replied: My age in years is not prime but odd and when reversed and added to my age you have a perfect square...
### Repetitiously
##### Stage: 3 Challenge Level:
The number 2.525252525252.... can be written as a fraction. What is the sum of the denominator and numerator?
### Summing Consecutive Numbers
##### Stage: 3 Challenge Level:
Many numbers can be expressed as the sum of two or more consecutive integers. For example, 15=7+8 and 10=1+2+3+4. Can you say which numbers can be expressed in this way?
### Sept 03
##### Stage: 3 Challenge Level:
What is the last digit of the number 1 / 5^903 ?
### Writ Large
##### Stage: 3 Challenge Level:
Suppose you had to begin the never ending task of writing out the natural numbers: 1, 2, 3, 4, 5.... and so on. What would be the 1000th digit you would write down.
### Whole Numbers Only
##### Stage: 3 Challenge Level:
Can you work out how many of each kind of pencil this student bought?
### Unlocking the Case
##### Stage: 2 Challenge Level:
A case is found with a combination lock. There is one clue about the number needed to open the case. Can you find the number and open the case?
### Guess the Dominoes
##### Stage: 1, 2 and 3 Challenge Level:
This task depends on learners sharing reasoning, listening to opinions, reflecting and pulling ideas together.
### Light the Lights Again
##### Stage: 2 Challenge Level:
Each light in this interactivity turns on according to a rule. What happens when you enter different numbers? Can you find the smallest number that lights up all four lights?
### Clever Carl
##### Stage: 2 and 3
What would you do if your teacher asked you add all the numbers from 1 to 100? Find out how Carl Gauss responded when he was asked to do just that.
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Factors the 96 space numbers that, once multiplied in pairs provide the product as 96. It has a total of 12 determinants of i m sorry 96 is the best factor and also the prime factors of 96 are 2 and also 3. The amount of all components of 96 is 252.
You are watching: What are all the factors of 96
Factors that 96: 1, 2, 3, 4, 6, 8, 12, 16, 24, 32, 48 and 96Negative determinants of 96: -1, -2, -3, -4, -6, -8, -12, -16, -24, -32, -48 and -96Prime components of 96: 2, 3Prime administer of 96: 2 × 2 × 2 × 2 × 2 × 3 = 25 × 3Sum of components of 96: 252
1 What are components of 96? 2 Factors that 96 by element Factorization 3 Factors the 96 in Pairs
## What are factors of 96?
We deserve to use department to inspect whether 2, 4, 6, 16, 24, and also 48 are factors of 96. Together they leave no remainder, castle are undoubtedly the determinants of 96.
Explore determinants using illustrations and interactive examples:
## How to calculation the components of 96?
There are two typical methods to find the determinants of 96:
Let us calculate the determinants of 96 using the division method. Let"s think about the numbers that can divide 96 without remainders. Us will begin with 1, then check 2, 3, 4, 5, 6, 7, 8, 9, etc. As much as 48 (which is exactly fifty percent of 96).
Division Result 96 ÷ 1 gives remainder 0, 1 is a factor 96 ÷ 2 gives remainder 0, 2 is a factor 96 ÷ 3 gives remainder 0, 3 is a factor 96 ÷ 4 gives remainder 0, 4 is a factor 96 ÷ 6 gives remainder 0, 6 is a factor 96 ÷ 8 gives remainder 0, 8 is a factor 96 ÷ 12 gives remainder 0, 12 is a factor 96 ÷ 16 gives remainder 0, 16 is a factor 96 ÷ 24 gives remainder 0, 24 is a factor 96 ÷ 32 gives remainder 0, 32 is a factor 96 ÷ 48 gives remainder 0, 48 is a factor 96 ÷ 96 gives remainder 0, 96 is a factor
## Factors the 96 by element Factorization
Factorization method to compose the number as a product that its factors. Prime Factorization describes finding which element numbers multiply with each other to type the initial number.
Step 1: Divide the number 96 v the smallest prime factor, say 2. 96 ÷ 2 = 48Step 2: Divide 48 again through 2. 48 ÷ 2 = 24Step 3: Divide 24 again by 2. 24 ÷ 2 = 12 and also continue this divisionStep 4: 12 ÷ 2 = 6Step 5: 6 ÷ 2 = 3Step 6: Now if we will certainly divide 3 through 2, we will acquire a spring number and also a element cannot it is in a fraction or a decimal number. So, let"s proceed with the following prime factor, to speak 3. 3 ÷ 3 = 1
The prime determinants of 96 thus obtained are created as 96 = 2 × 2 × 2 × 2 × 2 × 3 = 25 × 31, where 2 and 3 room the prime numbers.
Factor tree of 96
The same division process is best represented in the kind of a factor tree as offered below:A variable tree is a one-of-a-kind branching diagram whereby we discover the factors of a number, climate we find the determinants of those numbers, etc. Till we cannot aspect anymore.The end of a variable tree room all prime components of the original number.
There are countless ways of producing the factor tree the 96:
## Factors of 96 in Pairs
Pair factors of the number 96 are the totality numbers which when multiplied offer the initial number, i.e. 96. Pair factors might be either hopeful or negative. There are 12 pair determinants of 96 and also they are: (1,96), (-1,-96), (2,48), (-2,-48), (3,32), (-3,-32), (4,24), (-4,-24), (6,16), (-6,-16), (8,12), and (-8,-12).
Important Notes
96 is a composite number.2 and 3 room the just prime numbers that are determinants of 96.The number of factors of a given number is finite; 96 has actually a full of 12 factors.
See more: Lyrics To Joshua Fought The Battle Of Jericho, Joshua Fought The Battle Of Jericho
Tips and Tricks:
96 = 2 × 2 × 2 × 2 × 2 × 3 = 25 × 31To acquire the complete numbers of components of 96, simply add one to the exponent 5 and also 1 and also multiply their sums. (5 + 1) × (1 + 1) = 6 × 2 = 12Thus, 96 has actually 12 factors and they space 1, 2, 3, 4, 6, 8, 12, 16, 24, 32, 48, and 96.
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## Currently working through Apostol need a check on my logic
1. The problem statement, all variables and given/known data
Prove the following theorem.
If a≠0, then (a-1)-1=a
Theorems proven before/axioms I am allowed to use:
Existence of reciprocals axiom: there exist real numbers x and y where x≠0 such that xy=1
Possibility of division Thm: basically a-1=1/a
2. Relevant equations
I just need a check on my proof since I am self studying the book. Any logical mistakes on my proof or suggestions etc? This is my first time writing proofs.
3. The attempt at a solution
Assume (a-1)-1=a is true.
Then by the existence of reciprocals axiom there exists an x such that (a-1)-1x=1.
Hence,
(a-1)-1x=ax
ax=1
Now by the possibility of division theorem we have,
x=1/a=a-1
Finally since we assumed (a-1)-1=a was true it follows that:
(a-1)-1x=ax=aa-1=1
Q.E.D
It looks fine to me, I'm just a noob at proofs. Thanks.
PhysOrg.com science news on PhysOrg.com >> 'Whodunnit' of Irish potato famine solved>> The mammoth's lament: Study shows how cosmic impact sparked devastating climate change>> Curiosity Mars rover drills second rock target
Mentor
Quote by VectorField 1. The problem statement, all variables and given/known data Prove the following theorem. If a≠0, then (a-1)-1=a Theorems proven before/axioms I am allowed to use: Existence of reciprocals axiom: there exist real numbers x and y where x≠0 such that xy=1 Possibility of division Thm: basically a-1=1/a 2. Relevant equations I just need a check on my proof since I am self studying the book. Any logical mistakes on my proof or suggestions etc? This is my first time writing proofs. 3. The attempt at a solution Assume (a-1)-1=a is true. Then by the existence of reciprocals axiom there exists an x such that (a-1)-1x=1. Hence, (a-1)-1x=axax=1Now by the possibility of division theorem we have, x=1/a=a-1Finally since we assumed (a-1)-1=a was true it follows that: (a-1)-1x=ax=aa-1=1Q.E.D It looks fine to me, I'm just a noob at proofs. Thanks.
According to the first line of your proof,
Assume (a-1)-1=a is true.
you're assuming that the thing you're proving is true.
That's a definite No-No .
I suggest starting with:
a-1 has a reciprocal, (a-1)-1.
Therefore, (a-1)-1 a-1 = 1
...
Now, multiply that by a, with a being on the right.
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A275712 Decimal expansion of the sum of the alternating series of reciprocals of nonprime numbers. 7
8, 4, 8, 1, 3, 2, 2, 1, 1, 8, 7, 6, 9, 8 (list; constant; graph; refs; listen; history; text; internal format)
OFFSET 0,1 COMMENTS From Jon E. Schoenfield, Nov 10 2016: (Start) Let S(j) be the partial sum through the j-th term of the alternating series, i.e., S(j) = -Sum_{m=1..j} (-1)^m/A018252(m). The sequence of real values S(2*i-1) for i >= 1, i.e., of partial sums 1/1, 1/1 - 1/4 + 1/6, 1/1 - 1/4 + 1/6 - 1/8 + 1/9, ... (each of which ends with a positive term) will approach the limit from above, while the sequence of real values S(2*i) for i >= 1, i.e., of partial sums 1/1 - 1/4, 1/1 - 1/4 + 1/6 - 1/8, 1/1 - 1/4 + 1/6 - 1/8 + 1/9 - 1/10, ... (each of which ends with a negative term) will approach the limit from below. Let S'(j) = (S(j-1) + S(j))/2; equivalently, S'(j) = -(Sum_{m=1..j-1} (-1)^m/A018252(m) + (1/2)*(-1)^j/A018252(j)), so S'(j) can be viewed as an adjusted version of S(j), adjusted by using only half of the final term of S(j). At large values of j, successive values of S'(j) will fluctuate very little compared to the differences between successive values of S(j), because the averaging of successive values of S(j), which are above the limit at each odd value of j and below the limit by very nearly the same amount at each even value of j, causes the values of S'(j) to trace a path midway between that traced by the S(j) values for odd j and those for even j. Moreover, it can be observed (see the charts under Links) that the values of S'(j) themselves fall into three sharply distinct real-valued subsequences: one that converges toward the limit from above and consists of those values where both j and the j-th nonprime number (i.e., the reciprocal of the last term in S(j)) are even; one that converges toward the limit from below and consists of those values where j is odd and the j-th nonprime number is even; and one that stays very near the middle, converging even more rapidly toward the limit, and consisting of all those values where the j-th nonprime number is odd (regardless of the parity of j). The values in this last subsequence converge so rapidly that the first 1000 or so terms are sufficient to show that the limit is clearly 0.848132..., with the next digit very likely another 2. Using the first 10^7 terms, it becomes apparent ("zoomed in" more closely than in the 3-curve chart under Links) that the limit is 0.8481322118769887..., and having observed the behavior of this last subsequence out to nearly j=3*10^9, I am confident that the limit is 0.84813221187698878102544... But how many of these digits can be rigorously proved? (End) From Robert Price, Nov 13 2016: (Start) To support the comments above by Jon Schoenfield: The sum of an alternating arithmetic sequence whose terms are decreasing in absolute value will always converge. The method he has outlined, we might call the "Schoenfield Interpolation". To illustrate further and to confirm his method, the simpler sum: log(2) = 1/1 - 1/2 + 1/3 - 1/4 + ... can be calculated with the following Mathematica program. Note that after only 20,000 iterations, needing only a few seconds, the limit can be obtained to 13 digits. sum = 0; iter = 20000; For[i=1,i<=iter,i=i+2, sum = sum + 1/i; upper = sum; If[i>iter-10,avg1=(upper+lower)/2;Print[N[{upper,lower,avg1,(avg1+avg2)/2},15]]]; sum = sum - 1/(i+1); lower = sum; If[i>iter-10,avg2=(upper+lower)/2;Print[N[{upper,lower,avg2,(avg1+avg2)/2},15]]]; ]; N[Log[2],15] {0.693172191189447,0.693122168679318,0.693147179934382,0.693147181497164} {0.693172191189447,0.693122171181444,0.693147181185446,0.693147180559914} {0.693172188687571,0.693122171181444,0.693147179934508,0.693147180559977} {0.693172188687571,0.693122173683070,0.693147181185320,0.693147180559914} {0.693172186186196,0.693122173683070,0.693147179934633,0.693147180559977} {0.693172186186196,0.693122176184195,0.693147181185195,0.693147180559914} {0.693172183685320,0.693122176184195,0.693147179934758,0.693147180559977} {0.693172183685320,0.693122178684820,0.693147181185070,0.693147180559914} {0.693172181184945,0.693122178684820,0.693147179934883,0.693147180559977} {0.693172181184945,0.693122181184945,0.693147181184945,0.693147180559914} 0.693147180559945 (End) LINKS Jon E. Schoenfield, Chart showing the rapid convergence of the "adjusted sums" S'(j) (regardless of the parity of j or of the j-th nonprime) as compared to the unadjusted sums S(j) as described in Comments FORMULA -Sum_{k>=1} (-1)^k/A018252(k) = 1/1 - 1/4 + 1/6 - 1/8 + 1/9 - 1/10 + 1/12 - ... Equals 1 - A269229. EXAMPLE 0.84813221187698... MATHEMATICA NonPrime[n_] := FixedPoint[n + PrimePi@# &, n + PrimePi@n]; N[Sum[(-1)^n/NonPrime[n], {n, 1, 200}], 25] (* G. C. Greubel, Aug 06 2016 *) PROG (Sage) nonprime = (i for i in NN if i>0 and not i.is_prime()) s = RLF(0); s RealField(110)(s) for i in xrange(0, 50000000): s += (-1)^i / nonprime.next() print s # Terry D. Grant, Aug 06 2016 CROSSREFS Cf. A078437, A018252, A269229. Sequence in context: A144663 A197260 A155889 * A199266 A205383 A083948 Adjacent sequences: A275709 A275710 A275711 * A275713 A275714 A275715 KEYWORD nonn,more,cons AUTHOR Terry D. Grant, Aug 06 2016 EXTENSIONS Incorrect terms a(7)-a(10) deleted, and corrected terms a(7)-a(9) added, by Robert Price, Nov 09 2016 a(10)-a(13) from Robert Price, Nov 13 2016 STATUS approved
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New SAT Math Workbook
# New SAT Math Workbook
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Unformatted text preview: are 180° in a triangle. Since ∠FEG is the supplement of ∠HEG, ∠FEG = 70°. (C) The sum of the angles in a parallelogram is 360°. 12 x = 360° x = 30° Angle B = 5x = 5 · 30° = 150° 5. (A) The volume of a rectangular box is the product of its length, width, and height. Since 1 the height is 18 inches, or 1 feet, and the 2 length and width of the square base are the same, we have 1 x ⋅ x ⋅ 1 = 24 2 x 2 = 16 x=4 Angle O is a central angle equal to its arc, 100°. This leaves 80° for the other two angles. Since the triangle is isosceles (because the legs are both radii and therefore equal), angle ABO is 40°. 10. (A) d= = (5 − (3)) + (-5 - 1) 2 2 (8)2 + (-6)2 = 64 + 36 = 100 = 10 www.petersons.com 224 Chapter 13 Exercise 1 1. (C) Find the area in square feet and then convert to square yards by dividing by 9. Remember there are 9 square feet in one square yard. (18 · 20) ÷ 9 = 360 ÷ 9 = 40 square yards 2. (B) Area of parallelogram = b · h Exercise 2 1. (A) Area of equilateral triangle = s2 s2 3 4 Therefore, must equal 16 4 s 2 = 64 s=8 Perimeter is 8 + 8 + 8 = 24 2. (B) In 4 hours the hour hand moves through one-third of the circumference of the clock. C = 2π r = 2π ( 3) = 6π 1 ⋅ 6π = 2π 3 ( x + 7)( x − 7) = 15 x 2 − 49 = 15 x 2 = 64 x =8 Base = x + 7 = 15 3. 1 ⋅b⋅h 2 (D) Compare 2πr with 2π (r + 3). 3. (B) Area of triangle = 2π (r + 3) = 2πr + 6π Circumference was increased by 6π. Trying this with a numerical value for r will give the same result. 4. (E) In one revolution, the distance covered is equal to the circumference. C = 2πr = 2π (18) = 36π inches To change this to feet, divide by 12. 36π = 3π feet 12 Using one leg as base and the other as altitude, the area is 1 · 6 · 8 = 24. Using the hypotenuse 2 as base and the altitude to the hypotenuse will give the same area. 1 ⋅ 10 ⋅ h = 24 2 5h = 24 1 h = 4.8 ∴ ⋅ 10 ⋅ 4.8 = 24 2 4. (E) Area of rhombus = · product of 2 diagonals 1 1 Area = ( 4 x )( 6 x ) = 24 x 2 = 12 x 2 2 2 1 ( ) In 20 revolut...
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## This note was uploaded on 08/15/2010 for the course MATH a4d4 taught by Professor Colon during the Spring '10 term at Embry-Riddle FL/AZ.
Ask a homework question - tutors are online
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### Card Trick 2
Can you explain how this card trick works?
### Painting Cubes
Imagine you have six different colours of paint. You paint a cube using a different colour for each of the six faces. How many different cubes can be painted using the same set of six colours?
### Cube Paths
Given a 2 by 2 by 2 skeletal cube with one route `down' the cube. How many routes are there from A to B?
# Amazing Card Trick
##### Stage: 3 Challenge Level:
Felix, Matthew, Alice, Robert,Hayden, Jenna, Catherine, James, James, Nick, Kieran, Kayleigh, Bethany, Luke and Matthew, all from Cupernham School sent in explanations which involved a similar argument. Correct explanations were also recieved from Andrei of School 205 Bucharest, Sophia of the Maths Club at Stamford High School and Matthew of Finley Middle School .
If there a number of you from the same school, it is worth spending a little time sharing explanations and sending in a joint proposal as it will be unlikely that we shall be able to give credit to every individual's work in the future.
Here is an explanation:
This card trick has nothing to do with magic, just mathematical thinking. When you do this trick, the total amount of cards on the table is 33.
It works because there are 3 cards you put down in the first place, the cards you add to them to make 10 and the number of cards down the pack. These make a total of 33 if you add them together. The number you memorise is the 7th card into the original half of 26. That is the top 26+7 into the original half equal 33 so that is how it works.
eg. If you lay down the 9 diamonds, 3 hearts and 6 clubs, it would look like this:
9 3 6 (three cards on the table)
You then lay down 1, 7 and 4 cards respectively (making a total of a12 +3 = 15 cards).
1 7 4
You then count down the pack 9+3+6 cards = 18 cards.
That is 15 cards on the table plus 18 cards down the pack- making 33.
9 3 6
-----
1 7 4
9 3 6
The 2 last rows show the cards that make up the original card to 10. These cards add up to 30. The top row is the original three cards, so if you add them on, the final total is 33. This is always true, because the three cards are made up to ten every time. This is because you have to make up the numbers on the cards, then deal out the original number again when you have made the prediction. The trick is when you deal out the 26 cards at the beginning, you take note of the 7th card. This is the card you predict.
If you do this trick properly it can be very entertaining!
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# Analysing a small recursive algorithm
I need to calculate the complexity of func5, depending on variables $n, m$.
func4 is a function whose complexity is $\Theta(n+m)$
void func5(int a[], int n, int m, int b[])
{
if (n==0) { return; }
*b = func4(a,n,m);
func5(a+1,n-1,m,b+1);
}
I get an expression which looks like:
$$C_1*n + C_2*(nm+n^2) - C_2(1+2+3+..n)$$
$C_1$ is the operations done in each iteration of func5,
$C_2$ is the operations done in each call to func4,
and the substraction comes since func4 is receiving each time a smaller $n$ by one.
The answer says that complexity is $\Theta(n*m+n^2)$ but I don't understand how to find the constants leading to Big-Theta notation.
Thanks.
• You have not dealt with the recursion properly. See our reference question for how it works. Also, be mindful of Landau notation in two parameters; it's not clear what $\Theta(n + m)$ "simplifies" to in the case of $n=0$! Mar 9, 2016 at 1:14
Since $1+2+\cdots+n = \frac{n(n+1)}{2}$, your expression is equal to $$C_1 n + C_2 (nm+n^2) - C_2 \frac{n^2+n}{2} = C_2(nm+n^2/2) + (C_1-C_2/2)n.$$ This shows that for constant $m$ and large enough $n$, your expression is at least $(1/2-\epsilon)C_2(nm+n^2)$ (for any $\epsilon > 0$ and, if $C_1 \geq C_2/2$, even for $\epsilon = 0$), and at most $(1+\epsilon)C_2(nm+n^2)$ (for any $\epsilon > 0$ and, if $C_1 \leq C_2/2$, even for $\epsilon = 0$).
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Question 3 – Transfer Functions
Document Sample
``` ENGR4300 Test 3S Fall 2005
ENGR4300
Fall 2005
Test 3S
Name _______solution___________
Section________________________
Question 1 (25 points)___________
Question 2 (25 points) ___________
Question 3 (25 points)___________
Question 4 (25 points)___________
Total (100 points): ______________
Please do not write on the crib sheets.
On all questions: SHOW ALL WORK. BEGIN WITH FORMULAS, THEN
SUBSTITUTE VALUES AND UNITS. No credit will be given for
numbers that appear without justification.
1 of 12
ENGR4300 Test 3S Fall 2005
Question 1 – Astable Multivibrator (25 points)
The multivibrator above was built with the 555 timer chip shown. In the diagram, Vs is
the source voltage, Vc is the voltage across the capacitor C1, and VD is the voltage at the
discharge pin of the timer. You are given the following component values:
Test A: R1 = 4.7k ohms R2 = 6.8k ohms C1 = 0.68F Vs = 6V
Test B: R1 = 2.2k ohms R2 = 4.7k ohms C1 = 1.5F Vs = 12V
Test C: R1 = 6.8k ohms R2 = 8.6k ohms C1 = 0.47F Vs = 5V
1) Calculate the on time and off time of the multivibrator. (4 pt)
T1 0.693( R1 R2)C1 T1 = 0.693(4.7K+6.8K)(0.68) = 5.42ms
T 2 0.693( R2)C1 T2 = 0.693(6.8k)(0.68) = 3.20ms
Test A: Ton = 5.42ms Toff = 3.20ms
Test B: Ton = 7.17ms Toff = 4.89ms
Test C: Ton = 5.02ms Toff = 2.80ms
2) Calculate the frequency and period of the multivibrator. (4 pt)
T=Ton + Toff f=1/T T=5.42m+3.20m = 8.62ms f=1/8.62m=116Hz
Test A: T = 8.62ms f = 116 Hz
Test B: T = 12.06ms f = 82.9 Hz
Test C: T = 7.82ms f = 128 Hz
3) Calculate the duty cycle of the multivibrator. (2 pt)
DC = 100*Ton/T DC=100*5.42m/8.62m=62.9%
Test A: DC = 62.9%
Test B: DC = 59.5%
Test C: DC = 64.2%
2 of 12
ENGR4300 Test 3S Fall 2005
4) In theory, what are the maximum and minimum voltage across the capacitor, Vc? (2
pt)
Vmin = Vs/3 Vmax = 2*Vs/3 Vmin=6/3=2V Vmax = 2*6/3 = 4V
Test A: Vmin = 2V Vmax = 4V
Test B: Vmin = 4V Vmax = 8V
Test C: Vmin = 1.67V Vmax = 3.33V
5) When the transistor inside the 555 timer is closed, the output at pin 7 is grounded, and
therefore, equal to zero. When the transistor is open, the output at pin 7 can be found
using the voltage divider formed by R1 and R2. Find an expression for the voltage at pin
7 (when pin 7 is not grounded) in terms of the voltage across the capacitor, Vc, the source
voltage, Vs, and the two resistors R1 and R2. [Hint: Recall how you calculate the voltage
at the non-inverting input for a Schmitt trigger.] Do not substitute values. (3 pt)
R2
VD VS VC VC
R1 R 2
This can be simplified further, but it is not necessary.
6) Use the equation in 5) to find the maximum and minimum voltage at pin 7 when the
transistor is open. (2 pt)
VDlow = [6.8k/(6.8k+4.7k)](6-2) + 2 = 4.36V
VDhigh = [6.8k/(6.8k+4.7k)](6-4) + 4 = 5.18V
Test A: VDlow = 4.36V VDhigh = 5.18V
Test B: VDlow = 9.44V VDhigh = 10.72V
Test C: VDlow = 3.53V VDhigh = 4.27V
7) On the graph below, identify the outputs at pin (2, 6), 7 and 3 of the multivibrator.
Identify both parts of the cycle for each. (6 pt)
This shows the basic shape of the signals. The image on each test was different.
3 of 12
ENGR4300 Test 3S Fall 2005
Question 2 – Combinational Logic Circuits (25 points)
Test A and B:
1) You are given the following circuit
U11A
1
3 C 2
U3A
A 2
3
1
F
7432
7402
U13A
1
2
3
E
7486
U5A
U1A 1
1
3 2
3
G
B 2
D 7408
7400
a) Fill out the truth table below (10 pt)
A B C D E F G
0 0 0 1 1 0 1
0 1 1 1 0 0 0
1 0 1 1 0 0 0
1 1 1 0 1 0 0
b) What gate is the output at G equivalent to (circle one)? (1 pt)
AND NAND OR NOR XOR XNOR NOT
2) Fill out the truth table below to prove the following relationship: (14 pt)
if Q A B A B A B then Q A B
A B A B A B A B A B Q A+B
0 0 1 1 0 0 0 0 0
0 1 1 0 1 0 0 1 1
1 0 0 1 0 1 0 1 1
1 1 0 0 0 0 1 1 1
4 of 12
ENGR4300 Test 3S Fall 2005
Test C:
1) You are given the following circuit
U5A
1
3 C 2
U3A
A 2
3
1
F
7408
7402
U1A
1
2
3
E
7400
U5A
U13A 1
1
3 2
3
G
B 2
D 7408
7486
a) Fill out the truth table below (10 pt)
A B C D E F G
0 0 0 0 1 0 0
0 1 0 1 1 0 1
1 0 0 1 1 0 1
1 1 1 0 1 0 0
b) What gate is the output at G equivalent to (circle one)? (1 pt)
AND NAND OR NOR XOR XNOR NOT
2) Fill out the truth table below to prove the following relationship: (14 pt)
if Q A B A B A B then Q A B
A B A B A B A B A B Q A B
0 0 1 1 1 1 0 0 0
0 1 1 0 1 0 1 0 0
1 0 0 1 0 1 1 0 0
1 1 0 0 1 1 1 1 1
5 of 12
ENGR4300 Test 3S Fall 2005
Question 3 – Sequential Logic Gates (25 points)
Quiz A and B:
You are given the following circuit DSTM1 provides the clock for the counter. DSTM 1
provides the clock to the two flip flops. DSTM2 provides an initial reset pulse to all the
sequential chips. Therefore, you can assume that QA, QB, QC, QD, Q1, and Q2 are all
initially low.
U1A
DSTM1
CLK 1 2
7404
U3A U6A
1 3 1
A QA J
QB
4
5 12 Q
3 Q1
CLR
QC 6 U4A CLK 2
QD 1 2 4 Q
CLR
74393 K
7404 74107
2
13
U7A
1
J
12 Q
3 Q2
U5A CLK 2
1 2 4 Q
CLR
K
U2A 7404 74107
DSTM2
13
CLK 1 2
7404
1) Sketch the timing trace for each of the signals shown. (14 pt)
6 of 12
ENGR4300 Test 3S Fall 2005
Quiz C:
You are given the following circuit DSTM1 provides the clock for the counter. DSTM 1
provides the clock to the two flip flops. DSTM2 provides an initial reset pulse to all the
sequential chips. Therefore, you can assume that QA, QB, QC, QD, Q1, and Q2 are all
initially low.
U1A
DSTM1
CLK 1 2
7404
U3A U6A
1 3 1
A QA J
QB
4
5 12 Q
3 Q1
CLR
QC 6 U4A CLK 2
QD 1 2 4 Q
CLR
74393 K
7404 74107
2
13
U7A
1
J
12 Q
3 Q2
U5A CLK 2
1 2 4 Q
CLR
K
U2A 7404 74107
DSTM2
13
CLK 1 2
7404
1) Sketch the timing trace for each of the signals shown. (14 pt)
7 of 12
ENGR4300 Test 3S Fall 2005
2) We take the output of this circuit and use it as the input to the transistor circuit shown
below:
Quiz A Quiz B and C
Fill in the following table for all possible values of Q1 and Q2. (4 pt)
Quiz A Quiz B and C
Q1 Q2 Vout Q1 Q2 Vout
0V 0V 5V 0V 0V 5V
0V 5V 5V 0V 5V 0V
5V 0V 5V 5V 0V 0V
5V 5V 0V 5V 5V 0V
3) What are the values of the signals at Q1, Q2 and Vout after 8.9 milliseconds? (3 pt)
All Tests: Q1: 0V Q2: 0V Vout: 5V
8 of 12
ENGR4300 Test 3S Fall 2005
Quiz A and B
4) IC chips have a standard numbering convention that you should now know. Label pins
#1-8 on the IC below. (HINT: It doesn’t matter what the chip’s function is – just label
#1, #2, …). (1 pt)
5) Like op-amps, IC chips need two power connections in order to function. Label the
power (Vcc) and ground pins on the 7404 IC chip pictured below with the appropriate
voltage values. (1 pt)
6) The 7404 chip contains more than one inverter. How many does it contain? (1 pt)
6
7) You’ve primarily used two types of capacitors in the studio: electrolytic and ceramic
disc, each shown below:
Electrolytic: Ceramic Disc:
Which one is polarized (e.g. has a + and – lead) (circle one)? (1 pt)
neither electrolytic ceramic disc both
9 of 12
ENGR4300 Test 3S Fall 2005
Quiz C
4) IC chips have a standard numbering convention that you should now know. Label pins
#1-8 on the IC below. (HINT: It doesn’t matter what the chip’s function is – just label
#1, #2, …). (1 pt)
5) We’ve talked about the use of bypass capacitors when working with digital logic
chips. Which answer below best describes where they should be located and why we use
them? (1 pt)
a. Place one or more between digital logic inputs and ground to smooth the sharp
transients of digital switching signals that could otherwise corrupt proper chip
operation.
b. Place one or more in series with logic power (Vcc) where it enters the protoboard to
filter noise coming from the power supply.
c. Place one or more in parallel with Vcc and ground at each chip to filter noise.
d. Place only one between Vcc and high speed logic chip outputs to suppress output
oscillation (e.g. ringing).
6) What was the recommended range of bypass capacitor values given for the TTL logic
family used in the studio? (circle one) (1 pt)
a) 0.1pF – 0.5pF b) 0.01nF – 0.1nF c) 0.01uF – 0.1uF d) 0.1uF – 10uF
e) 10uF – 100uF f) I don’t know because I slept through that lecture.
7) You’ve primarily used two types of capacitors in the studio: electrolytic and ceramic
disc, each shown below:
Electrolytic: Ceramic Disc:
Which one is polarized (e.g. has a + and – lead) (circle one)? (1 pt)
neither electrolytic ceramic disc both
10 of 12
ENGR4300 Test 3S Fall 2005
Question 4 – Switching Circuits (25 points)
_
+
_
+
+
Quiz A: V1=4V V3= -10V V4=2V V5=8V R4=30k R5=2k
Quiz B: V1=6V V3= -8V V4=3V V5=10V R4=20k R5=1k
Quiz C: V1=5V V3= -7V V4=2V V5=9V R4=40k R5=2k
1) In the circuit shown, circle the model of a Schmitt trigger. (1 pt)
2) What is the reference voltage of the Schmitt trigger model? (1 pt)
Test A: Vref = 2V Test B: Vref = 3V Test C: Vref=2V
3) If the op-amp is putting out its maximum voltage at C, what is the voltage at point B,
the non-inverting input to the op-amp? (3 pt)
R2
R R (vout Vref ) Vref
v v+ = [2k/(2k+30k)](8-2) +2 = 2.375V
1 2
Test A: v+ = 2.38V Test B: v+ = 3.33V Test C: v+ = 2.33V
4) If the op-amp is putting out is minimum voltage at point C, what is the voltage at point
B, the non-inverting input to the op-amp? (3 pt)
R2
R R (vout Vref ) Vref
v v+ = [2k/(2k+30k)](-10 -2) +2 = 1.25V
1 2
Test A: v+ = 1.25V Test B: v+ = 2.48V Test C: v+ = 1.57V
11 of 12
ENGR4300 Test 3S Fall 2005
5) On the following plot, the input at point A is shown.
a) Mark the upper and lower thresholds of the hysteresis (2 pt), and
b) sketch and label the output at points B, C, D and E for the input shown. (12 pt)
Many of the signals will overlap. Use the PSpice labeling convention to identify the
signals by drawing unique symbols on important parts of each trace. Note that the input
has a little box. Use the following: B = + C = D = and E = .
This plot shows the basic shape of the signals for all three tests. Specific high and low
values are given below.
Test A: Signals change when input rises above 2.38V or falls below 1.25V
Signal B toggles between 1.25 and 2.38 volts. (H-L-H-L-H)
Signal C toggles between +8V and -10V (H-L-H-L-H)
Test B: Signals change when input rises above 3.33V or falls below 2.48V
Signal B toggles between 2.48 and 3.33 volts. (H-L-H-L-H)
Signal C toggles between +10V and -8V (H-L-H-L-H)
Test C: Signals change when input rises above 2.33V or falls below 1.57V
Signal B toggles between 1.57 and 2.33 volts. (H-L-H-L-H)
Signal C toggles between +9V and -7V (H-L-H-L-H)
All Tests: Signal D toggles between 0V and some voltage less than 5V (L-H-L-H-L)
(You do not know enough about diodes to know the exact voltage, so we
accepted anything in this range that was recognizable as not zero.)
Signal E is constant at 5V
8) Are the LEDs on or off at the following times (3 points)?
time LED1 LED2
0.3ms ON ON
0.8ms OFF ON
1.2ms ON ON
12 of 12
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CC-MAIN-2014-15
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## How much water should a 70 kg person drink?
60% of our body is water. For a person weighing 70kg, that means 42 kg or 42 litres!
How much water should I drink based on my weight in kg?
Water (in litre) to drink a day equals your weight (in kg) multiplied by 0.033. So, if you are 60 kg, you should drink about two litres of water every day.
How much water should I drink if I weigh 200 pounds?
Instead, think about how much you weigh and divide that number in half. That’s how many ounces of water you should drink per day. For instance, a person who is 200 pounds, should drink 100 oz. of water per day to be adequately hydrated.
### How do I know how much water I should drink?
How much water should you drink based on your weight?
1. Take your weight (in pounds) and divide that by 2.2.
2. Multiply that number depending on your age: If you’re younger than 30, multiply by 40.
3. Divide that sum by 28.3.
4. Your total is how many ounces of water you should drink each day.
How much water should I drink if I weigh 75 kg?
Water (in litres) to drink a day = Your Weight (in Kg) multiplied by 0.033. For example, if you are 60kg, you should drink about 2 litres of water every single day. At 90kg, you’ll around about 3 litres of water. All you need to do is multiply 0.033 to your weight in Kg.
Is 4 litres of water a day too much?
A total daily intake of around 2.7 liters (91 ounces) for women and 3.7 liters (125 ounces) for men can meet most adults’ needs (19). Depending on the other foods and beverages you consume, you may not need to drink 3 liters (100 ounces) of water per day to meet your fluid requirements.
#### How much water should a 75 kg man drink?
How much water should a 240 lb man drink?
How much water per day will help you lose weight?
Weight Ounces of Water Daily
220 pounds 148 ounces
230 pounds 154 ounces
240 pounds 161 ounces
250 pounds 168 ounces
How much water should a 280 pound person drink?
“It depends on your size and weight, and also on your activity level and where you live,” Nessler says. “In general, you should try to drink between half an ounce and an ounce of water for each pound you weigh, every day.” For example, if you weigh 150 pounds, that would be 75 to 150 ounces of water a day.
## Is it OK to drink 4 Litres of water a day?
Drinking enough water offers health benefits, however, drinking too much water, such as 3-4 liters of water, in a short period leads to water intoxication. This is induced by consuming four liters or more water in a few hours (rather than the entire day), lowering salt levels to unsafe levels.
How much water should you drink a day to lose belly fat?
Health experts suggest drinking about eight glasses of water (about 2 liters) per day but it can differ with every individual. If you are a gym regular and sweating out during workouts, your body would require more water.
How much water should a 64 kg person drink?
### How do you calculate water by body weight?
The easy way to calculate total body water is simply to multiply 0.6 times your weight in kilograms, since roughly 2/3 of your body weight is water.
How can drinking more water help you lose weight?
Drinking water helps boost your metabolism, cleanse your body of waste, and acts as an appetite suppressant. Also, drinking more water helps your body stop retaining water, leading you to drop those extra pounds of water weight.
What is the recommended amount of water to drink?
Most experts recommend drinking eight to nine cups of water per day. If you drink eight, eight ounce (the 8 x 8 rule) glasses of fluid each day you’ll generally be consuming the recommended daily fluid intake.
#### How much weight can you lose by drinking water?
Adequate water supply can help you in reducing body weight along with several other health benefits. In fact, based on a recent study, you can lose up to 10 pounds of body weight each year if you drink 1 gallon of water a day.
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https://thibaultlanxade.com/general/jalado-por-un-bloque-de-8-0-n-como-se-muestra-en-la-fi-gura-4-6a-un-bloque-de-20-n-se-desliza-hacia-la-derecha-con-velocidad-constante-calcule-c-entre-el-bloque-y-la-mesa-suponga-que-la-friccion
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Q:
# Pulled by an 8.0-N block, as shown in Figure 4-6a, a 20-N block slides to the right with constant velocity. Find c between the block and the table. Assume that the friction in the pulley is negligible.
Accepted Solution
A:
Unfortunately, without Figure 4-6a or any additional information, it is challenging to accurately determine the value of the coefficient of friction (c) between the block and the table. The coefficient of friction depends on various factors such as the nature of the surfaces in contact, the roughness of the surfaces, and the normal force acting on the block. However, I can provide you with a general understanding of the coefficient of friction and how it relates to the scenario you described. The coefficient of friction (c) is a dimensionless value that represents the ratio of the frictional force between two surfaces to the normal force pressing the surfaces together. It is typically denoted as either static friction (c_s) or kinetic friction (c_k), depending on whether the surfaces are at rest or in motion relative to each other. In your scenario, you mentioned that the 20-N block is being pulled by an 8.0-N force and sliding to the right with a constant velocity. This implies that the force of friction acting on the block is equal in magnitude but opposite in direction to the pulling force. Since the block is moving with a constant velocity, the force of friction must be equal to the applied force, balancing it out. In this case, the coefficient of kinetic friction (c_k) between the block and the table can be calculated using the equation: c_k = (force of kinetic friction) / (normal force) Since the normal force acting on the block is equal to its weight (20 N) in this scenario, the coefficient of kinetic friction (c_k) would be equal to the ratio of the applied force (8.0 N) to the weight of the block (20 N): c_k = 8.0 N / 20 N c_k = 0.4 Therefore, based on this simplified analysis, the coefficient of kinetic friction (c) between the block and the table would be approximately 0.4. However, please note that this value is an estimation based on limited information and assumptions, and the actual coefficient of friction could vary depending on the specific conditions and materials involved.
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https://interviewquestions.ap6am.com/tag/value-labs-aptitude/
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## Value Labs Aptitude Interview Questions & Answers
1. Question 1. A Train 110 M Long Is Running With A Speed Of 60 Km/hr. In What Time Will It Pass A Man Who Is Running At 6 Km/hr In The Direction Opposite To That In Which The Train Is Going ?
Speed of train relative to man = 60 + 6 = 66 km/hr.
= 66 * 5/18 = 55/3 m/sec.
Time taken to pass the men = 110 * 3/55 = 6 sec.
2. Question 2. Ram Bought 4 Dozen Apples At Rs. 12 Per Dozen And 2 Dozen Apples At Rs. 16 Per Dozen. He Sold All Of Them To Earn 20%. At What Price Per Dozen Did He Sell The Apples ?
6 dozen apples = Rs. (12 × 4 + 16 × 2) = Rs. 80
Gain = 20%
S.P = Rs. (120/100 × 80) = Rs. 96
Per dozen = Rs. (96/6) = Rs. 16
3. Aptitude Interview Questions
4. Question 3. In A Game Of 90 Points A Can Give B 15 Points And C 30 Points. How Many Points Can B Give C In A Game Of 100 Points ?
While A scores 90 points, B scores (90-15)=75 points and C scores (90-30)= 60 points
i.e., when B scores 75 points, C scores 60 points
=> When B scores 100 points, C scores 6075×100 = 80 points
i.e., in a game of 100 points, B can give C (100-80)=20 points
5. Question 4. 150 Men Consume 1050 Kg Of Rice In 30 Days. In How Many Days Will 70 Men Consume 980 Kg Of Rice?
Rate of consumption of each man = 1050/(150 * 30) = 7/30 kg/day
Let us say 70 men take x days to consume 150 kg.
Quantity consumed by each item in x days = (7x/30) kg
Quantity consumed by 70 men in x days = (7/30 x)(70) kg
= (7/30 x) * (70) = 960
x = 60 days.
6. Question 5. 22004 – 29303 + 101299 – 59532 = ?
22004 – 29303 + 101299 – 59532
123303 – 888.35 = 34468 = 34500
7. Wipro Aptitude Interview Questions
8. Question 6. A Alone Can Do A Piece Of Work In 6 Days And B Alone In 8 Days. A And B Undertook To Do It For Rs. 3200. With The Help Of C, They Completed The Work In 3 Days. How Much Is To Be Paid To C ?
C’s 1 day’s work = 1/3 – (1/6 + 1/8) = 1/3 – 7/24 = 1/24
A’s wages : B’s wages : C’s wages
1/6 : 1/8 : 1/24 = 4:3:1
C’s share = 1/8 * 3200 = Rs. 400
9. Question 7. In Digging A Pond 20 M * 10 M * 5 M The Volumes Of The Soil Extracted Will Be ?
20 * 10 * 5 = 1000
10. ABB Group Aptitude Interview Questions
11. Question 8. The Greatest Number Of Four Digits Which Is Divisible By 15, 25, 40 And 75 Is:
Greatest number of 4-digits is 9999.
L.C.M. of 15, 25, 40 and 75 is 600.
On dividing 9999 by 600, the remainder is 399.
Required number (9999 – 399) = 9600.
12. Question 9. The Product Of Two Numbers Is 4107. If The H.c.f. Of These Numbers Is 37, Then The Greater Number Is:
Let the numbers be 37a and 37b.
Then, 37a x 37b = 4107
ab = 3.
Now, co-primes with product 3 are (1, 3).
So, the required numbers are (37 x 1, 37 x 3) i.e., (37, 111).
Greater number = 111.
13. VMware Aptitude Interview Questions
14. Question 10. A Person Crosses A 600 M Long Street In 5 Minutes. What Is His Speed In Km Per Hour ?
Speed = 600/ 5 x 60 m/sec. = 2 m/sec. = 2 x 18/5km/hr =7.2 km/hr
15. Question 11. Worker A Takes 8 Hours To Do A Job. Worker B Takes 10 Hours To Do A Job. How Long Should It Take Both A And B, Working Together To Do The Same Job ?
first we need to calculate 1 hours work, then their collective work as A’s 1-hour work is 1/8
B’s 1-hour work is 1/10
(A+B)’s 1 hour work = 1/8 + 1/10= 9/40
So both will finish the work in 40/9 hours=4 4/9
16. Question 12. A And B Started A Business By Investing Rs.4000/- And Rs.5000/- Respectively. Find The A’s Share Out Of A Total Profit Of Rs.1800:
A = Rs.4000/-
B = Rs.5000/-
A share 4 parts & B share 5 parts
Total 9 parts —–> Rs.1800/-
—-> 1 part ——-> Rs.200/-
A share = 4 parts —–> Rs.800/-
17. Aptitude Interview Questions
18. Question 13. A Mixture Contains Alcohol And Water In The Ratio 4 : 3. If 5 Liters Of Water Is Added To The Mixture, The Ratio Becomes 4: 5. Find The Quantity Of Alcohol In The Given Mixture.
Let the quantity of alcohol and water be 4x litres and 3x litres respectively
4x/(3x+5) = 4/5
20x = 4(3x+5)
8x = 20
x = 2.5
Quantity of alcohol = (4 x 2.5) litres = 10 litres.
19. Question 14. A Boat Can Travel With A Speed Of 13 Km/hr In Still Water. If The Speed Of The Stream Is 4 Km/hr, Find The Time Taken By The Boat To Go 68 Km Downstream.
Speed downstream = (13 + 4) km/hr = 17 km/hr.
Time taken to travel 68 km downstream =68/17hrs = 4 hrs.
20. Question 15. Sam And Joan Are Playing A Tennis Match. If The Probability Of Sam’s Win Is 0.59, Then Find The Probability Of Joan’s Win.
Let event A = Sam wins and event B = Joan wins. Then,
P(A) = 0.59
Since if Sam wins, Joan cannot win and if Joan wins, Sam cannot win, so we can say that the events A and B are mutually exclusive. Other than these two events, there are
no any other possible outcomes. So,
P(A)+P(B) = 1
0.59+P(B) = 1
P(B) = 1-0.59 = 0.41
21. Question 16. The Ratio Between The Length And The Breadth Of A Rectangular Park Is 3: 2. If A Man Cycling Along The Boundary Of The Park At The Speed Of 12 Km/hr Completes One Round In 8 Minutes, Then The Area Of The Park (in Sq. M) Is:
Perimeter = Distance covered in 8 min. =12000x 8m = 1600 m.
Let length = 3x metres and breadth = 2x metres.
Then, 2(3x + 2x) = 1600 or x = 160.
Length = 480 m and Breadth = 320 m.
Area = (480 x 320) m2 = 153600 m2.
22. Question 17. The Average Of Runs Of A Cricket Player Of 10 Innings Was 32. How Many Runs Must He Make In His Next Innings So As To Increase His Average Of Runs By 4?
Average = total runs / no.of innings = 32
So, total = Average x no.of innings = 32 x 10 = 320.
Now increase in avg = 4runs. So, new avg = 32+4 = 36runs
Total runs = new avg x new no. of innings = 36 x 11 = 396
Runs made in the 11th inning = 396 – 320 = 76
23. Question 18. The Greatest Number Which On Dividing 1657 And 2037 Leaves Remainders 6 And 5 Respectively, Is:
Required number = H.C.F. of (1657 – 6) and (2037 – 5)
= H.C.F. of 1651 and 2032 = 127.
24. Wipro Aptitude Interview Questions
25. Question 19. Two Pipes A And B Can Fill A Tank In 20 And 30 Minutes Respectively. If Both The Pipes Are Used Together, Then How Long It Will Take To Fill The Tank ?
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# How do you calculate 60?
You have learned that to find 1% of a number means finding 1/100 of it. Similarly, finding 60% of a number means finding 60/100 (or 6/10) of it. 60% of \$700 → 60% × \$700. 60% of \$700 → 0.6 × \$700.
## How do you calculate 60 percent of a grade?
Assign 10 points to each weight. So, 60% = 600 points and 40% = 400 points. total grade = 54 + 28 =82.
## How do you calculate 60 percent of salary?
60% coverage Take your annual salary and divide by 52 (weeks). This is your weekly salary. 2. Multiply this number by either .
## What is a 4th of 40?
One-quarter of 40 is 10. Because one-quarter is equal to the fraction 1/4, we can find one-quarter of a number by finding 1/4 of that number, or by…
## How much will a 60 affect my grade?
If it is a test grade, which in my school they are worth 60%, then it will bring you down by pretty much a whole lot, probably about 7–10 points. If it is a 20% weight it will only bring it down by about 2–3 points.
## How do you find 4 percent of a number?
For this type of problem, you can simply divide the number that you want to turn into a percentage by the whole. So, using this example, you would divide 2 by 5. This equation would give you 0.4. You would then multiply 0.4 by 100 to get 40, or 40%.
## How do you find 3 percent of a number?
If you are required to convert a decimal number like 0.57 to a percentage, you are to simply multiply it by 100. That is, 0.57 x 100 = 57. Therefore, 0.57 as a percentage = 57% or 57 percent. Another example of converting a decimal to a percentage is 0.03 x 100 = 3% or 3 percent.
## How do you convert each part of the ratio to percentage?
So we know that to convert a ratio to percentage, we will first convert it to fraction as x:y=xy then we multiply it by 100 and put the percentage sign. Therefore, we can write 2 : 3 : 5 as 20%, 30%, 50% by substituting the value of x=10 in 2x, 3x, 5x. Now in (c) part we have the ratio as 1 : 4.
## What percent is 162 out of 200?
As per our online percentage calculator tool, 162 out of 200 is 81 percent. It is not hard to calculate percentage of 162 out of 200.
## What is a 18 out of 27?
Now we can see that our fraction is 66.666666666667/100, which means that 18/27 as a percentage is 66.6667%.
## What is 40 60 as a grade?
Now we can see that our fraction is 66.666666666667/100, which means that 40/60 as a percentage is 66.6667%.
## What is a 3rd of 30?
One third of thirty is the same as dividing 30 into three big ‘chunks’ and having one of them for yourself; 30 ÷ 3 is 10, so your ‘slice’ (or answer) is 10.
20% for a final exam is fairly standard. Usually most students need to get a high B or A on a 20% final to get an A in the class, but the weighted average will tell you what you need to get if you know your grades for everything else.
## Is 60 a good test score?
60% is usually considered to be a “passing” grade—usually a D. So it is not a “failing” grade. It means that the student has met the requirements of the course but below the average expectations of the course. A grade of 50% or lower indicates that the student has only fulfilled half of the expectations of the class.
## What is a C grade in college?
A+, A, A- indicates excellent performance. B+, B, B- indicates good performance. C+, C, C- indicates satisfactory performance. D+, D, D- indicates less than satisfactory performance. F indicates unsatisfactory performance (no credit: always include last date of attendance).
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# If $x={e}^{\mathrm{cos}2t}$ and $y={e}^{\mathrm{sin}2t}$ , prove that $\frac{dy}{dx}=-\frac{y\mathrm{log}x}{x\mathrm{log}y}$
Video Solution
Text Solution
Verified by Experts
## $\because x={e}^{\mathrm{cos}2t}$ and $y={e}^{\mathrm{sin}2t}$ $\therefore \frac{dx}{dt}=\frac{d}{dt}{e}^{\mathrm{cos}2t}={e}^{\mathrm{cos}2t}.\frac{d}{dt}\mathrm{cos}2t$ $={e}^{\mathrm{cos}2t}.\left(-\mathrm{sin}2t\right).\frac{d}{dt}\left(2t\right)$ $\frac{dx}{dt}=-2{e}^{\mathrm{cos}2t}.\mathrm{sin}2t\text{......}\left(i\right)$ and $\frac{dy}{dt}=\frac{d}{dt}{e}^{\mathrm{sin}2t}={e}^{\mathrm{sin}2t}.\frac{d}{dt}\mathrm{sin}2t$ $={e}^{\mathrm{sin}2t}\mathrm{cos}2t.\frac{d}{dt}2t$ $=2{e}^{\mathrm{sin}2t}.\mathrm{cos}2t\text{......}\left(ii\right)$ $\therefore \frac{dy}{dx}=\frac{dy/dt}{dx/dt}=\frac{2{e}^{\mathrm{sin}2t}.\mathrm{cos}2t}{-2{e}^{\mathrm{cos}2t}.\mathrm{sin}2t}\text{}$ $=\frac{{e}^{\mathrm{sin}2t}.\mathrm{cos}2t}{{e}^{\mathrm{cos}2t}.\mathrm{sin}2t}\text{.....}\left(iii\right)$ We know that , $\mathrm{log}x=\mathrm{cos}2t.\mathrm{log}e=\mathrm{cos}2t\text{}......\left(iv\right)$ and $\mathrm{log}y=\mathrm{sin}2t.\mathrm{log}e=\mathrm{sin}2t\text{.....}\left(v\right)$ $\therefore \frac{dy}{dx}=\frac{-y\mathrm{log}x}{x\mathrm{log}y}$ [using Eqs. (iv) and (v) in Eq. (iii) and $x=e\left(\mathrm{cos}2t\right),y={e}^{\mathrm{sin}2t}$]
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# Equation: How the FAA Estimates Space-Junk Risks
Space is full of junk. Beginning with the launch of Sputnik in 1957, humankind has boldly flung one heavy metallic gadget after another into orbit, often without much regard for the return trip. As a result, more than 16,000 uncontrollable objects, collectively weighing more than 6,000 tons, now circle the planet.
The problem: Whisper-thin traces of air gradually put the brakes on objects in low Earth orbit and create earthbound showers of hot metal bits. Between 10 and 40 percent of a crashing satellite can reach the ground as charred chunks ranging in size from a pebble to a cow. Some satellites come equipped with a rocket engine and propellant tank so controllers can aim their descent into an ocean instead of a neighborhood, but not all. To determine whether a satellite requires this kind of costly deorbiting system, the Federal Aviation Administration uses this equation to estimate the odds that a piece of it will someday hit someone on the ground. Care to calculate your chances?
Pf : Probability that a piece of space junk will survive reentry. For things like falling satellites, the equation is run separately for each chunk expected to result from its eventual breakup.
Ni : Average population density in area i. To simplify the analysis, Earth’s surface is divided into a grid of squares i, and the distribution is assumed to be uniform in each.
Pi : Probability that the object will hit area i—a function of latitude (lat) and the inclination (inc) of the object’s orbital path relative to the equator.
A : Casualty area. The bigger the object, the larger the area. Multiplying Ni by this tells how many people would get whacked if it falls in area i.
Ec : Expected casualties. It’s evaluated for every area i and every chunk from a given spacecraft. If the sum total exceeds .0001—i.e., a 1 in 10,000 chance of a single casualty—that craft requires a controlled deorbit.
Illustration: Josh Cochran
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# Does the function f(x)= -x^2+6x-1 have a minimum or maximum value?
Jan 28, 2017
The parabola will have a maximum value because the ${x}^{2}$ term is negative.
#### Explanation:
1. Because the function has the general form $f \left(x\right) = \textcolor{b l u e}{A} {x}^{2} + \textcolor{p u r p \le}{B} x + \textcolor{red}{C}$, we know the graph will be a parabola.
2. The sign of the ${x}^{2}$ term will tell us if the parabola opens up (like a $\cup$) or down (like a $\cap$):
If $A > 0$, opens up ($\cup$)
If $A < 0$, opens down ($\cap$)
In this case,
$f \left(x\right) = \textcolor{b l u e}{- \left(1\right)} {x}^{2} \textcolor{p u r p \le}{+ 6} x \textcolor{red}{- 1}$
$\textcolor{b l u e}{A = - 1}$ so the parabola will open "down" or $\cap$ which means the parabola will have a "peak" or maximum point.
graph{-x^2+6x-1 [-15, 15, -10, 10]}
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## The cascade model of mortality
It is often instructive to look at mortality models where the parameters have an underlying meaning in a process. One interesting example of this is the cascade model of mortality, described by Izsak & Gavrilov (1995). Under the cascade model an individual in state $$n$$ can do one of two things:
1. Move to state $$n+1$$ with intensity $$\lambda_0+n\lambda$$, or
2. Move to the dead state with intensity $$\mu_0+n\mu$$.
where $$\lambda_0>0$$ is the background rate of transition from one state to the next, $$\lambda>0$$ is the additional rate of transition, $$\mu_0>0$$ is the background rate of mortality and $$\mu>0$$ is the additional rate of mortality. This is illustrated in Figure 1.
Figure 1. The cascade model for mortality (for simplicity only transitions into and out of state $$n$$ are depicted).
As is clear from the multiplier $$n$$, the rate of transition from one state to the next becomes faster, and the rate of mortality increases as well. It is for this reason that the model is described as a cascade process. By letting $$n\to\infty$$, Izsak & Gavrilov (1995) got the following expression for the force of mortality:
$\mu(x) = \mu_0 + \frac{\mu\lambda_0\left(1-e^{-(\mu+\lambda)x}\right)}{\mu+\lambda e^{-(\mu+\lambda)x}}\qquad(1)$
where $$\mu(x)$$ is the force of mortality at age $$x$$. As noted in Richards (2008, Appendix 3), if we assume $$\mu_0>\frac{\lambda_0\mu}{\lambda}$$ we can make the following substitutions:
\eqalign{\epsilon &= \log\left(\mu_0-\frac{\mu\lambda_0}{\lambda}\right)\cr \beta &= \mu+\lambda \cr \rho &= -\log(\mu_0+\lambda_0) \cr \alpha &= \log \mu - \log \lambda + \log(\mu_0+\lambda_0)}\quad(2)
A spot of algebra then gives us this alternative expression for equation (1):
$\mu(x)=\frac{e^\epsilon+e^{\alpha+\beta x}}{1+e^{\alpha+\rho+\beta x}}\qquad(3)$
where $$\alpha, \beta, \rho$$ and $$\epsilon$$ are real-valued parameters to be estimated. The parameterisation in equation (3) is sometime called the Makeham-Beard model, and some estimated parameter values are given in Table 1.
Table 1. Estimated parameter values (MLEs) for 125,535 male pensioners in England & Wales (31,207 deaths).
ParameterEstimate
$$\hat\alpha$$ -13.2666
$$\hat\beta$$ 0.131902
$$\hat\epsilon$$ -5.07444
$$\hat\rho$$ 0.0931642
In a survey of sixteen models in Richards (2012), I identified the Makeham-Beard model as the best-fitting for a set of annuitant-mortality data. Subsequent practical work has confirmed the usefulness of this model for general pensioner mortality. In other words, the simple four-parameter cascade process is equivalent to one of the best, all-round mortality models for annuitant and pensioner mortality.
Izsak & Gavrilov (1995) observed that as $$x\to\infty$$ in equation (1), the limiting force of mortality is $$\mu_0+\lambda_0$$, i.e. that the limiting value for $$q_x$$ is less than 1 under the cascade model. This is echoed in Greenwood (1926), who suggested that there is indeed a limiting value for $$q_x$$, albeit one that is different for men and women. If we combine this insight with the substitutions in (2), we can see another interpretation of the so-called Beard parameter (after Robert Beard), $$\rho$$, in equation (3): $$e^{-\rho}$$ is the limiting force of mortality. This is illustrated in Figure 2.
Figure 2. Makeham-Beard force of mortality for males according to Table 1, together with limiting value.
References
Greenwood, M. (1926) The natural duration of cancer, Reports on Public Health and Medical Subjects, Her Majesty's Stationery Office, 33, 1–26.
Izsak, J. and Gavrilov, L. A. (1995) A typical interdisciplinary topic: questions of the mortality dynamics, Archives of Gerontology and Geriatrics, 20, 283–293.
Richards, S. J. (2008) Applying survival models to pensioner mortality data, British Actuarial Journal, 14(II), 257–326 (with discussion)
Richards, S. J. (2012) A handbook of parametric survival models for actuarial use, Scandinavian Actuarial Journal, 2012 (4), 233–257.
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A couple of integration problems
1. Apr 1, 2012
Cacophony
1. The problem statement, all variables and given/known data
a) S(4 is higher limit, 0 is lower limit) (x^4 - x^2 + 1)dx
b) S(pi is higher limit, -pi is lower limit) (cosx + sinx)dx
2. Relevant equations
The S is the integration sign
3. The attempt at a solution
a) = ((x^5)/5)-((x^3)/3)+x I(4 high, 0 low)
= (((4^5)/5)-((4^3)/3)+4)-(0)
Is this the final solution or is there another step i don't know about?
b) = (sinx + -cosx) dx I(pi high, -pi low)
= (sin(pi)-cos(pi))-(sin(-pi)-cos(-pi))
Is this the final solution?
2. Apr 1, 2012
LCKurtz
While your answers are technically correct, you should simplify them both. Put in the values of $\cos \pi$ and $\sin \pi$.
3. Apr 1, 2012
Cacophony
so basically:
(0-1)-(0-1)?
4. Apr 1, 2012
5. Apr 1, 2012
0 right
6. Apr 1, 2012
Cacophony
do I simplify the first one aswell? Cause someone said I didn't have to.
7. Apr 1, 2012
LCKurtz
0 is the correct answer, but I'm not sure you didn't make a couple of cancelling arithmetic mistakes along the way. In post #3 it appears you made the following step:
(sin(pi)-cos(pi))-(sin(-pi)-cos(-pi))
= (0-1)-(0-1)
The arithmetic in that step has two errors. With regard to your first one, I wouldn't consider it simplified until it is a single fraction reduced to lowest terms.
8. Apr 1, 2012
Cacophony
I'm not following. What do you mean reduced to lowest terms?
9. Apr 1, 2012
LCKurtz
(((4^5)/5)-((4^3)/3)+4)-(0)
I mean combine the three terms into a single fraction; get rid of all those parentheses.
A fraction is reduced to lowest terms when the numerator and denominator have no common factors. For example, you wouldn't leave an answer as $\frac{42}{30}$ when it could be reduced to $\frac{7}{5}$.
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## A community for students. Sign up today
Here's the question you clicked on:
## nikki156789 3 years ago 26. Calculate the perimeter of triangleDEF �������� a. 39.28 b. 40.32 c. 45.82 d. 72
• This Question is Closed
1. phaizzpham
is there like an image you could draw out?, it would make it simpler if there is an image
2. nikki156789
|dw:1332541848602:dw|
3. nikki156789
what does the degree?
4. nikki156789
ok?
5. nikki156789
myko that comes out to not an option
6. nikki156789
it comes out to 42
7. phaizzpham
this one is a tricky one let me try again
8. myko
|dw:1332542529410:dw| what this means?
9. phaizzpham
yeah that is throwing us off
10. nikki156789
idk which is why i dont understand it
11. myko
actualy, my bad, there is no solution i guess with the info you have
12. phaizzpham
r u positive that it is a t?
13. nikki156789
no its like an angle thing with a line through it.
14. phaizzpham
ahh thats means perpendicular i would think
15. myko
maybe it means that this top angle is half of the bottom ones?
16. nikki156789
idk which is why im having trouble doing this problem
17. phaizzpham
ive done that and i figured out the height which is six, but idk how to apply it to figure out the parameter
18. SUROJ
|dw:1332543779552:dw|
19. phaizzpham
i think the answer is 40.32
20. SUROJ
any reason behind that?
21. nikki156789
SUROJ be nice
22. nikki156789
how'd you get 40.32?
23. phaizzpham
|dw:1332543858044:dw| well i tried to find the height which is 12 using the h=root3*s/2 formula, then i cut the triangle in half, then applied the A^2+B^2=C^2 to find one of the sides which is 13.89 then added all of them together which is 2*13.89+14 which is 41.78 so it could be 40.32 lol
24. nikki156789
oh ok:)
25. phaizzpham
i could be wrong tho D:
26. nikki156789
:/
27. SUROJ
is that equilateral triangle?
28. phaizzpham
yeah it should be
29. nikki156789
i dont think so
30. nikki156789
no its not cuz only two sides are the same
31. phaizzpham
or an isolyze triangle ?
32. nikki156789
its an isosceles triangle
33. SUROJ
yes
34. phaizzpham
okay let me try again lol
35. nikki156789
ok
36. nikki156789
|dw:1332545361479:dw|
37. SUROJ
72
38. nikki156789
thats the 2nd triangle
39. nikki156789
so the perimeter of def is 72?
40. SUROJ
wait that is 42.8
41. SUROJ
yea the perimeter
42. SUROJ
sorry 45.8
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# Quick Review: Clocks and Calendars
Views:35426
In this article, you will learn the important points for solving questions based on Clocks and Calendars.
###### Calendars
• In an ordinary year, there are 365 days, which means 52 * 7 + 1, or 52 weeks and one day. This additional day, is called an odd day.
• Every 100th year starting from 1st AD, is a non-leap year, but every 4th century year is a leap year. So, any year divisible by 400 will be a leap year e.g.: 1200, 1600 and 2000. And the years 1800, 1900 will be non leap years as they are divisible by 100, but not 400.
• The concept of odd days is very important in calendars. In a century - i.e. 100 years, there will be 24 leap years and 76 non–leap years. This means that there will be 24 * 2 + 76 * 1 = 124 odd days. Since, 7 odd days make a week, to find out the net odd days, divide 124 by 7. The remainder is 5. This is the number of odd days in a century.
• 100 years give us 5 odd days as calculated above.
• 200 years give us 5 * 2 = 10 - 7 (one week) = 3 odd days.
• 300 years give us 5 * 3 = 15 - 14 (two weeks) = 1 odd day.
• 400 years give us {5 * 4 + 1 (leap century)} - 21} = 0 odd days.
• Now, if we start from 1st January 0001 AD; for 0 odd day, the day will be Sunday; for 1 odd day, the day will be Monday; for 2 odd days, it will be Tuesday; for 3 odd days, it will be Wednesday and so on.
###### Clocks
• A clock is a complete circle having 360 degrees. It is divided into 12 equal parts i.e. each part is 360/12 = 300. As the minute hand takes a complete round in one hour it covers 3600 in 60 min. In 1 min. it covers 360/60 = 60/ minute.
• Also, as the hour hand covers just one part out of the given 12 parts in one hour, this implies it covers 300 in 60 min. i.e. 1/20 per minute. Therefore, the relative speed of the minute hand is 6 - (1/2) = 5(1/2)degrees.
• Every hour, both the hands coincide once. In 12 hours, they will coincide 11 times. It happens due to only one such incident between 12 and 1'o clock.
• The hands are in the same straight line when they are coincident or opposite to each other.
• When the two hands are at a right angle, they are 15-minute spaces apart.
• In one hour, they will form two right angles and in 12 hours there are only 22 right angles. It happens due to right angles formed by the minute and hour hand at 3'o clock and 9'o clock.
• When the hands are in opposite directions, they are 30-minute spaces apart.
• If a clock indicates 9.15, when the correct time is 9, it is said to be 15 minutes too fast. On the other hand, if it indicates 8.45, when the correct time is 9, it is said to be 15 minutes too slow.
• If both the hour hand and minute hand move at their normal speeds, then both the hands meet after 65(5/11) minutes.
• 22 times in a day, the hands of a clock will be in a straight line but opposite in direction.
• 44 times in a day, the hands of a clock will be straight.
• 44 times in a day, the hands of a clock are at right angles.
• 22 times in a day, the hands of a clock coincide.
• When the minute hand is behind the hour hand, the angle between two hands at M minutes past H 'o clock will be 30(H-(M/5)) + (M/2)degree.
• When the minute hand is ahead of the hour hand, the angle between the two hands at M minutes past H 'o clock will be 30(H-(M/5)) - (M/2) degree.
Views:35426
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# 4 Results
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Subject:
Mathematics
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Provider:
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10/06/2016
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Equations and Inequalities
Type of Unit: Concept
Prior Knowledge
Students should be able to:
Add, subtract, multiply, and divide with whole numbers, fractions, and decimals.
Use the symbols <, >, and =.
Evaluate expressions for specific values of their variables.
Identify when two expressions are equivalent.
Simplify expressions using the distributive property and by combining like terms.
Use ratio and rate reasoning to solve real-world problems.
Order rational numbers.
Represent rational numbers on a number line.
Lesson Flow
In the exploratory lesson, students use a balance scale to find a counterfeit coin that weighs less than the genuine coins. Then continuing with a balance scale, students write mathematical equations and inequalities, identify numbers that are, or are not, solutions to an equation or an inequality, and learn how to use the addition and multiplication properties of equality to solve equations. Students then learn how to use equations to solve word problems, including word problems that can be solved by writing a proportion. Finally, students connect inequalities and their graphs to real-world situations.
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Algebra
Mathematics
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Lesson OverviewStudents apply the multiplication property of equality to solve equations.Key ConceptsIn the previous lesson, students solved equations of the form x + p = q using the addition property of equality. In this lesson, they will solve equations of the form px = q using the multiplication property of equality. They will multiply or divide both sides of an equation by the same number to obtain an equivalent equation.Since multiplication by a is equivalent to division by 1a, students will see that they may also divide both sides of the equation by the same number to get an equivalent equation. Students will also apply this property to solving a particular kind of equation, a proportion.Goals and Learning ObjectivesUse the multiplication property of equality to keep an equation balanced.Use the multiplication property of equality to solve equations of the form px = q for cases in which p, q, and x are all non-negative rational numbers.Use the multiplication property of equality to solve proportions.
Subject:
Algebra
Material Type:
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09/21/2015
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Students work in pairs to critique and improve their work on the Self Check from the previous lesson.Key ConceptsTo critique and improve the task from the Self Check and to complete a similar task with a partner, students use what they know about solving equations and relating the equations to real-world situations.Goals and Learning ObjectivesSolve equations using the addition or multiplication property of equality.Write word problems that match algebraic equations.Write equations to represent a mathematical situation.
Subject:
Algebra
Material Type:
Lesson Plan
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https://brainly.in/question/315289
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# Find the roots of the quadratic equation : 3x²- 2 √ 6x+2=0
1
by priyarakjoshi
2016-03-30T22:26:40+05:30
3x²-2√6x+2=0
a=3, b= -2√6, c=2
D=b²-4ac
D=(-2√6)² - 4(3×2)
D=24-24
D=0, I.e √D=0
So we get that D=0, so the equation has 2 equal roots.
x=-b±√D×1/2a
x=-(-2√6)±0×1/6
x=2√6/6
x=√6/3 or x=-√6/3
pls mark as brainliest if it helped you
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