url stringlengths 6 1.61k | fetch_time int64 1,368,856,904B 1,726,893,854B | content_mime_type stringclasses 3 values | warc_filename stringlengths 108 138 | warc_record_offset int32 9.6k 1.74B | warc_record_length int32 664 793k | text stringlengths 45 1.04M | token_count int32 22 711k | char_count int32 45 1.04M | metadata stringlengths 439 443 | score float64 2.52 5.09 | int_score int64 3 5 | crawl stringclasses 93 values | snapshot_type stringclasses 2 values | language stringclasses 1 value | language_score float64 0.06 1 |
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
http://mamicode.com/info-detail-1954065.html | 1,526,845,467,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794863684.0/warc/CC-MAIN-20180520190018-20180520205946-00028.warc.gz | 180,791,156 | 7,640 | # Piggy-Bank
TimeLimit: 2000/1000 MS (Java/Others) MemoryLimit: 65536/32768 K (Java/Others)
64-bit integer IO format:%I64d
Problem Description
Before ACM can do anything, a budget must be prepared and the necessary financial support obtained. The main income for this action comes from Irreversibly Bound Money (IBM). The idea behind is simple. Whenever some ACM member has any small money, he takes all the coins and throws them into a piggy-bank. You know that this process is irreversible, the coins cannot be removed without breaking the pig. After a sufficiently long time, there should be enough cash in the piggy-bank to pay everything that needs to be paid.
But there is a big problem with piggy-banks. It is not possible to determine how much money is inside. So we might break the pig into pieces only to find out that there is not enough money. Clearly, we want to avoid this unpleasant situation. The only possibility is to weigh the piggy-bank and try to guess how many coins are inside. Assume that we are able to determine the weight of the pig exactly and that we know the weights of all coins of a given currency. Then there is some minimum amount of money in the piggy-bank that we can guarantee. Your task is to find out this worst case and determine the minimum amount of cash inside the piggy-bank. We need your help. No more prematurely broken pigs!
Input
The input consists of T test cases. The number of them (T) is given on the first line of the input file. Each test case begins with a line containing two integers E and F. They indicate the weight of an empty pig and of the pig filled with coins. Both weights are given in grams. No pig will weigh more than 10 kg, that means 1 <= E <= F <= 10000. On the second line of each test case, there is an integer number N (1 <= N <= 500) that gives the number of various coins used in the given currency. Following this are exactly N lines, each specifying one coin type. These lines contain two integers each, Pand W (1 <= P <= 50000, 1 <= W <=10000). P is the value of the coin in monetary units, W is it‘s weight in grams.
Output
Print exactly one line of output for each test case. The line must contain the sentence "The minimum amount of money in the piggy-bank is X." where X is the minimum amount of money that can be achieved using coins with the given total weight. If the weight cannot be reached exactly, print a line "This is impossible.".
SampleInput
```3
10 110
2
1 1
30 50
10 110
2
1 1
50 30
1 6
2
10 3
20 4```
SampleOutput
```The minimum amount of money in the piggy-bank is 60.
The minimum amount of money in the piggy-bank is 100.
This is impossible.思路：裸的完全背包```
``` 1 #include<cstdio>
2 #include<cstring>
3 #include<cmath>
4 #include<algorithm>
5 using namespace std;
6 const int INF=0x3f3f3f3f;
7 const int maxn = 1000;
8 int weight[maxn], value[maxn];
9 int dp[510][10100];
10 int main()
11 {
12 int n;
13 scanf("%d",&n);
14 while(n--)
15 {
16 memset(weight,0,sizeof(weight));
17 memset(value,0,sizeof(value));
18 memset(dp,INF,sizeof(dp));
19 int v1,v2,v;
20 scanf("%d%d",&v1,&v2);
21 v=v2-v1;
22 int m;
23 scanf("%d",&m);
24 for(int i=1; i<=m; i++)
25 scanf("%d%d",&value[i],&weight[i]);
26 dp[0][0]=0;
27 for(int i=1; i<=m; i++)
28 {
29 for(int j=0; j<=v; j++)
30 dp[i][j]=dp[i-1][j];
31 for(int j=weight[i]; j<=v; j++)
32 dp[i][j]=min(dp[i][j], dp[i][j-weight[i]] + value[i] ) ;
33
34 }
35 if(dp[m][v]!=INF)
36 printf("The minimum amount of money in the piggy-bank is %d.\n",dp[m][v]);
37 else
38 printf("This is impossible.\n");
39 }
40 return 0;
41 }```
(0)
(0)
0条 | 1,036 | 3,806 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.953125 | 3 | CC-MAIN-2018-22 | latest | en | 0.931756 |
https://mcqslearn.com/sat/physics/quiz/quiz.php?page=20 | 1,721,511,131,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763517541.97/warc/CC-MAIN-20240720205244-20240720235244-00205.warc.gz | 349,637,361 | 16,702 | SAT Prep Courses
SAT Physics Certification Exam Tests
SAT Physics Practice Test 20
# Centre of Gravity Multiple Choice Questions (MCQs) PDF Download - 20
Books:
Apps:
The Centre of Gravity MCQ with Answers PDF (Centre of Gravity Multiple Choice Questions (MCQs) PDF e-Book) download Ch. 12-20 to prepare SAT Physics Practice Tests. Learn Turning Effects of Forces Test PDF, Centre of Gravity Multiple Choice Questions (MCQ Quiz) for online colleges enrolling. The Centre of Gravity MCQ App Download: Free certification app for latent heat, states of matter, centre of gravity test prep to learn distance learning courses.
The MCQ Quiz: In a meter rule, when the pivot is at the 50 cm mark, the ruler is; "Centre of Gravity" App (Play Store & App Store) with answers: Unbalanced; Balance; Vertical; At to-and-fro position; for online colleges enrolling. Practice Turning Effects of Forces Questions and Answers, Apple Book to download free sample for job placement test.
## Centre of Gravity MCQ Quiz with Answers : Test 20
MCQ 96:
In a meter rule, when the pivot is at the 50 cm mark, the ruler is
1. balance
2. unbalanced
3. vertical
4. at to-and-fro position
MCQ 97:
The assumption of tiny particles in continuous motion is known as
1. atomic model of matter
2. chemical model of matter
3. potential model of matter
4. kinetic model of matter
MCQ 98:
The unit of specific latent heat of fusion(lv) is
1. Joule
2. J Kg-1
3. kg m-3
4. Watt
MCQ 99:
The center of gravity of an object depends on the distribution of its
1. weight
2. temperature
3. magnitude
4. mass
MCQ 100:
The hotter the object, the amount of radiant heat emitted is
1. lesser
2. zero
3. greater
4. negative | 432 | 1,693 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.703125 | 3 | CC-MAIN-2024-30 | latest | en | 0.810924 |
https://www.padeepz.net/ma8151-notes-engineering-mathematics-1-regulation-2017-anna-university/ | 1,576,100,016,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575540533401.22/warc/CC-MAIN-20191211212657-20191212000657-00477.warc.gz | 810,038,484 | 14,063 | # MA8151 Notes Engineering Mathematics 1
## OUTCOMES : MA8151 Notes Engineering Mathematics 1
After completing this course, students should demonstrate competency in the following skills:
Use both the limit definition and rules of differentiation to differentiate functions.
Apply differentiation to solve maxima and minima problems.
Evaluate integrals both by using Riemann sums and by using the Fundamental Theorem of Calculus.
Apply integration to compute multiple integrals, area, volume, integrals in polar
coordinates, in addition to change of order and change of variables.
Evaluate integrals using techniques of integration, such as substitution, partial fractions and integration by parts.
Determine convergence/divergence of improper integrals and evaluate convergent improper integrals.
Apply various techniques in solving differential equations.
### TEXT BOOKS : MA8151 Notes Engineering Mathematics 1
1. Grewal B.S., “Higher Engineering Mathematics”, Khanna Publishers, New Delhi, 43rd Edition, 2014.
2. James Stewart, “Calculus: Early Transcendentals”, Cengage Learning, 7th Edition, New Delhi, 2015. [For Units I & III – Sections 1.1, 2.2, 2.3, 2.5, 2.7(Tangents problems only), 2.8,
3.1 to 3.6, 3.11, 4.1, 4.3, 5.1(Area problems only), 5.2, 5.3, 5.4 (excluding net change theorem), 5.5, 7.1 – 7.4 and 7.8].
#### REFERENCES : MA8151 Notes Engineering Mathematics 1
1. Anton, H, Bivens, I and Davis, S, “Calculus”, Wiley, 10th Edition, 2016.
2. Jain R.K. and Iyengar S.R.K., “Advanced Engineering Mathematics”, Narosa Publications, New Delhi, 3rd Edition, 2007.
3. Narayanan, S. and Manicavachagom Pillai, T. K., “Calculus” Volume I and II, S. Viswanathan Publishers Pvt. Ltd., Chennai, 2007.
4. Srimantha Pal and Bhunia, S.C, “Engineering Mathematics” Oxford University Press, 2015.
5. Weir, M.D and Joel Hass, “Thomas Calculus”, 12th Edition, Pearson India, 2016.
### Padeepz E-Learning Materials MA8151 Engineering Mathematics 1
Padeepz E-Learning Materials MA8151 Engineering Mathematics 1 we have provided the sample materials in this page. If you like the sample and want to buy the full subject the procedure is also provided in this page.
#### Partial derivatives
Partial Derivative Introduction:
A partial derivative of a function of several variables is the ordinary derivative with respect to one of the variables, when all the remaining variables are kept constant. Consider a function u=f(x,y). Here , u is the dependent variable and x & y are independent variables. The partial derivative of u=f(x,y) with respect to x is the ordinary derivative of u w.r.to x, keeping y constant. It is denoted by
Subject name Engineering Mathematics 1 Subject Code MA8151 Semester 1 Regulation 2017 Regulation
MA8151 Syllabus Engineering Mathematics 1
MA8151 Important Questions Engineering Mathematics 1
MA8151 Question Bank Engineering Mathematics 1
## 2 Replies to “MA8151 Notes Engineering Mathematics 1 Regulation 2017 Anna University”
1. Veerasundaram says: | 789 | 3,021 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.296875 | 3 | CC-MAIN-2019-51 | latest | en | 0.741352 |
http://2019price.info/geo/geo-coordinates-distance-java | 1,547,824,863,000,000,000 | text/html | crawl-data/CC-MAIN-2019-04/segments/1547583660175.18/warc/CC-MAIN-20190118151716-20190118173716-00105.warc.gz | 2,157,144 | 4,795 | # Geo Coordinates Distance Java
Price 2019 - Geo Coordinates Distance Java, Nab api documentation, Welcome to nab api. nab’s industry leading api capability is driving innovation and empowering developers, partners and customers. an important part of our journey is collaborating with you, so welcome aboard!. C# - calculating distance between two latitude and, I'm calculating the distance between two geocoordinates. i'm testing my app against 3-4 other apps. when i'm calculating distance, i tend to get an average of 3.3 miles for my calculation whereas other apps are getting 3.5 miles.. Android get address with street name, city for location, In this android tutorial, i will walk you through how to find the address based on the mobile location. we have got gps or network provider in the android device and we can use that to get the current location in terms of latitude and longitude. using the latitude and longitude we can get the address by google geocoding api..
## Geo Coordinates Distance Java - java - how to get mouse coordinates in java using eclipse [ with source code ]
how to install show box 5.05 on fire stick & fire tv or android devices (june 2018) - livy kodi - duration: 12:01. livy kodi 283,087 views ancient rome did not build this part 2 - world's largest stone columns - lost technology - baalbek - duration: 9:51. bright insight 1,049,588 views
Azimuth - wikipedia, Cartographical azimuth. the cartographical azimuth (in decimal degrees) can be calculated when the coordinates of 2 points are known in a flat plane (cartographical coordinates): = (−, −) remark that the reference axes are swapped relative to the (counterclockwise) mathematical polar coordinate system and that the azimuth is clockwise relative to the north.. Interactivate: activities - shodor, Build your own polygon and transform it in the cartesian coordinate system. experiment with reflections across any line, revolving around any line (which yields a 3-d image), rotations about any point, and translations in any direction.. Global positioning system - wikipedia, The global positioning system (gps), originally navstar gps, is a satellite-based radionavigation system owned by the united states government and operated by the united states air force. it is a global navigation satellite system that provides geolocation and time information to a gps receiver anywhere on or near the earth where there is an unobstructed line of sight to four or more gps .. Free gps software for java, symbian, palm, - maps-gps-info.com, A complete, up-to-date source for free gps software for java, symbian, palm, blackberry, iphone, psion and more. | 570 | 2,650 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.75 | 3 | CC-MAIN-2019-04 | longest | en | 0.838821 |
http://www.math.tamu.edu/~fnarc/m641/f15/m641f15_hw.html | 1,542,243,205,000,000,000 | text/html | crawl-data/CC-MAIN-2018-47/segments/1542039742322.51/warc/CC-MAIN-20181114232605-20181115014605-00144.warc.gz | 449,570,937 | 9,852 | Math 641-600 — Fall 2015
Assignments
Assignment 1 - Due Wednesday, September 9, 2015.
• Do the following problems.
1. Section 1.1: 3(c), 5, 7(a), 8
2. Section 1.4: 3
3. Let $U$ be a subspace of an inner product space $V$, with the inner product and norm being $\langle\cdot,\cdot \rangle$ and $\|\cdot\|$. Also, let $v$ be in $V$. (Do not assume that $U$ is finite dimensional or use arguments requiring a basis.)
1. Fix $v\in V$. Show that if there exists $p\in U$ such that $p$ satisfies either (i) $\min_{u\in U}\|v-u\| = \|v-p\|$ or (ii) $v-p\in U^\perp$, then it satisfies both (i) and (ii). Moreover, if $p$ exists, it is unique.
2. Suppose $p$ exists for every $v\in V$. Since $p$ is uniquely determined by $v$, we may define a map $P: V \to U$ via $Pv:=p$. Show that $P$ is a linear map and that $P$ satisfies $P^2 = P$. ($P$ is called an orthogonal projection. The vector $p$ is the orthogonal projection of $v$ onto $U$.)
3. If the projection $P$ exists, show that for all $w,z\in V$, $\langle Pw,z\rangle = \langle Pw,Pz\rangle= \langle w,Pz\rangle$. Use this to show that $U^\perp= \{w\in V\colon Pw=0\}$.
4. Suppose that the projection $P$ exists. Show that $V=U\oplus U^\perp$, where $\oplus$ indicates the direct sum of the two spaces.
4. Let $U$ and $V$ be as in the previous exercise. Suppose that $U$ is finite dimensional and that $B=\{u_1,u_2,\ldots,u_n\}$ is an ordered basis for $U$. In addition, let $G$ be the $n\times n$ matrix with entries $G_{jk}= \langle u_k,u_j\rangle$.
1. Let $v\in V$ and $d_k := \langle v,u_k\rangle$. Show that $p$ exists for every $v$ and is given by $p=\sum_j x_j u_j\in U$, where the $x_j$'s satisfy the normal equations, $d_k = \sum_j G_{kj}x_j$. (Hint: use exercise 3 above.)
2. Explain why the projection $P$ exists. Show that if B is orthonormal, then $Pv=\sum_j \langle v,u_j\rangle u_j$.
Assignment 2 - Due Wednesday, September 16, 2015.
• Read sections 2.1 and 2.2
• Do the following problems.
1. Section 1.2: 9, 10(a,b)
2. Section 1.3: 2(b)
3. Section 1.4: 4
4. Let V be an n dimensional vector space and suppose L:V→V is linear.
1. Let E={φ1,.., φ1} be a basis for V and let A be the matrix of L relative to E. Show that the characteristic polynomial pA(λ) := det(A − λ I) is independent of the choice of E, and so pL(λ) := pA(λ) is well defined, in the sense that it is independent of the choice of basis for V.
2. Use the previous part to show that Trace(L) := Trace(A) and det(L) := det(A) are also well defined,
3. Let $L:P_2\to P_2$ be given by $L(p)= \big((1-x^2)p'\big)' + 7p$. Find Trace(L) and det(L).
5. Let A and B be n×n matrices. Suppose that the range of B, Range(B), is an invariant subspace for A. Show that there is an n×n matrix X such that AB = BX.
6. (This is a generalization of Keener's problem 1.3.5.) Let $A$ be a self-adjoint matrix with eigenvalues $\lambda_1\ge \lambda_2,\ldots,\ge \lambda_n$. Show that for $2\le k < n$ we have $\max_U \sum_{j=1}^k \langle Au_j,u_j \rangle =\sum_{j=1}^k \lambda_j,$ where $U=\{u_1,\ldots,u_k\}$ is any o.n. set. (Hint: Put $A$ in diagonal form and use a judicious choice of $B$.)
7. Let U be a unitary, n×n matrix. Show that the following hold.
1. < Ux, Uy > = < x, y >
2. The eigenvalues of U all lie on the unit circle, |λ|=1.
3. Show that U is diagonalizable. (Hint: follow the proof for the self-adjoint case.)
8. Let $k(x,y) = x+ 3x^2y + xy^2$ and $\langle f,g\rangle=\int_{-1}^1 f(x)g(x)(1+x^2)dx$. Consider the operator $Lu=\int_{-1}^1 k(x,y) u(y)dy$. In the notes, we have shown that $L:P_2\to P_2$.
1. Relative to the inner product above, find $L^\ast$ and $\text{Null}(L^\ast)$.
2. Find a condition on $q\in P_2$ for which $Lp=q$ always has a solution. Is this different from what was in the notes?
Assignment 3 - Due Wednesday, September 23, 2015.
Assignment 4 - Due Friday, October 2, 2015.
Assignment 5 - Due Friday, October 9.
Assignment 6 - Due Friday, October 30.
Assignment 7 - Due Friday, November 6.
• Read sections 3.1 and 3.2.
• Do the following problems.
1. Section 2.2: 25(a,b), 26(b), 27(a)
2. Let $H_0$ be the set of all $f\in C^{(0)}[0,1]$ such that $f(0)=f(1)=0$ and that $f'$ is piecewise continuous. Show that $\langle f,g\rangle_{H_0} :=\int_0^1f'(x)g'(x)dx$ defines a real inner product on $H_0$.
3. We want to use the Galerkin method to numerically solve the boundary value problem (BVP): −u" = f(x), u(0) = u(1) = 0, f ∈ C[0,1]
1. Weak form of the problem. Let H0 be as in the previous problem. Suppose that $v\in H_0$. Multiply both sides of the eqaution above and use integration by parts to show that $\langle u,v\rangle_{H_0} = \langle f,v\rangle_{L^2[0,1]}$. This is called the weak'' form of the BVP.
2. Conversely, suppose that u ∈ H0 is also in C(2)[0,1] and that u satisfies
⟨u,v⟩H0 = ∫01 f(x) v(x) dx for all v ∈ H0.
Show that u satisfies the BVP.
3. Consider S0 := {s ∈S1/n(1,0) : s(0)=s(1)=0}. Show that S0 is spanned by φj(x) := N2(nx-j+1), j = 1 ... n-1. (Here, N2(x) is the linear B-spline.)
4. Show that the least-squares approximation s ∈ S0 to the solution u is given by s = ∑j αjφj(x), where the αj's satisfy Gα = β, with
βj = ⟨ y,φjH0 = ∫01 f(x) φj(x) dx, j=1 ... n-1 and Gkj = ⟨ φj, φkH0.
5. Show that Gkj = ⟨ φj, φkH0 is given by
Gj,j = 2n, j = 1 ... n-1
Gj,j-1 = - n, j = 2 ... n-1
Gj,j+1 = - n, j = 1 ... n-2
Gj,k = 0, all other possible k.
Assignment 8 - Due Friday, November 13.
Assignment 9 - Due Friday, November 20.
• Read sections 3.5 and 3.6.
• Do the following problems.
1. Section 3.4: 2(b)
2. Consider the real Hilbert space H0 := {v ∈ H1[0,1] : v(0) = v(1) = 0}, with the inner product < f,g >H0 = ∫01f ′(x) g ′(x)dx. Show that there is a function gy in H0 for which f(y) = < f,gy >H0. (gy is called a reproducing kernel and H0 is a reproducing kernel Hilbert space.) Hint: Use problem 4, HW3 to show that the Φy(f) := f(y) is a bounded linear functional on H0.
3. Let $L$ be a bounded operator on a Hilbert space $H$. Show that the closure of the the range of $L$ satisfies $\overline{R(L)} = N(L^\ast)^\perp$. (Hint: Follow the proof of the Fredholm alternative, which is just the special case where $R(L)$ is closed.)
4. A sequence {fn} in a Hilbert space H is said to be weakly convergent to f ∈ H if and only if lim n → ∞ < fn,g> = < f,g> for every g∈H. When this happens, we write f = w-lim fn. For example, if {φn} is any orthonormal sequence, then φn converges weakly to 0. You are given that every weakly convergent sequence is a bounded sequence (i.e. there is a constant C such that ||fn|| ≤ C for all n). Prove the following:
Let K be a compact linear operator on a Hilbert space H. If fn converges weakly to f, then Kfn converges to Kf — that is, lim n → ∞ || Kfn - Kf || = 0.
Hint: Suppose this doesn't happen, then there will be a subsequence of {fn}, say {fnk}, such that || Kfnk - Kf || ≥ ε for all k. Use this and the compactness of K to arrive at a contradiction. We remark that the converse is also true. If a bounded linear operator $K$ maps weakly convergent sequences into convergent sequences, then $K$ is compact.
5. Consider the Hilbert space $\ell^{\,2}$. Let $S=\{\{a_j\}_{j=1}^\infty \in \ell^{\,2}\colon \sum_{j=1}^\infty (1+j^2)\,|a_j|^2 < 1 \}$. Show that $S$ is a precompact subset of $\ell^{\,2}$.
6. Let L be a bounded self-adjoint linear operator on a Hilbert space $\mathcal H$. Show that these two formulas for $\|L\|$ are equivalent:
1. $\|L\| = \sup \{\|Lu\| : u \in {\mathcal H},\ \|u\| = 1\}$
2. $\|L\| = \sup \{|\langle Lu,u\rangle| : u\in {\mathcal H},\ \|u\|=1\}$
Assignment 10 - Due Monday, November 30.
Extra Problems - These are not to be handed in.
1. Let $L$ be in $\mathcal B (\mathcal H)$.
1. Show that $\|L^k\| \le \|L\|^k$, $k=2,3,\ldots$.
2. (We did this in class.) Let $|\lambda| \|L\|<1$. Show that $\big\|(I - \lambda L)^{-1} - \sum_{k=0}^{n-1}\lambda^k L^k\big\| \le \frac{|\lambda|^k \|L\|^k}{1 - |\lambda| \|L\|}.$
3. Let $L$ be as in problem 6, HW8. Estimate how many terms it would require to approximate $(I - \lambda L)^{-1}$ to within $10^{-8}$, if $|\lambda|\le 0.1$.
2. Use Newton's method (see text, problem 3.6.3) to approximate the cube root of 2. Show that the method converges.
3. Section 4.1: 6
4. Section 4.2: 1, 4, 8
5. Let $Lu=-u''$, $u(0)=0$, $u'(1)=2u(1)$.
1. Show that the Green's function for this problem is $G(x,y)=\left\{ \begin{array}{rl} -(2y-1)x, & 0 \le x < y \le 1\\ -(2x-1)y, & 0 \le y< x \le 1. \end{array} \right.$
2. Verify that $0$ is not an eigenvalue for $Kf(x) := \int_0^1G(x,y)f(y)dy$.
3. Show the orthonormal set of eigenfunctions for $L$ form a complete set in $L^2[0,1]$. (Hint: use tthe results from problem 4, HW10.
Updated 12/8/2015. | 3,140 | 8,666 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 1, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.0625 | 4 | CC-MAIN-2018-47 | latest | en | 0.775293 |
https://onlinejudge.org/board/viewtopic.php?f=5&t=8141&start=30 | 1,591,316,847,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590348492295.88/warc/CC-MAIN-20200604223445-20200605013445-00158.warc.gz | 469,841,558 | 12,440 | 406 - Prime Cuts
Moderator: Board moderators
Arm.Turbo
New poster
Posts: 21
Joined: Wed Aug 11, 2004 1:20 pm
At last AC. Thx to all =)
Rony
New poster
Posts: 16
Joined: Wed Jun 30, 2004 6:46 am
Location: Dhaka
Contact:
p-406
Hi,
Anyone can give me algorithm to find the middle and print the list .
Hi,
A1, I think you have'nt understand my question . For prime cuts problem I want to know the algo to find out the mid point and print the result because i have tried so many times but failed . There is a condition if list is
even then print C*2 number of prime numbers else (C*2)-1 from the middle . The given algo (By you ) will not solve this problem.
Thanks
Rony
Last edited by Rony on Sun Sep 26, 2004 10:43 am, edited 1 time in total.
A1
Experienced poster
Posts: 173
Joined: Wed Jan 28, 2004 3:34 pm
then :
Code: Select all
``````int lenthofarray; //some how you know it
float halfofarray;
int halfofarrayint;
halfofarray=lenthofarray/2.0;
halfofarrayint= halfofarray; //one kind of type casting
if(halfofarrayint == halfofarray) // then two middle
print -- array[halfofarrayint] and array[halfofarrayint+1];
else
print --array[halfofarrayint];``````
if you talk about prime cut then you have to do little more
Rony
New poster
Posts: 16
Joined: Wed Jun 30, 2004 6:46 am
Location: Dhaka
Contact:
A1 wrote:your question is not clear!
then :
Code: Select all
``````int lenthofarray; //some how you know it
float halfofarray;
int halfofarrayint;
halfofarray=lenthofarray/2.0;
halfofarrayint= halfofarray; //one kind of type casting
if(halfofarrayint == halfofarray) // then two middle
print -- array[halfofarrayint] and array[halfofarrayint+1];
else
print --array[halfofarrayint];``````
if you talk about prime cut then you have to do little more
Hi,
A1, I think you have'nt understand my question . For prime cuts problem I want to know the algo to find out the mid point and print the result because i have tried so many times but failed . There is a condition if list is
even then print C*2 number of prime numbers else (C*2)-1 from the middle . The given algo (By you ) will not solve this problem.
Thanks
Rony
A1
Experienced poster
Posts: 173
Joined: Wed Jan 28, 2004 3:34 pm
Ok Now I undersetand
Code: Select all
``````intger leanthofarray;
input: C;
if C*2>=leanthofarray
print every element of array.
else
{
if(leanthofarry is even)
{
start=(leanthofarry-(C*2))/2;
For i=(0+start) To i<(leanthofarray-start)
print array[i];
}
else
{
start=(leanthofarray-((C*2)-1))/2;
For i=(0+start) To i<(leanthofarray-start)
print array[i];
}
}``````
I think it will work
Rony
New poster
Posts: 16
Joined: Wed Jun 30, 2004 6:46 am
Location: Dhaka
Contact:
Hi,
Thanks a lot . I have got Acc with PE at last. You have done a lot for
me. Your algorithm is so easy which is very good . Carry on.
Who are u? Post some things about you and your UVA ID thats why
I can see your score. Again many many Thanks.
Regards
Rony
Ashkankhan
New poster
Posts: 12
Joined: Wed Oct 13, 2004 10:14 am
Location: Teh
Contact:
Is there any bodu here??????????
I delete it:D
Last edited by Ashkankhan on Mon Nov 01, 2004 9:50 am, edited 1 time in total.
Ashkankhan
New poster
Posts: 12
Joined: Wed Oct 13, 2004 10:14 am
Location: Teh
Contact:
Gods of algo helppppppppp
help me WA
[cpp]
#include <stdio.h>
int main()
{
int a[]={1,2,3,5,7,11,13,17,19,23,29,31,37,
41,43,47,53,59,61,67,71,73,79,83,89,97,
101,103,107,109,113,127,131,137,139,149,
151,157,163,167,173,179,181,191,193,197,
199,211,223,227,229,233,239,241,251,257,
263,269,271,277,281,283,293,307,311,313,
317,331,337,347,349,353,359,367,373,379,
383,389,397,401,409,419,421,431,433,439,
443,449,457,461,463,467,479,487,491,499,503,509,521,523,
541,547,557,563,569,571,577,587,593,599,601,607,613,617,
619,631,641,643,647,653,659,661,673,677,683,691,701,709,
719,727,733,739,743,751,757,761,769,773,787,797,809,811,
821,823,827,829,839,853,857,859,863,877,881,883,887,907,
911,919,929,937,941,947,953,967,971,977,983,991,997};
//printf("%d",a[84]);
int n,c,i=0,t;
while(scanf("%d %d",&n,&c) != EOF)
{
if(c!=0)
{
// if(n<=1000 && n>=1 && c<=n && c>=1 )
// {
i=0;
while(a<=n)
{
i++;
}
//printf("%d\n",i);
printf("%d %d:",n,c);
if(i%2==0)
{
t=(i-2*c)/2;
if(2*c-1>i)
{
for(int j=0;j<i;j++)
printf(" %d",a[j]);
}
else
for(int y=t;y<=t+(2*c-1);y++)
{
printf(" %d",a[y]);
}
}
else
{
t=(i-2*c-1)/2;
if(2*c>i)
{
for(int p=0;p<i;p++)
printf(" %d",a[p]);
}
else
for(int e=t+1;e<=t+(2*c-1);e++)
{
printf(" %d",a[e]);
}
}
printf("\n\n");
}
// }
}
return 0;
}
[/cpp]
sohel
Guru
Posts: 856
Joined: Thu Jan 30, 2003 5:50 am
Location: New York
... and which problem is this?
Ashkankhan
New poster
Posts: 12
Joined: Wed Oct 13, 2004 10:14 am
Location: Teh
Contact:
dear sohel
406:D
Sokar
New poster
Posts: 1
Joined: Tue Sep 07, 2004 9:45 pm
Ashkankhan wrote:help me WA
One problem I see with your solutions is it will not work for input that is larger than 997. Almost certainly the range of input will be more than 997?
Piers Kennedy
New poster
Posts: 3
Joined: Thu Apr 22, 2004 8:12 pm
406 input causing WA
Dear Ashkankhan,
My approach was very similar to yours. My code passed all the tests in the 406 threads but I still got WA. Sokar's suggestion is not why you are still getting WA (there is no input above 1000). For some reason for this problem you need to read all the input (and store) then print the output. As soon as I tried this I got AC.
Hope this helps.
A1
Experienced poster
Posts: 173
Joined: Wed Jan 28, 2004 3:34 pm
Very Small Problem
[c]i=0;
while(a<=n)
{
i++;
}
//printf("%d\n",i);
[/c]
Change it like this :
[c]i=0;
while(a<=n && a>0) //don't know why!!!!!!!
{
i++;
}
//printf("%d\n",i);
[/c]
you will get a nice AC
Ashkankhan
New poster
Posts: 12
Joined: Wed Oct 13, 2004 10:14 am
Location: Teh
Contact:
very very tnx A1.But I dint know why
frankhuhu
New poster
Posts: 30
Joined: Tue Jul 20, 2004 5:22 am
Contact:
406 TLE?? Need Help!!!
[cpp]
I think it's a simple question,but it's TLE,Why?
Here is my code.
#include <iostream.h>
bool p[1000010];
int prime[1010];
void Eratosthenes()
{
int i,j;
p[0] = p[1] = 0;
for (i=2; i<=1000000; i++) p=true;
for (i=2; i<=1000;)
{
for (j=i+i; j<=1000000; j+=i) p[j]=false;
for (i++; !p; i++);
}
}
void get_prime_table()
{
Eratosthenes();
prime[0]=1;
int i,j;
for (i=2,j=1;i<=1010;i++)
if (p==true) prime[j++]=i;
}
void solve(int n,int c)
{
get_prime_table();
int length,times;
int i;
for (i=0;prime<=n;i++);
length=i;
if (length%2==0) times=c*2;
else times=c*2-1;
if (times>length)
{
for (i=0;i<length-1;i++)
cout<<prime<<' ';
cout<<prime[length-1]<<endl;
}
else
{
if (length%2!=0)
{
for (i=(length-times+1)/2;i<(length+times-1)/2;i++)
cout<<prime<<' ';
cout<<prime[(length+times-1)/2]<<endl;
}
if (length%2==0)
{
for (i=(length-times)/2;i<(length+times-2)/2;i++)
cout<<prime<<' ';
cout<<prime[(length+times-2)/2]<<endl;
}
}
}
int main()
{
int N,C;
while (cin>>N>>C)
{
cout<<N<<' '<<C<<": ";
solve(N,C);
cout<<endl;
}
return 0;
}[/cpp] | 2,546 | 6,972 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3 | 3 | CC-MAIN-2020-24 | longest | en | 0.807186 |
https://community.qlik.com/t5/QlikView-App-Development/calculate-class/m-p/1401000/highlight/true | 1,571,143,997,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570986658566.9/warc/CC-MAIN-20191015104838-20191015132338-00224.warc.gz | 439,504,345 | 40,414 | # QlikView App Development
Discussion Board for collaboration related to QlikView App Development.
Contributor III
## calculate class
Hi everyone
I have two tables:
-First one contains information relative to the class:
For example: if the rank of my product’s sales of brand B1 in France <=2, then I will give him a class A;
The setting of class for each country/brand is different, and interval of rank is larger in Germany than France because we have more products to sale in Germany
--Second one is a straight table which contains information about sales.
In this table I should calculate the rank of sales per product/brand/county and his class according to his rank
In this example:
Product P1 of brand B1 in France has a class C because his sale’s rank is 6.
So for calculating class, I can use “if”:
if(country='France' and brand='B1', if(rank>6,'D', if(rank>4, 'C, if(rank>2, 'B','A'))), if(county='France' and brand='B2' ……..)
But I find my solution is very complicate, I want to find easier way to achieve it.
I can not achive it in script because rank is calculated in the straight table
But how?
Thanks
Tags (1)
4 Replies
Highlighted
Honored Contributor
## Re: calculate class
Hi Zhu,
try:
Rank:
FROM
class.xlsx
(ooxml, embedded labels, header is 1 lines, table is class);
Temp:
max(RankValue) as MaxRank;
PurgeChar(FieldValue('Rank',RecNo()),'<>=') as RankValue
AutoGenerate FieldValueCount('Rank');
Let vMaxRank = Peek('MaxRank',0,'Temp');
Class:
Brand,
Class,
Rank,
if(left(Rank,1) = '>', \$(vMaxRank),PurgeChar(Rank,'<>=')) as To
FROM
class.xlsx
(ooxml, embedded labels, header is 1 lines, table is class);
Class1:
country,
Brand,
Class,
To,
if(country=peek(country) and Brand=peek(Brand),Peek(To),0) +1 as From
Resident Class Order by country, Brand,To ;
Sales:
brand as Brand,
product,
sales,
Rank
FROM
class.xlsx
(ooxml, embedded labels, table is Sales);
Left Join(Class1)
IntervalMatch (Rank, country, Brand) LOAD From, To, country, Brand Resident Class1;
Left Join(Sales)
LOAD Rank, country, Brand, Class Resident Class1;
Drop Tables Class, Class1,Temp, Rank;
Giving this:
country Class sum(sales)
1058
FranceA429
FranceB310
FranceC216
FranceD103
Regards
Andrew
Honored Contributor
## Re: calculate class
Hi Zhu,
I deleted my original post and replaced it with another. This new script does not read the class from the sales table. It takes the class table and creates intervals of rank then uses the IntervalMatch function to form an association between these intervals and the rank given in the sales table. A join operation then adds the class field to the sales table.
Kind regards
Andrew
Contributor III
## Re: calculate class
Hi Andrew.
I think it was not clear in my original request. I have not field rank in y table salesin which I have only sales per product/brand/country, rank is an expression in my straight chart calculated according to my filters.
Xia
Contributor III
## Re: calculate class
Hi Andrew,
Some feedback for you.
As I said that I have not my rank in my file which is indicator calculated in live in my application.
So I used only firt part of your script to get the liste of my range of class with interval function.
And then I use concat to create two variables: one for class and another one for mapping like this
concat(Chr(39)&Classe&Chr(39),',', Key).
At the end, I create an expression in my straight table
=
Pick(
Match(
County&'|'&Brand&'|'&Rank(......... ),
\$(filedsmapping))+1,
'D',\$(class),
). | 896 | 3,550 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.796875 | 3 | CC-MAIN-2019-43 | latest | en | 0.837596 |
https://nivent.github.io/blog/interesting-equation-ii/ | 1,675,912,651,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764501066.53/warc/CC-MAIN-20230209014102-20230209044102-00756.warc.gz | 443,727,676 | 12,361 | # An Interesting Equation II
This will be another post analyzing a particular diophantine equation. Like the last time, this equation will be quadratic. Unlike last time, this time there will actually be some integral solutions. The equation in question is
where, of course, $x,y\in\Z$. In other words, we want to know which integers $y$ make $5y^2+4$ a perfect square. The answer may surprise you.
Note: I’m gonna use some facts about Dedekind domains that I haven’t proved anywhere on this blog (i.e. unique factorization of ideals into prime ideals). If I’m going to keep doing posts like this one, at some point I will write a post introducing the theory of Dedekind domains and proving that rings of integers are always Dedekind.
The first thing to realize is that working in $K=\Q(\sqrt 5)$ allows us to rewrite our equation as
where $\knorm:K\to\Q$ is the standard relative norm, so we are looking for elements of $K$ (integral coefficients and) with norm $4$. Since we only want elements with integer coefficients, we may be tempted to work in the ring $\Z[\sqrt 5]$. However, this ring isn’t super nice. Instead, the proper setting for this problem is the ring of integers $\ints K=\Z[\phi]$ where $\phi$’s satisfies $x^2-x-1=0$ (notice that $-\inv\phi$ also satisfies this polynomial).
Hence, our goal is to find elements of $\ints K$ with norm $4$. Well, technically our goal is to find elements of $\Z[\sqrt5]$ with norm $4$, but since $\Z[\sqrt 5]\subseteq\ints K$, our only concern is that $\ints K\sm\Z[\sqrt 5]$ might have some norm $4$ elements, but this will turn out to be a non-issue.
Fix $x,y\in\Z$ such that $x+y\phi$ has norm $4$. That is,
This shows that $x+y\phi$ is a factor of $4$, so the natural thing to do would be to invoke unique factorization somehow. We are in luck because $\ints K$ turns out to be a UFD. However, this is non-trivial to prove, so we won’t rely on this fact. We’ll instead rely on the slightly-easier-to-prove facts 1 about general Dedekind domains. In particular, that they recover unique factorizations on ideals. In terms of ideals, we have
Note that $(2)$ is prime as
This makes $(2)^2$ the (unique) factorization of $(4)$ into prime ideals. Because $x+y\phi$ and $x-y\inv\phi$ are not units (as they have non-unit norms), this means we must have
Thus, $x+y\phi=2u$ for some unit $u\in\units{\ints K}$ 2. Now, we have previously shown that $\units{\ints K}=\pm\eps^{\Z}$ for some fundamental unit $\eps\in\ints K$. It’s not too hard to show furthermore that one can take $\eps=\phi$ as such a fundamental unit. This means we have shown that every norm $4$ element of $\ints K$ is of the form $\pm2\phi^n$ for some $n\in\Z$.
We can actually say more. Because $\knorm(x+y\phi)=\knorm(2)$, we must have $\knorm(u)=1$; however, $\knorm(\phi)=-1$, so $u$ must be an even power of $\phi$. That is, the norm $4$ elements of $\ints K$ are exactly those of the form $\pm2\phi^{2n}$ for some $n\in\Z$.
Since we are looking for solutions to $x^2-5y^2=4$ and squaring erases signs, we’ll ignore the $\pm$, and define sequences $a_n,b_n\in\Z$ by $a_n+b_n\phi=2\phi^{2n}$. Using that $\phi^2=1+\phi$, we see that
so $a_{n+1}=a_n+b_n$ and $b_{n+1}=a_n+2b_n$. At this point, it is worthwhile to remark that $a_0=2$ and $b_0=0$, so induction shows that $a_n,b_n$ are both even for all $n$. This means that $a_n+b_n\phi\in2\ints K\subseteq\Z[\sqrt5]$, so the norm $4$ elements are really in 1-1 correspondence with the solutions we seek.
Now, we want “integers $y$ such that $5y^2+4$ is perfect” so we really only care about $b_n$. With this in mind, note that $a_n=b_{n+1}-2b_n$ so $a_{n+1}=b_{n+2}-2b_{n+1}$. Thus, $b_{n+2}-2b_{n+1}=b_{n+1}-b_n$ which we rearrange to read
We could get an explicit formula for this 3, but we’ll cheat a little instead. If you form a table of the first few values of $b_n$,
and stare at them for long enough, you may notice that it looks a lot like $b_n=2F_{2n}$ where $F_n$ is the $n$th Fibonacci number. Indeed, defining $\ast b_n:=2F_{2n}$ we see this sequence satisfies
Since we also have $\ast b_0=2F_0=0$ and $\ast b_1=2F_2=2$, this shows that $b_n=\ast b_n=2F_{2n}$. Returning to our original question (and noting that $\phi=(1+\sqrt5)/2$), we can write $a_n+b_n\phi=x_n+y_n\sqrt5$ where
Thus, the (positive) integers $y$ such that $5y^2+4$ exactly make up every other Fibonacci number! This is a very surprising fact on first glance, but it came fairly naturally from our analysis of the equation $x^2-5y^2=4$. The fibonacci numbers showed up since we worked in the ring $\ints K=\Z[\phi]$ where $\phi$ is the golden ratio, and it’s every other fibonacci number because $\knorm(\phi)=-1$ so you switch between $5y^2+4$ being a square and $5y^2-4$ being a square as you consider consecutive powers of $\phi$ (consecutive Fibonacci numbers).
1. that I’m still not proving here
2. We could have reached this conclusion earlier if we had used unique factorization at the level of elements, but the detour through ideals was a short one, so no biggy
3. using e.g. generating functions or linear algebra | 1,520 | 5,099 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.765625 | 4 | CC-MAIN-2023-06 | latest | en | 0.894485 |
https://www.math-edu-guide.com/CLASS-7-Important-Rules-Of-Exponents.html | 1,718,207,817,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861173.16/warc/CC-MAIN-20240612140424-20240612170424-00410.warc.gz | 793,773,054 | 6,415 | # CLASS-7IMPORTANT RULES OF EXPONENTS
IMPORTANT RULES OF EXPONENTS
1) In, a⁶ = a X a X a X a X a X a, we should read as ‘a’ to the power 6 (Six), base = a , index = 6 .
2) in, b⁸ = b X b X b X b X b X b X b X b, we should read as ‘b’ to the power 8 (Eight), base = b, index = 8.
3) in, y= y X y X y, we should read as ‘y’ to the power 3 (Three) or ‘y’ cubed (or cube), base = y, index = 3.
4) in, z² = z X z, we should read as ‘z’ to the power of 2 (Two) or ‘z’ square ( or squared), base = z, index = 2.
5) z¹ = z
6) (-1)ⁿ = 1, when index ‘n’ is even
7) (-1)ⁿ = - 1, when index ‘n’ is odd
1
8) Z X -------- = 1
Z
1
9) Z ÷ Z = Z X --------- = 1
Z
1
10) ---------- = a⁻ᵐ
aᵐ
This is called the Reciprocal of a power is written as a power with an index equal to the negative of the index of the power.
11) The laws of Multiplication – aᵐ X aⁿ = aᵐ⁺ⁿ, where ‘m’ & ‘n’ are numbers, and ‘a’ is a non-zero number.
1
12) The laws of division, aᵐ ÷ aⁿ = aᵐ X -------- = aᵐ⁻ⁿ,
aⁿ
where ‘m’ & ‘n’ are numbers, and ‘a’ is a non-zero number.
13) The laws of power, (aᵐ)ⁿ = aᵐⁿ, where ‘m’ & ‘n’ are numbers, and ‘a’ is a non-zero number.
14) where ‘a’ is non zero number but index is ‘0’ (zero), then a⁰ = 1
15) laws of th power of a product, ( a X b )ⁿ = aⁿ X bⁿ
aⁿ
and, (a / b)ⁿ = ---------- = aⁿ / bⁿ
bⁿ | 601 | 1,386 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.8125 | 4 | CC-MAIN-2024-26 | latest | en | 0.755324 |
http://scienceblogs.de/klausis-krypto-kolumne/2018/07/07/a-code-in-the-hollywood-movie-wanted/ | 1,550,377,893,000,000,000 | text/html | crawl-data/CC-MAIN-2019-09/segments/1550247481612.36/warc/CC-MAIN-20190217031053-20190217053053-00548.warc.gz | 245,756,814 | 25,213 | # A code in the Hollywood movie “Wanted”
The movie “Wanted” (starring Angelina Jolie and Morgan Freeman) contains a steganographic message. Can a reader decipher it?
Hiding a code in a movie, TV series or video clip is an interesting means of gaining publicity. Here are the examples I am aware of (if you know others please let me know):
• I have published several blog posts about the Fair Game code, which is still unsolved.
• In the TV series Breaking Bad, a character sequence can be seen that might represent a code.
• In the series Alias, every location name contains a marked letter. In the linear sequence read, these marked letters resulted a word, a hidden meaning of the episode.
• In the intro of the German TV film Im Namens meines Sohnes a message is hidden. It is easy to solve.
### The “Wanted” code
Earlier this year, blog reader Tony Patti informed me about another code hidden in a movie. This one is contained in Wanted, a 2008 thriller staring Angelina Jolie, Morgan Freeman, and James McAvoy. Here’s the trailer:
In the movie, a man named Sloan (Morgan Freeman) explains that 1000 years ago a clan of weavers discovered a mystical code hidden in a fabric. The code is based on mistakes in weave of the vertical threads. If the vertical thread is on top, it’s a “1”; if it’s below, it’s a “0”. Once decoded, the Binary Code reveals the name of the person that would die. Here is an example (thanks to Tony for the screenshot):
I don’t know the message that is encoded here. Can a reader find it? According to Tony, the Bacon cipher was used. The Bacon cipher encodes the alphabet with five binary digits (this picture is taken from Wikipedia):
Here’s another screenshot from the same movie Tony provided me:
As you see, the arm carrying a pistol bears a tattoo showing eight-bit sequences (also known as bytes in computer science). Does a reader know what these bytes mean?
I have searched for detailed information about these two codes online, but I didn’t find anything. If a reader knows more please let me know.
Further reading: The Top 50 unsolved encrypted messages: 29. The Pollaky cryptograms
## Subscribe to Blog via Email
Gib Deine E-Mail-Adresse an, um diesen Blog zu abonnieren und Benachrichtigungen über neue Beiträge via E-Mail zu erhalten.
## Kommentare (13)
1. #1 Magnus Ekhall
Sweden
7. Juli 2018
The message on the arm looks as if it could be ASCII. If so the three last characters are: MCSN
2. #2 Gerry and Andrea
7. Juli 2018
On other photos the ASCII codes on Angelina Jolie‘s arm are better visible. It has ten letters and starts with BPZM, then P or Q, then T or U, then – like Magnus said – MCSN.
3. #3 Richard SantaColoma
http://proto57.wordpress.com/
7. Juli 2018
There might be some confusion over Ms. Jolie’s arm tattoo, because she already has… in real life… several tattoos. So which ones are hers, and which ones were applied for her role in Wanted, are not clear.
But looking at the numbers on her arm, in the image you linked, Klaus, and the movie poster (which shows a slightly different angle), I come up with:
0100 0010
0101 0000
0101 1010
0100 1101
0101 0001
0101 0100(or 1)
0100 1101
0100 0011
0101 0011
0100 1110
But the thing is, the words at the top are a real tattoo! They read:
A PRAYER FOR THE WILD
AT HEART KEPT IN CAGES
… and then the binary numbers are stacked below those words, in the movie. So either the numbers are a later tattoo she really got, or added below the words only for the movie… I can’t find them in any pictures of her in “real life”, so I do think they were added.
By the way, “A PRAYER…” is “a line from Tennessee Williams’s 1941 play Stairs to the Roof adorning her arm: “A prayer for the wild at heart, kept in cages.””
4. #4 TWO
Uvalde, Texas
7. Juli 2018
ASCII is 7 bits
5. #5 TWO
7. Juli 2018
Maybe as simple as :
D B
E
E I
D M
???
A =1 B =2 and so on
6. #6 Gerry and Andrea
7. Juli 2018
According to an interview of Angelina Jolie with Wired „the binary codes on her arm are a part of the reading of the fabric of the Loom of Fate“. But according to two commenters on the forum http://www.vbaexpress.com/forum/showthread.php?19919-Solved-Binary-Tatoo BPZMPTMCSN stands for Brad Pitt an her children Zahara Marley, Pax Thien, Maddox Chivan and Shiloh Nouvel:
7. #7 Richard SantaColoma
http://proto57.wordpress.com/
7. Juli 2018
Gerry and Andrea: Well that cinches it then.. those are also her real tattoos.
And very interesting to me is that the movie promoters thought to use her genuine tattoos as a design feature in the poster, as an interesting feature… when it turns out, it has nothing to do with the movie at all!
I mean, as though Ms. Jolie was not interesting enough already…
8. #8 Gerry and Andrea
7. Juli 2018
Well, Angelina has many tattoos, but the binary codes as well as the words TOIL and TEARS on her arms were fake ones just for the film „Wanted“.
9. #9 TWO
7. Juli 2018
Sounds perfectly logical to me.
She has a real tattoo which is a list of coordinates and here is the explanation of that one :
N11° 33′ 00″ E104° 51′ 00″ : the place in Cambodia where Angelina Jolie’s eldest son Maddox was born.
N09° 02′ 00″ E038° 45′ 00″ : the place in Ethiopia were her daughter Zahara was born.
S22° 40′ 26″ E014° 31′ 40″ : the place in Namibia (Swakopmund) where Angelina gave birth to her biological daughter with Brad Pitt, Shiloh (Hebrew for “the peaceful one”), was born.
N10° 46′ 00″ E106° 41′ 40″ : the place in Vietnam where Angelina’s son Plax Then was born.
N43° 41′ 21″ E07° 14′ 28″
N43° 41′ 21″ E07° 14′ 28″:5 and 6 are a place in France for the twins Knox and Vivienne.
Angelina added a 7th line of coordinates for Brad Pitt’s place of birth in Oklahoma.
10. #10 schorsch
8. Juli 2018
the binary code reads indeed BPZMPTMCSN and stands for the Angelina Jolie-Brad Pitt family, resp. their children: Brad Pitt, Zahara Marley, Pax Thien, Maddox Chivan, Shiloh Nouvel
11. #11 schorsch
8. Juli 2018
@TWO:
You are right, ASCII-Code is a 7 bit code. But in these days, ASCII is rarely used in telegraphic applications; it’s more likely to be found encoded in computer memory. And since computer memory is usually organized in eight bit units, ASCII will usually be represented by 8 bit.
This creates redundancy and an useless overhead – in the present case not only the leading bit is redundant, but the three leading bits . I wouldn’t expect a cryptographer to create so much obvious redundancy, I wouldn’t expect her to use ASCII at all – unless her purpose is not concealment, but deception. A code which does not reveal it’s meaning at the first glance, but which is simple to read by the initiated…
And then it absolutely makes sence to use 8 bit encodes ASCII – the standard code for every Big Bang Theory-Nerd, Futurama-Freak or Simpsonmaniac.
12. #12 rolak
26. Oktober 2018
In the series Alias .. a hidden meaning of the episode
Der Satz klingt nach ‘jede Folge hat..’ – Beispiel? Liste? ⇒database?
Ist ja nicht so, daß beim Sichten einer Serie sekündlich unergründlich Archivare auf Monitore starren – es können also beim Gefundenen selbstverständlich Lücken mitgewachsen sein, zB bei
S01E14 – The Coup 1:42 G 16:6 E 28:23 E 37:43 W → GEEW
doch die Frage bleibt: Beispiel? Liste? ⇒database?
13. #13 rolak
9. November 2018 | 1,975 | 7,302 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.71875 | 3 | CC-MAIN-2019-09 | longest | en | 0.904472 |
https://byjus.com/worksheets/measuring-angles-worksheets/ | 1,653,732,644,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652663016373.86/warc/CC-MAIN-20220528093113-20220528123113-00030.warc.gz | 192,715,334 | 149,779 | # Measuring Angles Worksheet
Measuring angles is an essential part of the mathematics curriculum in higher classes, and it is better that kids have a basic understanding of them from kindergarten. The topic can be pretty tricky for kids. Before explaining more about measuring angles, it is important for parents to start with the basics of geometry. Parents can find a number of measuring angles worksheets on the BYJU’S site.
What is an Angle?
If a kindergarten kid is asked to define an angle, it might turn out to be difficult for him to answer. So to make the concept easier for them, try explaining to them with a diagram. Start by drawing two parallel lines intersecting each other. Now highlight the space between two given rays and mark them as angles. Make sure that before explaining what angles are, the kid is aware of the difference between lines and rays. Lines or Line-segments can be extended from both ends, whereas Ray can be extended in one direction only. These smaller concepts will be helpful in creating a base for bigger concepts like angles or degrees. Once the kid has a basic knowledge about angles, parents can easily access the BYJU’S measuring angles worksheets, and measuring and drawing angles worksheets for their practice. It is also important to teach the kids how to use a protractor as it will help them to measure the angles and determine their type. With the measuring angles worksheets free, children can get to know about the different types of angles.
• Right angle – Right angle or 90 degrees, forms at a point where a horizontal straight line meets a vertical straight line.
• Acute angle – An angle measuring between 0 to 89 degrees (i.e. less than 90); it is known as the acute angle.
• Obtuse angle – If the angle in the given rays measures somewhere between 90 to 180 degrees, it is an obtuse angle.
• Straight angle – If a straight line is drawn and measured, then the angle will measure 180 degrees and is called a straight angle.
Some students might find measuring angles confusing, so it is better for parents to start with the difference between rays and lines. Kids can be given worksheets that will help them to strengthen this concept. Once they have understood the basic idea, parents can talk to them about the various instruments in the geometry box and how to use them. This way, parents can build up the base for children to learn more about measuring angles. | 493 | 2,426 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.46875 | 4 | CC-MAIN-2022-21 | longest | en | 0.9403 |
https://stats.stackexchange.com/questions/232471/why-does-the-glm-function-does-not-return-an-r2-value | 1,642,918,733,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320304134.13/warc/CC-MAIN-20220123045449-20220123075449-00368.warc.gz | 584,223,087 | 34,313 | # Why does the glm function does not return an R^2 value?
The lm function in R retrieves an R^2 value.
The glm function, even if applied to a Gaussian family, does not retrieve an R^2 value.
What is/are the reason/reasons for this?
Thank you!
• This might help, or this or this Aug 30 '16 at 11:52
The glm function uses a maximum likelihood estimator (or restricted maximum likelihood). Maximum likelihood does not minimize the squared error (this is called [ordinary] least squares). Sometimes both estimators give the same results (in the linear/ordinary case for normal distributed error terms, see here) but this does not hold in general. Since the coefficient of determination $R^2$ is calculated by ordinary least-squares regression and not by maximum likelihood, there is no reason to display this measure.
PS: Also regard Nick Cox very valid comment below: $R^2$ may be also well-definied and interesting for GLM. My personal experience is that (as so often) some people like/accept it, while others do not.
• This is a little strong. For example, Zheng, B. and A. Agresti. 2000. Summarizing the predictive power of a generalized linear model. Statistics in Medicine 19: 1771–1781 argue cogently that the square of the correlation between predicted and observed is well-defined and often interesting and useful for GLMs. It's just that some of the interpretation of $R^2$ that goes with regression is irrelevant or inappropriate in a wider context. Aug 30 '16 at 12:02
• I also warn against conflating linear regression and OLS; for example, if regression were calculated by a general maximum likelihood routine, then $R^2$ wouldn't lose validity or value. Aug 30 '16 at 12:03 | 393 | 1,690 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.59375 | 3 | CC-MAIN-2022-05 | latest | en | 0.908129 |
https://www.teacherspayteachers.com/Browse/Search:volume%20of%20prisms,%20pyramids,%20cones | 1,558,835,266,000,000,000 | text/html | crawl-data/CC-MAIN-2019-22/segments/1558232258620.81/warc/CC-MAIN-20190526004917-20190526030917-00482.warc.gz | 964,299,461 | 54,205 | # Results forvolume of prisms, pyramids, cones
This foldable provides students with a total of 10 examples for finding the volume of the 3-D figures listed below: Prisms (Rectangular & Triangular) Pyramids (Rectangular & Triangular) Cylinders Cones Spheres This is a great way to keep students organized and provides them with a great s
Subjects:
Types:
\$4.00
55 Ratings
4.0
PDF (340.28 KB)
14 questions including: • Volume of simple right-geometric shapes, including: o Rectangular prisms o Triangular prisms o Trapezoids o Cylinders o Pyramids o Cones • Complex objects Includes six word problems. Answer key included.
Subjects:
Types:
\$1.00
not yet rated
N/A
PDF (401.88 KB)
Students will find the volume of 10 figures - cones, cylinders, spheres, hemisphere, prisms, and pyramids. Students will use the answer box to color the picture. An adorable safari leopard picture is the result! If you cut the page in half, it fits perfectly in an interactive notebook on facing p
Subjects:
Types:
\$2.00
not yet rated
N/A
PDF (3.34 MB)
This is a Jeopardy review game where students practice their skills solving area, volume and circumference. Problems include diagrams, word problems and thinking backwards. Students must know the formulas for the area of a square, rectangle, triangle and circle; circumference; volume of prisms, py
Subjects:
Types:
\$3.00
8 Ratings
4.0
PPT (598.5 KB)
***Note: this product is in Beta, so it is comparatively low-priced; I will be adding to it throughout the year, so buy now, pay a lower price, get free updates later (you will have to re-download)*** This is a bundle of 13 Units designed for a high school Geometry Class. Each unit plan includes Com
Subjects:
\$60.00
3 Ratings
4.0
ZIP (571.43 MB)
Included in this zip folder are 5 PDF files. 3 Are intro files and 2 are assignments. A brief description of each: Intro #1 is a 21 page file that introduces parallel and perpendicular slices of 3D figures. Intro #2 is a 41 page file that introduces volume of Prisms and Cylinders using Cavalieri
Subjects:
Types:
\$1.75
3 Ratings
3.3
ZIP (22.64 MB)
by
This is a visual organizer for surface are and volume of prisms, pyramids, cones, cylinders, and spheres. Teachers can add their own pictures or word problems (and add more rows if necessary), and students can use the organizers to calculate surface area and volume.
Subjects:
Types:
\$1.00
1 Rating
4.0
DOC (159 KB)
This is a set of 24 task cards covering volume of prisms, pyramids, cones, cylinders, and spheres! The cards can be used for student stations, partner activities, classroom warm-ups, etc. Student recording sheet and teacher answer key included! This product is sold for a DISCOUNTED price in the G
Subjects:
Also included in: Volume Activity Bundle
\$3.00
1 Rating
4.0
PDF (3.13 MB)
Included in this zip folder are 5 SMART Notebook files. 3 Are intro files and 2 are assignments. A brief description of each: Intro #1 is a 21 slide file that introduces parallel and perpendicular slices of 3D figures. Intro #2 is a 41 slide file that introduces volume of Prisms and Cylinders us
Subjects:
Types:
\$1.75
not yet rated
N/A
ZIP (19.73 MB)
Included in this zip folder are 5 Power Point files. 3 Are intro files and 2 are assignments. A brief description of each: Intro #1 is a 21 slide file that introduces parallel and perpendicular slices of 3D figures. Intro #2 is a 41 slide file that introduces volume of Prisms and Cylinders using
Subjects:
\$1.75
not yet rated
N/A
ZIP (14.97 MB)
This resources covers how perimeter, circumference, area, surface area and volume are affected when dimensional change occurs. Both proportional and non-proportional changes are practiced. You will find 2 pages of notes, 8 multi-task cards, 4 practice pages and a maze.This lesson is very detailed an
Subjects:
\$4.50
not yet rated
N/A
PDF (1.5 MB)
This resource provides academic vocabulary as well as examples to assist in finding the volume of prisms, pyramids, cones and cylinders.
Subjects:
Types:
\$1.50
not yet rated
N/A
DOCX (24.51 KB)
THREE fun activities to review finding the volume of prisms, pyramids, cones, cylinders, and spheres. See previews for more detail of each activity.
Subjects:
Types:
\$8.50
\$6.80
not yet rated
N/A
Bundle
My students love these review stations! Students will work in groups of 4 to solve the problems at 9 different stations. They will work the problems out on post-its and stick them to the answer board. Or even better, students can use the answer cards included in this product to slide into a pocke
Subjects:
Types:
\$3.00
not yet rated
N/A
PDF (32.03 MB)
Volume and Surface Area Coloring Activity (Prisms, Pyramids, Cylinders, Cones) This is a fun way for students to practice finding both volume and surface area of 3D figures. There are 12 problems total, 6 volume and 6 surface area. This includes rectangular prisms, cylinders, cones, and pyramids.
Subjects:
Types:
\$2.00
494 Ratings
4.0
PDF (1.85 MB)
Volume Task Cards (Prisms, Cylinders, Pyramids, and Cones) This activity includes 20 task cards in which students will practice finding the volume of the following figures: rectangular prism, triangular prism, trapezoidal prism, cylinders, square pyramids, rectangular pyramids, triangular pyramids,
Subjects:
Types:
Also included in: Volume and Surface Area Task Cards Bundle
\$3.00
90 Ratings
4.0
PDF (4.37 MB)
These three volume tasks will assess students knowledge of calculating the volume of spheres, cylinders, cones, and rectangular prisms. The activities are different levels of difficulty and can be used for differentiating, or throughout a geometry unit. You will receive PDFs of the 3 tasks. Answe
Subjects:
CCSS:
Also included in: Geometry Bundle, High School Resources
\$1.99
25 Ratings
4.0
PDF (671.93 KB)
Volume of Solids Worksheet Pages 1-2: Solids include: Prisms, Cylinders, Cones, Pyramids and Spheres; 12 Problems total Pages 3-4: Answer Key A nice worksheet that can be used for in class review, as homework, or an assessment. Surface Area and Volume Bundle
Subjects:
Types:
Also included in: Surface Area and Volume Bundle Geometry Practice Review
\$1.50
23 Ratings
4.0
PDF (733.67 KB)
Volume Interactive Notebook: Create this reference guide with your students! Volume of Spheres, Cones, Cylinders, Prisms, Pyramids...my 8th graders have tested and approved this resource! Please note: this product is included in a money saving bundle! Check it out: Voluminous Volume Growing Bundl
Subjects:
Types:
CCSS:
\$3.00
20 Ratings
3.9
PDF (1.82 MB)
This is an excerpt from my popular line of Bossy Brocci Math & Big Science workbooks on Amazon. Printing should be: LANDSCAPE and DOUBLE-SIDED, with the Flip being along the 'SHORT' edge or side Want MORE Power for your Dollar? Give Brocci Bundles a Try before you buy! To get the Bundles &
Subjects:
Types:
\$1.50
\$1.20
11 Ratings
4.0
PDF (436.15 KB)
This is an excerpt from my popular line of Bossy Brocci Math & Big Science workbooks on Amazon. Printing should be: LANDSCAPE and DOUBLE-SIDED, with the Flip being along the 'SHORT' edge or side Want MORE Power for your Dollar? Give Brocci Bundles a Try before you buy! To get the Bundles &
Subjects:
Types:
\$1.50
\$1.20
10 Ratings
4.0
PDF (394.38 KB)
This maze consists of 11 cones and pyramids that students must calculate the total and lateral surface area of. Not all boxes are used in the maze to prevent students from just guessing the correct route. In order to complete the maze students will have to calculate surface area 10 times. Importan
Subjects:
Types:
\$1.50
10 Ratings
4.0
PDF (3.66 MB)
This is an excerpt from my popular line of Bossy Brocci Math & Big Science workbooks on Amazon. Printing should be: LANDSCAPE and DOUBLE-SIDED, with the Flip being along the 'SHORT' edge or side Want MORE Power for your Dollar? Give Brocci Bundles a Try before you buy! To get the Bundles &
Subjects:
Types:
\$1.50
\$1.20
8 Ratings
4.0
PDF (395.78 KB)
This is an 21 slide power point lesson that has a couple of links to short video clips that shows the relationship between prisms and pyramids as well as the relationship between cylinders and cones. The power point lesson explains how to calculate the volume of a pyramid and cone. There are basic p
Subjects:
CCSS:
\$1.99
\$1.59
6 Ratings
4.0
PPTX (278.28 KB)
showing 1-24 of 559 results
Teachers Pay Teachers is an online marketplace where teachers buy and sell original educational materials. | 2,182 | 8,485 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.90625 | 3 | CC-MAIN-2019-22 | latest | en | 0.858973 |
http://forums.wolfram.com/mathgroup/archive/2011/Jun/msg00161.html | 1,591,354,115,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590348496026.74/warc/CC-MAIN-20200605080742-20200605110742-00009.warc.gz | 46,547,323 | 7,836 | Re: plotting contours on a sphere
• To: mathgroup at smc.vnet.net
• Subject: [mg119518] Re: plotting contours on a sphere
• From: Bob Hanlon <hanlonr at cox.net>
• Date: Tue, 7 Jun 2011 08:14:10 -0400 (EDT)
```My response only provided the replacement definition for p3. At a minimum, the OP's definition of f must also be used.
f[\[Theta]_, \[Phi]_] = Sin[\[Theta] + \[Phi]];
p3 = SphericalPlot3D[1, {\[Phi], 0, Pi}, {\[Theta], 0, 2 Pi},
ColorFunctionScaling -> False,
ColorFunction -> Function[{x, y, z, \[Theta], \[Phi], r},
ColorData["TemperatureMap"]
[Rescale[f[\[Phi], \[Theta]], {-1, 1}]]],
Mesh -> 9,
MeshFunctions -> {Function[{x, y, z, \[Theta], \[Phi], r},
f[\[Phi], \[Theta]]]},
BoundaryStyle -> Transparent]
\$Version
"8.0 for Mac OS X x86 (64-bit) (February 23, 2011)"
Bob Hanlon
---- "Nasser M. Abbasi" <nma at 12000.org> wrote:
=============
On 6/6/2011 3:23 AM, Bob Hanlon wrote:
> p3 = SphericalPlot3D[1, {\[Phi], 0, Pi}, {\[Theta], 0, 2 Pi},
> ColorFunctionScaling -> False,
> ColorFunction -> Function[{x, y, z, \[Theta], \[Phi], r},
> ColorData["TemperatureMap"]
> [Rescale[f[\[Phi], \[Theta]], {-1, 1}]]],
> Mesh -> 9,
> MeshFunctions ->
> {Function[{x, y, z, \[Theta], \[Phi], r},
> f[\[Phi], \[Theta]]]}]
>
>
> Bob Hanlon
>
Hi Bob;
what version of Mathematica did you use? on 8.0.1,
windows 7, I get an error message.
MeshFunctions::invmeshf: MeshFunctions->{At Line = 1, the input
was:,p3=SphericalPlot3D[1,{\[Phi],0,Pi},{\[Theta],0,2Pi},ColorFunctionScaling->False,
ColorFunction->Function[{x,y,z,\[Theta],\[Phi],r},
ColorData[TemperatureMap][Rescale[f[\[Phi],\[Theta]],{-1,1}]]],
Mesh->9,MeshFunctions->{Function[{x,y,z,\[Theta],\[Phi],r},f[\[Phi],
\[Theta]]]}],f[#5,#4]&} must be a pure function or a list of pure functions. >>
here is a screen shot also fyi
http://12000.org/tmp/june_5_2011/msg.png
--Nasser
--
Bob Hanlon
```
• Prev by Date: Re: Using Cases or Position to Find Strings in a List of Strings
• Next by Date: Re: Select positive solutions (depending on parameters)
• Previous by thread: Re: plotting contours on a sphere
• Next by thread: Mapping points | 736 | 2,151 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.59375 | 3 | CC-MAIN-2020-24 | latest | en | 0.590394 |
https://byjus.com/commerce/sandeep-garg-solutions-class-11-economics-chapter-2-tabular-presentation/ | 1,723,536,628,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641075627.80/warc/CC-MAIN-20240813075138-20240813105138-00606.warc.gz | 113,794,238 | 106,587 | # Sandeep Garg Solutions Class 11 Economics Chapter 2 - Tabular Presentation
Sandeep Garg Class 11 Economics Solutions Chapter 2 Tabular Presentation is illustrated by the professional economic educator from the contemporary edition of Sandeep Garg Economics Class 11 textbook solutions.
We at BYJU’S provide Sandeep Garg Economics class 11 Solutions to give a comprehensive insight into the subject to the students. These insights will be a valuable advantage to students while completing their homework or while studying for their exams.
There are numerous concepts in economics, but here we provide you with the solution from the Tabular Presentation, which will be convenient for the students to score well in the board exams.
## Sandeep Garg Solutions Class 11 – Chapter 2
Question 1
Show a blank table with the number of students giving JEE examination in terms of gender and stream (Science, Art, and Commerce).
Question 2
Arrange a blank table for a population of Mumbai according to gender, age, and marital status (Age should be below 30 years and above 30 years).
Question 3
In 2016, out of a total of 550 students in a college, 200 were for B.Com (H) and Eco(H) were remaining students. Make a table of the given information.
The given details can be outlined as follows.
In year 2015
Total number of students in a college =350
Total number of B.Com(H) students = 250
Total number of Eco(H) students = 350 – 250 = 100
In year 2016
Total number of students in a college = 550
Total number of Eco(H) students = 350
Total number of B.Com(H) students = 550 – 350 = 200
Distribution of Students in a College
Question 4
In a survey about the tea habit in two towns, the below-mentioned details were received:
(i) Town A: Total females population were 40%; where the tea drinkers were 45% and 20% were male non-tea drinkers.
(ii) Town B: Males population were 55%; 30% of males were non-tea drinkers and 15% were females, tea drinkers. Arrange the information in tabular form.
Percentage of tea drinkers and non-tea drinkers in City A and City B
Question 5
In a school, 620 parents applied for their child’s admission, out of which 130 are economically weaker section category. Here, 190 are self-employed, 40 parents are service class, and rest are of business class.
In the EWS division, 94 parents are service class. Suppose, if total business class applicants are 300, then give the information in a tabular form. | 577 | 2,449 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.59375 | 4 | CC-MAIN-2024-33 | latest | en | 0.944504 |
http://www.gurufocus.com/term/grossmargin/STEI/Gross%252BMargin/Stewart%2BEnterprises%252C%2BInc | 1,475,145,528,000,000,000 | text/html | crawl-data/CC-MAIN-2016-40/segments/1474738661781.17/warc/CC-MAIN-20160924173741-00283-ip-10-143-35-109.ec2.internal.warc.gz | 509,043,113 | 26,579 | Switch to:
Stewart Enterprises, Inc. (NAS:STEI)
Gross Margin
18.25% (As of Jul. 2013)
Gross Margin is calculated as gross profit divided by its revenue. Stewart Enterprises, Inc.'s gross profit for the three months ended in Jul. 2013 was \$23.2 Mil. Stewart Enterprises, Inc.'s revenue for the three months ended in Jul. 2013 was \$127.1 Mil. Therefore, Stewart Enterprises, Inc.'s Gross Margin for the quarter that ended in Jul. 2013 was 18.25%.
Stewart Enterprises, Inc. had a gross margin of 18.25% for the quarter that ended in Jul. 2013 => No sustainable competitive advantage
The 5-Year average Growth Rate of Gross Margin for Stewart Enterprises, Inc. was 0.00% per year.
Definition
Gross Margin is the percentage of Gross Profit out of sales or Revenue.
Stewart Enterprises, Inc.'s Gross Margin for the fiscal year that ended in Oct. 2012 is calculated as
Gross Margin (A: Oct. 2012 ) = Gross Profit (A: Oct. 2012 ) / Revenue (A: Oct. 2012 ) = 109.7 / 516.097 = (Revenue - Cost of Goods Sold) / Revenue = (516.097 - 406.415) / 516.097 = 21.25 %
Stewart Enterprises, Inc.'s Gross Margin for the quarter that ended in Jul. 2013 is calculated as
Gross Margin (Q: Jul. 2013 ) = Gross Profit (Q: Jul. 2013 ) / Revenue (Q: Jul. 2013 ) = 23.2 / 127.062 = (Revenue - Cost of Goods Sold) / Revenue = (127.062 - 103.867) / 127.062 = 18.25 %
* All numbers are in millions except for per share data and ratio. All numbers are in their own currency.
A positive Gross Profit is only the first step for a company to make a net profit. The gross profit needs to be big enough to also cover related labor, equipment, rental, marketing/advertising, research and development and a lot of other costs in selling the products.
Explanation
Warren Buffett believes that firms with excellent long term economics tend to have consistently higher margins.
Durable competitive advantage creates a high Gross Margin because of the freedom to price in excess of cost. Companies can be categorized by their Gross Margin
1. Greater than 40% = Durable competitive advantage
2. Less than 40% = Competition eroding margins
3. Less than 20% = no sustainable competitive advantage
Consistency of Gross Margin is key
Stewart Enterprises, Inc. had a gross margin of 18.25% for the quarter that ended in Jul. 2013 => No sustainable competitive advantage
Be Aware
If a company loses its competitive advantages, usually its gross margin declines well before its sales declines. Watching Gross Margin and Operating Margin closely helps avoid value trap situations.
Related Terms
Historical Data
* All numbers are in millions except for per share data and ratio. All numbers are in their own currency.
Stewart Enterprises, Inc. Annual Data
Oct04 Oct05 Oct06 Oct07 Oct08 Oct09 Oct10 Oct11 Oct12 Oct13 Gross Margin 23.30 20.76 22.29 21.50 19.12 18.01 19.35 19.59 21.25 21.68
Stewart Enterprises, Inc. Quarterly Data
Jul11 Oct11 Jan12 Apr12 Jul12 Oct12 Jan13 Apr13 Jul13 Oct13 Gross Margin 17.49 19.07 20.35 21.97 20.88 21.76 23.82 23.88 18.25 20.50
Get WordPress Plugins for easy affiliate links on Stock Tickers and Guru Names | Earn affiliate commissions by embedding GuruFocus Charts
GuruFocus Affiliate Program: Earn up to \$400 per referral. ( Learn More) | 857 | 3,252 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.59375 | 3 | CC-MAIN-2016-40 | longest | en | 0.915571 |
https://stonespounds.com/794-6-stones-in-stones-and-pounds | 1,713,826,125,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296818374.84/warc/CC-MAIN-20240422211055-20240423001055-00482.warc.gz | 480,337,766 | 6,529 | # 794.6 stones in stones and pounds
## Result
794.6 stones equals 794 stones and 8.4 pounds
You can also convert 794.6 stones to pounds.
## How to convert 794.6 stones to stones and pounds?
In order to convert 794.6 stones to stones and pounds we can take the decimal part of 794.6 stones and convert it into pounds. In this case we need to convert 0.6 stones to pounds.
We know that 1 stones equals 14 pounds, therefore to convert 0.6 stones to pounds we simply multiply 0.6 stones by 14 pounds:
0.6 stones × 14 pounds = 8.4 pounds
Finally, we can say that 794.6 stones in stones and pounds is equivalent to 794 stones and 8.4 pounds:
794.6 stones = 794 stones and 8.4 pounds
Seven hundred ninety-four point six stones is equal to seven hundred ninety-four stones and eight point four pounds.
## Conversion table
For quick reference purposes, below is the stones and stones to pounds conversion table:
stones(st) stones(st) pounds(lb)
795.6 stones 795 stones 8.4 pounds
796.6 stones 796 stones 8.4 pounds
797.6 stones 797 stones 8.4 pounds
798.6 stones 798 stones 8.4 pounds
799.6 stones 799 stones 8.4 pounds
800.6 stones 800 stones 8.4 pounds
801.6 stones 801 stones 8.4 pounds
802.6 stones 802 stones 8.4 pounds
803.6 stones 803 stones 8.4 pounds
804.6 stones 804 stones 8.4 pounds
## Units definitions
The units involved in this conversion are stones and pounds. This is how they are defined:
### Stones
The stone or stone weight (abbreviation: st.) is an English and imperial unit of mass now equal to 14 pounds (6.35029318 kg). England and other Germanic-speaking countries of northern Europe formerly used various standardised "stones" for trade, with their values ranging from about 5 to 40 local pounds (roughly 3 to 15 kg) depending on the location and objects weighed. The United Kingdom's imperial system adopted the wool stone of 14 pounds in 1835. With the advent of metrication, Europe's various "stones" were superseded by or adapted to the kilogram from the mid-19th century on. The stone continues in customary use in Britain and Ireland used for measuring body weight, but was prohibited for commercial use in the UK by the Weights and Measures Act of 1985.
### Pounds
The pound or pound-mass is a unit of mass used in the imperial, United States customary and other systems of measurement. A number of different definitions have been used; the most common today is the international avoirdupois pound, which is legally defined as exactly 0.45359237 kilograms, and which is divided into 16 avoirdupois ounces. The international standard symbol for the avoirdupois pound is lb; an alternative symbol is lbm (for most pound definitions), # (chiefly in the U.S.), and ℔ or ″̶ (specifically for the apothecaries' pound). The unit is descended from the Roman libra (hence the abbreviation "lb"). The English word pound is cognate with, among others, German Pfund, Dutch pond, and Swedish pund. All ultimately derive from a borrowing into Proto-Germanic of the Latin expression lībra pondō ("a pound by weight"), in which the word pondō is the ablative case of the Latin noun pondus ("weight"). Usage of the unqualified term pound reflects the historical conflation of mass and weight. | 811 | 3,218 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.984375 | 4 | CC-MAIN-2024-18 | latest | en | 0.849246 |
https://www.mql5.com/en/market/product/7798 | 1,542,498,579,000,000,000 | text/html | crawl-data/CC-MAIN-2018-47/segments/1542039743913.6/warc/CC-MAIN-20181117230600-20181118011945-00004.warc.gz | 934,646,735 | 15,237 | • Overview
• Reviews
# Analyst 2015
Method of trading and technical analysis using an analyst indicator
In any market, whether it is stocks, minerals (metals, raw materials) or currency pairs, we see a trade window with quotes (price) and trading time. The most important thing for us is to learn to analyze the market and find the entry points, observing all the trading rules, as well as to calculate trade exit points (profit) using only these data: price and time.
The Analyst indicator includes only the price and time data. Doing the math, and studying them on two principles, it compares the two results and determines the market mood at certain periods.
The most important thing in this process was to properly formulate the analysis rules for the analytical structure and the trading part of the system. I'll try to explain briefly how it works.
In real time we have the following:
1. When simultaneously using two methods for mathematic calculation of price movement over time, trading zones of several types appear, showing the market mood on a given interval of time.
2. Based on the principle (nature), trading zones are divided into several types, which we have tried to describe briefly and clearly for the trading terminal to understand them.
Buy areas are divided into:
• X areas (uncertainty and rising);
• fall down areas;
• growth areas;
• areas of sharp movements.
Sell areas are divided into:
• X areas (uncertainty and falling);
• growth areas;
• fall areas;
• reversal areas.
Brief description of these trade zones in real time greatly helps to select right trades in future. On the chart, you can see that after initiation of zones, they are all different, "thick", "thin", "medium", "high", "low", "wide", "narrow". And each has its own rules, principles and moods. We only need to understand their moods, behavior and their opportunities, of course.
3. Our next task was to define the moods and opportunities of the zones with high probability by using a certain mathematical algorithm in the calculations.
After examining the properties and setting in a certain calculation logic to study the "mood" of the patterns, we have managed to quite accurately describe the market mood and predict its movements. And now it really will help you improve the quality of your trading.
Each of you can backtest all the signals (they are not redrawn), and of all the recommendations of the analyst select only the ones you like the best.
### Parameters
• TimeFrame – timeframe. Possible values:
• Y5 – a five-year cycle;
• Y3 – a three-year cycle;
• Y – a year;
• Q – a quarter;
• MN1 – a month;
• W1 – a week;
• D1 – a day;
• H4 – 4 hours;
• H1 – 1 hour.
• Draw lines description – comments to the zones. true - enabled, false - disabled.
• Width – the width of timeframe dividing lines, 0 – do not show.
• Style – the style of timeframe dividing lines.
• Color – the color of timeframe dividing lines.
• M1 Color – color of growth zones.
• M2 Color – color of fall zones.
• М1 under M4 description – a comment for Sell areas (you can add your own comment).
• МS1 under MS4 description – a comment for Buy areas (you can add your own comment).
• М1 above M4 description – a comment for upper X areas (you can add your own comment).
• МS1 above MS4 description – a comment for upper X areas (you can add your own comment).
Pay special attention to the higher TF and absorption zones.
No reviews | 765 | 3,429 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.171875 | 3 | CC-MAIN-2018-47 | latest | en | 0.945593 |
http://teachhighschoolmath.blogspot.com/2011/11/sum-of-three.html | 1,369,102,880,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368699632815/warc/CC-MAIN-20130516102032-00029-ip-10-60-113-184.ec2.internal.warc.gz | 261,785,842 | 15,074 | ## Sunday, November 6, 2011
### Sum of Three
I got this activity from my brother-in-law and really like it. It is called "Sum of Three". In essence you have students in groups of three each doing a different problem. Then you have them add the answers together to be the sum of the three answers. They will bring this number up to you and ask if this is correct. If they are correct, then they get to move on to the next problem set. If they are incorrect, then they must decide how to go about getting the correct answer. This is the best part of the activity. When the students get the problem wrong, they automatically think it is somebody elses mistake. They then will systematically go through the process of how to do the problem with everyone. It is a great group activity. Here are my directions. I have given a couple examples as well.
1. Get your class into groups of three. I always do this on a random basis. Let's say you had 26 people in your class. Take 26 kids divided by 3 and you get 8 full groups of three. Count students one, two, three, and so on till eight, then start back at one, two, three and keep doing this until you run out of people. In this case, you will have 6 groups that have 3 in it and 2 groups that have 4 in it. Now the ones get together and the twos get together and so on.
2. Hand out a problem set A in paper form to each group. This will have three problems in it. A1, A2, and A3. Each student will work on a different problem. The group will then add the sum of all three answers.
3. When a group thinks it has a sum, they bring up THEIR PAPER to show you their answer. Don't have the students SAY the sum. This might give it away for the other students. If they are correct, give them the problem set B in paper form. If they are incorrect, then they must go back and find out where they made a mistake.
4. I have my students do all their work on a separate piece of paper to be turned in. This creates a little accountability as well as a place for them to work.
5. You will have to decide how many sets of problems you will want. I typically have 3 to 4 sets. I have the last set as extra difficult.
Here are a couple of examples of the "Sum of Three" activity.
This is a problem set for multiplying fractions in Intro to Algebra
This is the answer sheet that I use when the students come up to give me their SUM.
This is an Algebra "Sum of Three" problem set. I have the students add the y-intercepts of the line. PDF of the FRACTIONS "SUM OF THREE" ACTIVITY PDF of the EQUATION OF A LINE "SUM OF THREE" ACTIVITY | 659 | 2,613 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.8125 | 4 | CC-MAIN-2013-20 | longest | en | 0.93527 |
https://www.br.freelancer.com/projects/php-java/poker-browser-calculator/ | 1,503,520,348,000,000,000 | text/html | crawl-data/CC-MAIN-2017-34/segments/1502886123359.11/warc/CC-MAIN-20170823190745-20170823210745-00527.warc.gz | 879,091,901 | 27,058 | # Poker Browser Calculator
The idea of the project is to create something similar to:
The program/ application should be visible in a web browser, not download and install.
I don't mind the method of making it as long as it is viewable to my customers.
Option A: "0.1/0.25" Value: 17.5
Option B: "0.25/0.5" Value: 35
Option C: "0.5/\$1" Value: 70
Option D: "\$1/\$2" Value: 140
Option E: "\$2/\$4" Value: 210
Option F: "\$3/\$6" Value: 300
Option G: "\$5/10" Value: 300
Choice Menu 2: "Number of Tables"
Option A: "1 Table" (Multiply by 1)
Option B: 2 Tables (Multiply by 2)
Option C: 3 Tables (Multiply by 3)
Continue to:
Option K: 10 Tables (Multiply by 10)
Choice Menu 3: "Number of Players"
Option A: "6 Players" (Multiply By 15)
Option B: "10 Players" (Multiply By 12)
Choice Menu 4: "Hours of Play A Week"
Option A: "5 Hours" (Multiply By 5)
Option B: "10 Hours" (Multiply By 10)
Option C: "20 Hours" (Multiply By 20)
Option D: "30 Hours" (Multiply By 30)
Option E: "40 Hours" (Multiply By 40)
Option C: "60 Hours" (Multiply By 60)
Display answer in PGP's. I.e. 20,000 PGP's P/Week
Multiply by 52 and display 1,040,000 PGP's P/Year.
Next to the Per Year total, show an image relevant to the points. I.e. of points total is between 20,000-50,000 shows a hat, if it’s between 200,000-1,000,000 shows a car. I will give the corresponding Image's upon accepting.
Questions will be answered through PM.
Would like to get this project under way a.s.a.p.
Max 7 Day completion. Quite simple task so \$50 maximum. Will probably choose lowest bidder with good review(s)
Habilidades: Java, Javascript, PHP, Design de Sites | 571 | 1,840 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.015625 | 3 | CC-MAIN-2017-34 | longest | en | 0.781902 |
https://gpuzzles.com/mind-teasers/most-popular-logical-puzzle/ | 1,495,661,642,000,000,000 | text/html | crawl-data/CC-MAIN-2017-22/segments/1495463607862.71/warc/CC-MAIN-20170524211702-20170524231702-00349.warc.gz | 763,536,729 | 12,881 | • Views : 70k+
• Sol Viewed : 20k+
Mind Teasers : Most Popular Logical Puzzle
Difficulty Popularity
Outside a room there are three light switches. One of switch is connected to a light bulb inside the room.
Each of the three switches can be either 'ON' or 'OFF'.
You are allowed to set each switch the way you want it and then enter the room(note: you can enter the room only once)
Your task is to then determine which switch controls the bulb ??
Discussion
Suggestions
• Views : 50k+
• Sol Viewed : 20k+
Mind Teasers : Popular Deductive Logic Problem
Difficulty Popularity
Eight Brothers lives in an old house where there is no electricity and no computers or any any other gadget.
Brother-2: Playing Chess
Brother-3: Writing
Brother-4: making food for the family
Brother-5: sleeping and snoring
Brother-6: cleaning house
Brother-7: watering the plants
what is Brother-8 doing ?
• Views : 60k+
• Sol Viewed : 20k+
Mind Teasers : Trick Interesting Maths puzzle
Difficulty Popularity
There are three cars in a racing track. The track is made forming a perfect circle and is quite wide so that at one time, multiple cars can pass through it. The car which is leading is driving at 55 MPH and the slowest car is driving at 45 MPH. The car that is in middle of these two is driving in between the two speeds. For the time being you can say that the distance between the fastest car and the middle car is x miles and it is same between the middle car and the slowest car. Also, x is not equal to 0 or 1.
The car keeps running till the leading car catches up with the slowest car and then every car stops. Given the case, do you think that at any point, the distance between any two pairs will again become x miles?
• Views : 60k+
• Sol Viewed : 20k+
Mind Teasers : Short Story Mystery Puzzle
Difficulty Popularity
A Japanese ship was making its way through the Arabian Sea. The captain felt like taking a bath and thus went to the bathroom after removing his Rolex watch and gold bracelet. When he returned back from the bathroom, he found out that his valuables were missing.
He suspected five of his crew members. He called them in and asked all of them what they were doing for the last 15 minutes. They replied as follows:parachute
The French cook (with a butcher knife): I was in fridge room getting meat for cooking.
The German Engineer (with some tools in hand): I was working on the generator engine.
The Bangladeshi seaman (empty handed): I was on the mast correcting the flag which was upside down by mistake.
The American Radio officer (with his headphones in hand): I was messaging to company that we will reach the port in the next two days that is Friday morning at 1100 hrs.
The Indian navigation officer (empty handed): I was taking a nap before my night watch began.
Upon hearing all of them, the captain found out who was the thief. How did he know that? Who is the thief?
• Views : 80k+
• Sol Viewed : 20k+
Mind Teasers : Colditz Pascal Prison Escape Puzzle
Difficulty Popularity
Two world's famous prisoners 'Colditz' and 'Pascal' are locked in a cell.
They plan to escape from the cell.
They noticed there is an open window at 40 feet above the ground level.
Both of them tried very hard but are never able to reach there.
Then both of them decided to plan to escape by a tunnel and they start digging out.
After digging for just 5 days, Colditz and Pascal comes out with the much more easier plan than tunneling and they escaped.
what was the plan ?
• Views : 40k+
• Sol Viewed : 10k+
Mind Teasers : Popular Probability Brain Teaser
Difficulty Popularity
Kangwa, Rafael and Ferdinand plans for gun fighting.
They each get a gun and take turns shooting at each other until only one person is left.
History suggest
Kangwa hits his shot 1/3 of the time, gets to shoot first.
Rafael, hits his shot 2/3 of the time, gets to shoot next if still living.
Ferdinand having perfect record at shooting(100% accuracy) shoots last , if alive.
The cycle repeats. If you are Kangwa, where should you shoot first for the highest chance of survival?
• Views : 40k+
• Sol Viewed : 10k+
Mind Teasers : 3 Gallon Brain Teaser
Difficulty Popularity
You have been given three jars of 3 liters, 5 liters and 8 liters capacity out of which the 8 liters jar is filled completely with water. Now you have to use these three jars to divide the water into two parts of 4 liters each.
How can you do it making the least amount of transfers?
• Views : 80k+
• Sol Viewed : 20k+
Mind Teasers : Sherlock Holmes Murderer Mystery Puzzle
Difficulty Popularity
There are five people. One of them shot and killed one of the other five.
1. Dan ran in the NY City marathon yesterday with one of the innocent men.
2. Mike considered being a farmer before he moved to the city.
3. Jeff is a topnotch computer consultant and wants to install Ben's new computer next week.
4. The murderer had his leg amputated last month.
5. Ben met Jack for the first time six months ago.
6. Jack has been in seclusion since the crime.
7. Dan used to drink heavily.
8. Ben and Jeff built their last computers together.
9. The murderer is Jack's brother. They grew up together in Seattle.
Sherlock holmes arrived at the scene and immediately find the murderer.
who was the murderer ?
• Views : 50k+
• Sol Viewed : 20k+
Mind Teasers : Candle Trick Teaser Problem
Difficulty Popularity
Husband has prepared for a candle light dinner on the honeymoon for his wife. While they were having the dinner, a strong breeze flew through the open window and four candles out of ten were extinguished. After that, the husband closed the window.
How many candles were left in the end?
• Views : 80k+
• Sol Viewed : 20k+
Mind Teasers : Playing Cards Logic Riddle
Difficulty Popularity
Dr. Watson has a new card puzzle for Sherlock Holmes. He picks four cards out of the pack of 52 cards and lays them face down on the table. He offers four hints to Sherlock.
1) The left card can’t be greater than the one on right.
2) The difference between the first card and third card equals to eight.
3) None of the Ace is present
4) No face card has been included i.e. no queen king and jacks.
5) The difference between the second and fourth card is 7.
Sherlock smiles and tells all the four cards to Dr. Watson.
Can you tell the four cards?
• Views : 60k+
• Sol Viewed : 20k+
Mind Teasers : Wild West Cowboy Riddle
Difficulty Popularity
In the Wild West, you challenge two cowboys - Hunter Jack and Sharp Shooter Leo into a death match. They being better shooter than you readily accept your challenge. But they don't want to waste bullets and thus lay down a certain rules that are accepted by you as well. Here are the rules:
1) Everyone shoot in a given order till only one is left.
2) Everyone shoot only once when his turn arrive.
3) If any one of you is injured, the other two will finish him off together.
4) The worst shooter gets to shoot first (which is you) and the best one shoots at the last.
Now, what tactics will you use if you know that you hit every third shot of yours, Jack hits every second shot and Leo hits every shot?
Latest Puzzles
24 May
McDonald's Christmas Happy Meal Puzzle
On the occasion of Christmas, McDonald's...
23 May
Which mathematical symbols Replaces The Question Mark
Which mathematical symbols replaces the ...
22 May
What Is It six letter word. Riddle
It's a six letter word.
The first...
21 May
Trick Odd One Out Riddle
One of the below words is odd one out. C...
20 May
Shotgun Cant Kill Dean Winchester Riddle
In the bathroom, Dean Winchester shot hi...
19 May
Remove 2 Matchsticks 6-4=7 Equation Puzzle
Can you remove 2 matchsticks and make th...
18 May | 1,838 | 7,708 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.390625 | 3 | CC-MAIN-2017-22 | longest | en | 0.947729 |
https://fr.mathworks.com/matlabcentral/cody/problems/9-who-has-the-most-change/solutions/2019563 | 1,575,800,806,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575540508599.52/warc/CC-MAIN-20191208095535-20191208123535-00312.warc.gz | 388,821,520 | 15,951 | Cody
# Problem 9. Who Has the Most Change?
Solution 2019563
Submitted on 14 Nov 2019 at 5:14
This solution is locked. To view this solution, you need to provide a solution of the same size or smaller.
### Test Suite
Test Status Code Input and Output
1 Fail
a = [1 2 1 15]; b = 1; assert(isequal(most_change(a),b))
Assertion failed.
2 Fail
a = [ 1 2 1 15; 0 8 5 9]; b = 2; assert(isequal(most_change(a),b))
Matrix dimensions must agree. Error in most_change (line 3) [~,b] = max(a.*[.25; .10; .05; .01]); Error in Test2 (line 4) assert(isequal(most_change(a),b))
3 Fail
a = [ 1 22 1 15; 12 3 13 7; 10 8 23 99]; b = 3; assert(isequal(most_change(a),b))
Matrix dimensions must agree. Error in most_change (line 3) [~,b] = max(a.*[.25; .10; .05; .01]); Error in Test3 (line 5) assert(isequal(most_change(a),b))
4 Fail
a = [ 1 0 0 0; 0 0 0 24]; b = 1; assert(isequal(most_change(a),b))
Matrix dimensions must agree. Error in most_change (line 3) [~,b] = max(a.*[.25; .10; .05; .01]); Error in Test4 (line 3) assert(isequal(most_change(a),b))
5 Fail
a = [ 0 1 2 1; 0 2 1 1]; c = 1; assert(isequal(most_change(a),c))
Matrix dimensions must agree. Error in most_change (line 3) [~,b] = max(a.*[.25; .10; .05; .01]); Error in Test5 (line 3) assert(isequal(most_change(a),c))
6 Fail
% There is a lot of confusion about this problem. Watch this. a = [0 1 0 0; 0 0 1 0]; c = 2; assert(isequal(most_change(a),c)) % Now go back and read the problem description carefully.
Matrix dimensions must agree. Error in most_change (line 3) [~,b] = max(a.*[.25; .10; .05; .01]); Error in Test6 (line 4) assert(isequal(most_change(a),c))
7 Fail
a = [ 2 1 1 1; 1 2 1 1; 1 1 2 1; 1 1 1 2; 4 0 0 0]; c = 5; assert(isequal(most_change(a),c))
Matrix dimensions must agree. Error in most_change (line 3) [~,b] = max(a.*[.25; .10; .05; .01]); Error in Test7 (line 7) assert(isequal(most_change(a),c)) | 733 | 1,901 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.6875 | 4 | CC-MAIN-2019-51 | latest | en | 0.707289 |
https://www.mozaweb.com/en_US/Extra-3D_scenes-Time_zones-47119 | 1,568,608,139,000,000,000 | text/html | crawl-data/CC-MAIN-2019-39/segments/1568514572484.20/warc/CC-MAIN-20190916035549-20190916061549-00022.warc.gz | 948,277,006 | 15,985 | Quantity: 0
Total: 0,00
0
# Time zones
### Time zones
The Earth is divided into 24 time zones. Standard time is the time used within time zones.
Geography
Keywords
time zone, standard time, Greenwich, local time, International Date Line, UTC, Coordinated Universal Time, Earth, time difference, longitude, geography
Related items
### The Earth
• Prime Meridian
• International Date Line
The Earth is divided into 24 time zones.
Time zones were defined by geometrically dividing the Earth`s equator into 24 sections bordered by meridians, each 15° of longitude apart (because the Earth rotates 15° in 1 hour). Hence adjacent time zones have a time difference of exactly 1 hour, one hour less on the west, one hour more on the east.
Standard time is the time used in time zones. The local time at the meridian in the center of each time zone is defined as the standard time.
By convention, the starting point of the time zones is the Greenwich time zone (the section between the longitudes 7.5° W and 7.5° E).
The international date line is an imaginary line connecting the North Pole and the South Pole, its path roughly follows the 180° meridian. When crossing it from east to west we advance the calendar date by one day, when we cross in the opposite direction the calendar date to the west of the line is repeated.
### Standard time
• 0
• +1
• +2
• +3
• +4
• +5
• +6
• +7
• +8
• +9
• +10
• +11
• +12
• –11
• –10
• –9
• –8
• –7
• –6
• –5
• –4
• –3
• –2
• –1
• Greenwich Mean Time
The Earth is divided into 24 time zones.
Time zones were defined by geometrically dividing the Earth`s equator into 24 sections bordered by meridians, each 15° of longitude apart (because the Earth rotates 15° in 1 hour). Hence adjacent time zones have a time difference of exactly 1 hour, one hour less on the west, one hour more on the east.
Standard time is the time used in time zones. The local time at the meridian in the center of each time zone is defined as the standard time.
By convention, the starting point of the time zones is the Greenwich time zone (the section between the longitudes 7.5° W and 7.5° E).
The international date line is an imaginary line connecting the North Pole and the South Pole, its path roughly follows the 180° meridian. When crossing it from east to west we advance the calendar date by one day, when we cross in the opposite direction the calendar date to the west of the line is repeated.
### Time zone
• Prime Meridian
• International Date Line
The Earth is divided into 24 time zones.
Time zones were defined by geometrically dividing the Earth`s equator into 24 sections bordered by meridians, each 15° of longitude apart (because the Earth rotates 15° in 1 hour). Hence adjacent time zones have a time difference of exactly 1 hour, one hour less on the west, one hour more on the east.
Standard time is the time used in time zones. The local time at the meridian in the center of each time zone is defined as the standard time.
By convention, the starting point of the time zones is the Greenwich time zone (the section between the longitudes 7.5° W and 7.5° E).
The international date line is an imaginary line connecting the North Pole and the South Pole, its path roughly follows the 180° meridian. When crossing it from east to west we advance the calendar date by one day, when we cross in the opposite direction the calendar date to the west of the line is repeated.
### Day - night
• morning
• noon
• afternoon
• midnight
The Earth is divided into 24 time zones.
Time zones were defined by geometrically dividing the Earth`s equator into 24 sections bordered by meridians, each 15° of longitude apart (because the Earth rotates 15° in 1 hour). Hence adjacent time zones have a time difference of exactly 1 hour, one hour less on the west, one hour more on the east.
Standard time is the time used in time zones. The local time at the meridian in the center of each time zone is defined as the standard time.
By convention, the starting point of the time zones is the Greenwich time zone (the section between the longitudes 7.5° W and 7.5° E).
The international date line is an imaginary line connecting the North Pole and the South Pole, its path roughly follows the 180° meridian. When crossing it from east to west we advance the calendar date by one day, when we cross in the opposite direction the calendar date to the west of the line is repeated.
### Related items
#### Seas and bays
This animation demonstrates the most important seas and bays.
#### Uranus
Uranus is the 7th planet from the Sun, a gas giant.
#### Renewable sources of energy
Who wouldn’t be familiar with solar-powered calculators or windmills? Let’s have a look at the...
#### Jupiter
Jupiter is the largest planet of the Solar System, it has two and a half times the mass...
#### Sights of the world
A game about sights around the world.
#### The Sun
The diameter of the Sun is about 109 times that of the Earth. Most of its mass consists...
#### ISS
The International Space Station is a habitable satellite built with the cooperation of 16...
#### Formation of the Earth and Moon
This animation demonstrates how the Earth and the Moon were formed. | 1,251 | 5,197 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.671875 | 3 | CC-MAIN-2019-39 | longest | en | 0.859419 |
http://stackoverflow.com/questions/7829571/milliseconds-to-days | 1,394,615,732,000,000,000 | text/html | crawl-data/CC-MAIN-2014-10/segments/1394021585790/warc/CC-MAIN-20140305121305-00059-ip-10-183-142-35.ec2.internal.warc.gz | 176,498,895 | 15,610 | # milliseconds to days
i did some research, but still can't find how to get the days... Here is what I got:
int seconds = (int) (milliseconds / 1000) % 60 ;
int minutes = (int) ((milliseconds / (1000*60)) % 60);
int hours = (int) ((milliseconds / (1000*60*60)) % 24);
int days = ????? ;
-
from wikipedia: 86,400,000 (24×60×60×10×10×10) milliseconds — one day – amit Oct 19 '11 at 23:49
This isn't "I suck at math." This is pure "I'm lazy." – Zoltán Feb 11 at 10:02
If you don't have another time interval bigger than days:
int days = (int) (milliseconds / (1000*60*60*24));
If you have weeks too:
int days = (int) ((milliseconds / (1000*60*60*24)) % 7);
int weeks = (int) (milliseconds / (1000*60*60*24*7));
It's probably best to avoid using months and years if possible, as they don't have a well-defined fixed length. Strictly speaking neither do days: daylight saving means that days can have a length that is not 24 hours.
-
Don't forget about daylight savings changes – Mark Rotteveel Oct 20 '11 at 14:12
long days = TimeUnit.MILLISECONDS.toDays(milliseconds);
int days = (int) (milliseconds / 86 400 000 ) | 344 | 1,133 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.09375 | 3 | CC-MAIN-2014-10 | latest | en | 0.783247 |
https://teachchemistry.org/classroom-resources/discovering-equilibrium | 1,719,267,920,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198865490.6/warc/CC-MAIN-20240624214047-20240625004047-00054.warc.gz | 485,469,025 | 15,288 | # AACT Member-Only Content
You have to be an AACT member to access this content, but good news: anyone can join!
# Discovering Equilibrium Mark as Favorite (27 Favorites)
LESSON PLAN in Le Châtelier's Principle, Establishing Equilibrium, Equilibrium Constants, Reaction Quotient. Last updated March 25, 2020.
### Summary
In this lesson students manipulate sets of given conditions to discover what equilibrium is, and how the equilibrium is established from different starting conditions. Students then refer back to the activity as the foundation framework for the rest of Essential Knowledge 6.A, 6.B.1 and 6.B.2. This lesson focuses on both a qualitative and quantitative understanding of equilibrium.
High School (AP Chemistry)
### AP Chemistry Curriculum Framework
This lesson supports the following unit, topics and learning objectives:
• Unit 7: Equilibrium
• Topic 7.1: Introduction to Equilibrium
• TRA-6.A: Explain the relationship between the occurrence of a reversible chemical or physical process, and the establishment of equilibrium, to experimental observations.
• Topic 7.2: Direction of Reversible Reactions
• TRA-6.B: Explain the relationship between the direction in which a reversible reaction proceeds and the relative rates of the forward and reverse reactions.
• Topic 7.3: Reaction Quotient and Equilibrium Constant
• TRA-7.A: Represent the reaction quotient Qc or Qp, for a reversible reaction, and the corresponding equilibrium expressions Kc=Qc or Kp=Qp.
• Topic 7.4: Calculating the Equilibrium Constant
• TRA-7.B: Calculate Kc or Kp based on experimental observations of concentrations or pressures at equilibrium.
• Topic 7.5: Magnitude of the Equilibrium Constant
• TRA-7.C: Explain the relationship between very large or very small values of K and the relative concentrations of chemical species at equilibrium.
• Topic 7.7: Calculating Equilibrium Concentrations
• TRA-7.E: Identify the concentrations or partial pressures of chemical species at equilibrium based on the initial conditions and the equilibrium constant
• Topic 7.9: Introduction to Le Châtelier’s Principle
• TRA-8.A: Identify the response of a system at equilibrium to an external stress, using Le Châtelier's principle.
• Topic 7.10: Reaction Quotient and Le Châtelier’s Principle
• TRA-8.B: Explain the relationships between Q, K, and the direction in which a reversible reaction will proceed to reach equilibrium.
### Objectives
By the end of this lesson, students should be able to
• Understand the concept of dynamic equilibrium.
• Understand the meaning of K, and what its value signifies.
• Compare K to Q in order to predict which direction a system must proceed to reach equilibrium.
• Perform calculations involving K and manipulations of K.
• Use a graph to recognize the establishment of chemical equilibrium.
• Predict what will happen if a system at equilibrium is disturbed.
### Chemistry Topics
This lesson supports students’ understanding of
• Equilibrium
• Reversible Reactions
• Equilibrium Constant (K)
• Reaction Quotient (Q)
• ICE tables
• Le Chatelier’s Principle
### Time
Teacher Preparation: 5 minutes
Lesson:
• Introductory activity and discussion: 45 minutes
• Follow-up notes: several days depending on pace of class
### Materials
• Introductory activity:
• Large piece of paper or white board (to collect class data)
• Marker (to record class data)
• Optional:
• An optional version of this activity is included should you wish students to physically manipulate the reactions.
• In this case, you will need approximately 300 small manipulatives for a class of 30.
• These can be anything small: bingo chips, washers, pennies, paperclips, etc.
### Safety
• This activity does not require any special safety considerations.
### Teacher Notes
• The consideration of forward and reverse reactions and the meaning of K and Q serve to introduce the concepts in Big Idea 6. Please see the included document Teacher Background Notes for an overview.
• In this lesson an activity is provided to introduce the concept of equilibrium. This activity can be referred back to as the unit progresses (examples provided below in lesson outline).
• Teaching notes are provided which identify the relevant AP Chemistry Essential Knowledge standards (all of EK 6.A) for this topic.
• An Excel spreadsheet is provided that the teacher can manipulate to make different equilibrium graphs for graphical analysis practice problems (including disturbances to systems at equilibrium, covered subsequently in EK 6.B)
• A student note taker document and a teacher key for the note taker with teaching hints for a comprehensive coverage of EK 6.A are provided.
• Lesson Outline:
• Day 1: Introducing Equilibrium Activity (see #5 for optional version with manipulatives)
1. Before students come to class, copy the following class data table onto a large piece of paper or a white board.
1. Pass out the “Establishing Equilibrium” student worksheet. Have students read through the background notes and the sample procedure.Work through one to two rounds of the example activity to ensure students get the gist of what they are to do.
2. Assign each pair of students one of the groups (A – J) to collect data on.If you have more than 10 pairs, you can make up any reactant and product initial amounts.You can even let an extra group or two select their own initial amounts, so that the students can see that there is nothing “special” about the initial amounts.
3. As students complete their set of calculations, have them write their data on the class data table (see below for expected results). A copy of this data table has been provided on the student document as well.
1. Optional Manipulative Version: Students can be given physical “reactants” and “products” to manipulate instead of performing all of the work in their calculators.
1. Use the same group assignments (A – J starting conditions).
2. Use groups of three students instead of two (one to ‘work’ the reactants, one to ‘work’ the products, and one to record results).If this is done, be sure that groups H-J are also covered for discussion purposes.
3. Distribute small items to use for the reactants and products (examples given in material list).
4. To differentiate between “reactants” and “products”, have students use two pieces of paper, one labeled “reactants” and the other labeled “products”.Have students physically move manipulatives from one “reaction vessel” to the other.
5. Have students round to nearest whole number (since they cannot move a fraction of a chip).
6. The results will be very similar:
1. Once their data is recorded, have students begin working on the Analysis section of the worksheet. They can complete questions 1 – 5 using their data only, therefore do not need to wait for their classmates to finish.
2. Once all groups have recorded their data, review the Analysis questions before proceeding to the Class Discussion. Students should have recognized that when equilibrium is reached, no observable changes occur in the system, that reactant and product molecules are present, that the concentration of all species remains constant, and that a graph of concentration versus time reveals when equilibrium is established (EK 6.A.a and f).
3. The Class Discussion questions allow students to refine their understanding of the characteristics of a system at equilibrium, and how that equilibrium was achieved.This would also be an excellent time to introduce the concept of “equilibrium positions” (the actual concentrations of reactants and products at equilibrium, of which there are infinite possibilities) versus “equilibrium constant” (one value).
• Day 2 and additional class periods:
• The activity from Day 1 provides a foundation for the remainder of EK 6A, 6B.1 and 6B.2.Three documents have been provided for teacher use in covering these topics:
• Topics covered include:
• reversible reactions
• characteristics of equilibrium
• equilibrium constant/expression/calculations
• graphical analysis
• particulate diagrams
• the response of K to manipulating equations
• calculating Q
• predicting if a reaction system will proceed to the left or right to reach equilibrium
• Le Chatelier’s Principle
• Assessment/practice. The following FRQs provide excellent opportunities for student practice:
### Background
Many chemical and physical processes are reversible. One physical process that you are familiar with is the melting of ice. Ice melts when it is left in a glass on your kitchen counter. If you return the ice to the freezer, the process reverses, and the ice freezes. Many chemical reactions are also reversible. One example that you are sure to encounter in this course is the industrial production of ammonia via the Haber process:
N2 + 3 H2 ⇌ 2 NH3
An important biological example is the binding and releasing of oxygen gas by the hemoglobin molecule (Hb):
Hb(aq) + 4 O2(g) ⇌ Hb(O2)4(aq)
### Objectives
• Develop a definition for the term “dynamic equilibrium.”
• Recognize how to identify when a system which has reached equilibrium.
• Understand how equilibrium is established.
### Procedure
You will be simulating a reaction system which is proceeding towards equilibrium. The following is provided as an example of what you will be doing:
### Example
Take the reaction system R ⇌ P to equilibrium. The given set of conditions are:
Initial Amount of R = 30 moles
Initial Amount of P = 0 moles
Forward reaction: Each round, 30% of R reacts (R → P)
Reverse reaction: Each round, 25% of P reacts (P → R)
To calculate the final amount for each round:
Initial Amount - Amount Reacted + Amount Formed = Final Amount
Consider the reversible reaction R ⇌ P.
1. Obtain a set of starting conditions from your teacher. Record your set of conditions below:
• Initial amount R _______
• Initial amount P _______
• Forward reaction _______
• Reverse reaction _______
1. Perform several rounds of reactions. Round to two decimal places. Continue until the system reaches equilibrium. You will know when the system arrives at equilibrium.
2. Once equilibrium is reached, perform two more rounds (this will be helpful later when graphing your results).
3. Once you are done, record your data on the Class Data Table provided by your teacher.
### Analysis
1. Describe the changes in both the amounts of P and R from time “0” until equilibrium was reached. How did you know when you reached equilibrium?
2. Do the forward and reverse reactions actually cease? Why is equilibrium called dynamic equilibrium? What is happening on the molecular level?
1. How does the graph indicate when equilibrium has been reached?
2. For each round, the forward rate is the “amount of R reacted”, the reverse rate is the “amount of P reacted”.
1. Select several rounds for your reaction series calculate both the forward rate and the reverse rate at each of these time intervals.
Time Forward Rate Reverse Rate
2. What is happening to the rates? What happens to the rates at equilibrium?
### Class Discussion
Use the data from the Class Data Table to answer the following questions:
1. Consider the data from Groups A – G. At equilibrium, are the concentrations of R and P always the same? Provide an example to support your answer.
2. Consider the data from Groups A – G. Does the P/R ratio depend on the initial amounts of P and R? Provide an example to support your answer.
3. What happens to the P/R ratio if the system begins with only products instead of only reactants?
4. The P/R ratio can be called the “equilibrium constant.” Considering your answers to the questions above, why is this a valid term?
5. Consider the data from Groups A – J. Do the forward and reverse reaction percentages affect the P/R equilibrium ratio? Explain the mathematical relationship.
6. Consider the data from Group A and Group H.
1. Compare the P/R ratios. What do you observe?
2. The reaction from Group A is said to “favor the products”, while that from Group H is said to “favor the reactants.” What do you think this means?
3. Make an observation about the P/R ratio and whether the reactants or products are favored.
4. If a reaction system has a very large P/R ratio (or a very small one), what would this indicate?
5. Compare the graphs for each. What do you observe?
### Class Data Table
Use this table to copy down the results from each student group. | 2,646 | 12,392 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.9375 | 3 | CC-MAIN-2024-26 | latest | en | 0.837959 |
https://www.sdpuo.com/how-to-find-the-velocity/ | 1,718,677,078,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861746.4/warc/CC-MAIN-20240618011430-20240618041430-00425.warc.gz | 869,623,915 | 18,054 | June 18, 2024
This article provides a step-by-step guide to calculate velocity and its real-world applications in various fields such as physics, engineering, and sports. It also includes a list of common velocity problems with solutions, an interactive calculator, and infographics for velocity formulas and equations.
## Introduction
If you have ever been stuck in traffic and wondered how fast you were moving, you have pondered velocity. Velocity tells you the speed and direction of an object’s movement. Velocity comes in handy in various fields, from engineering to sports. In this article, you will learn what velocity is and how to calculate it.
## Step-by-Step Guide to Finding Velocity
Velocity is the rate at which an object changes its position, given in meters per second (m/s) or kilometers per hour (km/h). The basic formula to calculate velocity is velocity=(final position-initial position)/time.
Here are the five steps to finding velocity:
1. Determine the final and initial position of the object or point in question.
2. Determine the time it took for the object to move from initial point to the final point.
3. Calculate the change in position, which is the difference between final position and initial position.
4. Substitute the changes in position and time into the formula velocity=(final position-initial position)/time.
For example, suppose that a car travels 100 meters in 10 seconds. What is the car’s velocity? Using the formula, we can calculate the velocity to be (100m-0m)/10s=10m/s.
It is important to understand the different types of velocity, including instantaneous, average, and uniform velocity. Instantaneous velocity measures the velocity of an object at any given point of time. Average velocity, on the other hand, measures the average rate of change of position of an object over a period of time. Lastly, uniform velocity measures the constant rate of change of an object’s position over a period.
## Real-World Applications of Velocity
Velocity has practical applications in physics, engineering, sports, and other fields. One application, for instance, is in physics, where velocity is used to measure the speed and direction of an object’s movement. Engineers use velocity to calculate the performance of machines, such as determining how quickly a car will stop from a particular speed. Velocity comes in handy in sports, particularly in racing, where athletes must track their speed and make adjustments to improve their performance based on velocity.
Take, for example, a basketball player’s layup shot, which requires calculating the velocity of the ball to make the perfect shot. If the ball goes too fast, it will miss the hoop. But if the ball goes too slow, it will not make it through the hoop either. By understanding velocity, the basketball player can adjust the speed and placement of the ball to perfectly land the shot.
## Common Velocity Problems
Students often struggle with velocity problems, which require a good understanding of the fundamental principles of physics. However, with the right guidance, students will master the concept of velocity in no time. Here are some common velocity problems and how to solve them:
Problem 1: A race car travels 240 meters in 12 seconds. What is the car’s velocity?
Solution:
The given values are:
Initial position, x1 = 0
Final position, x2 = 240 m
Time, t = 12 s
Using the formula velocity=(final position-initial position)/time, we can calculate the velocity to be (240 m – 0 m)/12 s = 20 m/s.
Problem 2: Find the average velocity of a cyclist who cycles for 20 km in 2 hours.
Solution:
The given values are:
Initial position, x1 = 0
Final position, x2 = 20 km
Time, t = 2 hours
Using the formula velocity=(final position-initial position)/time, we can calculate the velocity to be (20 km – 0 km)/2 hours = 10 km/h.
## Interactive Calculator
Calculating velocity can be a bit daunting, especially for new learners. An interactive calculator or tool can help to simplify the process and provide hands-on experience. Here is a useful online tool to calculate velocity:
Speed Distance Time Calculator
This calculator allows you to perform calculations for various parameters, including speed, distance, and time. To use it, simply input the relevant values, and the calculator will output your results.
## Infographics for Velocity Formulas and Equations
Infographics are an excellent tool for illustrating and explaining formulas and equations for velocity. For instance, velocity-time graphs use different colors and graphics to help readers better understand the concepts. Here is an example of a velocity-time graph:
The graph shows the changes in velocity with respect to time. The slope of the graph at any given point represents the object’s acceleration or deceleration. By studying this graph, you can gain deeper insights into the principles of velocity and its practical applications.
## Conclusion
Velocity is a fundamental principle in physics, engineering, and sports, among other fields. By learning how to calculate velocity, you can gain valuable insights that will empower you in your studies and in your career. Remember, velocity is not just about speed; it also includes the direction of movement. Utilize the step-by-step guide, apply the formulas, and seek out tools like the interactive calculator and infographics to improve your understanding. With these resources, you will soon master velocity and be able to tackle velocity problems with greater confidence and understanding. | 1,120 | 5,544 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.65625 | 5 | CC-MAIN-2024-26 | latest | en | 0.926106 |
https://netsnvenworksess.web.app/185.html | 1,709,400,891,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947475833.51/warc/CC-MAIN-20240302152131-20240302182131-00091.warc.gz | 406,317,358 | 6,208 | # Ntransforming functions worksheet pdf
First there is an overview of how a, h, and k relate to transforming the parent quadratic function in vertex form, followed by several practice pr. Feel free to use a graphing calculator to check your answer, but you should be able to look at the. Flipped upside down as indicated by the negative in front. Only the first 51 of these values appear on the facing page the worksheet runs. Transformations of function worksheet for 11th grade. Sketch the graph of the parent function and the graph of the transformed function on the same set of axes. Gtcedk identifying types of functions graphically and analytically c. Use transformation of functions to move segments into new positions. Absolute value transformations displaying top 8 worksheets found for this concept some of the worksheets for this concept are graphing absolute value functions date period, absolute value transformations matchingmania, solving absolute value equations, name date, essential question lesson 1 absolute value functions, honors transforming absolute value.
In the following we will look at six of the most common graphs. Click the following links to download one step equations worksheets as pdf documents. Students formalize the period, frequency, phase shift, midline, and amplitude of a general sinusoidal function by understanding how the. Parent function worksheet 1 humble independent school. Worksheet transformations offunctions and theirgraphs 1. The parent graph yx logb passes through the points 1, 0 and b, 1 and has a vertical asymptote at x 0. The content you are trying to access requires a membership. Worksheets are transformations of graphs date period, graphical transformations of functions, transformation of functions work, graph the image of the figure using the transformation, transformations of functions name date, transformations of exponential functions work, exponential transformations work, linear transformation work 1. Tes global ltd is registered in england company no 02017289 with its registered office at 26 red lion square london wc1r 4hq. This activity can be used early in the year after students have studied linear functions.
Translatingshifts, vertical stretch or compression and reflection over the xaxis. Write an equation for the absolute function described. Identify the vertical displacement, amplitude, period, phase shift, domain and range. Translations and refl ections a family of functions is a group of functions with similar. Transformation of a quadratic function and identification of various types of shifts like horizontal shift, vertical shift and reflection are the major topics discussed in this unit of highschool function. Sketch the graphs of cosine and sine functions worksheet sketch the following functions over two cycles. A baseball is struck such that its position on a plane perpendicular to home plate can be described by the equation u. Free algebra 2 worksheets created with infinite algebra 2. This formula has direct application to the process of transforming probability density. Be able to graph the functions key points such as x and yintercepts are important and be able to identify the graphs. This lesson follows the pattern used in both algebra i and algebra ii. Transformation of functions worksheet teachers pay teachers. T3 scotland functions and transformations page 7 of 12 task 5 if you know the graph of y x2 can you draw quickly the graph of y x2. Graphing and finding properties of the root function and the reciprocal function.
The parent function y x stretched vertically by a factor of 2, shifted right 5 units and up 3 units. Using transformations to graph quadratic functions describe the following transformations on the function y x2. Amplitude and period for sine and cosine functions worksheet. The parent function y x reflected across the xaxis, then shifted left 2 units. The function fx 20x represents the daily rental fee for x days. Describe the transformations that have been applied to obtain the function from the given base function. Use the attached worksheets titled transformations of exponential functions to guide the lesson on transformations. Vanier college sec v mathematics department of mathematics 20101550 worksheet. The period of a function is the horizontal distance required for a complete cycle. As the students progress through the activity, check that they. The transformations of the letters in the words math rocks will be the answers to the worksheet. If a function contains more than one transformation it may be graphed using the following procedure.
Suppose x is a random variable whose probability density function is f x. Practice worksheets for graphing exponential functions and. Many of the equations which you will encounter are transformations of a few basic functions. Vertical translations a shift may be referred to as a translation. Here are your free resources for your lesson on parent functions and transformations worksheet, powerpoint guided notes, exit quiz, bell work, and more. Selection file type icon file name description size revision time user. Use your knowledge of the graph of the base function, and the transformations, to graph the function.
Using transformations to graph quadratic functions describe the following. Transformation worksheets contain skills on slides, flips, turns, translation, reflection and rotation of points and shapes. Functions, relations, and transformations 4 overview in discovering advanced algebra, students study mathematical functions modeling realworld problems. Describe how you can graph this function by transforming the graph of y 3. For a function that models a relationship between two quantities, interpret key features of graphs and tables in terms of the quantities, and sketch graphs showing key features given a verbal description of the relationship. Notice that all of the new functions in the chart differ from fx by some algebraic manipulation that happens after f plays its part as a function. Absolute value transformations worksheets kiddy math. Graphing logarithmic functions worksheet transformations of logarithmic functions. These graphs and their corresponding functions are. G h 4mja ednem wwxidt nhy 7iqnyf wian niot ree qa cl zg dedbmrnax d25. Displaying all worksheets related to transformation of functions. Graphing and finding properties of the root function and the reciprocal. You are transforming the original curve in each case.
On your calculator draw these two graphs and sketch two colours the results on the. Transformations of graphs and the modulus function pearson. Dec 22, 2009 this website and its content is subject to our terms and conditions. Submit an equation that will move the graph of the function yx2 right 4 units. N h pa gl 0l r cr ni8gkhmtvs4 zr vees oerrfvzerd y. Graphical transformations of functions in this section we will discuss how the graph of a function may be transformed either by shifting, stretching or compressing, or reflection. Graph transformations there are many times when youll know very well what the graph of a particular function looks like, and youll want to know what the graph of a very similar function looks like. This is our function this is our function vertically stretched this is our function vertically compressed this is our function horizontally compressed this is our function horizontally stretched this is our function reflected over the xaxis this is our function reflected over the yaxis this is our function with a horizontal shift right this is our function with.
Master graphing the root function, the reciprocal function and the asymptotes, the absolute value function, the quadratic function and lines. Transformations of function worksheet for 11th grade 1 for the curve y x 3 given in figure, draw. Modeling with functions mathematics vision project. For instance, here are the domains of the functions in example 3. Displaying all worksheets related to answers transformation of functions. Transformations of graphs date period kuta software llc. Transformation of exponential functions with answers. Using transformations, many other functions can be obtained from these parents functions. Graphing simple rational functions kuta software llc. Unit 7 transformations of functions emathinstruction. Answers transformation of functions worksheets lesson. Describe the transformations necessary to transform the graph of fx into that of gx. Stacked here is a vast collection of pdf worksheets on graphing quadratic functions to identify zeros, writing the quadratic function of the parabola, identifying properties of a parabola and more. Graph the following functions without using technology.
Free printable function worksheets pdf with answer keys on the domainrange, evaluating functions, composition of functions,1 to 1, and more. Basic sine function periodic functions definition, period, phase shift, amplitude, vertical shift. Transformation of exponential functions with answers displaying top 8 worksheets found for this concept some of the worksheets for this concept are transformations of exponential functions work, exponential transformations work, exponential functions date period, transformations of functions name date, transforming exponential and. Graphing functions using transformations tutoring and learning centre, george brown college 2014. A rational function is a function thatcan be written as a ratio of two polynomials. Then, discuss how to find the domain and range, intercepts, and intervals of increase and decrease. V o2u0 k1v38 qkxuqt oau lsuo3fqtpwta mrhex gl 6lqck. Absolute value vertical shift up 5, horizontal shift right 3. Mathematics linear 1ma0 transformation of graphs materials required for examination items included with question papers ruler graduated in centimetres and nil millimetres, protractor, compasses, pen, hb pencil, eraser. Suppose x is a random variable whose probability density function is fx.
Using degrees, find the amplitude and period of each function. The graph of y fx is transformed into the graph of y f x c by a stretch of scale factor c in the xdirection. Whether you want a homework, some cover work, or a lovely bit of extra practise, this is the. A periodic function is a function whose graph repeats itself identically from left to right. In addition, writing the coordinates of the transformed shapes and more are included. Transformation worksheets reflection, translation, rotation. Grab our free printable function worksheets featuring domain and range, function. Mathematicstranslations of linear functions tteaching suggestions his is a discovery activity that can be effectively done in groups. In this chapter, well discuss some ways to draw graphs in these circumstances. Describe the transformations required to obtain the trig function starting from the parent function. Then use the results of explorations 1 and 2 to compare the graph of k to the graph of f xx. Transformations can be combined within the same function so that one graph can be shifted, stretched, and reflected.
Functions, characteristics of functions, and terminology that students will use as they study functions during the remainder of the year are introduced in this activity. Introduce a new type of function, and then transform it. Worksheets are graph the image of the figure using the transformation, transformations of graphs date period, graphical transformations of functions, transformations of functions name date, transformations of exponential functions work, exponential transformations work, linear. Here, the abstract idea of a function grows out of students earlier experiences with linear equations and graphing. In these worksheets identify slides, flips and turns of the given figures. Solving linear equations worksheets pdf from transforming functions worksheet, source. Transforming the graph of the sine function engageny. Worksheet transformations offunctions and theirgraphs. Transformations of exponential functions worksheet. Remind students as they begin question 1 to use ordered pairs which include both positive and negative x values as they graph the lines.
Function transformations just like transformations in geometry, we can move and resize the graphs of functions let us start with a function, in this case it is fx x 2, but it could be anything. Transformations include translations left, right, up, and down, reflections across both the x and y axes, vertical stretching and. Infinite algebra 2 assignment 4 graphing functions by. By transforming a few reference points according to the rules in the last section, we will be able to graph variations of these functions. Absolute valuevertical shift up 5, horizontal shift right 3. Algebra function worksheets pdfs with answer keys on. Write the equation of the following functions, given the original function and the transformations performed. Steps for multiple transformations use the following order to graph a function involving more than one transformation. If you need to purchase a membership we offer yearly memberships for tutors and teachers and special bulk discounts for schools. Transformation of functions worksheets lesson worksheets. Combine transformations of graphs of linear functions.
287 797 1641 413 1485 530 428 409 833 1046 1124 363 1064 1340 829 86 1441 579 695 93 1010 1309 840 834 529 1328 488 399 1386 236 668 185 | 2,580 | 13,430 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.046875 | 3 | CC-MAIN-2024-10 | latest | en | 0.843487 |
https://studysoup.com/tsg/22447/calculus-early-transcendentals-1-edition-chapter-10-problem-20re | 1,606,924,246,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141711306.69/warc/CC-MAIN-20201202144450-20201202174450-00016.warc.gz | 492,436,171 | 11,190 | ×
×
# Areas of regions Find the area of the | Ch 10 - 20RE
ISBN: 9780321570567 2
## Solution for problem 20RE Chapter 10
Calculus: Early Transcendentals | 1st Edition
• Textbook Solutions
• 2901 Step-by-step solutions solved by professors and subject experts
• Get 24/7 help from StudySoup virtual teaching assistants
Calculus: Early Transcendentals | 1st Edition
4 5 1 327 Reviews
29
5
Problem 20RE
Areas of regions Find the area of the following regions. In each case, graph the curve(s) and shade the region in question.
The region inside the limaçon r = 2 + cos θ and outside the circle r = 2
Step-by-Step Solution:
Solution 20RE
Step 1:
Given the limaçon r = 2+ cos θ
The equation (up to translation and rotation) of a limaçon in polar coordinate has the form
Therefore in our question we have and
Here since the limocon is a simple closed curve.
Also given a circle
Step 2:
The graph of the curve is
Step 3 of 4
Step 4 of 4
#### Related chapters
Unlock Textbook Solution | 288 | 1,001 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.34375 | 3 | CC-MAIN-2020-50 | latest | en | 0.772907 |
https://lessonplanet.com/search?keywords=projectile+motion | 1,618,624,429,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038098638.52/warc/CC-MAIN-20210417011815-20210417041815-00032.warc.gz | 464,801,162 | 30,519 | ### We found143 reviewed resources for projectile motion
Videos (Over 2 Million Educational Videos Available)
5:25
Language and Creativity
8:55
The Circulatory System Part 1: The Heart
6:54
How to paint a watercolor and ink flower...
Other Resource Types ( 143 )
6 Items in Collection
Lesson Planet
#### The Physics of Fluid Mechanics
For Students 9th - 12th Standards
In this collection, Pascal’s law, Archimedes’ Principle, and Bernoulli’s Principle aid young engineers in understanding the physics of fluid mechanics. Armed with the knowledge of how forces are applied to fluids, scholars design...
Lesson Planet
#### Projectile Motion
For Teachers 9th - 12th
Here is a rare gem for physics teachers! An engaging collection of slides opens with a question about aiming a slingshot at a monkey as it lets go of the branch from which it hangs. Projectile motion concepts are presented with graphics,...
Lesson Planet
#### Projectile Motion Review
For Students 9th - 12th
Launching Fluffy, the Wonder Hamster, in a catapult might not be humane, but imagining the action to analyze projectile motion should be acceptable. This physics resource injects a little silliness, yet is quite serious about assessing...
Lesson Planet
#### Projectile Motion
For Teachers 9th - Higher Ed
A unique and useful presentation, this covers projectile motion for your general physics class. Many graphs display the vectors involved, and some even act as animations. Motion is broken down into its components and only then are the...
Lesson Planet
#### Projectile Motion Lab
For Teachers 9th - 12th
Employing their knowledge of projectile motion, physics masters form a hypothesis about how far a metal ball will fly after rolling down a ramp and off of a tabletop. Their goal is to predict correctly and catch the ball at the right...
Lesson Planet
#### Projectile Motion
For Students 9th - 12th
Hit a bullseye with an engaging presentation about the components of projectile motion. Viewers see how horizontal and vertical motion combine to create the parabolic path of projectile motion.
Lesson Planet
#### Projectile Motion Warm-up
For Teachers 9th - 12th
Three practice projectile motion problems are presented, each on an individual slide. During your projectile motion unit, you could display one of these each day at the beginning of class as a warm-up exercise. There are cute cartoon...
Lesson Planet
#### Projectile Motion Warm-up 3
For Teachers 9th - 12th
Four physics problems dealing with projectile motion are displayed for your class to solve, one on each slide. Use these to practice as a class, or as a pop quiz to assess their grasp on these concepts.
Lesson Planet
#### Projectile Motion and Quadratic Functions
For Teachers 9th - 12th Standards
Model the flight of a ball. Small groups collect time data for how long it takes to toss a ball between two people. The teacher leads a discussion about projectile motion to get the class thinking about how to calculate the equation of...
Lesson Planet
#### Exploring Projectile Motion
For Teachers 6th - 8th Standards
Live action Angry Birds is sure to be a popular STEM lesson plan. Young physicists differentiate between potential and kinetic energy, calculate speed based on distance and time, create graphs, and design and complete their own...
Lesson Planet
#### Projectile Motion
For Students 6th - 12th Standards
Have you ever wanted to shoot someone out of a canon? This simulation gives you that chance! In addition to a human, you can shoot a car, a piano, a cannon ball, and many other items. Scholars set the angle, initial speed, mass, air...
1:25
Lesson Planet
#### STEMbite: Projectile Motion of Abby's Arrow
For Students 7th - 12th
Projectile motion is seen in action as Andrew's niece, Abby, practices archery. This is a nice demonstration since most video clips that you can find online are actually just animations. That being said, there is nothing more to this...
4:16
Lesson Planet
#### Science of NFL Football: Projectile Motion and Parabolas
For Students 5th - 12th
Punt, pass, or kick your way to learning physics! An engaging lesson breaks down the mechanics of punting a football through an analysis of vectors. The ninth lesson of a video series explains the horizontal and vertical velocity...
Lesson Planet
#### Typical Numeric Questions for Physics I - Motion
For Students 9th - 12th
Seven pages with 33 multiple choice questions about motion make up this worksheet. Calculations must be performed in order to choose the correct answer. Physics majors will solve for angles, displacement, acceleration, speed, and time...
Lesson Planet
#### Projectile Motion Warm-up 2 - A Bit of Review
For Teachers Pre-K - 11th
Seven questions about projectile motion are posted for your physics class to review. The questions take up the first three slides, and answers are provided on the final two. You could use this as a review by discussing and working...
Lesson Planet
#### Projectile Motion
For Teachers 9th - 12th
Students observe projectile motion and calculate the speed of a baseball based on the time and distance traveled. They record the time, measure the distance, and draw the path of the ball's travel on a data table.
Lesson Planet
#### Parametric Projectile Motion
For Teachers 9th - 12th Standards
Learners relate projectile motion to parametric equations as they identify the velocity and angle of a projectile. They create a picture on the navigator to visualize the word problem.
Lesson Planet
#### Motion in Two Dimensions
For Teachers 9th - 12th
Two activities employ the use of kinematic equations for determining projectile ranges. They both require the use of a toy dart gun. Included in this resource are both the teacher's guide and student laboratory sheets. This is an...
Lesson Planet
#### Projectile Magic
For Teachers 9th - 11th Standards
What do the movies October Key and Harry Potter and the Sorcerer's Stone have in common? The fourth installment of a five-part module presents equations regarding projectile motion and how to rearrange them. Scholars view video clips...
Lesson Planet
#### Projectile Motion
For Teachers 9th - 12th
High schoolers determine the angle of launch that will give a projectile its greatest range. In this projectile motion activity, students recognize that the motion of a projectile results from the combined horizontal and vertical...
Lesson Planet
#### Projectile Motion Demonstration
For Teachers 9th - 11th
Students dicusss Newton's First Law of Motion. They calculate a projectile moving horizontally and vertically at different intervals. They particpate in an experiment in which the projectile is at an angle. They record and discuss the...
Lesson Planet
#### Magical Motion
For Teachers 9th - 11th Standards
Make solutions to projectile motion problems magically appear using equations. Pupils watch a clip from a Harry Potter movie and find the length of time it takes for a remembrall to fall into Harry's hands. They use a projectile motion...
Lesson Planet
#### Projectile and Circular Motion, Torque
For Teachers 11th - Higher Ed Standards
Introduce your young scholars to the concepts of circular motion, projectile motion, angular speed, simple harmonic motion, torque, center of mass, centripetal force, and Hooke's Law with a 86-slide presentation. The circular motion...
Lesson Planet
#### Satellite Motion
For Teachers 9th - 12th
Three, two, one, blastoff! Your physics, astronomy, or engineering class learns about satellite motion, both circular and elliptical, by viewing this set of slides. Explanatory graphics are included to further enhance learning. Several... | 1,616 | 7,671 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.78125 | 3 | CC-MAIN-2021-17 | latest | en | 0.866579 |
https://drbobsports.com/sports-betting-as-an-investment/ | 1,560,877,002,000,000,000 | text/html | crawl-data/CC-MAIN-2019-26/segments/1560627998808.17/warc/CC-MAIN-20190618163443-20190618185443-00209.warc.gz | 423,539,534 | 28,415 | # Section 1: Sports Betting as an Investment
## Making Money by Betting on Sports
Most people think that sports betting is about finding ‘sure things,’ but in reality such ‘locks’ are nothing more than gamblers’ fancy. Just as in real estate, currency, stocks, or any other speculative market, ‘sure things’ simply do not exist. As a professional sports bettor, my goal is to find and exploit many small edges over a long period of time to earn a compounding return. Winning 55% of games is very significant, and with very conservative bet sizing, you can grow your return very quickly. Investing \$10,000 into the stock market for a year and earning a 10% return is considered a great investment – but your return winning a modest 54% of your sports bets would trounce that return.
My picks have yielded a much higher risk adjusted return than the stock market. Obviously, the variance from season to season is formidable, but as anyone who had a significant amount invested in stocks or real estate in 2008 can tell you, such swings aren’t limited to sports. In the long run, my edge in what I do is far greater than the edge that you could hope to gain in any other speculative market.
## Juice, and the power of 55%
Even though most sports bettors are losers in their own right (as a whole, bettors actually win an average of only 48% of their bets – less than they would expect to win if they just flipped a coin for every game), their losses are compounded by the fact that the house takes a cut of winnings, also known as the ‘juice’ or ‘vig.’ Most sports books charge a 10% commission on wins, which means that a bettor must actually win 52.4% of his games just to break even. (Wagering \$100 per game, a bettor loses \$100 with a loss and wins \$90.91 with a win, so he must go 11-10 (11/21 = 52.38%) to break even).
In order to beat the juice and win in sports betting, a bettor must employ a disciplined approach in their analysis of each game using methods that have proven to be successful in the long run. I discuss my math models and analytical metrics in my Handicapping Methods essay, but you must realize that only the best and most knowledgeable handicappers can win more than 52.4% of their games. In their 2007 two page article about my handicapping success, the Wall Street Journal wrote, “…fewer than 100 people can sustain (win rates of 55%) over time. Most of them belong to professional betting syndicates that hire teams of statisticians, wager millions every week and keep their operations secret.”
Touts often claim to be able to hit 60% or higher, but as I explain in my essay on Bayesian Probability, anyone who tells you that their long term expected winning percentage is higher than 60% is deluding themselves. Ten or more years ago a sharp handicapper could win about 60% long term but those days are over, as odds makers have become more savvy in the past decade or so. For a bettor to claim a greater than 60% long term expected win percentage, that would be mean that Vegas would have to consistently release lines with egregious errors, and that simply just does not happen often enough nowadays for claims of a greater than 60% long term expected win percentage. Any short term win rates of around 60% or higher are simply due to blind, short-term luck. For instance, last year (2016) in the first season using a new NFL play-by-play model, Dr. Bob Sports’ NFL Best Bet sides were an incredible 66-26 (71.7%), but that record was enhanced by winning a very large percentage of close games (31-12 on Best Bets decided by 7 points or less) rather than splitting the close ones. It still would have been a great season on NFL Best Bet sides (62%) if the close games had been 50% but I still can’t expect the new model to win 60%-plus on sides based on that one season – although the play-by-play model back-tested at a very profitable 56% winners.
I often hear amateur gamblers erroneously claim that winning 55% of games isn’t even enough to beat juice. As demonstrated above, a bettor only needs to win 52.4% to break even, and a 55% bettor will be very profitable in the long run if they pursue an optimal money management strategy.
Of course, as in any game of chance, there is variability in the actual results and just because you have won 55% in the past and expect to win 55% in the future doesn’t mean that you’re going to win 55% this upcoming season. There is variance in sports betting, as there is in most investments, and I calculate the standard deviation to figure out how much of my bankroll I can safely wager on each game during the season to accommodate potential negative swings while having very little chance of exhausting my bankroll. I have extensively quantified the variance that exists in sports betting, and use mathematical formulas to dictate the exact optimal amount to invest so as to maximize the ratio of profits to variance.
My long term percentage on College Football Best Bets is 56% (1290-1017-40 over 29 years) and the new NFL play-by play model was 100-69 (59.2%) in 2016. However, despite being a combined 148-107 (58.0%) on Football Best Bets, college and NFL, in 2016, I will continue to use 55% winners as a realistic goal going forward. If I expect 55% on 200 Football Best Bets (I had 255 last year, which was higher than expected) then the expected profit at -110 odds would be 200 x (0.55 – (1.1 x 0.45), which is +11.0 units (or +22.0 Stars if my average Best Bet is rated 2-Stars). The Kelly Criterion recommends a wager of 3.4% of your bankroll for a wager with a 55% chance of winning and odds of -110. However, the Kelly formula assumes sequential betting and sports betting usually involves simultaneous betting, which is part of the reason behind using some fraction of full Kelly to reduce risk. If I play 2.0% of my initial bankroll per bet, or 1.0% per Star, (i.e. flat betting) then my expected return during football season (5 months) is 22.0%. Adjusting your bankroll after each week rather than flat betting will increase your expected return, as explained in the KC simulation section of my money management section.
A basketball season with 53.5% winners (my career percentage is 53.9%) on 500 bets would on average yield +11.75 units ( (500*.535) – (500*.465)*1.1 ), or +23.5 Stars if my average Best Bet is rated 2-Stars. Using a conservative 1.6% of bankroll per bet (full Kelly at 53.5% at -110 odds is 2.35% of bankroll), or 0.8% per Star, results in an expected return of 18.8%. So, despite a lower overall winning percentage and smaller average wager size, a season’s worth of basketball wagers is fairly comparable to a season of football because there are so many more Best Bets in basketball season.
The NBA Guru Basketball service has achieved even higher returns in the 5 seasons that the Guru has been with Dr Bob Sports. The NBA Guru has an incredible record of 647-532-20 (54.9%) on his Best Bets over 5 seasons and 1366-1118-41 on a Star Basis for +136.0 Stars (with an extra -0.2 for added juice), which is an average of +27.2 Stars per season. You can risk more of your bankroll per play with the NBA Guru because he has a higher win percentage and fewer plays. I recommend 2.0% of your bankroll per play, or 1.0% per Star on NBA Guru Best Bets.
## Money Management
Money Management is as critical to a sports investor as picking winners. I have devoted many hours of careful analysis and math to optimal money management systems, which I have painstakingly outlined in my Money Management articles. Sports betting is more high risk (higher volatility and standard deviation of return) than stocks, but also results in a higher return if you follow a proven long term winning handicapper (of which there are very few).
My Money Management articles outline how to adjust your bet sizing based on your goals (expected return vs. probability of positive returns), your investment length (one season or many), your growth preference (flat or compounding), your risk tolerance (high or low) and the proportion of your overall bankroll which is made up by sports betting.
It is always better to set conservative expectations to avoid over betting.
## Factoring in the Cost of my Service
The cost of my College football service is \$895, the cost of the NFL service is \$995 (\$1,595 for both services), my Basketball service is \$895 (\$2,195 for all Football and Dr. Bob’s Basketball service), and the NBA Guru subscription is well worth the \$1495 given how profitable he’s been (\$3,295 for all Football and all Basketball, including the NBA Guru). You must factor in that cost when calculating your expected return on investment (ROI). As explained above, winning 55% on the Football Best Bets and 53.5% on my Basketball Best Bets would yield an expected profit of +45.5 Stars and let’s assume the NBA Guru profits +27.2 Stars as well (he’s averaged +27.2 Stars per season). Let’s say you decide to play 1.0% of your initial bankroll per star on the Football Best Bets and NBA Guru Best Bets and 0.8% per star on the Basketball Best Bets, as in the example above. Doing so would have an expected total return 68.0% per year based on flat-betting using your initial bankroll. Using an optimal betting strategy, as explained in the advanced money management section, would yield even higher long term returns while protecting the downside risk in the inevitable negative variance seasons that plague even the best long term handicappers.
If you had \$20,000 that you could comfortably afford to risk as your sports wagering bankroll and \$3,295 went to pay for the all Football and all Basketball service, then you would have \$16,705 left for wagering. As explained above the expected return on the combined Dr Bob Football and Basketball and NBA Guru Basketball services is +68.0% per year (using a less optimal flat betting approach), which would result in a return wagering profit of +\$11,359 on the \$16,705 initial bankroll. The overall profit, after factoring in the cost of the services, would be \$8,064 ((\$16,705 x 0.68) – \$3,295 = +\$8,064), which is a very good 40.3% expected return on your \$20,000. That percentage return is higher for higher bankrolls and lower for lower bankrolls since the cost of service becomes a smaller percentage of higher bankrolls and a higher percentage of smaller bankrolls. If you want to subscribe to the all Football and all Basketball package you would need a total of at least \$4,846 to invest to expect a positive return after factoring in the cost of the service. The calculations above are based on expected results based on long term records and some years are better and some years are worse.
## What is a Point spread?
Before I delve into rigorous explanations of how a bettor can gain an advantage against the point spread, it is important to understand what the spread actually represents. Point spreads were invented to keep bettors interested in games between teams of different talent levels – if a perennial powerhouse like Alabama plays a mid level team such as South Alabama, you’ll find very few people willing to bet on which team will win the game since Bama would be such a prohibitive favorite. However, most are willing to bet on whether Alabama will ‘cover the point spread’ and win by a certain number of points. If the point spread is 21.5, then the Crimson Tide must win by 22 or more points for their side of the bet to cover, while South Alabama must either win outright or lose by 21 points or less to cover their side. Point spreads are designed so that the probability of each outcome is roughly equal, and are generally set so as to approximate the median score differential between the two teams at the given site of the game.
However, skewed public perception, results-oriented analysis, and unsound metrics result in point spreads that are often biased one way or another. While the casual bettor does not possess the capacity to exploit these advantages, I have used mathematical models, situational analysis, significant trends and quantitative player analysis that are far more complex and accurate than anything else on the market to gain an advantage, which is why I have won 53% to 56% of my Best Bets (depending on the sport) over the last 29 years.
## How are the lines set?
While the odds makers do to try approximate the median margin of victory between two teams, they also try to reduce their exposure to risk by setting lines such that the public money will fall evenly on both sides of a game, so that they can offset the bets against each other and earn a profit on the juice (cut of winnings taken by the house, explained below) without exposing themselves to large potential losses. Thus, odds makers are often in the business of gauging public perception rather than team performance, and therefore the betting public actually sets the line. In more recent years, the betting public has had less influence on the odds than professional betting syndicates or sharp money has had, but there is still value to be found – although in different ways than in previous decades. If Georgia is 4 points better than Georgia Tech according to my advanced metrics and analysis, but the aggregate public perception is that Georgia is 7 points better than Georgia Tech, then the posted point spread is likely to be closer to 6 or 7 points (public perception) than it will be to 4 points (the realistic difference between the teams). This makes my job as a professional handicapper much less daunting; not only can I exploit lines where the odds makers are off, but I can also exploit the uniformed opinions of the general betting public, and more recently take advantage of betting syndicates and ‘quants’ that rely more and more on algorithms but can overlook some of the hidden value in changes in team personnel or lineups and in the particular match up between two teams.
## Isn’t gambling risky?
I don’t believe that the term ‘gambling’ applies to what I do. I sell information to subscribers, with which they can take positive expectation positions in uncertain markets. With correct financial optimization and bankroll management, long term risks are nominal compared to the risks of investing in other, more conventional markets. Just as a single stock may go up or down in a day, any one team may win or lose a given game. But as long as the investor maintains a long-term perspective, understands variance, and doesn’t over-extend themselves or bet more than they can easily handle, risk can be highly mitigated, and they can earn a very attractive risk adjusted return. | 3,282 | 14,612 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2019-26 | longest | en | 0.976791 |
https://www.coursehero.com/file/5904601/161-1-Class14-No-vioice/ | 1,521,689,511,000,000,000 | text/html | crawl-data/CC-MAIN-2018-13/segments/1521257647758.97/warc/CC-MAIN-20180322014514-20180322034514-00662.warc.gz | 772,323,539 | 182,305 | {[ promptMessage ]}
Bookmark it
{[ promptMessage ]}
161_1_Class14_No_vioice
# 161_1_Class14_No_vioice - EE161 Electromagnetic Waves...
This preview shows pages 1–7. Sign up to view the full content.
EE161 Electromagnetic Waves Spring, 2010 Prof. Y. Ethan Wang Electrical Engineering Dept. UCLA
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Lesson 14 Numerical Methods for Electromagnetics Finite Difference Time-Domain (FDTD)Method 2-D FDTD Formulation Parallel Plate Waveguide Example
Numerical Methods Finding the electromagnetic field distribution is a 3-D space is equivalent to solve Maxwell’s equations, which can be done numerically using computers Integral equations Maxwell’s equations Differential equations + Boundary conditions Linear equations or Matrix equations Time-domain approaches: solving for the time-varying field variable Frequency-domain approaches: solving for the phasor of the field FDTD, Time-domain finite element, Time-domain integral equation Finite element method, Method of Moments, mode matching Numerical methods obtain the field value at certain spatial points, time steps or the coefficients of basis functions upon which the field is expanded
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Finite Difference Time Domain (FDTD) •Emulate the electromagnetic waves in time-domain •From the curl equations of Maxwell equations •First proposed by Yee in 1966 Didn’t receive much attention until 1980’s No need to invert a big matrix!!! The result of FDTD method is the sampled electromagnetic field values on those spatial nodes and at different time steps FDTD methods are carried out based on a discretized time-space, e.g. time steps and spatial nodes on rectangular meshes.
3-D Formulation of FDTD H E E × = + t ε σ E H × -∇ = t μ Based on the curl equations: Using central difference operator to approximate any differential operator For example, if E (N) is to represent the electric field at time step N and E (N+1) represent the field at time step N+1, the time differential operator at the time step N+1/2 would now be, ) ( 2 ) ( ) 1 ( ) 2 / 1 ( t O t E E t E N N N + - 2245 + + Time-step size Infinite precision can be obtained if the time- step and grid size approach to zero
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
3-D Formulation of FDTD - = y E z E t H z y x μ 1 The time-domain curl equations lead to: - = z E
This is the end of the preview. Sign up to access the rest of the document.
{[ snackBarMessage ]}
### Page1 / 16
161_1_Class14_No_vioice - EE161 Electromagnetic Waves...
This preview shows document pages 1 - 7. Sign up to view the full document.
View Full Document
Ask a homework question - tutors are online | 656 | 2,856 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.078125 | 3 | CC-MAIN-2018-13 | latest | en | 0.794282 |
http://www.solutioninn.com/look-ahead-at-the-scenario-described-in-prob-2773-notice | 1,508,417,415,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187823284.50/warc/CC-MAIN-20171019122155-20171019142155-00199.warc.gz | 585,136,531 | 8,249 | Look ahead at the scenario described in Prob 27 7 3 Notice
Look ahead at the scenario described in Prob. 27.7-3. Notice the steady trend upward in the number of applications over the past three years—from 4,600 to 5,300 to 6,000. Suppose now that the admissions office of Ivy College had been able to foresee this kind of trend and so had decided to use exponential smoothing with trend to do the forecasting. Suppose also that the initial estimates just over three years ago had been expected value = 3,900 and trend = 700. Then, with any values of the smoothing constants, the forecasts obtained by this forecasting method would have been exactly correct for all three years.
Illustrate this fact by doing the calculations to obtain these forecasts when the smoothing constant is α = 0.25 and the trend smoothing constant is β = 0.25. (Use hand calculations rather than an Excel template.)
Membership | 197 | 902 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.96875 | 3 | CC-MAIN-2017-43 | latest | en | 0.959087 |
https://ccssmathanswers.com/180-days-of-math-for-first-grade-day-108-answers-key/ | 1,726,402,938,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651630.14/warc/CC-MAIN-20240915120545-20240915150545-00866.warc.gz | 132,552,850 | 58,372 | # 180 Days of Math for First Grade Day 108 Answers Key
By accessing our 180 Days of Math for First Grade Answers Key Day 108 regularly, students can get better problem-solving skills.
## 180 Days of Math for First Grade Answers Key Day 108
Directions: Solve each problem.
Question 1.
Draw 5 squares. Draw a circle around the second square.
Explanation:
Drawn 5 squares and drawn a circle around the
second square as shown above.
Question 2.
7 + 10 =
17,
Explanation:
Given to add 10 to 7 we will get
7
+10
17 .
Question 3.
6,
Explanation:
Given to subtract 7 from 13 we will get
13
– 7
6 .
Question 4.
Complete the pattern.
_____ 58 59 60 61
57,
Explanation:
If we see the numbers the number are
increasing by 1, the third number 59 is
1 more than the second number 58,
the fourth number 60 is 1 more than
third number 59 and fifth number 61 is 1
more than the fourth number 60,
So the first missing number in the pattern is
58 – 1 = 57.
Question 5.
Will the solid roll?
Circle:
yes
no
Explanation:
The given solid will not roll,
so circled no as shown above.
Question 6.
How many cubes make this solid?
_________ cubes
4 cubes,
Explanation:
4 cubes make the given solid.
Question 7.
Number of Home Runs
Marc Ramon Jamai 7 4 15
Sammy hit two more home runs than Ramon. How many home runs did Sammy hit?
6 home runs,
Explanation:
Given list of number of home runs made in that
Sammy hit two more home runs than Ramon.
Ramon has hit 4 runs, So more home runs did Sammy hit
than Ramon are 4 + 2 = 6 home runs.
Question 8.
There are 7 wheels. There is 1 tricycle. The rest are wagons. How many wagons are there? | 451 | 1,630 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.5 | 4 | CC-MAIN-2024-38 | latest | en | 0.88909 |
https://stemgeeks.net/computerscience/@armoredbanana/making-a-script-to-get-updates-when-your-voting-power-is-100percent-part-04 | 1,696,113,502,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510730.6/warc/CC-MAIN-20230930213821-20231001003821-00409.warc.gz | 584,095,463 | 82,729 | # Making a script to get updates when your voting power is 100% Part 04
in #computerscience2 years ago
Hello everyone I hope you are all doing well!
Sorry about not posting for a while, I got quite busy with the school as you may know yourself.
In the last post we were able to grab a users information from https://hivestats.io/@username
By using my code you should of seen the following info displayed:
``````python output
---------------
User: armoredbanana
Vote Power: 98.00%
Time untill full: 2 hours 24 minutes
Effective HP: 21.56 HP
Actual HP: 11.5 HP
---------------
``````
As you can see we were able to get the Time Until Full in a string format. This is not very useful if we want to check back in that given time.
In today's post you should see how we convert the Full Hive Power from a string into a integer that will tell you exactly how many minutes you need to wait. This will be useful in the next tutorial where you will learn how to send an email to yourself when your hive power is full!
Lets get into it!
(before I dive right into it I want to mention I cleaned up my code a bit and made it object oriented. You can find the code at this github repo:
Converting String to Integer
This can be a daunting task, but with a few different if statements it wont be long until we have our integer. Here is the function I created.
``````python code for converting string to integer
# gets the number of minutes left for the user to have full hive power again
def get_time(self):
string = str(self.time_to_100.text)
temp = string.replace(" ", "")
self.total_minute_value = 0
if (temp.find("Full") == 0):
self.total_minute_value = 0
# Getting the number of minutes remaining based on the day value
if (temp.find("day") != -1):
day_index = temp.find("day")
day_value = int(temp[day_index-1])
minute_day_value = day_value * 1440
self.total_minute_value += minute_day_value
# Getting the number of minutes remaining based on the hour value
if (temp.find("hour") != -1):
hour_index = temp.find("hour")
if (temp[hour_index-2].isdigit()):
hour_value = int(temp[hour_index-2] + temp[hour_index-1])
minute_hour_value = hour_value * 60
self.total_minute_value += minute_hour_value
elif (temp[hour_index-1].isdigit()):
hour_value = int(temp[hour_index-1])
minute_hour_value = hour_value * 60
self.total_minute_value += minute_hour_value
# Getting the number of minutes remaining based on the minute value
if (temp.find("minute") != -1):
minute_index = temp.find("minute")
if (temp[minute_index-2].isdigit()):
minute_value = int(temp[minute_index-2] + temp[minute_index-1])
self.total_minute_value += minute_value
elif (temp[minute_index-1].isdigit()):
minute_value = int(temp[minute_index-1])
self.total_minute_value += minute_value
return self.total_minute_value
``````
Now this looks like a lot so let me break it down for you.
1. First we need to load our string (time_to_100) into a string.
2. Now this string contains spaces, which we can replace with a great function in python. "string_variable.replace(" ", ""). Using this function we replace every space with a no space.
3. Next we check if its full. In python you can find a list of strings (aka words) by using the string_variable.find() function. If our hive power is full, on hive stats it displays it as "Full" so we search for that word. IF that word is present the .find("Full") function returns the index of the first letter in our string which will be 0.
Thats why we check if temp.find("Full") == 0. If so know we are full and we make the total_minute_value = 0
4. IF its not full we need to keep on checking. IF the word we are searching for does not exist in our string, then the .find() function will return a -1 value. So IF temp.find("day") != -1 (temp.find("day") IS NOT EQUAL to -1) then we know the word day is in our string. We are able to easily convert the string into a integer day_value = int(temp[day_index-1]) which we then must convert to minutes. 1 day has 1440 minutes which is an easy multiplication.
5. We must keep doing this for hours and minutes as well. The tricky part with hours and minutes is that it can be more than one integer. We can get around this by checking if there are two numbers by using the .isdigit() function. IF the index of our string is a digit, the .isdigit() function will return TRUE. IF its not a digit, it will return FALSE. Through this we can establish how many digits we are dealing with and accommodate our if statements as seen in the code below:
``````python code for converting hours into minutes
# Getting the number of minutes remaining based on the hour value
if (temp.find("hour") != -1):
hour_index = temp.find("hour")
if (temp[hour_index-2].isdigit()):
hour_value = int(temp[hour_index-2] + temp[hour_index-1])
minute_hour_value = hour_value * 60
self.total_minute_value += minute_hour_value
elif (temp[hour_index-1].isdigit()):
hour_value = int(temp[hour_index-1])
minute_hour_value = hour_value * 60
self.total_minute_value += minute_hour_value
``````
At the end of all of this, we load all of our minutes into a total_minute_value variable to easily know how many minutes we have to wait until our hive power is full.
Now when you call the function wait_time = object.get_time() it will return an integer value that represents the amount of minutes we need to wait until our hive power is full again. As I mentioned earlier this can be very helpful when we will send ourselves an email to go like again!
``````python code
import updater_mod
import multiprocessing
import time
import schedule
import smtplib, ssl
#100%_hivepower_0%
# object armoredbanana
user1.get_user_data()
check_backc_time_user_1 = user1.get_time()
print(user1)
user2.get_user_data()
user2.get_time()
print(user2)
# object bhanutejap
user3.get_user_data()
user3.get_time()
print(user3)
``````
updater_mod is the code I posted above at the very beginning. This is our class that we create our objects from. If you are not familiar with object oriented programming, I strongly recommend that you learn about that first before anything else as it makes replicating users in this case really easy!
In the code I created 3 different objects called user1, user2, user3 to show the simplicity of getting multiple users information using object oriented programming or OOP for short.
``````python output of the code above
---------------
User: armoredbanana
Vote Power: 98.36%
Time untill full: 1 hour 58 minutes
Effective HP: 21.56 HP
Actual HP: 11.5 HP
Full Hive Power in: 118 minutes
---------------
---------------
Vote Power: 94.59%
Time untill full: 6 hours 29 minutes
Effective HP: 104,775.67 HP
Actual HP: 5,496.05 HP
Full Hive Power in: 389 minutes
---------------
---------------
User: bhanutejap
Vote Power: 60.34%
Time untill full: 1 day 23 hours 35 minutes
Effective HP: 0 HP
Actual HP: 0 HP
Full Hive Power in: 2855 minutes
---------------
``````
Here you can see @armoredbana, @trostparadox,and @bhanutejap
info. It displays it to us now nicely in minutes which again is very useful for our next tutorial.
Troubles
I had some issues with getting the gecodriver in my path, but it was easily solved by the following line of code in the updater_mod.py
``````python code in updater_mod.py
self.driver = webdriver.Firefox(executable_path = r'C:\Users\thoma\Documents\Blockchain\webscraping\geckodriver.exe', options=options) #this is the path where my geckodriver exists
``````
I also had quite a few errors when converting hours and days because sometimes there are 12 hours remaining but other times there are only 8 hours remaining. That one singe digit makes a big difference.
Sort:
Hey everyone thanks for reading my post! I forgot to put in the github repo link so here it is: https://github.com/Thomas-J-Kidd/Hive-Power-Checker.git
Thank you again to everyone!
I enjoy learning from you! Keep posting and sharing. The Blockchain is such a great tool.
Thanks a lot! And yes its a great place! Check out my next post coming later today! | 1,994 | 8,009 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2023-40 | latest | en | 0.826709 |
https://www.coursehero.com/file/230892/hw7sol/ | 1,519,355,718,000,000,000 | text/html | crawl-data/CC-MAIN-2018-09/segments/1518891814311.76/warc/CC-MAIN-20180223015726-20180223035726-00717.warc.gz | 839,592,287 | 26,442 | {[ promptMessage ]}
Bookmark it
{[ promptMessage ]}
hw7sol
# hw7sol - (~ 5 x(Has(Joe,x ^ Dog(x => Scary(Joe d Either...
This preview shows page 1. Sign up to view the full content.
CS4600 - Introduction to Intelligent Systems Homework 6 - Sample Solution Write the following sentences in first-order logic using these literals: Has(Joe,x), Dog(x), Cat(x), Cute(x), Scary(x). a) Joe has a cute dog. 5 x (Has(Joe,x) ^ Dog(x) ^ Cute(x)) b) All of Joe’s dogs are cute. 2200 x ((Has(Joe,x) ^ Dog(x)) => Cute(x)) c) Unless Joe has a dog, he is scary.
This is the end of the preview. Sign up to access the rest of the document.
Unformatted text preview: (~ 5 x (Has(Joe,x) ^ Dog(x))) => Scary(Joe) d) Either Joe has at least one cat and at least one dog or he is scary (but not both at the same time). 5 x,y (Has(Joe,x) ^ Has(Joe,y) ^ Dog(x) ^ Cat(y)) <=> ~Scary(Joe) e) Not all dogs are both scary and cute. ~ 2200 x (Dog(x) => (Scary(x) ^ Cute(x)))...
View Full Document
{[ snackBarMessage ]}
Ask a homework question - tutors are online | 322 | 1,039 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.109375 | 3 | CC-MAIN-2018-09 | latest | en | 0.69675 |
https://gamedev.stackexchange.com/questions/20934/how-to-create-adjustable-formula-for-rpg-level-up-requirements/55458#55458 | 1,638,836,690,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964363327.64/warc/CC-MAIN-20211206224536-20211207014536-00543.warc.gz | 324,089,576 | 37,479 | # How to create adjustable formula for RPG level up requirements?
I'm trying to create a formula that can be modified simply by changing two values: number_of_levels, and last_level_experience. This is to enable people modding the game to change the levelling requirements.
I've got it so that I can specify the number of XP needed for the last level up, but I want to be able to control the XP needed for the first level up, which in this case can differ wildly. For example, if I have 40 levels, and 1,000,000 XP for the last level, the first level up requirement is then 625. But if I change the levels to 80, the first level up becomes 156. In both cases, the last level needs 1,000,000.
There must be some way to get the computer to work out a suitable curve given just these two basic values.
#include <iostream>
int main()
{
int levels = 40;
if (levels < 2) levels = 2;
int experience_for_last_level = 1e6;
float fraction = 1.0 / levels;
{
int i = 0;
float fraction_counter = fraction;
int counter = levels;
int total = 0;
for (i = 1; i <= levels; ++i, fraction_counter += fraction, --counter)
{
int a = static_cast<int>(fraction_counter * experience_for_last_level / counter);
std::cout <<"Level "<<i<<": "<<a<<" ("<<counter<<")"<<"\n";
total += a;
}
std::cout << "\nTotal Exp: " << total;
}
}
Output:
Level 1: 625 (40) Level 15: 14423 (26) Level 29: 60416 (12)
Level 2: 1282 (39) Level 16: 16000 (25) Level 30: 68181 (11)
Level 3: 1973 (38) Level 17: 17708 (24) Level 31: 77499 (10)
Level 4: 2702 (37) Level 18: 19565 (23) Level 32: 88888 (9)
Level 5: 3472 (36) Level 19: 21590 (22) Level 33: 103124 (8)
Level 6: 4285 (35) Level 20: 23809 (21) Level 34: 121428 (7)
Level 7: 5147 (34) Level 21: 26250 (20) Level 35: 145833 (6)
Level 8: 6060 (33) Level 22: 28947 (19) Level 36: 179999 (5)
Level 9: 7031 (32) Level 23: 31944 (18) Level 37: 231249 (4)
Level 10: 8064 (31) Level 24: 35294 (17) Level 38: 316666 (3)
Level 11: 9166 (30) Level 25: 39062 (16) Level 39: 487499 (2)
Level 12: 10344 (29) Level 26: 43333 (15) Level 40: 999999 (1)
Level 13: 11607 (28) Level 27: 48214 (14)
Level 14: 12962 (27) Level 28: 53846 (13)
• The fundamental problem is that there are infinitely many XP level curves that would end with the last level requiring that much XP. You have not constrained the dimensions of the problem, because you have not stated how you want the XP to change from level to level. Do you want an exponential growth curve? A parabolic growth curve? A linear one? Your problem is unsolvable in its current state. Personally, if I were modding the game, I'd want more control over the XP curve than just last level number and last level XP. I'd want to control the actual curve itself. Dec 12 '11 at 8:02
• I can allow modders to control levelling via a script. Dec 12 '11 at 8:23
Though there are infinitely many ways to choose them, it is common for leveling curves to follow a power rule such as the following one:
f(level) == A * exp(B * level)
The major advantage of this formula can be easily explained: for a given rule, there is a fixed value N such that each level costs N percent more than the previous one.
f(1) - f(0) == experience_for_first_level
f(levels) - f(levels - 1) == experience_for_last_level
Two equations, two unknowns. This looks good. Simple maths give A and B:
B = log(experience_for_last_level / experience_for_first_level) / (levels - 1);
A = experience_for_first_level / (exp(B) - 1);
Resulting in the following code:
#include <cmath>
#include <iostream>
int main(void)
{
int levels = 40;
int xp_for_first_level = 1000;
int xp_for_last_level = 1000000;
double B = log((double)xp_for_last_level / xp_for_first_level) / (levels - 1);
double A = (double)xp_for_first_level / (exp(B) - 1.0);
for (int i = 1; i <= levels; i++)
{
int old_xp = round(A * exp(B * (i - 1)));
int new_xp = round(A * exp(B * i));
std::cout << i << " " << (new_xp - old_xp) << std::endl;
}
}
And the following output:
1 1000 9 4125 17 17012 25 70170 33 289427
2 1193 10 4924 18 20309 26 83768 34 345511
3 1425 11 5878 19 24245 27 100000 35 412462
4 1702 12 7017 20 28943 28 119378 36 492389
5 2031 13 8377 21 34551 29 142510 37 587801
6 2424 14 10000 22 41246 30 170125 38 701704
7 2894 15 11938 23 49239 31 203092 39 837678
8 3455 16 14251 24 58780 32 242446 40 1000000
• If only all answers were this well planned and thought out.
– Nate
Dec 12 '11 at 23:07
• The curve here is much more palatable. Dec 22 '11 at 22:26
• Good answer. This may be a stupid question, but how do you calculate N that you described above? What if you wanted to make N the pluggable variable? Let me know if I should ask a separate question for this. Sep 5 '13 at 17:30
• @tieTYT the relationship between N and B is exp(B) = 1 + N, or B = log(1 + N). So if you want each level to require e.g. 15% more than the previous one, you’ll need B = log(1 + 0.15) = 0.13976. Sep 6 '13 at 9:23
Don't forget to round the numbers after you figured out your curve. It doesn't make much sense to tell the player he needs 119,378 experience points to reach the next level — because the person would always understand it as "roughly 120,000". Thus you will be better off doing the rounding yourself, and presenting "clean" results to your players. For example the following code (which extends upon the Sam Hocevar's) will attempt to round up to ≈2.2 significant digits (obviously that constant can be tweaked as you want):
from math import exp, log
levels = 40
xp_for_first_level = 1000
xp_for_last_level = 1000000
B = log(1.0 * xp_for_last_level / xp_for_first_level) / (levels - 1)
A = 1.0 * xp_for_first_level / (exp(B) - 1.0)
def xp_for_level(i):
x = int(A * exp(B * i))
y = 10**int(log(x) / log(10) - 2.2)
return int(x / y) * y
for i in range(1, levels+1):
print( "%d: %d" % (i, xp_for_level(i) - xp_for_level(i-1)) )
The output is:
1: 1000 9: 4200 17: 17100 25: 70000 33: 287000
2: 1190 10: 4900 18: 20300 26: 84000 34: 340000
3: 1420 11: 5900 19: 24200 27: 100000 35: 420000
4: 1710 12: 7000 20: 28700 28: 119000 36: 490000
5: 2030 13: 8400 21: 34000 29: 142000 37: 590000
6: 2420 14: 10000 22: 42000 30: 171000 38: 700000
7: 2870 15: 11900 23: 49000 31: 203000 39: 840000
8: 3400 16: 14200 24: 59000 32: 242000 40: 1000000 | 2,301 | 6,807 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.671875 | 3 | CC-MAIN-2021-49 | latest | en | 0.725741 |
https://prewar.mgcc.info/d-type/?rdp_we_resource=https%3A%2F%2Fen.wikipedia.org%2Fwiki%2FElectrical_reactance | 1,657,056,905,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656104628307.87/warc/CC-MAIN-20220705205356-20220705235356-00207.warc.gz | 503,554,302 | 41,476 | In electrical circuits, reactance is the opposition presented to alternating current by inductance or capacitance.[1] Greater reactance gives smaller current for the same applied voltage. Reactance is similar to resistance in this respect, but differs in that reactance does not lead to dissipation of electrical energy as heat. Instead, energy is momentarily stored in the reactance, and a quarter-cycle later returned to the circuit, whereas a resistance continuously loses energy.
Reactance is used to compute amplitude and phase changes of sinusoidal alternating current (AC) going through a circuit element. Like resistance, reactance is measured in ohms, with positive values indicating inductive reactance and negative indicating capacitive reactance. It is denoted by the symbol . An ideal resistor has zero reactance, whereas ideal inductors and capacitors have zero resistance. As frequency increases, inductive reactance increases and capacitive reactance decreases.
Comparison to resistance
Reactance is similar to resistance in that larger reactance leads to smaller currents for the same applied voltage. Further, a circuit made entirely of elements that have only reactance (and no resistance) can be treated the same way as a circuit made entirely of resistances. These same techniques can also be used to combine elements with reactance with elements with resistance but complex numbers are typically needed. This is treated below in the section on impedance.
There are several important differences between reactance and resistance, though. First, reactance changes the phase so that the current through the element is shifted by a quarter of a cycle relative to the phase of the voltage applied across the element. Second, power is not dissipated in a purely reactive element but is stored instead. Third, reactances can be negative so that they can 'cancel' each other out. Finally, the main circuit elements that have reactance (capacitors and inductors) have a frequency dependent reactance, unlike resistors which have the same resistance for all frequencies, at least in the ideal case.
The term reactance was first suggested by French engineer M. Hospitalier in L'Industrie Electrique on 10 May 1893. It was officially adopted by the American Institute of Electrical Engineers in May 1894.[2]
Capacitive reactance
A capacitor consists of two conductors separated by an insulator, also known as a dielectric.
Capacitive reactance is an opposition to the change of voltage across an element. Capacitive reactance is inversely proportional to the signal frequency (or angular frequency ) and the capacitance .[3]
There are two choices in the literature for defining reactance for a capacitor. One is to use a uniform notion of reactance as the imaginary part of impedance, in which case the reactance of a capacitor is the negative number,[3][4][5]
.
Another choice is to define capacitive reactance as a positive number,[6][7][8]
.
In this case however one needs to remember to add a negative sign for the impedance of a capacitor, i.e. .
At , the magnitude of the capacitor's reactance is infinite, behaving like an open circuit (preventing any current from flowing through the dielectric). As frequency increases, the magnitude of reactance decreases, allowing more current to flow. As approaches , the capacitor's reactance approaches , behaving like a short circuit.
The application of a DC voltage across a capacitor causes positive charge to accumulate on one side and negative charge to accumulate on the other side; the electric field due to the accumulated charge is the source of the opposition to the current. When the potential associated with the charge exactly balances the applied voltage, the current goes to zero.
Driven by an AC supply (ideal AC current source), a capacitor will only accumulate a limited amount of charge before the potential difference changes polarity and the charge is returned to the source. The higher the frequency, the less charge will accumulate and the smaller the opposition to the current.
Inductive reactance
Inductive reactance is a property exhibited by an inductor, and inductive reactance exists based on the fact that an electric current produces a magnetic field around it. In the context of an AC circuit (although this concept applies any time current is changing), this magnetic field is constantly changing as a result of current that oscillates back and forth. It is this change in magnetic field that induces another electric current to flow in the same wire (counter-EMF), in a direction such as to oppose the flow of the current originally responsible for producing the magnetic field (known as Lenz's Law). Hence, inductive reactance is an opposition to the change of current through an element.
For an ideal inductor in an AC circuit, the inhibitive effect on change in current flow results in a delay, or a phase shift, of the alternating current with respect to alternating voltage. Specifically, an ideal inductor (with no resistance) will cause the current to lag the voltage by a quarter cycle, or 90°.
In electric power systems, inductive reactance (and capacitive reactance, however inductive reactance is more common) can limit the power capacity of an AC transmission line, because power is not completely transferred when voltage and current are out-of-phase (detailed above). That is, current will flow for an out-of-phase system, however real power at certain times will not be transferred, because there will be points during which instantaneous current is positive while instantaneous voltage is negative, or vice versa, implying negative power transfer. Hence, real work is not performed when power transfer is "negative". However, current still flows even when a system is out-of-phase, which causes transmission lines to heat up due to current flow. Consequently, transmission lines can only heat up so much (or else they would physically sag too much, due to the heat expanding the metal transmission lines), so transmission line operators have a "ceiling" on the amount of current that can flow through a given line, and excessive inductive reactance can limit the power capacity of a line. Power providers utilize capacitors to shift the phase and minimize the losses, based on usage patterns.
Inductive reactance is proportional to the sinusoidal signal frequency and the inductance , which depends on the physical shape of the inductor:
.
The average current flowing through an inductance in series with a sinusoidal AC voltage source of RMS amplitude and frequency is equal to:
Because a square wave has multiple amplitudes at sinusoidal harmonics, the average current flowing through an inductance in series with a square wave AC voltage source of RMS amplitude and frequency is equal to:
making it appear as if the inductive reactance to a square wave was about 19% smaller than the reactance to the AC sine wave.
Any conductor of finite dimensions has inductance; the inductance is made larger by the multiple turns in an electromagnetic coil. Faraday's law of electromagnetic induction gives the counter-emf (voltage opposing current) due to a rate-of-change of magnetic flux density through a current loop.
For an inductor consisting of a coil with loops this gives:
.
The counter-emf is the source of the opposition to current flow. A constant direct current has a zero rate-of-change, and sees an inductor as a short-circuit (it is typically made from a material with a low resistivity). An alternating current has a time-averaged rate-of-change that is proportional to frequency, this causes the increase in inductive reactance with frequency.
Impedance
Both reactance and resistance are components of impedance .
where:
• is the complex impedance, measured in ohms;
• is the resistance, measured in ohms. It is the real part of the impedance:
• is the reactance, measured in ohms. It is the imaginary part of the impedance:
• is the square root of minus one, usually represented by in non-electrical formulas. is used so as not to confuse the imaginary unit with current, commonly represented by .
When both a capacitor and an inductor are placed in series in a circuit, their contributions to the total circuit impedance are opposite. Capacitive reactance and inductive reactance contribute to the total reactance as follows:
where:
• is the inductive reactance, measured in ohms;
• is the capacitive reactance, measured in ohms;
• is the angular frequency, times the frequency in Hz.
Hence:[5]
• if , the total reactance is said to be inductive;
• if , then the impedance is purely resistive;
• if , the total reactance is said to be capacitive.
Note however that if and are assumed both positive by definition, then the intermediary formula changes to a difference:[7]
but the ultimate value is the same.
Phase relationship
The phase of the voltage across a purely reactive device (i.e. with zero parasitic resistance) lags the current by radians for a capacitive reactance and leads the current by radians for an inductive reactance. Without knowledge of both the resistance and reactance the relationship between voltage and current cannot be determined.
The origin of the different signs for capacitive and inductive reactance is the phase factor in the impedance.
For a reactive component the sinusoidal voltage across the component is in quadrature (a phase difference) with the sinusoidal current through the component. The component alternately absorbs energy from the circuit and then returns energy to the circuit, thus a pure reactance does not dissipate power. | 1,934 | 9,637 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.875 | 3 | CC-MAIN-2022-27 | latest | en | 0.938782 |
https://brainiak.in/521/r-perform-operations-971-107-98-104-05-10-7-10-c-4-11-10-3-8-1-10-4-d-2-12-106-3-5-105 | 1,670,248,661,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446711017.45/warc/CC-MAIN-20221205132617-20221205162617-00477.warc.gz | 173,553,849 | 9,386 | # R. Perform the following operations : a. 3.971 × 107 + 1.98 × 104 ; b. 1.05 × 10-4 - 9.7 × 10-5; c. 4.11 × 10-3 + 8.1×10-4; d. 2.12 × 106 - 3.5 × 105
more_vert
4) Solve the following problems
R. Perform the following operations :
a. 3.971 × 107 + 1.98 × 104 ;
b. 1.05 × 10-4 - 9.7 × 10-5;
c. 4.11 × 10-3 + 8.1×10-4;
d. 2.12 × 106 - 3.5 × 105
more_vert
verified
A) 3.971 × 107 + 1.98 × 104
= 3.971 × 107 + 0.00198 × 107
= (3.971 + 0.00198) × 107
= 3.97298 × 107
B) 1.05 × 10−4 − 9.7 × 10−5
= 10.5 × 10−5 − 9.7 × 10−5
= (10.5 − 9.7) × 10−5
= 0.80 × 10−5
= 8.0 × 10−6
C) 4.11 × 10−3 + 8.1 ×10−4
= 41.1 × 10−4 + 8.1 × 10−4
= (41.1 + 8.1) × 10−4
= 49.2 × 10−4
= 4.92 × 10−3
D) 2.12 × 106 − 3.5 × 105
= 21.2 × 105 − 3.5 × 105
= (21.2 − 3.5) × 105
= 17.7 × 105
= 1.77 × 106 | 493 | 788 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.90625 | 3 | CC-MAIN-2022-49 | latest | en | 0.44124 |
https://www.experts-exchange.com/questions/25009355/Networkdays-Alternative-with-Time.html | 1,529,490,831,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267863516.21/warc/CC-MAIN-20180620085406-20180620105406-00111.warc.gz | 797,606,607 | 17,280 | # Networkdays Alternative, with Time
Merry Christmas All!
I'm trying to track down an alternative to the NETWORKDAYS formula because for some bizarre reason we aren't allowed to install the Analysis Toolpak at work.
I've got this formula, which works pretty good....
=SUMPRODUCT(--(COUNTIF(HolidayList,ROW(INDIRECT(INT(E3)&":"&TODAY())))=0), --(WEEKDAY(ROW(INDIRECT(INT(E3)&":"&TODAY())),3)<5))
...but it only counts whole days. What I'm after is something that takes the time into account as well. Any ideas?
Cheers,
Wayne
LVL 48
###### Who is Participating?
Commented:
Well, I think you can just use your existing formula with an adjustment for the times. That formula is effectively counting from 12:00 AM on the date in E3 up to the end of today so you can just subtract those times, i.e.
=SUMPRODUCT(--(COUNTIF(HolidayList,ROW(INDIRECT(INT(E3)&":"&TODAY())))=0),--(WEEKDAY(ROW(INDIRECT(INT(E3)&":"&TODAY())),3)<5))-MOD(E3,1)+MOD(NOW(),1)-1
format result cell as [h]:mm
That works assuming today isn't a weekend or holiday. If you want to allow that possibility change to
=SUMPRODUCT(--(COUNTIF(HolidayList,ROW(INDIRECT(INT(E3)&":"&TODAY())))=0),--(WEEKDAY(ROW(INDIRECT(INT(E3)&":"&TODAY())),3)<5))-MOD(E3,1)+(MOD(NOW(),1)-1)*(WEEKDAY(TODAY(),3)<5)*(COUNTIF(HolidayList,TODAY())=0)
regards, barry
0
Commented:
Hello Wayne,
So given that formula I take it you need to account for holidays? Are you counting all hours of the working days or do you just want to count between specific times?
You have start date and time in E3 I presume so you are taking that time into account but not the end time (just TODAY()?)
Could the start time/date be on a holiday or weekend?
regards, barry
0
Author Commented:
Hi Barry,
Yes, I need to account for holidays. I basically need to discount the entire 24 hours of any holiday or weekend days.
I had the TODAY() function in the formula because it will fail if I use NOW() because it needs an integer, but I will need to use NOW() in the final formula.
The start time will never be on a weekend or holiday.
Cheers,
Wayne
0
Author Commented:
Excellent! Thanks Barry!
0
Question has a verified solution.
Are you are experiencing a similar issue? Get a personalized answer when you ask a related question.
Have a better answer? Share it in a comment. | 624 | 2,303 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3 | 3 | CC-MAIN-2018-26 | latest | en | 0.881027 |
http://slideplayer.com/slide/4828508/ | 1,566,063,228,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027313436.2/warc/CC-MAIN-20190817164742-20190817190742-00373.warc.gz | 174,843,383 | 20,135 | The University of Texas at Austin, CS 395T, Spring 2008, Prof. William H. Press 1 Computational Statistics with Application to Bioinformatics Prof. William.
Presentation on theme: "The University of Texas at Austin, CS 395T, Spring 2008, Prof. William H. Press 1 Computational Statistics with Application to Bioinformatics Prof. William."— Presentation transcript:
The University of Texas at Austin, CS 395T, Spring 2008, Prof. William H. Press 1 Computational Statistics with Application to Bioinformatics Prof. William H. Press Spring Term, 2008 The University of Texas at Austin Unit 17: Classifier Performance: ROC, Precision-Recall, and All That
The University of Texas at Austin, CS 395T, Spring 2008, Prof. William H. Press 2 Unit 17: Classifier Performance: ROC, Precision-Recall, and All That (Summary) The performance of a classifier is a 2x2 contingency table –the “confusion matrix” of TP, FP, FN, TN Most classifiers can be varied from “conservative” to “liberal” –call a larger number of TPs at the expense of also a larger number of FPs –it’s a one-parameter curve so one classifier might dominate another or there might be no clear ordering between them There is a thicket of terminology –TPR, FPR, PPV, NPV, FDR, accuracy –sensitivity, specificity –precision, recall ROC plots TPR (y-axis) as a function of FPR (x-axis) –0,0 monotonically to 1,1 –in practice, convex because can be trivially upgraded to its convex hull –but can be misleading when the numbers of actual P’s and N’s are very different Precision-Recall plots are designed to be useful in just that case –can go back and forth between Precision-Recall and ROC curves –if a classifier dominates in one, it dominates in the other PPV vs. NPV is another equivalent performance measure
The University of Texas at Austin, CS 395T, Spring 2008, Prof. William H. Press 3 A (binary) classifier classifies data points as + or − If we also know the true classification, the performance of the classifier is a 2x2 contingency table, in this application usually called a confusion matrix. good! bad! (Type I error) bad! (Type II error) As we saw, this kind of table has many other uses: treatment vs. outcome, clinical test vs. diagnosis, etc.
The University of Texas at Austin, CS 395T, Spring 2008, Prof. William H. Press 4 Most classifiers have a “knob” or threshold that you can adjust: How certain do they have to be before they classify a “+”? To get more TP’s, you have to let in some FP’s! Notice there is just one free parameter, think of it as TP, since FP(TP) = [given by algorithm] TP + FN = P (fixed number of actual positives, column marginal) FP + TN = N (fixed number of actual negatives, column marginal) So all scalar measures of performance are functions of one free parameter (i.e., curves). And the points on any such curve are in 1-to-1 correspondence with those on any other such curve. If you ranked some classifiers by how good they are, you might get a different rankings at different points on the scale. On the other hand, one classifier might dominate another at all points on the scale. more conservative more liberal TP FP FN TN Cartoon, not literal:
The University of Texas at Austin, CS 395T, Spring 2008, Prof. William H. Press 5 Terminology used to measure the performance of classifiers Different combinations of ratios have been given various names. All vary between 0 and 1. A performance curve picks one as the independent variable and looks at another as the dependent variable. Dark color is numerator, dark and light color is denominator. Blue parameters: 1 is good. Red: 0 is good. “one minus”
The University of Texas at Austin, CS 395T, Spring 2008, Prof. William H. Press 6 ROC (“Receiver Operating Characteristic”) curves plot TPR vs. FPR as the classifier goes from “conservative” to “liberal” blue dominates red and green neither red nor green dominate the other You could get the best of the red and green curves by making a hybrid or “Frankenstein” classifier that switches between strategies at the cross-over points.
The University of Texas at Austin, CS 395T, Spring 2008, Prof. William H. Press 7 ROC curves can always be “upgraded” to their convex hull by replacing any concave portions by a random sample List points classified as + by B but not A. Start up the curve to A. When you reach A, start adding a fraction of them (increasing from 0 to 1) randomly, until you reach B. Continue on the curve from B. Using data with known “ground truth” answers, you can find what “knob” settings correspond to A and B. Then you can apply the convex classifier to cases where you don’t know the answers.
The University of Texas at Austin, CS 395T, Spring 2008, Prof. William H. Press 8 Since ROC curves don’t explicitly show any dependence on the constant P/N (ratio of actual + to – in the sample) they can be misleading if you care about FP versus TP lam = (0:0.01:1); fpr =.05 + 0.2 * lam; tpr = 1 - (.05 + 0.2*(1-lam)); fpr(1) = 0; fpr(end) = 1; tpr(1) = 0; tpr(end) = 1; plot(fpr,tpr) Suppose you have a test for Alzheimer’s whose false positive rate can be varied from 5% to 25% as the false negative rate varies from 25% to 5% (suppose linear dependences on both): FPR = 0.15, TPR=0.85 Now suppose you try the test on a population of 10,000 people, 1% of whom actually are Alzheimer’s positive: FP swamps TP by ~17:1. You’ll be telling 17 people that they might have Alzheimer’s for every one who actually does. It is unlikely that your test will be used. In a case like this, ROC, while correct, somewhat misses the point.
The University of Texas at Austin, CS 395T, Spring 2008, Prof. William H. Press 9 Precision-Recall curves overcome this issue by comparing TP with FN and FP prec = tpr*100./(tpr*100+fpr*9900); prec(1) = prec(2); % fix up 0/0 reca = tpr; plot(reca,prec) Continue our toy example: note that P and N now enter never better than ~0.13 0.01 By the way, this shape “cliff” is what the ROC convexity constraint looks like in a Precision-Recall plot. It’s not very intuitive.
The University of Texas at Austin, CS 395T, Spring 2008, Prof. William H. Press 10 For fixed marginals P,N the points on the ROC curve are in 1-to-1 correspondence with the points on the Precision-Recall curve. That is, both display the same information. You can go back and forth. pre, rec from TPR, FPR TPR, FPR from pre, rec pre = TPRP TPRP + FPRN rec = TPR rec ( 1 ¡ pre ) pre P N = FPR It immediately follows that if one curve dominates another in ROC space, it also dominates in Precision-Recall space. (Because a crossing in one implies a crossing in the other, by the above equations.) But for curves that cross, the metrics in one space don’t easily map to the other. For example, people sometimes use “area under the ROC curve”. This doesn’t correspond to “area under the Precision-Recall curve”, or to anything simple.
The University of Texas at Austin, CS 395T, Spring 2008, Prof. William H. Press 11 One also sees used PPV and NPV (more often as a pair of numbers than as a curve) PPV: given a positive test, how often does the patient have the disease. NPV: given a negative test, how often is the patient disease-free. PPV = 0.054 NPV = 0.998
The University of Texas at Austin, CS 395T, Spring 2008, Prof. William H. Press 12 It’s easy to get from PPV,NPV to ROC or vice versa. Or, for that matter, any other of the parameterizations. In Mathematica, for example:
Similar presentations | 1,897 | 7,429 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.640625 | 3 | CC-MAIN-2019-35 | latest | en | 0.831013 |
https://www.aqua-calc.com/convert/density/pennyweight-per-us-gallon-to-pound-per-cubic-foot | 1,563,563,040,000,000,000 | text/html | crawl-data/CC-MAIN-2019-30/segments/1563195526337.45/warc/CC-MAIN-20190719182214-20190719204214-00462.warc.gz | 621,315,470 | 9,839 | # Convert pennyweights per gallon to pounds per (cubic foot)
## dwt/gal to lb/ft³ (dwt:pennyweight, gal:gallon, lb:pound, ft:foot)
### pennyweights per gallon equal to pounds per cubic foot conversion results
• 1
through
20
pennyweights per gallon
• 1 dwt/gal = 0.025647495 lb/ft³
• 2 dwt/gal = 0.051294991 lb/ft³
• 3 dwt/gal = 0.076942486 lb/ft³
• 4 dwt/gal = 0.102589981 lb/ft³
• 5 dwt/gal = 0.128237477 lb/ft³
• 6 dwt/gal = 0.153884972 lb/ft³
• 7 dwt/gal = 0.179532467 lb/ft³
• 8 dwt/gal = 0.205179963 lb/ft³
• 9 dwt/gal = 0.230827458 lb/ft³
• 10 dwt/gal = 0.256474953 lb/ft³
• 11 dwt/gal = 0.282122449 lb/ft³
• 12 dwt/gal = 0.307769944 lb/ft³
• 13 dwt/gal = 0.333417439 lb/ft³
• 14 dwt/gal = 0.359064935 lb/ft³
• 15 dwt/gal = 0.38471243 lb/ft³
• 16 dwt/gal = 0.410359925 lb/ft³
• 17 dwt/gal = 0.436007421 lb/ft³
• 18 dwt/gal = 0.461654916 lb/ft³
• 19 dwt/gal = 0.487302411 lb/ft³
• 20 dwt/gal = 0.512949907 lb/ft³
• 21
through
40
pennyweights per gallon
• 21 dwt/gal = 0.538597402 lb/ft³
• 22 dwt/gal = 0.564244897 lb/ft³
• 23 dwt/gal = 0.589892393 lb/ft³
• 24 dwt/gal = 0.615539888 lb/ft³
• 25 dwt/gal = 0.641187384 lb/ft³
• 26 dwt/gal = 0.666834879 lb/ft³
• 27 dwt/gal = 0.692482374 lb/ft³
• 28 dwt/gal = 0.71812987 lb/ft³
• 29 dwt/gal = 0.743777365 lb/ft³
• 30 dwt/gal = 0.76942486 lb/ft³
• 31 dwt/gal = 0.795072356 lb/ft³
• 32 dwt/gal = 0.820719851 lb/ft³
• 33 dwt/gal = 0.846367346 lb/ft³
• 34 dwt/gal = 0.872014842 lb/ft³
• 35 dwt/gal = 0.897662337 lb/ft³
• 36 dwt/gal = 0.923309832 lb/ft³
• 37 dwt/gal = 0.948957328 lb/ft³
• 38 dwt/gal = 0.974604823 lb/ft³
• 39 dwt/gal = 1.000252318 lb/ft³
• 40 dwt/gal = 1.025899814 lb/ft³
• 41
through
60
pennyweights per gallon
• 41 dwt/gal = 1.051547309 lb/ft³
• 42 dwt/gal = 1.077194804 lb/ft³
• 43 dwt/gal = 1.1028423 lb/ft³
• 44 dwt/gal = 1.128489795 lb/ft³
• 45 dwt/gal = 1.15413729 lb/ft³
• 46 dwt/gal = 1.179784786 lb/ft³
• 47 dwt/gal = 1.205432281 lb/ft³
• 48 dwt/gal = 1.231079776 lb/ft³
• 49 dwt/gal = 1.256727272 lb/ft³
• 50 dwt/gal = 1.282374767 lb/ft³
• 51 dwt/gal = 1.308022262 lb/ft³
• 52 dwt/gal = 1.333669758 lb/ft³
• 53 dwt/gal = 1.359317253 lb/ft³
• 54 dwt/gal = 1.384964748 lb/ft³
• 55 dwt/gal = 1.410612244 lb/ft³
• 56 dwt/gal = 1.436259739 lb/ft³
• 57 dwt/gal = 1.461907234 lb/ft³
• 58 dwt/gal = 1.48755473 lb/ft³
• 59 dwt/gal = 1.513202225 lb/ft³
• 60 dwt/gal = 1.53884972 lb/ft³
• 61
through
80
pennyweights per gallon
• 61 dwt/gal = 1.564497216 lb/ft³
• 62 dwt/gal = 1.590144711 lb/ft³
• 63 dwt/gal = 1.615792206 lb/ft³
• 64 dwt/gal = 1.641439702 lb/ft³
• 65 dwt/gal = 1.667087197 lb/ft³
• 66 dwt/gal = 1.692734692 lb/ft³
• 67 dwt/gal = 1.718382188 lb/ft³
• 68 dwt/gal = 1.744029683 lb/ft³
• 69 dwt/gal = 1.769677178 lb/ft³
• 70 dwt/gal = 1.795324674 lb/ft³
• 71 dwt/gal = 1.820972169 lb/ft³
• 72 dwt/gal = 1.846619664 lb/ft³
• 73 dwt/gal = 1.87226716 lb/ft³
• 74 dwt/gal = 1.897914655 lb/ft³
• 75 dwt/gal = 1.923562151 lb/ft³
• 76 dwt/gal = 1.949209646 lb/ft³
• 77 dwt/gal = 1.974857141 lb/ft³
• 78 dwt/gal = 2.000504637 lb/ft³
• 79 dwt/gal = 2.026152132 lb/ft³
• 80 dwt/gal = 2.051799627 lb/ft³
• 81
through
100
pennyweights per gallon
• 81 dwt/gal = 2.077447123 lb/ft³
• 82 dwt/gal = 2.103094618 lb/ft³
• 83 dwt/gal = 2.128742113 lb/ft³
• 84 dwt/gal = 2.154389609 lb/ft³
• 85 dwt/gal = 2.180037104 lb/ft³
• 86 dwt/gal = 2.205684599 lb/ft³
• 87 dwt/gal = 2.231332095 lb/ft³
• 88 dwt/gal = 2.25697959 lb/ft³
• 89 dwt/gal = 2.282627085 lb/ft³
• 90 dwt/gal = 2.308274581 lb/ft³
• 91 dwt/gal = 2.333922076 lb/ft³
• 92 dwt/gal = 2.359569571 lb/ft³
• 93 dwt/gal = 2.385217067 lb/ft³
• 94 dwt/gal = 2.410864562 lb/ft³
• 95 dwt/gal = 2.436512057 lb/ft³
• 96 dwt/gal = 2.462159553 lb/ft³
• 97 dwt/gal = 2.487807048 lb/ft³
• 98 dwt/gal = 2.513454543 lb/ft³
• 99 dwt/gal = 2.539102039 lb/ft³
• 100 dwt/gal = 2.564749534 lb/ft³
• dwt/gal stands for dwt/US gal
• pennyweight per gallon stands for pennyweight per US gallon
• pennyweights per gallon stands for pennyweights per US gallon
#### Foods, Nutrients and Calories
ORCHARD FUSION, ORGANIC 100% JUICE SMOOTHIE, UPC: 070893031740 contain(s) 51 calories per 100 grams or ≈3.527 ounces [ calories | price ]
Foods high in Phenylalanine
#### Gravels, Substances and Oils
CaribSea, Freshwater, Super Naturals, Jelly Beans weighs 1 585.83 kg/m³ (99.00013 lb/ft³) with specific gravity of 1.58583 relative to pure water. Calculate how much of this gravel is required to attain a specific depth in a cylindricalquarter cylindrical or in a rectangular shaped aquarium or pond [ weight to volume | volume to weight | price | density ]
Hydrogen bromide [HBr] weighs 3.6452 kg/m³ (0.00211 oz/in³) [ weight to volume | volume to weight | price | mole to volume and weight | density ]
Volume to weightweight to volume and cost conversions for Refrigerant R-404A, liquid (R404A) with temperature in the range of -51.12°C (-60.016°F) to 68.34°C (155.012°F)
#### Weights and Measurements
newton per square decimeter is a derived metric measurement unit of pressure applied by force of one newton on a surface of one square decimeter.
The pressure forces on any solid surface by liquid or gas, do not apply to any single point on that surface. Instead, these forces are spread equally along the whole surface.
Convert stone per square millimeter to troy pound per square micron or convert between all units of surface density measurement
#### Calculators
Electricity cost calculator per hours, days, weeks, months and years | 2,445 | 5,444 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.671875 | 3 | CC-MAIN-2019-30 | latest | en | 0.342769 |
https://mathemorsels.wordpress.com/tag/puzzles/ | 1,532,070,340,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676591543.63/warc/CC-MAIN-20180720061052-20180720081052-00068.warc.gz | 706,732,462 | 8,342 | # Playtime
What if I told you that one of the top mathematicians of today makes discoveries in his field through valuable playtime? It’s true! And he is not the only one. Just think about why children play. For a child, playtime is work. It is their way of figuring out how the world works as they […]
# 111111111×111111111
Why do so many people enjoy mathematics? Is it because they like talking in some other language and sounding like an alien? No, it’s because they like to solve puzzles and discover patterns. Here is one fun pattern to ponder for today. 1 x 1 = 1 11 x 11 = 121 111 x 111 […] | 164 | 608 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2018-30 | latest | en | 0.950702 |
https://www.physicsforums.com/threads/newtons-iii-law-help-me-understand.58704/ | 1,505,958,807,000,000,000 | text/html | crawl-data/CC-MAIN-2017-39/segments/1505818687592.20/warc/CC-MAIN-20170921011035-20170921031035-00530.warc.gz | 810,396,095 | 15,646 | Newton's III Law :- Help me Understand!
1. Jan 5, 2005
Mo
Hello all, need a little bit of help understanding Newton's third Law.
It states that:
" While body A exerts a force on body B, body B exerts an equal and opposite force on body A"
Okay, i can understand that.I also understand that the force are acting upon different bodies.
so (and heres the question..) if i push my desklamp across the .. desk.. then i am exerting a force in one direction on the lamp, and it is exerting an equal and opposite force on my hand.
If so, why can i still get the desklamp to move. In my textbook it tells me that the forces do not cancel each other out, because they act on different bodies, but .. still ... i obviously don't understand it all that well.
Okay the next question from the book was
Explain why forces occur only in pairs"
Is it simply enough to say: Whenever one body exerts a force on a second body, the second body always eerts a force on the first body, hence forces occur only in pairs.
Help + Advice please!
Regards,
Mo
2. Jan 5, 2005
dextercioby
The 3 laws of Newton do not act independent of each other.U'll have to use them at the same time.The answer to your 'dilemma' is that,according to the second law,to each force corresponds an acceleration.In your case,the hand that pushes the lamp gives it an acceleration equal to the ratio between the force and the lamps mass.The total force acting on the lamp is just the force u're using to push it along the table.Of course,nthere's the friction force,too,but that,if the coefficent is small,doesn't alter the results significantly.
Daniel.
3. Jan 5, 2005
Mo
Thanks for your reply, but im still not 100% , am i right in saying that the reason that my hand never acclerated off in the opposite direction (after all the lamp is exerting a force on my hand in the opposite direction) is because of the ratio between force and mass (making it a very small acceleration?)
4. Jan 5, 2005
dextercioby
It's not only your hand,yer whole body should be accelerating in the opposite direction with a tiny acceleration (due to the big ratio of masses:yours & the lamps).You don't move,becuase of the friction between your feet and and the floor,or between your butt and the chair your butt uses to relax...
Daniel.
5. Jan 5, 2005
Mo
Thanks for your help :)
Regards.
Mo
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook | 593 | 2,447 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.703125 | 4 | CC-MAIN-2017-39 | longest | en | 0.927358 |
http://mathoverflow.net/questions/62532/k-structures-on-k-vector-spaces/62549 | 1,469,626,509,000,000,000 | text/html | crawl-data/CC-MAIN-2016-30/segments/1469257826907.66/warc/CC-MAIN-20160723071026-00319-ip-10-185-27-174.ec2.internal.warc.gz | 166,893,820 | 17,391 | # $k$ structures on $K$ vector spaces
The following statement is made in Borel's Linear Algebraic groups, section 11 on $k$ structures.
Let $V$ and $W$ be $K$ vector spaces with $k$ structures. If $f:V\to W$ is $K$ linear, then $f$ is said to be defined over $k$ if $f(V_k)\subset W_k$ and these are elements of $Hom_K(V,W)_k\subset Hom_K(V,W)$. This is a $k$ structure if $W$ is finite dimensional.
The author seems to be making no assumption on the dimension of $V$ (which is the source of my problem). If we allow $V$ to be infinite dimensional then it seems to me that it is incorrect that $Hom_K(V,W)_k$ is a $k$ structure on $Hom_K(V,W)$ as is given by the following example. Let $V_k=\oplus_{i\geq0} ke_i$ and take $K$ to be an extension which is not a finite $k$ vector space, and define $f\in Hom_K(V,K)$ by choosing $f(e_i)$ which are linearly independent over $k$. We will never be able to write such an element as $\sum f_i\otimes \alpha_i$ with $f_i\in Hom_K(V,K)_k$.
So my question is : Do we need to assume both $V,W$ to be finite dimensional $K$ vector spaces for $Hom_K(V,W)$ to have a $k$ structure?
Also, what are the other references for $k$ structures and rationality properties?
-
It would help to formulate your main question explicitly, since what you've written down in the third paragraph looks confusing at first sight. The final question is easier to answer. Borel used ad hoc methods in the 1960s Columbia lectures to avoid explicit use of schemes. (The section here is AG.11.) Similarly, Springer's later textbook has a somewhat ad hoc treatment of rationality questions starting with Chapter 11. The recent Cambridge book Pseudo-reductive Groups by Conrad-Gabber-Prasad may be the best modern reference for rationality properties. – Jim Humphreys Apr 21 '11 at 14:40
Edited the question. Thanks for the reference. – Rex Apr 21 '11 at 16:39
I added a couple of tags and also replaced my earlier unhelpful answer. – Jim Humphreys Apr 23 '11 at 13:38
The statement appears to be wrong. What you need is for $V$ to be finitely generated. It is a general theorem of commutative algebra that, if $R\rightarrow S$ is a flat map of commutative rings, and $M$ and $N$ are $R$-modules with $M$ finitely presented over $R$, then the natural map $$Hom_R(M,N)\rightarrow Hom_S(S\otimes_RM,S\otimes_RN)$$ induces an isomorphism $$Hom_R(M,N)\otimes_RS\rightarrow Hom_S(S\otimes_RM,S\otimes_RN).$$ See Eisenbud, Proposition 2.10.
If we take $R=k$ and $S=K$, you get the result about $k$-structures as long as $V$ is finitely generated.
-
To replace my earlier offhand dismissal of the basic question here, it may be useful to add some comments on where things actually go wrong in Section 11.1 of Borel's Chapter AG. As Rex suggests in the question, there is an overstated claim: ... this is even a $k$-structure provided that $W$ is finite dimensional. In particular, when $W=K$, we have a $k$-structure on the dual $V^*$ of $V$. The problem here is that $W$ needs to be finite dimensional over $k$ (not just over $K$), or in other words $K/k$ needs to be a finite field extension. Only then for example can you concretely describe an arbitrary linear functional $f:V \rightarrow K$ in terms of a collection of linear functionals $V_k \rightarrow k$. (Here the dimension of $V$ is irrelevant, however.) So the answer to the original question is that $V$ can be arbitrary while $W$ should be finite dimensional over $K$ along with $[K:k]<\infty$.
Probably the stronger Borel/Bass version just quoted is never actually needed or referred to later on, though to verify that would require some checking. In this subject it's fairly unnatural to work with dual spaces of infinite dimensional vector spaces. However, polynomial algebras (infinite dimensional over some base field) do play a major role in the traditional foundations of algebraic geometry.
Anyway, Chapter AG is telegraphic in style, with few worked-out details and with only broad references to the literature then available such as Mumford's "red book". So the non-obvious points do need checking. It's worth recalling the context from which Borel's second edition (1991) arose. He was invited to give a short lecture course at Columbia in Spring 1968, an "interesting" time there as indicated in the footnote to the original Introduction of the W.A. Benjamin lecture notes published in 1969. In that Introduction, reproduced in the second edition, Borel points out that Bass wrote up the lecture notes and was largely responsible for assembling the ad hoc background material in AG. Later circumstances didn't favor careful refinement by Borel of the original lecture notes; he mainly expanded and revised some later parts. He had not originally planned to do a second edition, which motivated me to write my own book. But his later lectures on the structure theory led him to streamline some arguments and made it natural to add some further topics.
-
I love your statement that «However, polynomial algebras do play a major role in the traditional foundations of algebraic geometry.» :) – Mariano Suárez-Alvarez Apr 25 '11 at 16:06
@Mariano: That's obviously an understatement, but I wanted to emphasize that it isn't enough (for rationality questions) to limit everything to finite dimensional vector spaces $V$. At the same time, the full linear dual of an infinite dimensional space doesn't seem to come up in this context. Though it needs to be checked that Borel's later treatment doesn't rely on the overstatement in AG.11. – Jim Humphreys Apr 25 '11 at 20:46
Thanks a lot..! – Rex Apr 28 '11 at 9:41 | 1,426 | 5,600 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.765625 | 3 | CC-MAIN-2016-30 | latest | en | 0.906682 |
http://inotivity.com/2019/04/19/ | 1,561,524,921,000,000,000 | text/html | crawl-data/CC-MAIN-2019-26/segments/1560628000164.31/warc/CC-MAIN-20190626033520-20190626055520-00463.warc.gz | 88,471,945 | 18,772 | # Inotivity Inotivity I want worksheet and I want it now.
## Printable Math Worksheets Addition And Subtraction
By Ysabel Gay. Worksheet. Published at Friday, April 19th, 2019 22:42:12 PM.
Developmentally Appropriate Activities, There are many active, and far more interesting, ways for children to begin understanding words and numbers than via worksheets (Mason, 1986). A classroom with a developmentally appropriate curriculum is a print-rich environment. The walls are covered with signs naming objects, stories children have dictated, lists of...
## Multiplication And Division Word Problems 3rd Grade
By Delit Roussel. Worksheet. Published at Friday, April 19th, 2019 22:10:59 PM.
Worksheets, workbooks, and printables. Is there a place for them in the early childhood setting? Today I’m explaining why I think worksheets are not appropriate for young children. Welcome to the Child Led Environments Series where we are exploring how to set up and cultivate an environment conducive to child-led...
## Multiplying Decimals By 10 100 And 1000 Worksheet
By Pierrette Louis. Worksheet. Published at Friday, April 19th, 2019 21:52:48 PM.
Do you have students with great artistic abilities? Let them shine as they complete worksheets by asking them to draw or complete art projects. They could label items or talk about other ways that the art applies to the subject. Technology Usage, Rather than completing a worksheet, have students complete...
## Factoring 2 Factoring By Grouping Worksheet Answer Key
By Pierrette Louis. Worksheet. Published at Tuesday, April 30th, 2019 23:21:16 PM.
Worksheets are Too Abstract, Young children are still in Piaget’s Preoperational Stage, which means they need symbols to represent objects. These young children cannot think abstractly. For example, they need a ball in their hands to understand what a ball is. Seeing the word ball on a worksheet or sometimes...
## Distributive Property Of Multiplication Worksheets
By Pierrette Louis. Worksheet. Published at Tuesday, April 30th, 2019 12:35:17 PM.
I’ve had numerous people ask me recently to teach them how to make worksheets for their classroom and products for TeachersPayTeachers. Being a visual learner myself, I decided to lay it out step-by-step in the hopes of making it as simple as possible! Yes, there is a LOT of information...
### Graphing Quadratic Functions Worksheet Answers Algebra 2
By Pierrette Louis. Worksheet. Published at Tuesday, April 30th, 2019 21:52:08 PM.
Insert any shapes and lines. You will use the same steps as making a border, but I will show you how to format it so it is the perfect size and in the right place. There is nothing that bugs me more than when a shape is not perfect! This...
#### Free Printable 5th Grade Multiplication Worksheets
By Pierrette Louis. Worksheet. Published at Tuesday, April 30th, 2019 14:55:44 PM.
School should be a welcoming, peaceful place for children – an environment to which children come eager to see what challenging, stimulating, and fun activities are in store. Children know they may not succeed at everything they try, but also know they will be valued for who they are. Children’s...
##### Add Subtract Multiply Divide Decimals Worksheet
By Pierrette Louis. Worksheet. Published at Tuesday, April 30th, 2019 14:32:29 PM.
Now is the time to add some clipart images to your worksheets to give them some personality. You can find tons of clipart on the Internet, but you have to be sure to check out each site’s policies before just using any clipart. Perhaps your school already has a license...
###### Super Teacher Worksheets Login And Password 2016
By Pierrette Louis. Worksheet. Published at Tuesday, April 30th, 2019 12:09:35 PM.
Worksheets Do Not Provide Real, Meaningful Experiences. I go into a lot of detail about meaningful experiences in this post. Basically, a child needs to have a reason for learning the concept. Completing a worksheet is not a good enough reason for a child. Providing activities that connect to real...
## The Cell Cycle Pogil Answer Key Extension Questions
By Pierrette Louis. Worksheet. Published at Wednesday, May 01st, 2019 00:42:06 AM.
Of course, you will need to make multiple copies of the worksheet for your students. Consider printing one copy of the worksheet at home or in your classroom, and using the color copier in your school office, if one is available. If your worksheet has a lot of color on...
Any content, trademark’s, or other material that might be found on the Inotivity website that is not Inotivity’s property remains the copyright of its respective owner/s. In no way does Inotivity claim ownership or responsibility for such items, and you should seek legal consent for any use of such materials from its owner. | 1,068 | 4,771 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.5625 | 3 | CC-MAIN-2019-26 | longest | en | 0.914463 |
https://docs.nvidia.com/deeplearning/modulus/_modules/modulus/utils/sympy/numpy_printer.html | 1,679,432,154,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296943746.73/warc/CC-MAIN-20230321193811-20230321223811-00748.warc.gz | 253,743,203 | 34,846 | # Source code for modulus.utils.sympy.numpy_printer
"""
Helper functions for converting sympy equations to numpy
"""
import types
import inspect
import numpy as np
import symengine as se
import sympy as sp
NP_LAMBDA_STORE = {}
[docs]def np_lambdify(f, r):
"""
generates a numpy function from a sympy equation
Parameters
----------
f : Sympy Exp, float, int, bool or list of the previous
the equation to convert to a numpy function.
If float, int, or bool this gets converted
to a constant function of value f. If f is a list
then output for each element in list is is
concatenated on axis -1.
r : list, dict
A list of the arguments for f. If dict then
the keys of the dict are used.
Returns
-------
np_f : numpy function
"""
# possibly lambdify list of f
if not isinstance(f, list):
f = [f]
# convert r to a list if dictionary
# break up any tuples to elements in list
if isinstance(r, dict):
r = list(r.keys())
no_tuple_r = []
for key in r:
if isinstance(key, tuple):
for k in key:
no_tuple_r.append(k)
else:
no_tuple_r.append(key)
# lambidfy all functions in list
lambdify_f = []
for f_i in f:
# check if already a numpy function
if isinstance(f_i, types.FunctionType):
# add r inputs to function
args = inspect.getargspec(f_i).args
def lambdify_f_i(**x):
return f_i(**{key: x[key] for key in args})
else:
# check if already lambdified equation
if (f_i, tuple(no_tuple_r)) in NP_LAMBDA_STORE.keys():
lambdify_f_i = NP_LAMBDA_STORE[(f_i, tuple(no_tuple_r))]
else: # if not lambdify it
try:
if not isinstance(f_i, bool):
f_i = float(f_i)
except:
pass
if isinstance(f_i, (float, int)): # constant function
def loop_lambda(constant):
return (
lambda **x: np.zeros_like(next(iter(x.items()))[1])
+ constant
)
lambdify_f_i = loop_lambda(f_i)
elif type(f_i) in [
type((se.Symbol("x") > 0).subs(se.Symbol("x"), 1)),
type((se.Symbol("x") > 0).subs(se.Symbol("x"), -1)),
bool,
]: # TODO hacky sympy boolian check
def loop_lambda(constant):
if constant:
return lambda **x: np.ones_like(
next(iter(x.items()))[1], dtype=bool
)
else:
return lambda **x: np.zeros_like(
next(iter(x.items()))[1], dtype=bool
)
lambdify_f_i = loop_lambda(f_i)
else:
try: # first try to compile with Symengine
kk = []
for k in no_tuple_r:
if isinstance(k, str):
kk.append(se.Symbol(k))
else:
kk.append(k)
kk = [se.Symbol(name) for name in sorted([x.name for x in kk])]
se_lambdify_f_i = se.lambdify(kk, [f_i], backend="llvm")
def lambdify_f_i(**x):
if len(x) == 1:
v = list(x.values())[0]
else:
v = np.stack(
[v for v in dict(sorted(x.items())).values()],
axis=-1,
)
out = se_lambdify_f_i(v)
if isinstance(out, list):
out = np.concatenate(out, axis=-1)
return out
except: # fall back on older SymPy compile
sp_lambdify_f_i = sp.lambdify(
[k for k in no_tuple_r], f_i, [NP_SYMPY_PRINTER, "numpy"]
)
def lambdify_f_i(**x):
v = sp_lambdify_f_i(**x)
if isinstance(v, list):
v = np.concatenate(v, axis=-1)
return v
# add new lambdified function to dictionary
NP_LAMBDA_STORE[(f_i, tuple(no_tuple_r))] = lambdify_f_i
# add new list of lambda functions
lambdify_f.append(lambdify_f_i)
# construct master lambda function for all
def loop_grouped_lambda(lambdify_f):
def grouped_lambda(**invar):
output = []
for lambdify_f_i in lambdify_f:
output.append(lambdify_f_i(**invar))
return np.concatenate(output, axis=-1)
return grouped_lambda
return loop_grouped_lambda(lambdify_f)def _xor_np(x):
return np.logical_xor(x)
def _min_np(x):
return_value = x[0]
for value in x:
return_value = np.minimum(return_value, value)
return return_value
def _max_np(x):
return_value = x[0]
for value in x:
return_value = np.maximum(return_value, value)
return return_value
def _heaviside_np(x):
return np.heaviside(x, 0)
def _equal_np(x, y):
return np.isclose(x, y)
NP_SYMPY_PRINTER = {
"amin": _min_np,
"amax": _max_np,
"Heaviside": _heaviside_np,
"equal": _equal_np,
"Xor": _xor_np,
}
SYMENGINE_BLACKLIST = [sp.Heaviside, sp.DiracDelta] | 1,174 | 3,912 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.59375 | 3 | CC-MAIN-2023-14 | latest | en | 0.439136 |
https://www.perlmonks.org/index.pl/?node_id=3333;parent=573138 | 1,713,840,125,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296818452.78/warc/CC-MAIN-20240423002028-20240423032028-00353.warc.gz | 845,501,105 | 8,502 | There's more than one way to do things PerlMonks
### comment on
Need Help??
All,
If you have ever felt that people debating the big-oh notation of some algorithm sound like they are speaking a foreign language, then this tutorial is for you. You may have even decided to educate yourself by checking out the Wikipedia entry and have been convinced it was a foreign language. You are not alone.
You may have already read An informal introduction to O(N) notation by dws which is excellent. This tutorial will repeat much of the same information in a much more elementary manner as well as go into detail about how useful or not the notation is. It is likely I won't be completely accurate about a number of things in the tutorial. My hope is that by the end, your understanding of the topic will be sufficient enough that you can understand the corrections others make (as I am sure they will).
#### What Is The Big-O
This tutorial covers the Big-O as it relates to computer science. If you were thinking of something else (perhaps Fridays in the Chatterbox), you can stop reading now. Simply put, it describes how the algorithm scales (performs) in the worst case scenario as it is is run with more input. Since my simple explanation may not be simple enough - let me give an example. If we have a sub that searches an array item by item looking for a given element, the scenario that the Big-O describes is when the target element is last (or not present at all). This particular algorithm is O(N) so the same algorithm working on an array with 25 elements should take approximately 5 times longer than an array with 5 elements.
It is easy to lose sight of the fact that there is more to consider about an algorithm other than how fast it runs. The Big-O can also be used to describe other behavior such as memory consumption. We often optimize by trading memory for time. You may need to choose a slower algorithm because it also consumes less of a resource that you need to be frugal with.
#### What The Big-O Is Not
Constants: The Big-O is not concerned with factors that do not change as the input increases. Let me give an example that may be suprising. Let's say we have an algorithm that needs to compare every element in an array to every other element in the array. A simple implementation may look like:
```for my \$i (0 .. \$#array) {
for my \$j (0 .. \$#array) {
next if \$j == \$i;
# Compare \$i, \$j
}
}
This is O(N^2). After a little bit of testing we decide that this is far too slow, so we make a little optimization.
```for my \$i (0 .. \$#array - 1) {
for my \$j (\$i + 1 .. \$#array) {
# Compare \$i, \$j
}
}
We have just cut our run time in half - YAY! Guess what, the Big-O has stayed the same O(N^2). This is because N^2 / 2 only has one variable part. The divided by 2 remains the same (constant) regardless of the input size. There are valid mathematical reasons for doing this but it can be frustrating to see two algorithms with the exact same Big-O that results in one running twice as fast as the other.
Implementation Details: The Big-O is an uncaring cold-hearted jerk. It does not care if you can't afford to buy the extra RAM needed for your problem and have to resort to tying your hash to disk. You are on your own. It also doesn't care that the data structure you would need to implement to achieve O(Log Log N) is so complex you will never be able to maintain it. In a nutshell, the Big-O lives in the land of theory and doesn't care very much about the real world.
#### What The Big-O Is Good For
The good news is that the Big-O belongs to an entire family of notation. This tutorial will not cover it but family members include describing the average and best cases. It also serves as a good indicator of what algorithm to use once you take your individual circumstances into consideration. Let me give a contrived example:
Let's consider using cacheing as an optimization. In theory, the Big-O is going to ignore it saying your input is all different and you will never benefit from it. In reality, you test it and discover that you have a 60% hit rate. You do a little more experimenting and discover that the input size required for a more complex algorithm to be faster is larger than your real maximum input size. This all despite the more complex alorithm having a more favorable Big-O.
In a nutshell, the Big-O of a given algorithm combined with the specific problem knowledge is a great way to choose the best algorithm for your situation.
#### What Do Those Symbols Mean?
So by this point you should realize that Big-O (theory) without context (real world) is not very useful. You are now armed with the knowledge necessary to start using Big-O as the mercenary it is. Ok Big-O, what exactly do you mean that algorithm is O(N Log N)? I am going to duck at this point and suggest you read the node by dws or the Wikipedia entry I linked to earlier. You may now be wondering if Big-O is really inanimate, perhaps even an abstract concept and not at all real as I have made it out to be. If so, how then can you determine the Big-O of a given algorithm? Analysis of algorithms is not for the faint of heart, so I must once again duck.
I have not really added anything to any of the other links I referenced. I do hope however that I have put it in plain enough english to be understood by even the most extreme novice. I welcome those more knowledgeable than myself to add corrections as well as provide additional content. I would only ask that you do so in the same spirit of this tutorial (understandable by non-CS majors).
Update: Removed an incorrect analogy regarding slopes thanks to blokhead.
Also see Sorting a list of IP addresses (aka Why I hate Big O) by jeffa
Cheers - L~R
In reply to Big-O Notation - What is it good for? by Limbic~Region
Title:
Use: <p> text here (a paragraph) </p>
and: <code> code here </code>
to format your post; it's "PerlMonks-approved HTML":
• Are you posting in the right place? Check out Where do I post X? to know for sure.
• Posts may use any of the Perl Monks Approved HTML tags. Currently these include the following:
<code> <a> <b> <big> <blockquote> <br /> <dd> <dl> <dt> <em> <font> <h1> <h2> <h3> <h4> <h5> <h6> <hr /> <i> <li> <nbsp> <ol> <p> <small> <strike> <strong> <sub> <sup> <table> <td> <th> <tr> <tt> <u> <ul>
• Snippets of code should be wrapped in <code> tags not <pre> tags. In fact, <pre> tags should generally be avoided. If they must be used, extreme care should be taken to ensure that their contents do not have long lines (<70 chars), in order to prevent horizontal scrolling (and possible janitor intervention).
• Want more info? How to link or How to display code and escape characters are good places to start.
Create A New User
Domain Nodelet?
Chatterbox?
and the web crawler heard nothing...
How do I use this?Last hourOther CB clients
Other Users?
Others avoiding work at the Monastery: (4)
As of 2024-04-23 02:39 GMT
Sections?
Information?
Find Nodes?
Leftovers?
Voting Booth?
No recent polls found | 1,667 | 7,045 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.203125 | 3 | CC-MAIN-2024-18 | latest | en | 0.961352 |
https://www.unitconverters.net/power/kilovolt-ampere-to-joule-second.htm | 1,726,552,448,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651739.72/warc/CC-MAIN-20240917040428-20240917070428-00399.warc.gz | 939,134,531 | 3,570 | Home / Power Conversion / Convert Kilovolt Ampere to Joule/second
# Convert Kilovolt Ampere to Joule/second
Please provide values below to convert kilovolt ampere [kV*A] to joule/second [J/s], or vice versa.
From: kilovolt ampere To: joule/second
### Kilovolt Ampere to Joule/second Conversion Table
Kilovolt Ampere [kV*A]Joule/second [J/s]
0.01 kV*A10 J/s
0.1 kV*A100 J/s
1 kV*A1000 J/s
2 kV*A2000 J/s
3 kV*A3000 J/s
5 kV*A5000 J/s
10 kV*A10000 J/s
20 kV*A20000 J/s
50 kV*A50000 J/s
100 kV*A100000 J/s
1000 kV*A1000000 J/s
### How to Convert Kilovolt Ampere to Joule/second
1 kV*A = 1000 J/s
1 J/s = 0.001 kV*A
Example: convert 15 kV*A to J/s:
15 kV*A = 15 × 1000 J/s = 15000 J/s | 280 | 689 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.4375 | 3 | CC-MAIN-2024-38 | latest | en | 0.42037 |
https://www.analystforum.com/t/quant-questions/1963 | 1,713,049,118,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296816853.44/warc/CC-MAIN-20240413211215-20240414001215-00058.warc.gz | 593,960,360 | 8,869 | Quant Questions
John has just performed a hypothesis test and has calculated the p-value to be 0.13, which of the following is most likely to be true? A: Do not reject the null hypothesis at the 10% significance level, and do not reject the null hypothesis at the 5% significance level. B: Do not reject the null hypothesis at the 10% significance level, and reject the null hypothesis at the 5%significance level. C: Reject the null hypothesis at the 10% significance level, and do not reject the null hypothesis at the 5%significance level. D: Reject the null hypothesis at the 10% significance level, and reject the null hypothesis at the 5%significance level. If John tosses a two sided coin 6 times, which of the following is closest to the probability of obtaining exactly three heads? A: 0.02 B: 0.16 C: 0.31 D: 1.88 In testing the significance of the correlation coefficient of 0.4, assuming that there are 90 observations, which of the following is most likely to be true? A: A t-statistic of 5.09 and 88 degrees of freedom B: A t-statistic of 5.09 and 89 degrees of freedom C: A t-statistic of 4.09 and 88 degrees of freedom D: A t-statistic of 4.09 and 89 degrees of freedom Which of the folowing is least likely to be required in order to perform an F-test? A: The sum of the squared errors B: The total number of observations C: The mean regression sum D: The total number of parameters to be estimated Which of the following best describe observations that significantly reduce what would otherwise be a high correlation? A: Outliers B: Non-linear relations C: Independent variables D: Spurious correlations
If John tosses a two sided coin 6 times, which of the following is closest to the probability of obtaining exactly three heads? A: 0.02 B: 0.16 C: 0.31 D: 1.88 6c3 * .5^3 * /5^3 = .3125 Choice C In testing the significance of the correlation coefficient of 0.4, assuming that there are 90 observations, which of the following is most likely to be true? A: A t-statistic of 5.09 and 88 degrees of freedom B: A t-statistic of 5.09 and 89 degrees of freedom C: A t-statistic of 4.09 and 88 degrees of freedom D: A t-statistic of 4.09 and 89 degrees of freedom tstat = r * sqrt (n-2) / sqrt ( 1-r^2) = .4 * sqrt(88) / sqrt ( 1 - .4^2) = 4.094 therefore Choice: C --> t-statistic of 4.09 and 88 dof.
John has just performed a hypothesis test and has calculated the p-value to be 0.13, which of the following is most likely to be true? A: Do not reject the null hypothesis at the 10% significance level, and do not reject the null hypothesis at the 5% significance level. B: Do not reject the null hypothesis at the 10% significance level, and reject the null hypothesis at the 5%significance level. C: Reject the null hypothesis at the 10% significance level, and do not reject the null hypothesis at the 5%significance level. D: Reject the null hypothesis at the 10% significance level, and reject the null hypothesis at the 5%significance level. Given that p-value is the smallest level of significance at which the Null hyp cannot be rejected Choice A – do not reject at 10% and do not reject at 5% seems to be the choice. Which of the folowing is least likely to be required in order to perform an F-test? A: The sum of the squared errors B: The total number of observations C: The mean regression sum D: The total number of parameters to be estimated For a F test => There should be 2 parameters? I am guessing D would be not required. Which of the following best describe observations that significantly reduce what would otherwise be a high correlation? A: Outliers B: Non-linear relations C: Independent variables D: Spurious correlations B : Non-linear relations.
1. ?? Don’t know what is a p-value 2. C 6c3 * 0.5^3 * 0.5^3 = 0.3125 3. C t-statistic = r * sqrt (n-2) / sqrt ( 1-r^2) = 0.4 * sqrt(90 -2) / sqrt ( 1 - 0.4^2) = 4.094 DoF = n - 2 = 90 - 2 = 88 4. ?? F-test related to Regression??? Is this ANOVA?? do we have it for L1?? 5. A The presence of Outliers (extreme observations, either +ve or -ve) reduces the correlation coefficient. - Dinesh S
CPK and dinesh, could u guys pls explain how do u derive the answer for that coin tossing question in detail? Which LOS is this and any reading material that may help? I am totally confused. Thanks in advance.
Coin tossing is the binomial probability distribution and for the correlation coefficient – this article outlines correlation: http://irp.savstate.edu/irp/glossary/correlation.html and these were the two lines there with my comments preceded by CPK below: (Maybe Joey can help identify the right answer) If the relationship is curvilinear, the “r” will give false and misleading readings that substantially underestimate the relationship. CPK: Based on the above – a non-linear relationship would reduce a otherwise STRONG Relationship The easy way to test and see whether the relationship is linear is to plot a scatter diagram and see if the “points” scatter in a more or less linear direction. On a scatter diagram, the coefficient measures the slope of the general pattern of points plotted and the width of the ellipse that encloses those points. The width of the ellipse indicates the extent of the relationship and hence, the magnitude, or absolute value of the coefficient. Some analysts advise removing any “outlier” cases from consideration and treat them a priori as aberrations so that they do not bias the relationship remaining among the more “normal” cases. CPK: OUTLIER biases the relation. It could end up being a HIGHER CORRELATION if there are outliers. And Dinesh p-value is the smallest level of significance at which the Null hyp cannot be rejected HTH CPK
First three are handled above, yes? Which of the folowing is least likely to be required in order to perform an F-test? A: The sum of the squared errors B: The total number of observations C: The mean regression sum D: The total number of parameters to be estimated A, B, and D are almost certainly required. I don’t even know what C is, so it’s not required. Which of the following best describe observations that significantly reduce what would otherwise be a high correlation? A: Outliers B: Non-linear relations C: Independent variables D: Spurious correlations Answer has got to be B because that certainly lowers the correlation but A isn’t a bad answer either. Outliers can either increase or decrease the correlation depending on what kind of outlier we are talking about. C and D are definitely out.
These are the answers guys: 1) A The p-value is the smallest value of alpha for which the null hypothesis is rejected. If the p-value is greater than or equal to the significance level , the null hypothesis is not rejected. If the p-value is smaller than the significance level, the null hypothesis is rejected. As the p-value in this question is 0.13, the null hypothesis is not rejected at the 5% or 10% significance level. 2) C Using the binomial distribution: Probability of obtaining exactly three heads = 6 C 3 x 0.5^3 x 0.5^3 (Where 6 C 3 is number of combinations drawing 3 from 6) Probability of obtaining exactly three heads = 20 x 0.125 x 0.125 = 0.31 3) C When testing the correlation coefficient: Degrees of freedom = n-2 = 90-2 = 88 T-statistic = [r(n-2)^1/2]/[(1-r)^2] = [0.4(90-2)^1/2]/[(1-0.4)^2] = 4.09 4) C The F-test uses the regression sum of squares, not the mean regression sum. 5) A Outliers are observations that would significantly reduce what would otherwise be a high correlation.
They are wrong on #5. Outliers have an ambiguous effect on correlation. Non-linearity always reduces correlation.
Thanks cpk, so the p-value is the smallest value of alpha for which the null hypothesis is rejected. ------- FAIL TO REJECT-------|-------REJECTION REGION------- -------|---------------|----------|--------------------------------------- ------0.05----------0.10-----0.13------------------------------------ Since our alpha1 is 0.05 (5%) and alpha2 is 0.10 (10%) are below 0.13 (our p-value), we do not reject the Null. Any significance level beyond 0.13, the Null is rejected. So answer ‘A’ makes sense. maparam, Question 2 is a BRV, where I labeled getting a HEAD as the success probability ‘p’ n tries = 3 total tries = 6 n - x = 6 -3 = 3 P(Head) = 1/2 = 0.5 P(Tail) = 1/2 = 0.5 ---------------------------------------------------------------------------------------- P(X = x) = (number of ways to choose x from n) * §^x * (1- p)^(n-x) ---------------------------------------------------------------------------------------- It’s so intuitive, than to remember this messy formula Joey, Non-linear relations are sure to reduce the correlation, but what they probably want us to assume, here, for this question, that we are in a capsule of L1 and we know no more than Linear Regression, probably a similar question on L2 (where we have the non-linear regression stuff) would surely be an answer of B. Guys, and I still don’t understand, how is a F-Test related to Linear Regression… Schweser notes never talk of this… is there something that I am missing? - Dinesh S
joey – since the question specifically asks Which of the following best describe observations that significantly reduce what would otherwise be a high correlation? A: Outliers B: Non-linear relations C: Independent variables D: Spurious correlations which best describes – the observations – would Outliers be the answer – because nothing else corresponds to Observations themselves
That’s a point. It’s hard to take the grammar of these very seriously though. So Dinesh - The idea is that in linear regression you just check to see if the slope is significant using a t-test. In every other kind of regression (e.g., multivariate regression, polynomial regression) you need a test that asks whether or not the whole regression model is significant or not. You use an F-statistic to do this in most cases (assume normal errors, blah, blah).
JoeyDVivre Wrote: ------------------------------------------------------- > So Dinesh - The idea is that in linear regression > you just check to see if the slope is significant > using a t-test. In every other kind of regression > (e.g., multivariate regression, polynomial > regression) you need a test that asks whether or > not the whole regression model is significant or > not. You use an F-statistic to do this in most > cases (assume normal errors, blah, blah). thanks Joey, so Schweser has just explained us the t-test for detecting the significance of the slope/intersept of the regresion line, but they have never talked about significance of the complete model. I’ll need to check if this and ANOVA tables are there for the L1 LOS. - Dinesh S
I don’t think it’s part of the L1 LOS and I think they have even phased it out on the LII LOS. Don’t take my word for it though.
maybe I am missing something here guys… but in question 2 how did you get 20 = 6 c 3 Thanks
there are 6C3 ways to choose exacly 3 heads from a possibility of 6 trials (6 heads) - Dinesh S
Dinesh… thanks… I understand that… but I was wondering if there is a formula for this or is this common sense that I am not picking up on… Thanks
There is a formula available too… ---------------------------------------------------------------------------------------- P(X = x) = (number of ways to choose x from n) * §^x * (1- p)^(n-x) ---------------------------------------------------------------------------------------- but hope you don’t use it too much… it’ better to understand ‘why’ than to plug-and-chug. - Dinesh S
I used 6C3 / 2^6 But I realise this only works because the probabilities are equal. I don’t get how the formula above works
Consider something which is binomial probability of success = .2 Probability of failure = .8 Now you have 3 chances to repeat the experiment. Your experiment outcomes are SSS SSF SFS FSS SFF FSF SFF FFF 3 Successes: SSS = 1 * .2 ^ 3 = 3C0 (0.2^3) * (0.8 ^ 0) Whereever the S appears – there is only 1 way that you can get 3 S Possibility of 2 Successes SSF = 3 * (.2 ^ 2) * (0.8 ^ 1) = 3C2 * (.2 ^ 2) * (0.8 ^ 1) This needs to be multiplied by 3 because SSF, SFS and FSS are 3 distinct possibilities. and so on. | 3,010 | 12,163 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.34375 | 3 | CC-MAIN-2024-18 | latest | en | 0.913271 |
http://www.map.mpim-bonn.mpg.de/index.php?title=Some_calculations_involving_configuration_spaces_of_distinct_points&diff=next&oldid=15002 | 1,695,375,773,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233506339.10/warc/CC-MAIN-20230922070214-20230922100214-00855.warc.gz | 68,616,364 | 8,236 | # Some calculations involving configuration spaces of distinct points
(Difference between revisions)
## 1 Introduction
‘The complement of the diagonal’ and ‘the Gauss map’ ideas play a great role in different branches of mathematics. The Haefliger-Wu invariant is a manifestation of these ideas in the theory of embeddings. The complement to the diagonal idea originated from two celebrated theorems: the Lefschetz Fixed Point Theorem and the Borsuk-Ulam Antipodes Theorem.
## 2 Construction and examples
For a manifold $X$$ {{Stub}} == Introduction == ; ‘The complement of the diagonal’ and ‘the Gauss map’ ideas play a great role in different branches of mathematics. The Haefliger-Wu invariant is a manifestation of these ideas in the theory of embeddings. The complement to the diagonal idea originated from two celebrated theorems: the Lefschetz Fixed Point Theorem and the Borsuk-Ulam Antipodes Theorem == Construction and examples == ; For a manifold X, \widetilde X denotes ''the deleted product'' of X, i.e. X^2 minus an open tubular neighborhood of the diagonal. It is a manifold with boundary and has the standard free involution. {{beginthm|Definition}}[of the Haefliger-Wu invariant \alpha] \label{DefHaef} The Haefliger-Wu invariant \alpha:\mathrm{Emb}^{k}N\to \pi_{\mathrm{eq}}^{k-1}(\widetilde{N}) is induced by the Gauss map, also denoted by \alpha. The Gauss map assigns to an individual embedding f:N\to\R^{k} an equivariant map \widetilde{N}\to S^{k-1} defined by the formula (x,y)\mapsto \frac{f(x)-f(y)} {\|f(x)-f(y)\|}, \quad (x,y)\in\widetilde{N}\subset N\times N. {{beginthm|Theorem}} The Haefliger-Wu invariant \alpha:\mathrm{Emb}^m N\to\pi^{m-1}_{\mathrm{eq}}( \widetilde N) is one-to-one for m\ge 3n+4. {{endthm}} == Invariants == ; ... == Classification/Characterization == ; ... == Further discussion == ; ... == References == {{#RefList:}} [[Category:Manifolds]]X$, $\widetilde X$$\widetilde X$ denotes the deleted product of $X$$X$, i.e. $X^2$$X^2$ minus an open tubular neighborhood of the diagonal. It is a manifold with boundary and has the standard free involution.
Definition 2.1.[of the Haefliger-Wu invariant $\alpha$$\alpha$]
The Haefliger-Wu invariant $\alpha:\mathrm{Emb}^{k}N\to \pi_{\mathrm{eq}}^{k-1}(\widetilde{N})$$\alpha:\mathrm{Emb}^{k}N\to \pi_{\mathrm{eq}}^{k-1}(\widetilde{N})$ is induced by the Gauss map, also denoted by $\alpha$$\alpha$. The Gauss map assigns to an individual embedding $f:N\to\R^{k}$$f:N\to\R^{k}$ an equivariant map $\widetilde{N}\to S^{k-1}$$\widetilde{N}\to S^{k-1}$ defined by the formula
$\displaystyle (x,y)\mapsto \frac{f(x)-f(y)} {\|f(x)-f(y)\|}, \quad (x,y)\in\widetilde{N}\subset N\times N.$
Theorem 2.2. The Haefliger-Wu invariant $\alpha:\mathrm{Emb}^m N\to\pi^{m-1}_{\mathrm{eq}}( \widetilde N)$$\alpha:\mathrm{Emb}^m N\to\pi^{m-1}_{\mathrm{eq}}( \widetilde N)$ is one-to-one for $2m\ge 3n+4$$2m\ge 3n+4$.
...
...
... | 918 | 2,915 | {"found_math": true, "script_math_tex": 11, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 12, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.734375 | 3 | CC-MAIN-2023-40 | latest | en | 0.764292 |
https://www.semanticscholar.org/paper/CHAPTER-3-Convergence-of-Random-Variables-/763b7d2f4eaac249a60ebb70249cde480b4a9533 | 1,513,242,779,000,000,000 | text/html | crawl-data/CC-MAIN-2017-51/segments/1512948542031.37/warc/CC-MAIN-20171214074533-20171214094533-00399.warc.gz | 800,872,876 | 42,440 | # CHAPTER 3 Convergence of Random Variables
• Published 2002
#### Abstract
In probability and statistics, it is often necessary to consider the distribution of a random variable that is itself a function of several random variables, for example, Y = g(X1, · · · , Xn); a simple example is the sample mean of random variables X1, · · · , Xn. Unfortunately, finding the distribution exactly is often very difficult or very time-consuming even if the joint distribution of the random variables is known exactly. In other cases, we may have only partial information about the joint distribution of X1, · · · , Xn in which case it is impossible to determine the distribution of Y . However, when n is large, it may be possible to obtain approximations to the distribution of Y even when only partial information about X1, · · · , Xn is available; in many cases, these approximations can be remarkably accurate. The standard approach to approximating a distribution function is to consider the distribution function as part of an infinite sequence of distribution functions; we then try to find a “limiting” distribution for the sequence and use that limiting distribution to approximate the distribution of the random variable in question. This approach, of course, is very common in mathematics. For example, if n is large compared to x, one might approximate (1 + x/n)n by exp(x) since
### Cite this paper
@inproceedings{2002CHAPTER3C, title={CHAPTER 3 Convergence of Random Variables}, author={}, year={2002} } | 327 | 1,513 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2017-51 | latest | en | 0.895359 |
https://scicomp.stackexchange.com/questions/2360/how-can-i-derive-a-bound-on-the-spurious-oscillations-in-the-numerical-solution/2361 | 1,582,845,405,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875146907.86/warc/CC-MAIN-20200227221724-20200228011724-00353.warc.gz | 471,809,567 | 35,859 | # How can I derive a bound on the spurious oscillations in the numerical solution of the 1D advection equation?
$\frac{\partial u}{\partial t} + c\frac{\partial u}{\partial x} = 0$ in $\Omega=[0,1]$
$u(0,t)=u(1,t)$
$u(x,0)=g(x)$
where $g(x)$ has a jump discontinuity at $x^*\in (0,1)$.
It is my understanding that for linear finite difference schemes of higher than first order, spurious oscillations occur near the discontinuity as it is advected over time, resulting in a distortion of the solution from its expected wave shape. According to wikipedia explanation, it seems that these oscillations typically occur when a discontinuous function is approximated with a finite fourier series.
For some reason, I can't seem to grasp how a finite fourier series can be observed in the solution of this PDE. In particular, how can I estimate a bound on the "over-shoot" analytically?
The first order upwind method is monotone; it does not introduce spurious oscillations. But it is only first order accurate, resulting in so much numerical diffusion as to be unusable for many purposes. Godunov's Theorem states that linear spatial discretizations of higher than first order cannot be monotone. To rigorously control oscillations, we use Total Variation Diminishing (TVD) schemes. TVD methods are typically limited to second order accuracy. For higher order, we must either relax our request, leading to Total Variation Bounded (TVB) methods like (Weighted) Essential Non-Oscillatory ((W)ENO), or we must relax the definition of TVD to "maximum-principle preserving" or similar, where the initial extrema are in terms of an initial reconstructed solution, resulting in special limiting schemes.
• My apologies... for some reason, I got the impression that this was true for the first order scheme as well. I edited the question to reflect this comment. – Paul May 28 '12 at 6:37
Linear finite difference discretization of a 1D problem with periodic boundaries leads to a discretization of the form $$U^{n+1} = LU^n$$
where $L$ is a circulant matrix. The eigenvectors of any circulant matrix are discrete Fourier modes $$v_j = \exp(ijh\xi)$$ (here $h$ is the grid spacing and $\xi$ is the wavenumber, which ranges from zero up to the highest wavenumber representable on the grid). These eigenvectors form a basis for all functions that can be represented on the grid. If you express the solution in terms of these discrete Fourier modes, then the numerical method is diagonalized, i.e. each Fourier component is multiplied by a (generally complex) scalar factor at each step. The scalar factor is often referred to as the amplification factor, and what I have just described is known as von Neumann analysis. It is analogous to Fourier analysis of linear PDEs, in which one uses a Fourier basis, to "diagonalize" the linear differential operators.
You can find nice explanations, for instance, in the text of Strikwerda or LeVeque.
• I'm familiar with von neumann analysis. But can I really use this analysis to derive a bound on the spurious oscillations? – Paul May 28 '12 at 6:22
• I was responding mainly to your statement I can't seem to grasp how a finite fourier series can be observed in the solution of this PDE. But yes, you can get such bounds from this analysis. For instance, you could look at the worst case secenario in which all modes constructively interfere. However, this is likely to be a very pessimistic bound. In practice, I haven't seen anyone derive bounds other than TVD or TVB (which are quite strong and don't hold for linear schemes). – David Ketcheson May 28 '12 at 6:25
• You could probably get a more interesting bound by looking at the dispersion relation for the highest wavenumber modes. But I've never seen it done. – David Ketcheson May 28 '12 at 6:28
Not all spurious oscillations are Gibbs phenomena. They look similar, but there are Gibbs oscillations for all finite Fourier approximations of discontinuous functions (they just get smaller as you add more terms). Whereas, there are non-oscillatory representations of discontinuous functions resulting from the solution of finite difference approximations to PDEs that don't require infinite series.
Bathe (Inf–sup testing of upwind methods, PDF) has a paper on this for finite element methods (convection-diffusion, IIRC) in 1-D that involves computing the constant for the $\inf$-$\sup$ condition and relating that to oscillations. You might gain some insight from that.
• This is a useful paper, but note that inf-sup stability does not provide strong control of oscillations. No amount of inf-sup stability can provide a TVD method, for example. And in light of Godunov's Theorem, it makes no sense to go looking for linear spatial discretizations if we intend to have non-oscillatory solutions of greater than first order. Note that the Peclet number appears in all the methods in this paper, and the methods degrade to first order accuracy as $\mathrm{Pe}\to \infty$, while also not being TVD. – Jed Brown May 27 '12 at 19:00
• These are all true statements. It only really applies to convection-diffusion problems. – Bill Barth May 27 '12 at 21:59
As for your last question about the connection between finite Fourier series and finite element approximation: In general, if you try to project a function with a jump onto a finite dimensional space whose basis functions are continuous, you get Gibbs phenomenon. This is true if the basis is a finite Fourier series (where the basis functions are the sines and cosines) or if the basis are the usual finite element hat functions -- it's a property of the projection plus the unsuitability of the basis functions.
• I'm happy to be proved wrong, since I'm clearly out of practice, but I'm not buying your comment about projections onto hat functions without further qualification. My quick computation using my old 1-D MATLAB code from my first year FEM class shows that the projection of the step function onto $H_0^1$ using hat functions is non-oscillatory. Do you have an example that can show what I'm missing? – Bill Barth May 27 '12 at 22:27
• Never mind. Old code is old. I can reproduce oscillations. Previous comment retracted. – Bill Barth May 27 '12 at 22:45
• I'm glad I could help :-) – Wolfgang Bangerth May 29 '12 at 17:33
One approach is via the equivalent equation, that is, the differential equation to which your discrete method gives the closest aproximation. This is never the differential equation that you intended to solve. Then you look at the asymptotic solution of the equivalent equation, for a step function as initial data. Look at Bouche, D., Bonnaud, G. and Ramos, D., 2003. Comparison of numerical schemes for solving the advection equation. Applied mathematics letters, 16(2), pp.147-154. | 1,588 | 6,777 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.171875 | 3 | CC-MAIN-2020-10 | latest | en | 0.923726 |
https://priorprobability.com/2014/06/27/lies-damned-lies-and-statistics/ | 1,686,229,528,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224654871.97/warc/CC-MAIN-20230608103815-20230608133815-00240.warc.gz | 481,633,052 | 27,716 | ## Lies, damned lies, and … statistics
Note: this is the fourth part of our review of the paper “Bayesian reasoning in science” by Colin Howson and Peter Urbach. (The fifth and final installment of our review shall appear on 28 June.)
Let us return to Howson and Urbach’s Bayesian paper today. After presenting Bayes’s rule and the Bayesian approach to truth on pp. 371-372 of their paper (see our Bayesian blog posts of 25-26 June for reference), Howson and Urbach concede that the Bayesian approach to truth “has been widely criticized because it is based on personal, hence subjective, probabilities [cf. the problem of priors we talked about in our post of 26 June titled “Beliefs are like gambles”]. Scientific inference, critics say, should be perfectly objective.” Howson and Urbach thus spend the rest of their paper comparing and contrasting the Bayesian approach to truth with its leading challenger, what they refer to as the “classical statistical inference” model, an alternative approach to truth associated with the work of such giants as R. A. Fisher, Jerzy Neyman, and Egon Pearson (all of whom Howson & Urbach lump together as “classical statisticians”).
In brief, Howson and Urbach begin the second part of their paper by noting that the “classical” or non-Bayesian approach to truth “has two principal parts, the first relating to the testing of hypothesis (using significance tests) and the second to estimating the values of unknown parameters.” (In this post, we shall focus on Howson and Urbach’s critique of Fisherian hypothesis testing and the related idea of “significance”.) The authors then take a simple example to illustrate the Fisherian approach: an experimenter tossing a coin 20 times and counting the number of times the coin lands “heads” in order to test whether the coin is fair or not. “There are 21 possibilities,” they write, “ranging from no heads and 20 tails to 20 heads and no tails.” But how does the experimenter in this simple example know whether the coin is fair, i.e. how does he actually “test” his hypothesis in this case? If he is a Fisherian, he must perform a secondary “significance test”; that is, he must now proceed to “test” his results from the 20 previous coin tosses (though not the coin itself).
You will find the splendid details of Howson and Urbach’s critique of significance testing on pp. 372-373 of their paper, but their main point, as we understand it, is this: whether the experimenter’s coin-toss results in the example above are “significant” in a statistical sense at some predetermined level (such as 0.05) tells us nothing about the actual coin being tested! Why? Because a significance test is not a direct test of truth; it is simply a secondary or subsidiary test of one’s experimental data. (By way of analogy, consider the difference between a historical or legal investigation into the actual contents of a document versus an investigation of the way in which that document was made.) There is thus no necessary or logical relation between the “significance” of a given statistical test and the truth of the hypothesis being tested.
Worse yet, Howson and Urbach note that significance results are easy to manipulate and are super-sensitive to experimental design. In particular, they present this additional critique of significance testing on p. 373 of their paper — the stopping-rule problem:
In our earlier example, it was assumed … that because the coin was tossed 20 times, all of the possible outcomes would exhibit [some combination of] 20 heads and/or tails. But these are the possible outcomes only if the experimenter has a premeditated plan to throw the coin 20 times. Had the plan been to stop the experiment when, say six heads appeared, he could have got just the result he did, but with a different list of unrealized, possible outcomes.
So what? Here’s what:
Because significance is calculated by reference to these [unrealized, possible] outcomes, a result could be significant if the experimenter had had one plan (or stopping rule in mind), but not significant if it was another.
In short, in the eloquent words of Howson and Urbach: “This dependence of significance tests … on the subjective, possibly unconscious intentions of the experimenter is an astonishing thing to discover at the heart of supposedly objective methodologies. It is also a most inappropriate thing to find any methodology, for the plausibility, or cognitive value, of a hypothesis … should not depend on the experimenter’s mind.” (Ouch!)
But hold on in a minute … what about the problem of subjective priors (which we noted in our post “Beliefs are like gambles” below)? Does the subjective Bayesian approach to truth fare any better than standard Fisherian methods? Stay tuned … | 1,038 | 4,768 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2023-23 | latest | en | 0.928122 |
http://stackoverflow.com/questions/15825766/how-to-find-width-of-1d-discrete-gaussian-kernel-for-a-certain-sigma/15825933 | 1,411,121,772,000,000,000 | text/html | crawl-data/CC-MAIN-2014-41/segments/1410657131238.51/warc/CC-MAIN-20140914011211-00037-ip-10-196-40-205.us-west-1.compute.internal.warc.gz | 260,541,363 | 14,991 | # How to find width of (1D discrete) Gaussian Kernel for a certain sigma?
Is there a rule of thumb or mathematical equation that tells me how wide my (1D discrete) Gaussian Kernel should be for a certain sigma?
Lets say, I choose a sigma of 1.87, should my kernel be 5 values/steps/pixels wide, 7 oder rather 25 in order to perform standardized image smoothing? Thank you.
-
Pick a threshold that you consider ignorable, like `T = 0.01`. Then solve `exp(-x^2/(2s^2)) / sqrt(2pi s^2) < T`:
`|x| < s sqrt(-2 ln(sqrt(2pi s^2) T))`
The right hand side gives you the width. For `s = 1.87` and `T = 0.01`, this gives you a width of 4 pixels. | 195 | 639 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.046875 | 3 | CC-MAIN-2014-41 | latest | en | 0.790285 |
https://www.aqua-calc.com/convert/density/tonne-per-metric-cup-to-pound-per-cubic-foot | 1,563,728,798,000,000,000 | text/html | crawl-data/CC-MAIN-2019-30/segments/1563195527089.77/warc/CC-MAIN-20190721164644-20190721190644-00139.warc.gz | 619,532,567 | 10,341 | # Convert tonnes per (metric cup) to pounds per (cubic foot)
## t/metric cup (t/metric c) to lb/ft³ (t:tonne, c:cup, lb:pound, ft:foot)
### tonnes per metric cup equal to pounds per cubic foot conversion results
• 1
through
20
tonnes per metric cup
• 1 t/metric c = 249 711.842 lb/ft³
• 2 t/metric c = 499 423.684 lb/ft³
• 3 t/metric c = 749 135.526 lb/ft³
• 4 t/metric c = 998 847.368 lb/ft³
• 5 t/metric c = 1 248 559.21 lb/ft³
• 6 t/metric c = 1 498 271.052 lb/ft³
• 7 t/metric c = 1 747 982.894 lb/ft³
• 8 t/metric c = 1 997 694.736 lb/ft³
• 9 t/metric c = 2 247 406.578 lb/ft³
• 10 t/metric c = 2 497 118.42 lb/ft³
• 11 t/metric c = 2 746 830.262 lb/ft³
• 12 t/metric c = 2 996 542.104 lb/ft³
• 13 t/metric c = 3 246 253.946 lb/ft³
• 14 t/metric c = 3 495 965.788 lb/ft³
• 15 t/metric c = 3 745 677.63 lb/ft³
• 16 t/metric c = 3 995 389.472 lb/ft³
• 17 t/metric c = 4 245 101.314 lb/ft³
• 18 t/metric c = 4 494 813.156 lb/ft³
• 19 t/metric c = 4 744 524.998 lb/ft³
• 20 t/metric c = 4 994 236.84 lb/ft³
• 21
through
40
tonnes per metric cup
• 21 t/metric c = 5 243 948.682 lb/ft³
• 22 t/metric c = 5 493 660.524 lb/ft³
• 23 t/metric c = 5 743 372.366 lb/ft³
• 24 t/metric c = 5 993 084.208 lb/ft³
• 25 t/metric c = 6 242 796.05 lb/ft³
• 26 t/metric c = 6 492 507.892 lb/ft³
• 27 t/metric c = 6 742 219.734 lb/ft³
• 28 t/metric c = 6 991 931.576 lb/ft³
• 29 t/metric c = 7 241 643.418 lb/ft³
• 30 t/metric c = 7 491 355.26 lb/ft³
• 31 t/metric c = 7 741 067.102 lb/ft³
• 32 t/metric c = 7 990 778.944 lb/ft³
• 33 t/metric c = 8 240 490.786 lb/ft³
• 34 t/metric c = 8 490 202.628 lb/ft³
• 35 t/metric c = 8 739 914.47 lb/ft³
• 36 t/metric c = 8 989 626.312 lb/ft³
• 37 t/metric c = 9 239 338.154 lb/ft³
• 38 t/metric c = 9 489 049.996 lb/ft³
• 39 t/metric c = 9 738 761.838 lb/ft³
• 40 t/metric c = 9 988 473.68 lb/ft³
• 41
through
60
tonnes per metric cup
• 41 t/metric c = 10 238 185.522 lb/ft³
• 42 t/metric c = 10 487 897.364 lb/ft³
• 43 t/metric c = 10 737 609.206 lb/ft³
• 44 t/metric c = 10 987 321.048 lb/ft³
• 45 t/metric c = 11 237 032.89 lb/ft³
• 46 t/metric c = 11 486 744.732 lb/ft³
• 47 t/metric c = 11 736 456.574 lb/ft³
• 48 t/metric c = 11 986 168.416 lb/ft³
• 49 t/metric c = 12 235 880.258 lb/ft³
• 50 t/metric c = 12 485 592.1 lb/ft³
• 51 t/metric c = 12 735 303.942 lb/ft³
• 52 t/metric c = 12 985 015.784 lb/ft³
• 53 t/metric c = 13 234 727.626 lb/ft³
• 54 t/metric c = 13 484 439.468 lb/ft³
• 55 t/metric c = 13 734 151.31 lb/ft³
• 56 t/metric c = 13 983 863.152 lb/ft³
• 57 t/metric c = 14 233 574.994 lb/ft³
• 58 t/metric c = 14 483 286.836 lb/ft³
• 59 t/metric c = 14 732 998.678 lb/ft³
• 60 t/metric c = 14 982 710.52 lb/ft³
• 61
through
80
tonnes per metric cup
• 61 t/metric c = 15 232 422.362 lb/ft³
• 62 t/metric c = 15 482 134.204 lb/ft³
• 63 t/metric c = 15 731 846.046 lb/ft³
• 64 t/metric c = 15 981 557.888 lb/ft³
• 65 t/metric c = 16 231 269.73 lb/ft³
• 66 t/metric c = 16 480 981.572 lb/ft³
• 67 t/metric c = 16 730 693.414 lb/ft³
• 68 t/metric c = 16 980 405.256 lb/ft³
• 69 t/metric c = 17 230 117.098 lb/ft³
• 70 t/metric c = 17 479 828.94 lb/ft³
• 71 t/metric c = 17 729 540.782 lb/ft³
• 72 t/metric c = 17 979 252.624 lb/ft³
• 73 t/metric c = 18 228 964.466 lb/ft³
• 74 t/metric c = 18 478 676.308 lb/ft³
• 75 t/metric c = 18 728 388.15 lb/ft³
• 76 t/metric c = 18 978 099.992 lb/ft³
• 77 t/metric c = 19 227 811.834 lb/ft³
• 78 t/metric c = 19 477 523.676 lb/ft³
• 79 t/metric c = 19 727 235.518 lb/ft³
• 80 t/metric c = 19 976 947.36 lb/ft³
• 81
through
100
tonnes per metric cup
• 81 t/metric c = 20 226 659.202 lb/ft³
• 82 t/metric c = 20 476 371.044 lb/ft³
• 83 t/metric c = 20 726 082.886 lb/ft³
• 84 t/metric c = 20 975 794.728 lb/ft³
• 85 t/metric c = 21 225 506.57 lb/ft³
• 86 t/metric c = 21 475 218.412 lb/ft³
• 87 t/metric c = 21 724 930.254 lb/ft³
• 88 t/metric c = 21 974 642.096 lb/ft³
• 89 t/metric c = 22 224 353.938 lb/ft³
• 90 t/metric c = 22 474 065.78 lb/ft³
• 91 t/metric c = 22 723 777.622 lb/ft³
• 92 t/metric c = 22 973 489.464 lb/ft³
• 93 t/metric c = 23 223 201.306 lb/ft³
• 94 t/metric c = 23 472 913.148 lb/ft³
• 95 t/metric c = 23 722 624.99 lb/ft³
• 96 t/metric c = 23 972 336.832 lb/ft³
• 97 t/metric c = 24 222 048.674 lb/ft³
• 98 t/metric c = 24 471 760.516 lb/ft³
• 99 t/metric c = 24 721 472.358 lb/ft³
• 100 t/metric c = 24 971 184.2 lb/ft³
#### Foods, Nutrients and Calories
NATURAL SLICED ALMONDS, UPC: 041512083013 weigh(s) 63.4 gram per (metric cup) or 2.12 ounce per (US cup), and contain(s) 633 calories per 100 grams or ≈3.527 ounces [ calories | weight to volume | volume to weight | price | density ]
Foods high in Vitamin D and Recommended Dietary Allowances (RDAs) for Vitamin D
#### Gravels, Substances and Oils
CaribSea, Marine, Aragonite, Special Coarse Aragonite weighs 1 153.3 kg/m³ (71.99817 lb/ft³) with specific gravity of 1.1533 relative to pure water. Calculate how much of this gravel is required to attain a specific depth in a cylindricalquarter cylindrical or in a rectangular shaped aquarium or pond [ weight to volume | volume to weight | price | density ]
Sodium Aluminate, powder [NaAlO2] weighs 1 550 kg/m³ (96.76334 lb/ft³) [ weight to volume | volume to weight | price | mole to volume and weight | density ]
Volume to weightweight to volume and cost conversions for Refrigerant R-508B, liquid (R508B) with temperature in the range of -106.67°C (-160.006°F) to -6.65°C (20.03°F)
#### Weights and Measurements
A troy ounce is a troy weight measurement unit
Scientifically, the weight of a body is the gravitational force acting on the body, and directed towards the center of gravity with magnitude of the force depending on the mass of the body.
Convert troy pound per cubic millimeter to long ton per liter or convert between all units of density measurement
#### Calculators
Volume of a rectangular box calculator. Definition and formulas. | 2,627 | 5,877 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.0625 | 3 | CC-MAIN-2019-30 | latest | en | 0.243736 |
https://www.forbes.com/2009/02/06/math-archimedes-churchill-opinions-contributors_0207_joseph_tartakovsky.html | 1,656,149,817,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103034877.9/warc/CC-MAIN-20220625065404-20220625095404-00351.warc.gz | 851,604,072 | 83,951 | BETA
This is a BETA experience. You may opt-out by clicking here
# Math Wrath
At the tender age of 8, I concluded that, among the varied destinies shimmering before me, being a profound mathematical genius was not one of them. I won't have a number named after me, like Signor Fibonacci, or propose a problem to perplex the generations, like Monsieur Fermat. Chances are I won't even get a dinner tip right.
This is because in second grade, I caught a flu and missed the week in which multiplication was taught. When I returned, the class had progressed to long division, which, without a solid grasp of multiplication, is a futile art. That year, I renounced the hateful science altogether.
I believed my misfortune was to have been born in an age when mathematics is so advanced that it requires things like supercomputers and long division. Then I learned what sort of people great mathematicians are. I learned, in short, about the sort of person I might have become.
Archimedes (c. 287 B.C. - c. 212 B.C.) lived in Syracuse when Rome besieged it, and the invading general ordered that the city's most famous native be protected. The soldier sent for him found the man lost in thought, etching symbols in the sand. When the Greek gestured absently for the centurion to wait, murmuring, "Don't disturb my circles," the furious soldier forgot his manners (and his orders). With his long sword he performed what later geometers would call segmentation on Archimedes's head, perhaps making literal the thinker's famed absentmindedness. The episode reveals an important truth, or axiom, or even postulate, about mathematicians: An abstract art produces an abstracted mind.
Karl Friedrich Gauss (1777-1855), perhaps history's greatest mathematician, was one day wrapped in a particularly sticky problem. A messenger broke in to relate that his wife was dying. Gauss, absorbed, replied, "Tell her to wait a minute until I've finished."
Those with number-lovers in the family learn that mathematicians are often as emotive as an abacus. Peter Gustav Lejeune Dirichlet (1805 -1859), inventor of the "formal" definition of a function, so hated writing letters that many friends never once received one from him. It was only on the occasion of the birth of his first child that he was moved to communicate; overwhelmed with emotion, he wired his father-in-law: "2 + 1 = 3."
But worse even than abstraction is the mathematician's instinctive pedantry. Charles Babbage (1792-1871), the British inventor of the "Analytical Engine," a precursor to the computer, objected to these lines of Tennyson: "Every moment dies a man, / Every moment one is born." If that were accurate, he reasoned, the "population of the world would be at a standstill." He telegrammed Tennyson a fix: "Every moment dies a man, / Every moment one and one-sixteenth is born."
For more on this topic, see
Is math taught, or only inflicted? John von Neumann (1903-1957), who devised the notion of a zero-sum game, admitted that "[i]n mathematics you don't understand things, you just get used to them." Well, I refuse. Why teach math in the age of the calculator? The device is available everywhere, from cellphones to fashionable watches.
Once a visitor to the Indian prodigy Srinivasa Ramanujan (1887-1920) noted that his cab number, 1729, seemed "rather a dull one." "No," replied Ramanujan, "it is a very interesting number. It is the smallest number expressible as the sum of two cubes in two different ways." He did that in his head. So what? Give me two minutes and my calculator watch, and I'll do the same without exerting any little gray cells.
Is my existence blighted by an inability to add 15% to a check? Do I suffer because I can't compute how many blocks intervene between 16th and 31st Streets? I'm not a supercomputer. The parents of Hungarian mathematician Paul Erdos (1913-1996) had him amuse house guests by calculating how many seconds they had lived. Mine didn't, but do you think I feel neglected? Far from it.
Our numerals are imported from the Middle East. As the war on terror progresses, it is crucial to reduce our dependence on the region. Numerals are abstractions, and abstractions lose wars.
Benjamin Franklin and Winston Churchill were the two most productive men of their ages. "I acquired fair writing pretty soon," wrote Franklin, "but I fail'd in the arithmetic and made no progress in it"--which practically proves that a life of numbers and a life of industry are incompatible. In his memoir My Early Life, Churchill describes struggling with math as he tried (repeatedly) to pass into the prestigious Royal Military Academy at Sandhurst. Speaking of fractions, cosines and quadratic equations, he wrote:
"I have never met any of these creatures since. With my third and successful examination, they passed away like the phantasmagoria of a fevered dream. I am assured that they are most helpful in engineering, astronomy and things like that. It is very important to build bridges and canals and to comprehend all the stresses and potentialities of matter, to say nothing of counting all the stars and even universes and measuring how far off they are, and foretelling eclipses, the arrival of comets and such like. I am very glad there are quite a number of people born with a gift and a liking for all of this; like great chess-players who play 16 games at once blindfold and die quite soon of epilepsy. Serve them right!"
He quit math forever in 1894, at age 20. (I was precocious in this respect.) And yet he had a fairly successful career nonetheless. The immortal example of Churchill, savior of the West, proves not only that math was unnecessary to save civilization, but that, as it nearly kept him out of the academy, math positively endangers it.
The old toast of the Mathematical Society of England was, "Pure mathematics; may it never be of use to any man!" Take out the "pure," whatever that means, and raise your glass.
Joseph Tartakovsky is a contributing editor of the Claremont Review of Books and a J.D. candidate at Fordham Law School. | 1,376 | 6,075 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.65625 | 3 | CC-MAIN-2022-27 | latest | en | 0.973367 |
https://community.alteryx.com/t5/Weekly-Challenge/Challenge-46-Formatting-a-Date-from-an-Integer/td-p/40308/page/52 | 1,695,548,253,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233506632.31/warc/CC-MAIN-20230924091344-20230924121344-00641.warc.gz | 202,842,671 | 47,342 | Happy 8th birthday to the Maveryx Community! Take a walk down memory lane in our birthday blog, and don't miss out on the awesome birthday present that all Maveryx Community members get to take advantage of!
alteryx Community
# Weekly Challenge
Solve the challenge, share your solution and summit the ranks of our Community!
Also available in | Français | Português | Español | 日本語
###### IDEAS WANTED
Want to get involved? We're always looking for ideas and content for Weekly Challenges.
## Challenge #46: Formatting a Date from an Integer
8 - Asteroid
Solution.
9 - Comet
Solution Attached
8 - Asteroid
Here is my solution
Formula used:
Spoiler
if StartsWith([date], "1")
then ReplaceFirst([date], "1", "20")
else ReplaceFirst([date], "0", "19")
endif
8 - Asteroid
Pretty simple one.
12 - Quasar
my solution
7 - Meteor
Solution for the Challenge #46: Formatting a Date from an Integer
7 - Meteor
Here is my solution
7 - Meteor
Here is my solution for this problem. I used wayy too many formulas.
7 - Meteor
Completed
7 - Meteor
only one tool solution 🙂 | 268 | 1,083 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.765625 | 3 | CC-MAIN-2023-40 | latest | en | 0.758604 |
http://www.rhlschool.com/math4n25.htm | 1,708,732,088,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474470.37/warc/CC-MAIN-20240223221041-20240224011041-00750.warc.gz | 54,651,061 | 1,734 | Name_______________________________________________Date_________________________
Mathematics Problem Solving
Volume 4, Number 25, March 22, 1999
www.rhlschool.com
Those Surveyed
1. One hundred adults answered questions in a telephone survey. There were an equal number of men and women surveyed. Only 20%, (1/5) of the men admitted that they had ever eaten Toasty Ouchy Cereal. Half of the women said that they had eaten that cereal.
Put a T in front of every statement that is true.
Put an M in front of every statement that might be true.
Put an F in front of every statement that is false.
1. ____ Of those surveyed, fifteen more women than men said that they had ever eaten Toasty Ouchy Cereal.
2. ____ Of those surveyed, fifteen more women than men had ever eaten Toasty Ouchy Cereal.
3. ____ Of those surveyed, fifteen more men than women said that they had ever eaten Toasty Ouchy Cereal.
4. ____ Sixty-five of the people surveyed did not say that they had ever eaten Toasty Ouchy Cereal.
5. ____ In the country where the survey was taken, Toasty Ouchy Cereal is eaten by more women than men.
6. ____ In the country where the survey was taken, Toasty Ouchy Cereal is eaten by more men than women.
rhlschool.com | 299 | 1,227 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.28125 | 3 | CC-MAIN-2024-10 | latest | en | 0.975927 |
https://www.clutchprep.com/chemistry/practice-problems/123013/consider-the-signs-for-hrxn-and-srxn-for-several-different-reactions-in-which-ca | 1,618,188,286,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038065903.7/warc/CC-MAIN-20210411233715-20210412023715-00490.warc.gz | 820,570,338 | 30,626 | Spontaneous Reaction Video Lessons
Concept
# Problem: Consider the signs for ΔHrxn and ΔSrxn for several different reactions. In which case is the reaction spontaneous at all temperatures?a) ΔHrxn < 0; ΔSrxn < 0b) ΔHrxn > 0; ΔSrxn > 0c) ΔHrxn < 0; ΔSrxn > 0d) ΔHrxn > 0; ΔSrxn < 0
###### FREE Expert Solution
Recall that when ΔH and ΔS are given, we can determine the spontaneity of the reaction based on their signs
We can use the Punnett square shown below:
82% (301 ratings)
###### Problem Details
Consider the signs for ΔHrxn and ΔSrxn for several different reactions. In which case is the reaction spontaneous at all temperatures?
a) ΔHrxn < 0; ΔSrxn < 0
b) ΔHrxn > 0; ΔSrxn > 0
c) ΔHrxn < 0; ΔSrxn > 0
d) ΔHrxn > 0; ΔSrxn < 0 | 279 | 761 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.8125 | 3 | CC-MAIN-2021-17 | latest | en | 0.710753 |
https://docs.lammps.org/stable/fix_accelerate_cos.html | 1,726,537,992,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651722.42/warc/CC-MAIN-20240917004428-20240917034428-00388.warc.gz | 190,243,629 | 7,941 | $$\renewcommand{\AA}{\text{Å}}$$
# fix accelerate/cos command
## Syntax
fix ID group-ID accelerate value
• ID, group-ID are documented in fix command
• accelerate/cos = style name of this fix command
• value = amplitude of acceleration (in unit of velocity/time)
## Examples
fix 1 all accelerate/cos 2.0e-7
## Description
Give each atom a acceleration in x-direction based on its z coordinate. The acceleration is a periodic function along the z-direction:
$a_{x}(z) = A \cos \left(\frac{2 \pi z}{l_{z}}\right)$
where $$A$$ is the acceleration amplitude, $$l_z$$ is the $$z$$-length of the simulation box. At steady state, the acceleration generates a velocity profile:
$v_{x}(z) = V \cos \left(\frac{2 \pi z}{l_{z}}\right)$
The generated velocity amplitude $$V$$ is related to the shear viscosity $$\eta$$ by:
$V = \frac{A \rho}{\eta}\left(\frac{l_{z}}{2 \pi}\right)^{2}$
and it can be obtained from ensemble average of the velocity profile:
$V = \frac{\sum\limits_i 2 m_{i} v_{i, x} \cos \left(\frac{2 \pi z_i}{l_{z}}\right)}{\sum\limits_i m_{i}},$
where $$m_i$$, $$v_{i,x}$$, and $$z_i$$ are the mass, $$x$$-component velocity, and $$z$$-coordinate of a particle, respectively.
The velocity amplitude $$V$$ can be calculated with compute viscosity/cos, which enables viscosity calculation with periodic perturbation method, as described by Hess. Because the applied acceleration drives the system away from equilibration, the calculated shear viscosity is lower than the intrinsic viscosity due to the shear-thinning effect. Extrapolation to zero acceleration should generally be performed to predict the zero-shear viscosity. As the shear stress decreases, the signal-to-noise ratio decreases rapidly, and the simulation time must be extended accordingly to get converged results.
In order to get meaningful results, the group ID of this fix should be all.
## Restart, fix_modify, output, run start/stop, minimize info
No information about this fix is written to binary restart files. None of the fix_modify options are relevant to this fix. No global or per-atom quantities are stored by this fix for access by various output commands. No parameter of this fix can be used with the start/stop keywords of the run command.
This fix is not invoked during energy minimization.
## Restrictions
This fix is part of the MISC package. It is only enabled if LAMMPS was built with that package. See the Build package page for more info.
Since this fix depends on the $$z$$-coordinate of atoms, it cannot be used in 2d simulations.
## Default
none
(Hess) Hess, B. Journal of Chemical Physics 2002, 116 (1), 209–217. | 682 | 2,647 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.96875 | 3 | CC-MAIN-2024-38 | latest | en | 0.807634 |
https://www.physicsforums.com/threads/conveyor-speed-distance-calculations.582817/ | 1,527,157,811,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794866201.72/warc/CC-MAIN-20180524092814-20180524112814-00380.warc.gz | 805,389,667 | 16,155 | # Conveyor Speed / Distance calculations
1. Mar 1, 2012
### Paulvp1968
I would like the formulae for calculating the distance created bewteen products transferring from one conveyor to another with a line of products moving at x M/min transferring onto a conveyor running y M/min. I need to be able to calculate what the gap will be between the products as this is critical to the design of the system. Urgent Help required PLEASE!!!!
2. Mar 1, 2012
### tvavanasd
A very simplified approximation of product gap can be calculated if you assume infinite product acceleration from one conveyor to the other. If you are interested, I will help you find the solution, but I can't just give it to you.
Vx = upstream conveyor speed (m/min)
Vy = downstream conveyor speed (m/min)
L = product length (m)
G = gap between products (m)
Note that the units of the conveyor speed must be the same, and the gap units will be the same as those for the product length. Please use lower case "m" for metre.
Question:
If we assume that the product will accelerate from speed Vx to speed Vy at the instant that the leading edge of the product contacts conveyor y, and both products start fully on conveyor x, how much time will pass before the leading edge of the trailing product will contact conveyor y?
Last edited: Mar 1, 2012
3. Mar 1, 2012
### Paulvp1968
Assume both products are touching on conveyor x and based upon a product width of 1m moving at 5m/min then 12 seconds would be the time elapsed for the leading edge of the trailing product to contact conveyor y
4. Mar 1, 2012
### tvavanasd
FYI, I just changed the conveyor speed variables names to Vx and Vy.
So the time is obviously the product length divided by upstream speed Vx, or:
t = L/Vx
a distance:
D = Vy * t
(Yes, I did just introduce variables t and D, but they are only temporary.)
Making sense?
I'm trying to move this along quickly...
G = ?
5. Mar 1, 2012
### Paulvp1968
So D=2m so if i am understanding this G=D-L = 1m and that G is proportional to the original gap
6. Mar 1, 2012
### tvavanasd
Yes, G = D - L
Sorry, what was Vy? It must be 15 m/min, although I don't think you mentioned it.
You can either use L = actual product length (assuming no gap) or Leff = effective product length which is equal to product length + initial gap.
Yes, the final gap is proportional to product length (or effective product length).
So what is G in terms for L, Vx and Vy? I thought that was what you wanted.
7. Mar 2, 2012
### Paulvp1968
So if G=D-L and we have the following form the calculations performed above
t= L/Vx = 0.2min
in the same time period "t" the leading edge of the product will move Vy*t to give product D = 10*0.2 = 2m. Therefore with G=D-L then G=2-1 so the gap between the products is 1m.
Calculated in terms for L, Vx and Vy, I can see the following G= Vy/VxL-L am I correct?
8. Mar 2, 2012
### tvavanasd
Not quite, it should be:
G = L * (Vy / Vx) - L
or more simply:
G = L * [(Vy / Vx) - 1]
To check:
If Vy = Vx, G should be 0.
Looks like I got D and G mixed up previously when I indicated that Vy = 15 m/min (I was wrong).
Last edited: Mar 2, 2012
9. Mar 5, 2012
### Paulvp1968
Many thanks for your asistance. Much appreciated. | 878 | 3,235 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.703125 | 4 | CC-MAIN-2018-22 | latest | en | 0.918879 |
https://samairey.com/sales-tax-and-discount-worksheet-with-answers/ | 1,660,215,900,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882571284.54/warc/CC-MAIN-20220811103305-20220811133305-00028.warc.gz | 482,621,756 | 15,639 | Sales Tax And Discount Worksheet With Answers. 3% 18) original price of a book: What is the sale price of the dress?
In a grocery store, a \$21 case of toilet paper is labeled, get a 20% discount. what is the discount? What is the sale price of the case of soda? The total bill is 86 50.
### A Machine Is Sold At Rs 66000 After A 10% Discount And 10% Vat On The Marked Price.
Sales tax and discount worksheet. 17) the sales tax in texas is 8.25% and an item costs \$400. Task cards (5 problem task cards and 1 answer key card)each card measures 3.67 (width) by 4.25 (height) inches.
### In A Department Store, A \$50 Dress Is Marked, Save 25%. What Is The Discount?
Find the amount of discount, sale price, sales tax, and total cost for each item. Markup discount and tax worksheet answers. 2nd grade getting to know you worksheet.
### Calculating Purchase Price Accounting For Discounts And Sales Tax.
Discount commission grade 6 displaying top 8 worksheets found for this concept. \$4.00 at 25% off basketball: The bubbles allow heir answers and results in a more confident student.
### Tax Tip And Discount Worksheet 371.
0.10 x \$15.00 = \$1.50 the sale price is calculated as follows: Convert the percent to decimal. What is the sale price of the dress?
### 48% 15) Original Price Of A Cd:
In a department store, a \$40 dress is marked, save 25%. what is the discount? 21 posts related to sales tax and discount worksheet 7th grade answer key. What is the sale price of the case of soda?
Posted in printable worksheets | 387 | 1,535 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.078125 | 3 | CC-MAIN-2022-33 | latest | en | 0.890174 |
https://decaforum.decawave.com/t/the-question-is-about-the-calibration-of-the-antenna-delay-and-the-creation-of-a-reference-device/11013 | 1,631,985,639,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780056548.77/warc/CC-MAIN-20210918154248-20210918184248-00269.warc.gz | 256,586,621 | 5,235 | # The question is about the calibration of the antenna delay and the creation of a reference device
To increase the accuracy in the local positioning system we are creating, we want to calibrate the antenna delay. Also, to calibrate the antenna delay, we want to create a reference device.
When creating a reference device, we use the methodology described in APS 014 paragraph 2.2.1. That is, we take three devices that we place in the corners of an equilateral triangle. The length of the sides of the triangle is known. Next, we compose a Euclidean matrix with these known actual distances.
Then we set the antenna delay in the devices to 0 and make a measurement between the devices using the TWR method, recording the measured values.
Then we have a question that we haven’t been able to answer yet. In APS 014, table No. 3 is presented as an example, which we are focusing on. In this table, it is not clear to us how the values [email protected], PRF=16MHz, Channel=were determined2 and [email protected],914 m, PRF=16MHz, Channel=2, which affect the values of distances entered in the second Euclidean matrix.
Please explain how the values in the example in APS 014 were obtained, and how we should determine the RangeBias values when creating a reference device.
This number is an RSSI based correction factor, see section 3 of APS011.
The correct correction factor for your system will depend on the transmit power, the distance used and the antenna gain.
While you can use values in table 2 of APS011 my advice would be to characterise this correction factor for your system and use your own correction table.
This has the advantage that you can then define which range / RSSI you want to call a correction of 0. You can then make this zero point such that it matches the range you are using for your delay calibrations.
We know that this correction factor is RSSI based and depends on the transmit power, the distance used and the antenna gain. We had studied APS011 also. The question is that there are no RangeBias values in the table 2 APS011 that are given in table 3 in example in APS014. I would like to have explanations of how the values appeared in table 3 APS011 when [email protected], PRF=16MHz, Channel=2 and [email protected], PRF=16MHz, Channel=2. And we do not understand why in Table 3 APS014 two RangeBias values are subtracted from the measured distance value with an antenna delay equal to 0. It is not clear to us how the numbers +0.13 and -0.07 turned out in table 3 and why both of these values are taken into account when getting the values entered into the matrix.
Do you suggest, that we use our own values instead of the correction coefficients given in the table? Can you then tell us how to create our own correction table yourself?
How I created my own was I collected large amounts of data of UWB measured range vs truth between 8 different units and found the delay calibration values and RSSI correction factors that resulted in minimum average error for that data set.
I then assume the same RSSI correction factors for all units and so future units only need the antenna delay calibrated against a known unit. | 699 | 3,165 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.15625 | 3 | CC-MAIN-2021-39 | latest | en | 0.918754 |
https://www.edaboard.com/showthread.php?291593-dc-to-dc-signal-booster-for-analog-to-digital-convertor | 1,555,848,562,000,000,000 | text/html | crawl-data/CC-MAIN-2019-18/segments/1555578531462.21/warc/CC-MAIN-20190421120136-20190421141855-00053.warc.gz | 681,903,277 | 16,615 | # dc to dc signal booster for analog to digital convertor
1. ## dc to dc signal booster for analog to digital convertor
I would like to digitalise an analog DC ammeter of 0-750A range using pic16f876a. The voltage across the shunt resistor of ammeter ranges from 0-60mv DC. If I connect this shunt voltage to the input pin of 10 bit Analog to digital convertor of pic then the resolution will be very low. With 5v as reference, 1 division is equal to 4.8mv. So by converting this value to ampere range it will be 4.8*12.5= 60A. So the ampere changes by 60A for each division. ie 60A,120A....etc. I would like to boost the input DC value of 0-60mv to some what near the reference voltage, ie 0-5V. Any body have idea of doing this DC to DC signal boosting without loading the input signal. Please help me ...
Regards
sajeev
•
2. ## Re: dc to dc signal booster for analog to digital convertor
Use an op amp to gain to the voltage from 0-.06 to 0-5.0
--[[ ]]-- | 272 | 964 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.59375 | 3 | CC-MAIN-2019-18 | latest | en | 0.81852 |
https://www.allaboutcircuits.com/tools/lorentz-force-calculator | 1,695,909,762,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510412.43/warc/CC-MAIN-20230928130936-20230928160936-00706.warc.gz | 703,887,294 | 17,712 | # Lorentz Force Calculator
## This calculator will determine the combined Lorentz force exerted on a charged particle moving through both an electric field and a magnetic field.
### Lorentz Force Calculator
Enter any 5 of the 6 variables and the calculator will provide the missing value.
In this Lorentz force calculator, enter any five of the six variables and this calculator will provide the value for the missing parameter. The six variables in the Lorentz formula are the:
• Lorentz force
• Particle charge
• Particle velocity
• Electric field strength
• Magnetic field strength
• Relative angle of the velocity vector to the magnetic field vector
### Lorentz Force Definition—What is Lorentz Force?
The Lorentz force is the combined force exerted on a particle with charge q moving with velocity v through both a magnetic field B and an electric field E. Let's first examine the effects of the electric and magnetic fields independently.
### Lorentz Force Electric Field Component
The electric field imparts a force on the charged particle that is independent of the velocity vector of the particle. The direction of the force is a function of the polarity of the charged particle. A positively charged particle will be accelerated in the same linear orientation as the electric field, whereas a negatively charged particle will be accelerated in a direction opposite to the electric field. These effects can be seen in Figure 1.
### The Magnetic Field and Lorentz Force
The magnetic field imparts a force on the charged particle that is a function of both the polarity of the charge and the velocity of the particle. This is a cross-product such that the force is perpendicular to both the velocity vector and the magnetic field vector (Figure 2).
##### Figure 2. The Lorentz force acting on a charged particle moving through a magnetic field
Due to this, the particle will curve perpendicularly to both the instantaneous velocity vector and the magnetic field, as shown in Figure 3.
### How to Calculate Lorenz Force—the Lorentz Force Equation
To determine the total force exerted on a charged particle moving through both a magnetic field and an electric field, we can sum the forces exerted by both individually. This is expressed using the following Lorentz force equation:
$$F_l = qE + q v \times B$$
##### Equation 1. Lorentz force equation
Where:
• Fl = the Lorentz force expressed in Newtons
• q = the total charge of the particle expressed in coulombs
• E = The electric field expressed in V/m (or equivalently N/C)
• v = The velocity of the charged particle expressed in m/s
• B = The magnetic field expressed in teslas
As can be seen from the equation, the Lorentz force is the sum of the force exhibited by the electric field on the particle (qE) and the force exhibited by the magnetic field on the particle (qv X B). If you need to find the force exhibited by only one field, the other field’s strength can be set to 0.
The contribution of the magnetic field to the total Lorentz force is a cross product, meaning that it is dependent on the angle of the magnetic field relative to the particle’s velocity. Based on this, we can further simplify the Lorentz force formula to the following:
$$F_l = qE + q v B \sin(\theta)$$
##### Equation 2. Simplified Lorentz force equation
Where:
• $$\theta$$ = The angle between the particle’s velocity vector and the magnetic field vector, expressed in radians
For this calculator, the angle is limited to:
$$-\frac{\pi}{2} \le \theta \le \frac{\pi}{2}$$
Or, equivalently, if the angle is expressed in degrees:
$$-90\le \theta \le 90$$
### Lorentz Force Applications
Understanding this force is important in analyzing or designing applications such as motors, sensors, speakers, and imaging. Other applications of the Lorentz force include:
• Cyclotrons and other particle accelerators
• Bubble chamber for tracking the trajectories of charged particles. | 874 | 3,946 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.953125 | 4 | CC-MAIN-2023-40 | longest | en | 0.858562 |
https://educatorpages.com/site/mrspeitz/pages/103933 | 1,638,839,992,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964363327.64/warc/CC-MAIN-20211206224536-20211207014536-00214.warc.gz | 294,132,216 | 8,020 | Chapter 12 Study Guide
Lesson 1
Words to Know: position, motion, distance, speed, average speed, velocity, acceleration
Ideas to Understand:
1. How is motion measured?
2. How can position be described using other objects?
3. What is the difference between speed and velocity?
4. What are examples of acceleration?
Lesson 2
Words to Know: force, friction, drag force, gravity, net force, balanced forces, unbalanced forces, inertia
Ideas to Understand
1. How does force affect moving objects?
2. What are forces that can be exerted?
3. How are combined effects of many forces measured?
4. What are some examples of inertia?
Lesson 3
Words to Know: momentum
Ideas to Understand:
1. What is necessary to accelerate an object?
2. What is the relationship between the variables in the second law of motion?
3. What is the third law of motion?
What happens to mass and weight in space? | 206 | 892 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.6875 | 3 | CC-MAIN-2021-49 | latest | en | 0.923862 |
https://www.jiskha.com/users/jason | 1,534,876,337,000,000,000 | text/html | crawl-data/CC-MAIN-2018-34/segments/1534221218391.84/warc/CC-MAIN-20180821171406-20180821191406-00063.warc.gz | 935,482,042 | 3,557 | # jason
1. ## Vectors
Determine if the line through A(1, 2, 3) and B(4, 8, -3) and the line through C(3, 5, 3) and D(0, 2, 4) intersect. Having a hard time figuring out this problem.
2. ## Chemistry
A weak acid with an initial pH of 3.2 was titrated with a strong base. 15 mL of 0.1 M NaOH was added to the acid to reach the equivalence point at a pH of 8.6. What would you expect the approximate pH of the analyte to be after the first 5 mL of 0.1 M NaOH
3. ## Chemistry
What is the pH of a 0.25 M acetic acid solution if the Ka of acetic acid is 1.8 x 10-5? 2.67 0.60 5.34 11.33 I got A, 2.67. I think this is correct as well since acetic acid is a weaker acid and the ph # would make sense.
4. ## Social Studies
8.Which of the following best describes changes in the population of Washington over the last 200 years? Cultural diversity has increased due to migration and immigration. A.The Native American population has not changed locations. B.Immigration from
5. ## social studies
Describe four measures that can be taken to improve or develop human resources in ghana
1. ## Vectors
Steve, I still don't understand how those values for t and s are the same, when I do substitute the numbers in, it gives me different values.
2. ## Vectors
Also, I think you've got the wrong answer posted. I don't know how you got -1/3, but when I use both equations for the third equation, I get different results.
3. ## Vectors
Does my answer have to be expressed as a point, or just what the parameters are that satisfy the condition that they make both lines intersect?
4. ## Social Studies
thank you so muchhhhh!!!!!!
5. ## Social Studies
A.Cultural diversity has increased due to migration and immigration. B.The Native American population has not changed locations. C..Immigration from Pacific Rim countries has decreased due to a lack of C.work. D.African Americans arrived in the state as | 505 | 1,898 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.6875 | 3 | CC-MAIN-2018-34 | latest | en | 0.940895 |
https://socratic.org/questions/how-do-you-graph-y-2sqrtx-3-and-how-is-it-different-from-the-parent-function#153676 | 1,656,532,356,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103642979.38/warc/CC-MAIN-20220629180939-20220629210939-00665.warc.gz | 579,170,162 | 6,026 | # How do you graph y=-2sqrtx-3 and how is it different from the parent function?
Jun 19, 2015
See below
#### Explanation:
The parent function is $y = \sqrt{x}$
graph{sqrtx [-4.07, 11.74, -4.17, 3.73]}
Working outwards from that:
Multiplying by 2 stretches the graph vertically
$y = 2 \sqrt{x}$
graph{2sqrtx [-4.02, 13.766, -2.07, 6.814]}
Then the $-$ reflects the graph across the $x$-axis. (It makes the $y$'s negative)
$y = - 2 \sqrt{x}$
graph{-2sqrtx [-2.88, 17.13, -8.22, 1.775]}
Finally, subtracting $3$ translates everything down $3$ (subtracts $3$ from the most recent $y$-values.
$y = - 2 \sqrt{x} - 3$
graph{y=-2sqrtx-3 [-2.07, 20.44, -11.14, 0.105]} | 261 | 666 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 11, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.40625 | 4 | CC-MAIN-2022-27 | latest | en | 0.716403 |
http://www.godlikeproductions.com/forum1/message2067516/reply34795662/copyright | 1,498,262,029,000,000,000 | text/html | crawl-data/CC-MAIN-2017-26/segments/1498128320201.43/warc/CC-MAIN-20170623220935-20170624000935-00441.warc.gz | 529,067,635 | 5,238 | Users Online Now: 2,000 (Who's On?) Visitors Today: 2,692,603 Pageviews Today: 3,288,963 Threads Today: 649 Posts Today: 11,239 07:53 PM
Back to Forum Back to Thread
Message Subject This is Huge: 25,920 Minutes from December 3, 2012, 11:11am to December 21, 2012, 11:11am.((THE TIME HAS COME)))
Poster Handle Anonymous Coward
Post Content
The majick of 9 (and 18 and 108)
The center point of 12-3-12(9) and 12-21-12(9) would be 12-12-12(9).
25,920=18=9
25,920/2=12,960=18=9
25,920 divided by 24 = 1080 miles which is the radius of the Moon."
1080p(pi in the eye/I) provides the clearest picture...LOL, no coincidences.
"DIAMETER OF THE SUN
Masonic number 33.333333333 x 25,920 = 864,000 which equals the Diameter of the Sun in miles."
864,000=18=9
Quoting: Anonymous Coward 28466199
"In this fascinating 3 dimensional model of 'SRI CHAKRA SPACE FRAMETM' (SCSF), 9pyramids are intertwined in an enticing manner forming a configuration of 45 projected pyramids."
45 projected pyramids 4+5= 9 | 337 | 1,000 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.953125 | 3 | CC-MAIN-2017-26 | latest | en | 0.784943 |
https://metanumbers.com/557187 | 1,623,618,919,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623487610841.7/warc/CC-MAIN-20210613192529-20210613222529-00128.warc.gz | 369,600,526 | 7,567 | ## 557187
557,187 (five hundred fifty-seven thousand one hundred eighty-seven) is an odd six-digits composite number following 557186 and preceding 557188. In scientific notation, it is written as 5.57187 × 105. The sum of its digits is 33. It has a total of 3 prime factors and 8 positive divisors. There are 366,600 positive integers (up to 557187) that are relatively prime to 557187.
## Basic properties
• Is Prime? No
• Number parity Odd
• Number length 6
• Sum of Digits 33
• Digital Root 6
## Name
Short name 557 thousand 187 five hundred fifty-seven thousand one hundred eighty-seven
## Notation
Scientific notation 5.57187 × 105 557.187 × 103
## Prime Factorization of 557187
Prime Factorization 3 × 79 × 2351
Composite number
Distinct Factors Total Factors Radical ω(n) 3 Total number of distinct prime factors Ω(n) 3 Total number of prime factors rad(n) 557187 Product of the distinct prime numbers λ(n) -1 Returns the parity of Ω(n), such that λ(n) = (-1)Ω(n) μ(n) -1 Returns: 1, if n has an even number of prime factors (and is square free) −1, if n has an odd number of prime factors (and is square free) 0, if n has a squared prime factor Λ(n) 0 Returns log(p) if n is a power pk of any prime p (for any k >= 1), else returns 0
The prime factorization of 557,187 is 3 × 79 × 2351. Since it has a total of 3 prime factors, 557,187 is a composite number.
## Divisors of 557187
8 divisors
Even divisors 0 8 4 4
Total Divisors Sum of Divisors Aliquot Sum τ(n) 8 Total number of the positive divisors of n σ(n) 752640 Sum of all the positive divisors of n s(n) 195453 Sum of the proper positive divisors of n A(n) 94080 Returns the sum of divisors (σ(n)) divided by the total number of divisors (τ(n)) G(n) 746.45 Returns the nth root of the product of n divisors H(n) 5.92248 Returns the total number of divisors (τ(n)) divided by the sum of the reciprocal of each divisors
The number 557,187 can be divided by 8 positive divisors (out of which 0 are even, and 8 are odd). The sum of these divisors (counting 557,187) is 752,640, the average is 94,080.
## Other Arithmetic Functions (n = 557187)
1 φ(n) n
Euler Totient Carmichael Lambda Prime Pi φ(n) 366600 Total number of positive integers not greater than n that are coprime to n λ(n) 91650 Smallest positive number such that aλ(n) ≡ 1 (mod n) for all a coprime to n π(n) ≈ 45754 Total number of primes less than or equal to n r2(n) 0 The number of ways n can be represented as the sum of 2 squares
There are 366,600 positive integers (less than 557,187) that are coprime with 557,187. And there are approximately 45,754 prime numbers less than or equal to 557,187.
## Divisibility of 557187
m n mod m 2 3 4 5 6 7 8 9 1 0 3 2 3 1 3 6
The number 557,187 is divisible by 3.
## Classification of 557187
• Arithmetic
• Deficient
### Expressible via specific sums
• Polite
• Non-hypotenuse
• Square Free
### Other numbers
• LucasCarmichael
• Sphenic
## Base conversion (557187)
Base System Value
2 Binary 10001000000010000011
3 Ternary 1001022022120
4 Quaternary 2020002003
5 Quinary 120312222
6 Senary 15535323
8 Octal 2100203
10 Decimal 557187
12 Duodecimal 22a543
20 Vigesimal 39cj7
36 Base36 bxxf
## Basic calculations (n = 557187)
### Multiplication
n×i
n×2 1114374 1671561 2228748 2785935
### Division
ni
n⁄2 278594 185729 139297 111437
### Exponentiation
ni
n2 310457352969 172982801128738203 96383768012518253114961 53703782547591007898365774707
### Nth Root
i√n
2√n 746.45 82.2875 27.3212 14.0994
## 557187 as geometric shapes
### Circle
Diameter 1.11437e+06 3.50091e+06 9.75331e+11
### Sphere
Volume 7.24589e+17 3.90132e+12 3.50091e+06
### Square
Length = n
Perimeter 2.22875e+06 3.10457e+11 787981
### Cube
Length = n
Surface area 1.86274e+12 1.72983e+17 965076
### Equilateral Triangle
Length = n
Perimeter 1.67156e+06 1.34432e+11 482538
### Triangular Pyramid
Length = n
Surface area 5.37728e+11 2.03862e+16 454941 | 1,299 | 3,950 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.578125 | 4 | CC-MAIN-2021-25 | latest | en | 0.805469 |
https://www.assignguru.com/mcqs/1/715724391/ | 1,702,226,185,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679102612.80/warc/CC-MAIN-20231210155147-20231210185147-00146.warc.gz | 696,822,820 | 6,475 | #### motion
Newton's first law of motion
an object's motion will not change unless an unbalanced force acts on the object
distance
How far an object moves
Displacement
Distance and direction of an object's change in position from the starting point.
speed
The distance an object travels per unit of time
Velocity
the speed and direction of a moving object
Acceleration
The rate at which velocity changes
force
a push or pull on an object
inertia
The tendency of an object to resist a change in motion
constant
neither slows down or stops
instantaneous speed
the speed of an object at one instant of time
negative acceleration
slowing down | 141 | 660 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.734375 | 3 | CC-MAIN-2023-50 | latest | en | 0.815033 |
https://www.techylib.com/en/view/parakeetincurable/maths_in_english_statistics_and_its_applications | 1,527,115,020,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794865830.35/warc/CC-MAIN-20180523215608-20180523235608-00323.warc.gz | 846,564,864 | 11,693 | # Maths in English: statistics and its applications
Software and s/w Development
Dec 13, 2013 (4 years and 6 months ago)
102 views
Rosa Mª Avila
Laia Bayarri
Cristina Lopera
4r B
Ivón Cardenas
Maths in English: statistics
and its applications
Artificial vision
Índex
1
-
Artificial vision
1.1 Introduction to artificial vision
1.2 Augmented reality
1.3 Stereo vision and multi
-
cam projections
2
-
Virtual worlds
2.1 Second life
2.2 Open wonderland
2.3 Explain what is Open Scene graphs
1.
The artificial vision
1.1 Introduction to artificial vision
Is the computer vision is methods for acquiring, processing,
analyzing, and understanding images and, in general, high
-
dimensional data from the real world in order to produce
numerical or symbolic information.
A theme in the development of this field has been to
duplicate the abilities of human vision by electronically
perceiving and understanding an image.
Example
Some applications
Controlling processes, e.g., an industrial robot.
Navigation, e.g., by an autonomous vehicle or mobile robot.
1.2 Augmented reality
OpenCV:
APPLICATIONS
2D and 3D feature toolkit Egomotion
estimation .
1.3 Stereo Vision and multi
-
cam
projections
3D
reconstruction
In
computer vision
and
computer graphics,
3D reconstruction
is the
process of capturing the shape and appearance of real objects. This
process can be accomplished either by active or passive methods.
Pro Forma
Pro FORMA (Probabilistic Feature
-
based On
-
line Rapid Model
Acquisition) The invention allows the dimensions of the object read
only designated using any ordinary webcam, allowing you to rotate
and move any object in front of the camera and be scanned in real
time. Advanced mathematical calculations performed to generate
three
-
dimensional views and so ignore any other object than the
scanned object.
Example
2. Virtual worlds
2.1 Second life
Second Life is an online virtual world developed by Linden Lab.
It was launched on June 23, 2003.
Residents can explore the world meet other residents, socialize,
participate in individual and group activities, and create and
trade virtual property and services with one another. Second Life
is intended for people aged 16 and over.
The software is a three
-
dimensional modeling tool based on
simple geometric shapes that allows residents to build virtual
objects. There is also a procedural scripting language.
Some features and examples
-
Club
Penguin
-
Habbo
-
Smeet
-
Second Life
2.2 Open
wonderland
Examples
2.3 Explain what is Open Scene graphs
The Open Scene Graph is an open source high performance 3D graphics
toolkit, used by application developers in fields such as visual simulation,
games, virtual reality, scientific visualization and modeling. Written entirely
in Standard C++ and OpenGL it runs on all Windows platforms, OSX,
GNU/Linux, IRIX, Solaris, HP
-
Ux
, AIX
and Free BSD
operating systems.
The Open Scene Graph is now well established as the world leading scene
graph technology, used widely in the vis
-
sim, space, scientific, oil
-
gas, games
and virtual reality industries.
Example | 731 | 3,139 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.875 | 3 | CC-MAIN-2018-22 | latest | en | 0.851751 |
https://www.coursehero.com/file/5923147/Homework-3-Solution/ | 1,490,607,243,000,000,000 | text/html | crawl-data/CC-MAIN-2017-13/segments/1490218189466.30/warc/CC-MAIN-20170322212949-00395-ip-10-233-31-227.ec2.internal.warc.gz | 905,389,747 | 87,863 | Homework 3 Solution
# Homework 3 Solution - MEEN 260 Introduction to Engineering...
This preview shows pages 1–2. Sign up to view the full content.
MEEN 260 Introduction to Engineering Experimentation Homework 3: Confidence Intervals Assigned: Thursday, 5 Feb. 2009 Due: Thursday, 12 th February 2009, 5:00pm Review the following concepts discussed in lecture: 1) Confidence Intervals based on Samples (Chapter 3, Sections 3.6) Homework Problems: Problem 1) 400 observations of engine power are made with hp x 300 = and hp s x 15 ± = . Assuming a 95% confidence level, determine: a) the expected range of the population mean b) the upper bound on the population mean Problem 2) 40 observations of torque are made with Nm x 300 = and Nm s x 15 ± = . Assuming a 95% confidence level, determine: a) the expected range of the population mean, b) the upper bound on the population mean. c) Compare the results with Problem 1 to determine the effect of sample size. Problem 3) A certain type of carbon nanotube is manufactured to a nominal dimension of 120 μm. A sample of six nanotubes was selected and the following measurements were made: 118μm 126 μm 112μm 130μm 121μm 118μm Determine the upper and lower tolerance values within the population mean may be expected to fall with a significance level of 20%. Problem 4) Assume your friend used the same data given in Problem 3, and calculated the confidence interval as: 120 μm ± 8.491 μm What confidence level did your friend assume?
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
This is the end of the preview. Sign up to access the rest of the document.
## This note was uploaded on 07/28/2010 for the course MEEN 260 taught by Professor Langari during the Fall '08 term at Texas A&M.
### Page1 / 4
Homework 3 Solution - MEEN 260 Introduction to Engineering...
This preview shows document pages 1 - 2. Sign up to view the full document.
View Full Document
Ask a homework question - tutors are online | 499 | 2,014 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5 | 4 | CC-MAIN-2017-13 | longest | en | 0.875822 |
https://mail.python.org/pipermail/python-ideas/2011-September/011526.html | 1,607,049,854,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141733120.84/warc/CC-MAIN-20201204010410-20201204040410-00141.warc.gz | 386,907,166 | 3,217 | # [Python-ideas] Proposal: add a calculator statistics module
Massimo Di Pierro massimo.dipierro at gmail.com
Tue Sep 13 03:14:47 CEST 2011
```You only need:
def E(f,data): return sum(f(x) for x in data)/len(data)
Than you can compute ANY expectation value
data = range(0,10)
average = E(lambda x:x, data)
variance = E(lambda x:(x-mu)**2,data)
skewness = E(lambda x:(x-mu)**3,data)/variance**(2.0/3.0)
X = [random() for i in range(N)]
Y = [random() for i in range(N)]
XY = [(X[i],Y[i]) for i in range(N)]
covariance = E[lambda x,y: x*y, XY] - E(lambda x,y:x, XY)*E(lambda x,y:y, XY)
Hope it makes sense.
etc.etc.
Massimo
On Sep 12, 2011, at 8:00 PM, Steven D'Aprano wrote:
> I propose adding a basic calculator statistics module to the standard library, similar to the sorts of functions you would get on a scientific calculator:
>
> mean (average)
> variance (population and sample)
> standard deviation (population and sample)
> correlation coefficient
>
> and similar. I am volunteering to provide, and support, this module, written in pure Python so other implementations will be able to use it.
>
> Simple calculator-style statistics seem to me to be a fairly obvious "battery" to be included, more useful in practice than some functions already available such as factorial and the hyperbolic functions.
>
> The lack of a standard solution leads people who need basic stats to roll their own. This seems seductively simple, as the basic stats formulae are quite simple. Unfortunately doing it *correctly* is much harder than it seems. Variance, in particular, is prone to serious inaccuracies. Here is the most obvious algorithm, using the so-called "computational formula for the variance":
>
>
> def variance(data):
> # σ2 = 1/n**2 * (n*Σ(x**2) - (Σx)**2)
> n = len(data)
> s1 = sum(x**2 for x in data)
> s2 = sum(data)
> return (n*s1 - s2**2)/(n*n)
>
> Many stats text books recommend this as the best way to calculate variance, advice which makes sense when you're talking about hand calculations of small numbers of moderate sized data, but not for floating point. It appears to work:
>
> >>> data = [1, 2, 4, 5, 8]
> >>> variance(data) # exact value = 6
> 6.0
>
> but unfortunately it is numerically unstable. Shifting all the data points by a constant amount shouldn't change the variance, but it does:
>
> >>> data = [x+1e12 for x in data]
> >>> variance(data)
> 171798691.84
>
> Even worse, variance should never be negative:
>
> >>> variance(data*100)
> -1266637395.197952
>
> Note that using math.fsum instead of the built-in sum does not fix the numeric instability problem, and it adds the additional problem that it coerces the data points to float. (If you use Decimal, this may not be what you want.)
>
> Here is an example of published code which suffers from exactly this problem:
>
> https://bitbucket.org/larsyencken/simplestats/src/c42e048a6625/src/basic.py
>
> and here is an example on StackOverflow. Note the most popular answer given is to use the Computational Formula, which is the wrong answer.
>
> http://stackoverflow.com/questions/2341340/calculate-mean-and-variance-with-one-iteration
>
> I would like to add a module to the standard library to solve these sorts of simple stats problems the right way, once and for all.
>
> Thoughts, comments, objections or words of encouragement are welcome.
>
>
>
> --
> Steven
>
> _______________________________________________
> Python-ideas mailing list
> Python-ideas at python.org
> http://mail.python.org/mailman/listinfo/python-ideas
``` | 928 | 3,549 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.296875 | 3 | CC-MAIN-2020-50 | latest | en | 0.832092 |
https://sites.google.com/view/cristian-ortiz/udea2019groupoids | 1,568,904,336,000,000,000 | text/html | crawl-data/CC-MAIN-2019-39/segments/1568514573533.49/warc/CC-MAIN-20190919142838-20190919164838-00369.warc.gz | 682,513,905 | 14,798 | # Lie groupoids and applications
## (Graduate Course at UdeA - Medellin)
### Why Lie groupoids?
Lie groupoids appeared first in Ehresmann's global formulation of PDE's and nowadays play an important role in Differential Geometry, Foliation Theory, Poisson Geometry, Noncommutative Geometry, Index Theory, Orbifolds and Stacks and Generalized Geometry, among others. The infinitesimal counterpart of a Lie groupoid is given by a Lie algebroid, an object that unifies Lie algebras, regular foliations but also provides an appropriate setting to study singular geometric structures, e.g. Poisson structures.
### When and Where?
Every Tuesday and Thursday from 10am to 12pm. Room 5-118.
### Office Hours
Every Monday from 10:00 to 11:00 at Room 4-331A. Students are very welcome to ask questions about any subject related to the course. Please feel free to ask your questions, I will do my best to discuss all issues that might help us solve your doubts, I will be happy to do it. If you would like to talk about Math in general, I am also available and looking forward to discuss.
### Pre-requisites
Differentiable manifolds.
### Contents
This course is an introduction to Lie groupoids and Lie algebroids, paying special attention to geometric structures on Lie groupoids and their connections with Symplectic Geometry. The course is organized as follows:
1. Lie groupoids and Lie algebroids: Lie groupoids, main concepts and examples; Lie algebroids, main concepts and examples; the Lie functor and the integrability problem.
2. Vector bundles over Lie groupoids and Lie algebroids: Vector bundles over Lie groupoids (VB-groupoids), connections with representations up to homotopy, infinitesimal description (VB-algebroids).
3. Geometric structures on Lie groupoids: Symplectic groupoids and Poisson structures; Riemannian groupoids; Linearization Theorem.
4. Lie groupoids and singular spaces: Morita equivalence; Lie groupoids and orbifolds; Lie groupoids and quotient stacks; Symplectic groupoids vs Symplectic structures on quotient stacks.
### Notes
I will provide notes containing an expanded version of the material presented during the lectures. These lecture notes are very much in progress and the file will be updated every week. Students can require access to the lecture notes here.
### Homeworks
• Homework 1 can be found here.
### References
The following are notes and books that I recommend as very good sources of study for the first part of the course:
1. M. Cranic and R. Fernandes, "Lectures on the integrability of Lie brackets". Available here.
2. K. Mackenzie, "General theory of Lie groupoids and Lie algebroids", London Math. Soc. Lecture Notes Series 213.
3. I. Moerdijk and J. Mcrun, "Introduction to Foliations and Lie Groupoids". Cambridge University Press.
For the second part of the course, I will follow the papers:
1. Gracia-Saz, A., Mehta, R., “Lie algebroid structures on double vector bundles and representation theory of Lie algebroids”, Advances in Mathematics 223, 4, 1236-1275 (2010). Available here.
2. Gracia-Saz, A., Mehta, R., “VB-groupoids and representation theory of groupoids”, Journal of Symplectic Geometry 15 (3) 741-783 (2017). Available here.
but also you are encouraged to read
3. Arias Abad, C., Crainic, M., "Representations to homotopy of groupoids and the Bott spectral sequence", Adv. Math., 248, 416-452 (2013).
4. Arias Abad, C., Crainic, M., "Representations up to homotopy of Lie algebroids", Crelle, 2012 (663), 91-126 (2012)
where general representations up to homotopy are introduced and studied in detail.
For the third part of the course, the following references are recommended:
1. J.P. Dufour and N.T. Zung, "Poisson structures and their normal forms", Progress in Mathematics, Birkhauser
2. M. del Hoyo and R. Fernandes, "Riemannian metrics on Lie groupoids", Journal für die reine und angewandte Mathematik (Crelle) 735, 143-173 (2018). Available here.
For the last part of the course, the following survey is highly recommended:
1. M. del Hoyo, "Lie groupoids and differentiable stacks", Portugaliae Mathematica, 70 (2) 161-209 (2013). Available here. | 1,023 | 4,154 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.578125 | 3 | CC-MAIN-2019-39 | latest | en | 0.867867 |
https://www.physicsforums.com/threads/newtons-third-law-forces.470873/ | 1,701,967,336,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100677.45/warc/CC-MAIN-20231207153748-20231207183748-00696.warc.gz | 1,032,098,164 | 15,016 | # Newton's third law + forces
• furor celtica
In summary, three identical boxes with weight W are stacked against a vertical wall. The lowermost box is in contact with the wall while the other two boxes are positioned in a certain way. When a horizontal force of magnitude P is applied to the middle box, it can move if P>2µW. However, for sliding to occur between the upper and middle boxes, the force must be greater than 4µW due to the lower frictional force between them. The question is whether the middle and upper boxes will move together or not. Drawing a free body diagram and solving for accelerations will help determine this. Care must be taken when considering the normal forces acting on the middle box.
## Homework Statement
three identical boxes, each of weight W, are to be stacked one on top of another against a vertical wall. The lowermost box is in contact with the wall, and the other two boxes are positioned as shown in the diagram. The middle box is pushed into position by the application of a horizontal force of magnitude P.
Show that if P>4µW (where µ is coefficent of friction between any two boxes), sliding takes place between the upper and middle boxes (i.e. friction is not strong enough for the upper box to follow the middle box in its movement).
## The Attempt at a Solution
I don't understand why it is 4µW, shouldn't it be 3µW or 2µW? I'm not sure.
Firstly the middle box must be able to move, i.e. P>(friction from lowermost box), P>(normal reaction) x µ, P> 2µW. However, the frictional force between the upper and middle boxes is only Wµ, os shouldn't the 2µW force be enough? Or 3µW if both frictional forces are added, I'm not even sure, but any way i can't get to 4µW.
#### Attachments
• Untitled.png
441 bytes · Views: 400
The question is if the middle or upper boxes move together or not if a force P>4µW is applied. Just draw the free body diagram and solve for the accelerations of both blocks. Take care to the normal forces acting on the middle box.
ehild | 486 | 2,014 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4 | 4 | CC-MAIN-2023-50 | latest | en | 0.925355 |
https://examsdaily.in/sameer-scientist-syllabus-pdf-download-exam-pattern | 1,701,976,535,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100686.78/warc/CC-MAIN-20231207185656-20231207215656-00218.warc.gz | 282,018,667 | 21,813 | # SAMEER Scientist C Syllabus 2021 PDF – Download Exam Pattern Here!!!!
0
SAMEER Scientist C Syllabus 2021 PDF – Download Exam Pattern Here!!!!. Society for Applied Microwave Electronics Engineering & Research has uploaded the Syllabus and Exam Pattern for the Post of Scientist C & B. So, Candidates can download the SAMEER Scientist Exam Pattern & Syllabus 2021 PDF in our blog & Start your Preparation now. Stay Connected with us for further updates
## SAMEER Syllabus 2021:
Name of the Board Society for Applied Microwave Electronics Engineering & Research Name of the Post Scientist C & B Exam Date Announce Soon Status Syllabus Available
### SAMEER Scientist Exam Pattern 2021:
Subject Name Duration Type Electronics & Communication Engineering 3 Hours Objective Type Physics Atmospheric Sciences
### SAMEER Scientist Exam Syllabus 2021:
#### Electronics & Communication Engineering:
Section 1: Engineering Mathematics
Linear Algebra: Vector space, basis, linear dependence and independence, matrix algebra, eigenvalues and eigen vectors, rank, solution of linear equations – existence and uniqueness.
Calculus: Mean value theorems, theorems of integral calculus, evaluation of definite and improper integrals, partial derivatives, maxima and minima, multiple integrals, line, surface and volume integrals, Taylor series.
Differential Equations: First order equations (linear and nonlinear), higher order linear differential equations, Cauchy’s and Euler’s equations, methods of solution using variation of parameters, complementary function and particular integral, partial differential equations, variable separable method, initial and boundary value problems.
Vector Analysis: Vectors in plane and space, vector operations, gradient, divergence and curl, Gauss’s, Green’s and Stoke’s theorems.
Complex Analysis: Analytic functions, Cauchy’s integral theorem, Cauchy’s integral formula; Taylor’s and Laurent’s series, residue theorem.
Numerical Methods: Solution of nonlinear equations, single and multi-step methods for differential equations, convergence criteria.
Probability and Statistics: Mean, median, mode and standard deviation; combinatorial probability, probability distribution functions – binomial, Poisson, exponential and normal; Joint and conditional probability; Correlation and regression analysis.
Section 2: Networks, Signals and Systems
Network solution methods: nodal and mesh analysis; Network theorems: superposition, Thevenin and Norton’s, maximum power transfer; Wye‐Delta transformation; Steady state sinusoidal analysis using phasors; Time domain analysis of simple linear circuits; Solution of network equations using Laplace transform; Frequency domain analysis of RLC circuits;
Linear 2‐port network parameters: driving point and transfer functions; State equations for networks.
Continuous-time signals: Fourier series and Fourier transform representations, sampling theorem and applications; Discrete-time signals: discrete-time Fourier transform (DTFT), DFT, FFT, Z-transform, interpolation of discrete-time signals; LTI systems: definition and properties, causality, stability, impulse response, convolution, poles and zeros, parallel and cascade structure, frequency response, group delay, phase delay, digital filter design techniques.
Section 3: Electronic Devices
Energy bands in intrinsic and extrinsic silicon; Carrier transport: diffusion current, drift current, mobility and resistivity; Generation and recombination of carriers; Poisson and continuity equations; P-N junction, Zener diode, BJT, MOS capacitor, MOSFET, LED, photo diode and solar cell; Integrated circuit fabrication process: oxidation, diffusion, ion implantation, photolithography and twin-tub CMOS process.
Section 4: Analog Circuits
Small signal equivalent circuits of diodes, BJTs and MOSFETs; Simple diode circuits: clipping, clamping and rectifiers; Single-stage BJT and MOSFET amplifiers: biasing, bias stability, mid- frequency small signal analysis and frequency response; BJT and MOSFET amplifiers: multi-stage, differential, feedback, power and operational; Simple op-amp circuits; Active filters; Sinusoidal oscillators: criterion for oscillation, single-transistor and op- amp configurations; Function generators, wave-shaping circuits and 555 timers; Voltage reference circuits; Power supplies: ripple removal and regulation.
Section 5: Digital Circuits
Number systems; Combinatorial circuits: Boolean algebra, minimization of functions using Boolean identities and Karnaugh map, logic gates and their static CMOS implementations, arithmetic circuits, code converters, multiplexers, decoders and PLAs; Sequential circuits: latches and flip‐flops, counters, shift‐registers and finite state machines; Data converters: sample and hold circuits, ADCs and DACs; Semiconductor memories: ROM, SRAM, DRAM; 8-bit microprocessor (8085): architecture, programming, memory and I/O interfacing.
Section 6: Control Systems
Basic control system components; Feedback principle; Transfer function; Block diagram representation; Signal flow graph; Transient and steady-state analysis of LTI systems; Frequency response; Routh- Hurwitz and Nyquist stability criteria; Bode and root-locus plots; Lag, lead and lag-lead compensation; State variable model and solution of state equation of LTI systems.
Section 7: Communications
Random processes: autocorrelation and power spectral density, properties of white noise, filtering of random signals through LTI systems; Analog communications: amplitude modulation and demodulation, angle modulation and demodulation, spectra of AM and FM, superheterodyne receivers, circuits for analog communications; Information theory: entropy, mutual information and channel capacity theorem; Digital communications: PCM, DPCM, digital modulation schemes, amplitude, phase and frequency shift keying (ASK, PSK, FSK), QAM, MAP and ML decoding, matched filter receiver, calculation of bandwidth, SNR and BER for digital modulation; Fundamentals of error correction, Hamming codes; Timing and frequency synchronization, inter-symbol interference and its mitigation; Basics of TDMA, FDMA and CDMA.
Section 8: Electromagnetics
Electrostatics; Maxwell’s equations: differential and integral forms and their interpretation, boundary conditions, wave equation, Poynting vector; Plane waves and properties: reflection and refraction, polarization, phase and group velocity, propagation through various media, skin depth; Transmission lines: equations, characteristic impedance, impedance matching, impedance transformation, S- parameters, Smith chart; Waveguides: modes, boundary conditions, cut-off frequencies, dispersion relations; Antennas: antenna types, radiation pattern, gain and directivity, return loss, antenna arrays; Basics of radar; Light propagation in optical fibers.
#### Physics:
Section 1: Mathematical Physics
Linear vector space: basis, orthogonality and completeness; matrices; vector calculus; linear differential equations; elements of complex analysis: Cauchy- Riemann conditions, Cauchy’s theorems, singularities, residue theorem and applications; Laplace transforms, Fourier analysis; elementary ideas about tensors: covariant and contravariant tensor, Levi-Civita and Christoffel symbols.
Section 2: Classical Mechanics
D’Alembert’s principle, cyclic coordinates, variational principle, Lagrange’s equation of motion, central force and scattering problems, rigid body motion; small oscillations, Hamilton’s formalisms; Poisson bracket; special theory of relativity: Lorentz transformations, relativistic kinematics, mass‐energy equivalence.
Section 3: Electromagnetic Theory
Solutions of electrostatic and magnetostatic problems including boundary value problems; dielectrics and conductors; Maxwell’s equations; scalar and vector potentials; Coulomb and Lorentz gauges; Electromagnetic waves and their reflection, refraction, interference, diffraction and polarization; Poynting vector, Poynting theorem, energy and momentum of electromagnetic waves; radiation from a moving charge.
Section 4: Quantum Mechanics
Postulates of quantum mechanics; uncertainty principle; Schrodinger equation; one-, two- and three- dimensional potential problems; particle in a box, transmission through one dimensional potential barriers, harmonic oscillator, hydrogen atom; linear vectors and operators in Hilbert space; angular momentum and spin; addition of angular momenta; time independent perturbation theory; elementary scattering theory.
Section 5: Thermodynamics and Statistical Physics
Laws of thermodynamics; macrostates and microstates; phase space; ensembles; partition function, free energy, calculation of thermodynamic quantities; classical and quantum statistics; degenerate Fermi gas; black body radiation and Planck’s distribution law; Bose‐Einstein condensation; first and second order phase transitions, phase equilibria, criticalpoint.
Section 6: Atomic and Molecular Physics
Spectra of one‐ and many‐electron atoms; LS and jj coupling; hyperfine structure; Zeeman and Stark effects; electric dipole transitions and selection rules; rotational and vibrational spectra of diatomic molecules; electronic transition in diatomic molecules, Franck‐Condon principle; Raman effect; NMR, ESR, X-ray spectra; lasers: Einstein coefficients, population inversion, two and three level systems.
Section 7: Solid State Physics & Electronics
Elements of crystallography; diffraction methods for structure determination; bonding in solids; lattice vibrations and thermal properties of solids; free electron theory; band theory of solids: nearly free electron and tight binding models; metals, semiconductors and insulators; conductivity, mobility and effective mass; optical, dielectric and magnetic properties of solids; elements of superconductivity: Type-I and Type II superconductors, Meissner effect, London equation. Semiconductor devices: diodes, Bipolar Junction Transistors, Field Effect Transistors; operational amplifiers: negative feedback circuits, active filters and oscillators; regulated power supplies; basic digital logic circuits, sequential circuits, flip‐flops, counters, registers, A/D and D/A conversion.
Section 8: Nuclear and Particle Physics
Nuclear radii and charge distributions, nuclear binding energy, Electric and magnetic moments; nuclear models, liquid drop model: semi‐empirical mass formula, Fermi gas model of nucleus, nuclear shell model; nuclear force and two nucleon problem; alpha decay, beta‐decay, electromagnetic transitions in nuclei; Rutherford scattering, nuclear reactions, conservation laws; fission and fusion; particle accelerators and detectors; elementary particles, photons, baryons, mesons and leptons; quark model.
#### Atmospheric Sciences:
Section A: Atmospheric Science
Fundamental of Meteorology, Thermal structure of the atmosphere and its composition, Radiation Balance and Laws, Wind Belts, Monsoon, Climate. Atmospheric Thermodynamics. Hydrostatic equilibrium and: Hydrostatic equation, variation of pressure with height, geopotential, Tropical convection. Atmospheric Electricity. Cloud Physics. Observation Techniques of the Atmosepheric Properties.
Fundamental equations. Pressure, gravity, centripetal and Corolis forces, continuity equation in Cartesian and isobaric coordinates, Scale analysis, inertial flow, geostrophic and gradient winds, thermal wind, vorticity. Atmospheric turbulence, baroclinic instabiltiy. Atmosphreric Waves.
Tropical meteorology: Trade wind inversion, ITCZ; monsoon trough tropical cyclones, their structure and development theory; monsoon depressions; Climate variability and forcings; Madden-Julian oscillation (MJO), ENSO, QBO (quasi-biennial oscillation) and sunspot cycles. Primitive equations of Numerical Weather Prediction. General Circulation and Climate Modelling.
Synoptic weather forecasting, prediction of weather elements such as rain, maximum and minimum temperature and fog. Data Assimilation.
Section B: Ocean Sciences
Seawater Properties, T-S diagrams, Ocean Observations, Ocean Tide and Waves and their properties. Coastal processes and Estuary Dynamics. coastal zone management. Wind Driven Circulation: Ekman, Sverdrup, Stommel and Munk theories, Inertial currents; geostrophic motion; barotropic and baroclinic conditions; Oceanic eddies. Global conveyor belt circulation. Subtropical gyres; Western boundary currents; equatorial current systems; Current System in the Indian Ocean.
Momentum equation, mass conservation, vorticity. Ocean and Wave Modeling, Ocean State Forecasting. Data Assimilation. Ocean Turbulence.
Chemical Property of seawater, major and minor elements, their behavior and chemical exchanges across interfaces and residence times in seawater, Element chemistry in atypical conditions-estuaries, Biochemical cycling of nutrients, trace metals and organic matter. Air-sea exchange of important biogenic dissolved gases; carbon dioxide-carbonate system; alkalinity and control of pH; biological pump. Marine Pollution. Primary and secondary production; factors controlling phytoplankton and zooplankton abundance and diversity; nekton and fisheries oceanography.
SAMEER Scientist Exam Atmospheric Science Syllabus 2021 PDF
SAMEER Scientist Exam Electronics and Communication Engineering Syllabus 2021 PDF
SAMEER Scientist Exam Physics Syllabus 2021 PDF
Official Site | 2,715 | 13,407 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.203125 | 3 | CC-MAIN-2023-50 | latest | en | 0.73161 |
https://www.exceldemy.com/excel-find-average-of-specific-cells/ | 1,725,711,659,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700650826.4/warc/CC-MAIN-20240907095856-20240907125856-00478.warc.gz | 734,111,475 | 53,086 | # How to Find Average of Specific Cells in Excel (3 Handy Ways)
## Method 1 – Using the AVERAGEIF Function to Find the Average of Specific Cells in Excel
### 1.1 Using AVERAGEIF and Comparison Operator
Find the average of the Physics scores that are greater than or equal to 75 using the comparison operator.
Steps:
• Go to the C16 cell and enter the following formula.
`=AVERAGEIF(D5:D14,">="&75)`
Cells D5:D14 represent the marks in Physics while the “>=”&75 specify the criterion which is greater than or equal to 75.
The results should look like the image given below.
### 1.2 Applying AVERAGEIF to Match Specific Text
Steps:
• Select cell C16 and enter the following formula.
`=AVERAGEIF(C5:C14,"*Beta*",D5:D14)`
Ranges C5:C14 and D5:D14 refer to the Section and Physics columns respectively. Meanwhile, the “*Beta*” represents the criteria to match. As a note, the asterisk (*) character before and after Beta indicates an exact match.
The results should look like the picture given below.
### 1.3 Using AVERAGEIF with Single Criteria
Steps:
• Select cell C17 and enter the following formula.
`=AVERAGEIF(C5:C14,C16,D5:D14)`
Ranges C5:C14 and D5:D14 represent the Section and Physics columns respectively. Cell C16 points to Section Alpha which is given criterion.
The output should look like shown in the image below.
## Method 2 – Using AVERAGEIFS Function
### 2.1 Employing AVERAGEIFS with Double Criteria
Steps:
• Select cell C18 and enter the following formula.
`=AVERAGEIFS(D5:D14,C5:C14,C17,D5:D14,">="&85)`
Formula Breakdown:
• AVERAGEIFS(D5:D14,C5:C14,C17,D5:D14,”>=”&85) → finds average for the cells specified by a given set of conditions or criteria. D5:D14 is the average_range argument which is the Physics column. C5:C14 is the criteri_range1 argument which refers to the Section column and the C17 is the criteria1 argument which is Section Gamma. Following this, D5:D14 is the criteri_range2 argument which refers to the Physics column, and the “>=”&85 is the criteria2 argument which represents the values greater than and equal to 85.
• Output88.0
The results should look like the image below.
### 2.2 Combining AVERAGE, LARGE, and SMALL Functions to Calculate Top and Bottom 3 Averages
Steps:
• Select cell C16 and enter the following
`=AVERAGE(LARGE(D5:D14,{1,2,3}))`
Formula Breakdown:
• LARGE(D5:D14,{1,2,3}) → returns the nth largest value in a dataset. Here, range D5:D14 represents the Physics column. {1,2,3} refers to the 3 of the largest values in the Physics column.
• Output89, 88, 87
• AVERAGE(LARGE(D5:D14,{1,2,3})) → becomes
• AVERAGE(89, 88, 87) → returns the average of the arguments. Here, the values of 89, 88, and 87 are summed and divided by 3 to return their respective average.
• Output88.0
• Select cell C17 and enter in the following formula.
`=AVERAGE(SMALL(D5:D15,{1,2,3}))`
Formula Breakdown:
• SMALL(D5:D14,{1,2,3}) → returns the nth smallest value in a dataset. Here, range D5:D14 represents the Physics column. {1,2,3} refers to the 3 of the smallest values in the Physics column.
• Output60, 62, 73
• AVERAGE(SMALL(D5:D14,{1,2,3})) → becomes
• AVERAGE(60, 62, 73) → returns the average of the arguments. Here, the values of 60, 62, and 73 are summed and divided by 3 to return their respective average.
• Output65.0
## Method 3 – Using DAVERAGE Function to Find Average of Specific Cells in Excel
Steps:
• Select cell D17 and enter the following formula.
`=DAVERAGE(B4:E14,B17,C16:C17)`
Formula Breakdown:
• DAVERAGE(B4:E14, B17, C16:C17) → averages the values in a database that match the specified conditions. B4:E14 is the database argument that represents all the cells in the dataset. B17 is the field argument, which refers to the Chemistry subject. Lastly, the C16:C17 is the criteria argument, which is Section Alpha.
• Output81.5
## Calculate the Average of Only Cells with Values in Excel
Steps:
• Select cell C16 and enter the following formula.
`=AVERAGEA(D5:D14)`
Range D5:D14 represents the scores in Physics.
## Related Articles
Get FREE Advanced Excel Exercises with Solutions! | 1,105 | 4,091 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.703125 | 4 | CC-MAIN-2024-38 | latest | en | 0.823528 |
http://dongtaikeji.cn/terms/v/velocity.asp | 1,597,349,644,000,000,000 | text/html | crawl-data/CC-MAIN-2020-34/segments/1596439739073.12/warc/CC-MAIN-20200813191256-20200813221256-00244.warc.gz | 25,179,940 | 27,448 | ### What Is the Velocity of Money?
The velocity of money is a measurement of the rate at which money is exchanged in an economy. It is the number of times that money moves from one entity to another. It also refers to how much a unit of currency is used in a given period of time. Simply put, it's the rate at which consumers and businesses in an economy collectively spend money. The velocity of money is usually measured as a ratio of gross domestic product (GDP)聽to a country's M1 or M2 money supply.
The velocity of money is important for measuring the rate at which money in circulation is being used for purchasing goods and services. It is used to help economists and investors gauge the health and vitality of an economy. High money velocity is usually associated with a healthy, expanding economy. Low money velocity is usually associated with recessions and contractions.
1:29
### Understanding the Velocity of Money
Velocity of money is a metric calculated by economists. It shows the rate at which money is being transacted for goods and services in an economy. While it is not necessarily a key economic indicator, it can be followed alongside other key indicators that help determine economic health like GDP, unemployment, and inflation. GDP and the money supply are the two components of the velocity of money formula.
Economies聽that exhibit a higher velocity of money relative to others tend to be more developed. The velocity of money is also known to fluctuate with business cycles. When an economy is in an expansion, consumers and businesses tend to more readily spend money causing the velocity of money to increase. When an economy is contracting, consumers and businesses are usually more reluctant to spend and the velocity of money is lower.
Since the velocity of money is typically correlated with business cycles, it can also be correlated with key indicators. Therefore, the velocity of money will usually rise with GDP and inflation. Alternatively, it is usually expected to fall when key economic indicators like GDP and inflation are falling in a contracting economy.
### Key Takeaways
• Velocity of money is a measurement of the rate at which money is exchanged in an economy.
• The velocity of money equation divides GDP by money supply.
• The velocity of money formula shows the rate at which one unit of money supply currency is being transacted for goods and services in an economy.
• The velocity of money is typically higher in expanding economies and lower in contracting economies.
### Example of Velocity of Money
Consider an economy consisting of two individuals, A and B, who have \$100 each. A buys a car from B for \$100. Then B purchases a home from A for \$100. B has kids and enlists A's help in adding new construction to his home. For his efforts, B pays A \$100. A also sells a car he owns to B for \$100. Thus, both parties in the economy have made transactions worth \$400, even though they only possessed \$100 each. In this economy, the velocity of money would be two resulting from the \$400 in transactions divided by the \$200 in money supply. This multiplication in the value of goods and services exchanged is made possible through the velocity of money in an economy.
### The Velocity of Money Formula
While the above provides a simplified example of the velocity of money, the velocity of money is used on a much larger scale as a measure of transactional activity for an entire country鈥檚 population. In general, this measure can be thought of as the turnover of the money supply for an entire economy.
For this application, economists typically use GDP and either M1 or M2 for the money supply. Therefore, the velocity of money equation is written as GDP divided by money supply.
Velocity of money formula:
Velocity of Money = GDP / Money Supply
GDP is usually used as the numerator in the velocity of money formula though gross national product (GNP) may also be used as well. GDP represents the total amount of goods and services in an economy that are available for purchase. In the denominator, economists will typically identify money velocity for both M1 and M2.
M1 is defined by the Federal Reserve as the sum of all currency held by the public and聽transaction deposits聽at depository institutions. M2 is a broader measure of money supply, adding in savings deposits,聽time deposits,聽and real money market mutual funds.
The St. Louis Federal Reserve tracks the quarterly velocity of money using both M1 and M2.
### Velocity of Money and the Economy
There are differing views among economists as to whether the velocity of money is a useful indicator of the health of an economy or, more specifically, inflationary pressures. The "monetarists" who subscribe to the聽quantity theory of money聽argue that money velocity should be stable absent changing expectations, but a change in money supply can alter expectations and therefore money velocity and inflation. For example, an increase in the money supply should theoretically lead to a commensurate increase in GDP prices because there is more money chasing the same level of goods and services in the economy. The opposite should happen with a decrease in money supply. Critics, on the other hand, argue that in the聽short term, the velocity of money is highly variable, and prices are resistant to change, resulting in a weak and indirect link between money supply and inflation.
Empirically, data suggests that the velocity of money is indeed variable. Moreover, the relationship between money velocity and inflation is also variable. For example, from 1959 through the end of 2007, the velocity of M2 money stock averaged approximately 1.9x with a maximum of 2.198x in 1997 and a minimum of 1.653x in 1964. Since 2007, the velocity of money has fallen dramatically as the Federal Reserve greatly expanded its balance sheet in an effort to combat the聽global financial crisis聽and deflationary pressures. As of the second quarter of 2019, the M2 velocity of money was 1.457. Money velocity has been increasing from its most recent trough of 1.432 in the second quarter of 2017. The second quarter of 2017 also shows the lowest reading of M2 money velocity in history.
1:23 | 1,273 | 6,212 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2020-34 | latest | en | 0.964303 |
http://stackoverflow.com/questions/20055594/recursive-function-for-counting-divisible-digits-in-c | 1,467,330,111,000,000,000 | text/html | crawl-data/CC-MAIN-2016-26/segments/1466783399425.79/warc/CC-MAIN-20160624154959-00152-ip-10-164-35-72.ec2.internal.warc.gz | 292,021,375 | 16,731 | # Recursive function for counting divisible digits in C
I need to write a recursive function which accepts a number, and it needs to check whether each number is divisible with the previous number. For example, if the input is 63542, result should be 2, while only 6 is divisible by 3, and 4 is divisible by 2. These are the only numbers that are divisible with the their previous number. I have the following code, but it returns every time one more. For the above example it should return 2 but it returns 3.
``````#include <stdio.h>
int deliv(int num)
{
int temp = num%100;
int counter = 0;
if(num == 0)
{
return 0;
}
else if((temp/10)%(num%10) == 0)
{
counter++;
return counter + deliv(num/10);
}
else return counter + deliv(num/10);
}
int main()
{
int result = deliv(63542);
printf("%d\n", result);
return 0;
}
``````
-
What about numbers like 8421. Does that return "2" for the pairing of 8,4 and 2,1. Or does it return "3" to include "4,2" as well? – selbie Nov 18 '13 at 19:03
@selbie it should return 3. – user1726549 Nov 18 '13 at 19:04
Also - your program will crash if "num" passed into deliv is a multiple of 10. Try passing in "5000" instead of "63542" and see what happens. – selbie Nov 18 '13 at 19:05
@selbie any idea how to solve it? – user1726549 Nov 18 '13 at 19:07
You want to change your test condition from `if(num == 0)` to `if(num < 10 )` because if num is a single digit number, temp becomes 0 since `temp = num/10`, which is divisible by `num%10`.
Also, add this condition to avoid crashing when two consecutive digits are 0.
``````if(num < 10)
{
return 0;
}
//To avoid crash due to 2 zeroes
else if(temp==0)
{
return counter + deliv(num/10);
}
else if((temp/10)%(num%10) == 0)
{
counter++;
return counter + deliv(num/10);
}
else return counter + deliv(num/10);
``````
-
But it will never be less than 0. – user1726549 Nov 18 '13 at 19:05
Sorry, a typo, that was meant to be 10, thanks. – Shubham Nov 18 '13 at 19:06
Okay, solved it, thanks! – user1726549 Nov 18 '13 at 19:07 | 630 | 2,011 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.078125 | 3 | CC-MAIN-2016-26 | latest | en | 0.801908 |
https://www.physicsforums.com/threads/kinetic-friction.187726/page-2 | 1,623,783,262,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623487621519.32/warc/CC-MAIN-20210615180356-20210615210356-00520.warc.gz | 789,585,252 | 13,716 | # Kinetic friction
I contacted my TA about the problem. The formula was simply V^2 / 2*g*coef=s
Thank you for all your help!
yep vf^2=2as
v^2/(2a)=s
a=g*Uk
v^2/(2*g*uk)=s
yes but you should've let Cassie do it or done it though pm's :s.
Also your signs are wrong, gravity (if we're going negative y is down) *.44 will give a -acceleration which opposes the motion to the right (+x side of the x axis).
and the velocity given is initial not final.
Oh oops. Alright I see | 146 | 474 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.734375 | 3 | CC-MAIN-2021-25 | latest | en | 0.901837 |
http://www.physicsforums.com/printthread.php?t=192465 | 1,408,765,145,000,000,000 | text/html | crawl-data/CC-MAIN-2014-35/segments/1408500825010.41/warc/CC-MAIN-20140820021345-00409-ip-10-180-136-8.ec2.internal.warc.gz | 541,713,244 | 3,609 | Physics Forums (http://www.physicsforums.com/index.php)
- - A good math course? (http://www.physicsforums.com/showthread.php?t=192465)
Jwink3101 Oct19-07 04:32 PM
A good math course?
Hello. This question may be hard to desribe for people who do not go to my school and know the courses but i will try. I am going to include course descriptions from the coursebook below.
Anyway, i am a sophmore physics major at Carnegie Mellon University. Let me add that i am doing an applied physics track and an engineering studies minor. I have always liked engineering and may end up there but i wanted to do undergrad in physics regardless.
Anyway, as part of the physics curriculum, i take a course called Physical Analysis and later Mathematical Methods of Physics. They cover almost all of the math needed post-calculus for physics. The thing is, i need to take another math course. I could take Diff EQ but Phys Analysis covers that. I could take Matrix Algebra but Math Mathods should cover that. Most other courses require me to have a pre-req that i do not want (an intro to mathematical proofs). Does anybody have suggestions? I was thinking of partial differential equations but i am not sure.
Here are descriptions for the courses in the physics department where i will be learning math:
Physical Analysis: This course aims to develop analytical skills and mathematical
modeling skills across a broad spectrum of physical phenomena,
stressing analogies in behavior of a wide variety of systems.
Specific topics include dimensional analysis and scaling in
physical phenomena, exponential growth and decay, the harmonic
oscillator with damping and driving forces, linear approximations
of nonlinear systems, coupled oscillators, and wave motion.
Necessary mathematical techniques, including differential
equations, complex exponential functions, matrix algebra, and
elementary Fourier series, are introduced as needed.
Math Methods: This course introduces, in the context of physical systems, a
variety of mathematical tools and techniques that will be needed
for later courses in the physics curriculum. Topics will include,
linear algebra, vector calculus with physical application, Fourier
series and integrals, partial differential equations and boundary
value problems. The techniques taught here are useful in more
advanced courses such as Physical Mechanics, Electricity and
Magnetism, and Advanced Quantum Physics. transformations, four-vectors, invariants, and applications to
particle mechanics.
Anyhelp would be great
bravernix Oct19-07 06:38 PM
I always thought a differential geometry class would be interesting, if one is offered. Or, how about complex analysis? PDEs may be good, but I have been told by many people that graduate E&M (say, using Jackson) will teach you all the PDE techniques you need.
ytoruno Oct21-07 03:20 AM
this all depends on wether you were taught calculus with proofs or you took a real analysis class, which is usually not a physics requirement. If you're good with proofs take a Diff geometry class. From the physics point of view i would suggest a class on complex variables. A math topic i personally find profoundly interesting is formal language theory, you won't regret it taking that one.
javierR Oct21-07 04:57 AM
A course in PDEs is ultimately essential in applied physics/engineering. Complex analysis as well. These two should be covered a bit in math methods courses in the physics dept., but if you have to take another math course, these are good. I don't know about CM, but in my second year of college I took a semester of ordinary differential equations and a semester of linear algebra, the latter introducing the basis of proofs. These were requirements for a physics major in addition to the mathematical techniques taught within the physics department. Then courses in the math department beyond these require better understanding of proofs, though useful calculational techniques are still present for partial differential equations and complex analysis. Good luck.
ice109 Oct21-07 10:48 AM
past ordinary/partial differential equations and linear algebra all the classes that aren't applied math classes require proofs.
PowerIso Oct21-07 12:19 PM
Hehe, even my applied math classes require proofs.
All times are GMT -5. The time now is 10:39 PM. | 894 | 4,337 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.703125 | 3 | CC-MAIN-2014-35 | latest | en | 0.92731 |
http://www.enotes.com/homework-help/what-resonant-frequency-10-mh-inductor-0-005-uf-452208 | 1,462,349,432,000,000,000 | text/html | crawl-data/CC-MAIN-2016-18/segments/1461860122533.7/warc/CC-MAIN-20160428161522-00103-ip-10-239-7-51.ec2.internal.warc.gz | 498,365,201 | 11,431 | # What is the resonant frequency of a 10-mH inductor and a 0.005-uF capacitor?They showed me how they got their answer fr=(1/6.28(sq rt of LC) ) but when I do it I don't get their exact answer which is 22,500Hz or 22.5kHz
Posted on
The resonant frequency of a series or parallel alternating LC circuit is the same. It can be computed from the resonance condition which is (for a series circuit) the equality of voltage drops on the inductor and capacitor or (for a parallel circuit) the equality of currents through the inductor and capacitor. This equality reduces, in turn, to the condition of their reactances being equal. X(L)= X(C)
`omega*L = 1/(omega*C)` where `omega =2*pi*F`
This leads to a value for the resonant frequency of
`F = 1/(2*pi*sqrt(L*C))`
For the values given in text of L=10 mH and C=0.005 microF=5 nF (and having `pi` correct up to 25 decimals) one has
`F =1/(2*pi*sqrt(0.01*5*10^(-9))) =22507.9 Hz=22508 Hz =22.5 kHz`
Sources: | 284 | 958 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.296875 | 3 | CC-MAIN-2016-18 | longest | en | 0.928922 |
https://www.varsitytutors.com/ssat_middle_level_math-help/numbers-and-operations?page=2 | 1,585,869,642,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585370508367.57/warc/CC-MAIN-20200402204908-20200402234908-00008.warc.gz | 1,192,156,837 | 47,464 | # SSAT Middle Level Math : Numbers and Operations
## Example Questions
### Example Question #671 : Ssat Middle Level Quantitative (Math)
Express this ratio in simplest form:
Explanation:
A ratio involving decimals can be simplified as follows:
First, move the decimal point over a common number of places in each number so that both numbers become whole. In this case, it would be two places:
(Note the addition of a zero at the end of the first number.)
Now, rewrite as a fraction and divide both numbers by :
The ratio, simplified, is .
### Example Question #12 : Numbers And Operations
In Mrs. Jones' class, the ratio of boys to girls is 3:2. If there are 8 girls in the class, how many total students are in the class?
Explanation:
Using the ratio, we find that there are 12 boys in the class .
Summing 12 and 8 gives the total number of 20 students.
### Example Question #13 : Numbers And Operations
A 16-ounce bottle of Charlie's Fizzy Fizz Root Beer costs 89 cents. Give the price per ounce to the nearest tenth of a cent.
Explanation:
Divide 89 by 16:
The soda costs about 5.6 cents per ounce.
### Example Question #11 : Numbers And Operations
A batting average is defined as the ratio of hits to turns at bat. It is expressed as a decimal rounded to the nearest thousand.
In 2013, Yadier Molina's batting average was 0.319. If he maintains this same batting average over the 2014 baseball season, how many turns at bat should he expect to need to achieve 50 hits? (Nearest whole number)
Explanation:
Since a batting average is a ratio of hits to turns at bat,
Let be the number of turns at bat Molina needs to get 50 hits. Then:
Solve for :
Molina should need 157 turns at bat,
### Example Question #11 : Numbers And Operations
Carlos owns a hamburger and hot-dog stand. When the fridge in the stand is full, there are 16 hamburgers and hot dogs total. If the ratio of hamburgers to hot dogs is always 5:3, how many hot dogs are in the cart when it is full?
Explanation:
When solving a ratio problem, first add each part of the ratio:
So, one part in this ratio is equal to 8. The number of hamburgers and hot dogs in the fridge is currently only the amount it can hold. To figure out how much it can hold total, we can multiply by the denominator of the fraction since the numerator is 1.
64 is the total amount of hamburgers and hot dogs that can be held in the fridge. Divide 8 into 64 in order to figure out how many parts fit into this number.
This number tells us how many times we need to multiply the ratio by in order to figure out how many hamburgers or hot dogs can be in the fridge at once. Since we are trying to figure out how many hot dogs are in the fridge when it is full, we use the second number, 3. We use the number 3 because, since the ratio is hamburgers to hotdogs, the second number represents the number of hot dogs.
This gives you your answer. There are 24 hot dogs in the fridge when it is full.
### Example Question #672 : Ssat Middle Level Quantitative (Math)
There are girls and boys in Miss Bailey's class. What is the ratio of girls to boys?
Explanation:
The ratio of girls to boys in Miss Bailey's class is , which can be reduced to when both the numerator and denominator are divided by .
### Example Question #11 : How To Find A Ratio
In a library, there are children's books and young adult books. What is the ratio of young adult books to children's books?
Explanation:
The ratio of young adult books to children's books is to , or , which can be reduced to .
### Example Question #11 : Numbers And Operations
There are male and female employees at a company. What is the ratio of females to males?
Explanation:
The ratio of females to males is to , or , which can be reduced to .
### Example Question #19 : Numbers And Operations
One dollar is equal to about 116.76 Japanese yen. For how many yen should a tourist to Japan be able to exchange $2,500 at that rate (nearest whole yen)? Possible Answers: Correct answer: Explanation: One dollar is equal to 116.76 yen, so multiply$2,500 by this conversion rate to get
yen.
### Example Question #12 : Numbers And Operations
Sharon is having a birthday party. So far, she has invited twenty of her friends from school, nine of whom are girls. She wants to make the ratio of boys to girls at the party two to one. If she decides to add her two cousins, both girls, to the guest list, and no other girls, how many more boys does she need to invite? | 1,096 | 4,505 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.53125 | 5 | CC-MAIN-2020-16 | latest | en | 0.888167 |
https://www.studiestoday.com/question-paper-mathematics-cbse-class-4-mathematics-question-paper-set-r-312980.html | 1,726,830,703,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700652246.93/warc/CC-MAIN-20240920090502-20240920120502-00507.warc.gz | 903,206,680 | 23,086 | # CBSE Class 4 Mathematics Question Paper Set R
Read and download CBSE Class 4 Mathematics Question Paper Set R designed as per the latest question paper pattern and Class 4 examination guidelines issued by CBSE, NCERT and KVS. The past year Question Papers for Class 4 Mathematics have been provided with solutions which will help students to assess their performance and find out topics in Mathematics Class 4 which they need to improve to get better marks in Class 4 exams. After solving these last year papers also refer to solved Sample Papers for Class 4 Mathematics available on our website to build strong understanding of the subject
## Mathematics Question Paper Class 4
Students can refer to the below Class 4 Mathematics Question Paper designed to help students understand the pattern of questions that will be asked in Class 4 exams. Please download CBSE Class 4 Mathematics Question Paper Set R
### Mathematics Class 4 Question Paper
A. Choose the correct options.
1. 66521 x 10 = 66521 665210 66520
2. 7 x 600 = 4200 4300 4400
3. 81 ÷ 9 = 90 8 9
4. 7th multiple of 8 = 56 57 58
5. A prime number has only ___ factors. 1 unlimited 2
B. Fill in the blanks.
1. When 0 is divided by any number, the quotient is always ________.
2. The product of 200 and 30 is _____________________.
3. A number which has more than two factors is called a __________ number.
4. The product of a number and __________ is the number itself.
5. All _____________ numbers are divisible by 2.
C. Solve the following.
1. Find the product of 826 and 314.
2. Find the quotient and the remainder for 3165 ÷ 24.
3. What will be the dividend if divisor is 6, quotient is 9, and remainder is 4?
4. Find if 6 is a factor of 984.
5. Find the first four multiples of 1
D. Read the following questions properly. Identify the operation and solve .
1. There are 1409 students in a school. If each student pays a monthly fee of R.O. 35, how much does the school collect as fees every month?
2. The product of two numbers is 2756. If one of them is 4, find the other.
3. Find all the factors of 24.
1. Tony is 9 years old. His father is 4 times his age. If his grandfather is 30 years older than his father, what is his grandfather’s age?
2. Find the missing number if 24 x 2 = 4 x
## More Study Material
### Previous Year Question Paper Mathematics Class 4
We hope you liked the above provided CBSE Class 4 Mathematics Question Paper Set R. To get an understanding of the type of questions which were asked in previous year exams, it is important for Class 4 students to refer to CBSE Class 4 Mathematics Question Paper Set R that we have provided above. Students can download all the Question Papers for Class 4 Mathematics which were issued in the exams so that you can practise them and solve all types of questions which have been asked till now. If you do CBSE Class 4 Mathematics Question Paper Set R for Class 4 Mathematics you will easily understand the way examination question papers are set by teachers.
### Class 4 Mathematics Previous Year Question Papers
The past year question papers CBSE Class 4 Mathematics Question Paper Set R covers all the topics which have been suggested by CBSE in their Class 4 Mathematics syllabus for the current year. By doing all CBSE Class 4 Mathematics Question Paper Set R PYP (previous year papers) you will understand the regular questions and MCQ questions for Class 4 Mathematics which are always asked and learn them so that you can get full marks in those questions.
We have provided links above of CBSE Class 4 Mathematics Question Paper Set R in Pdf format so that you can just click the links and easily download the papers with answers. On studiestoday, we have also provided a download and print option for all CBSE Class 4 Mathematics last year papers and Sample Papers for Class 4 Mathematics to make it more easy for students. You can download CBSE Class 4 Mathematics Previous Year Question Papers and Class 4 Mathematics Sample Papers in PDF absolutely free of cost.
#### Mathematics Previous Year Question Papers Class 4 with Solutions
We have provided CBSE Class 4 Mathematics Question Paper Set R with solutions. You will be able to understand the type of answers which you should write in Class 4 Mathematics exams to score good marks. You should attempt all the last year question paper for Class 4 and Class 4 Mathematics MCQ Test in examination conditions at home and then compare their answers with the solutions provided by our teachers. This way you will be able to identify your weak areas and also understand how the teachers of Class 4 Mathematics set their questions in the exams.
#### CBSE Class 4 Mathematics Previous Year Question Papers with solutions
We have provided the last 10 years Class 4 Mathematics Question Papers. Students who want to practice question papers from all previous years can easily get them on studiestoday. We are the leading website in India from which you can download all the latest Class 4 Mathematics study material for free as per the CBSE syllabus for Class 4 Mathematics for the current academic year.
Where can I download CBSE Class 4 Mathematics Question Paper Set R in Pdf
You can download CBSE Class 4 Mathematics Question Paper Set R from StudiesToday.com
Can I download the previous year Question Papers of Class 4 Mathematics in Pdf
Is the Class 4 Mathematics Question Paper available for the latest session
Yes, the CBSE Class 4 Mathematics Question Paper Set R issued for Class 4 Mathematics have been made available here for latest academic session
How can I download and print CBSE Class 4 Mathematics Question Paper Set R
You can easily access the link above and download CBSE Class 4 Mathematics Question Paper Set R and save on your computer of mobile
Is there any charge for CBSE Class 4 Mathematics Question Paper Set R
There is no charge for CBSE Class 4 Mathematics Question Paper Set R you can download everything free
How can I improve my scores by solving CBSE Class 4 Mathematics Question Paper Set R
Regular practice of sample question paper given on studiestoday for CBSE Class 4 Mathematics Question Paper Set R can help you to score better marks in exams
Are there any websites that offer free CBSE Class 4 Mathematics Question Paper Set R
Yes, studiestoday.com provides all latest CBSE Class 4 Mathematics Question Papers with answers based on the latest format issued for current academic session
Can model paper CBSE Class 4 Mathematics Question Paper Set R be accessed on mobile devices
Yes, studiestoday provides CBSE Class 4 Mathematics Question Paper Set R in Pdf and can be accessed on smartphones and tablets.
Are mock CBSE Class 4 Mathematics Question Paper Set R available in multiple languages
Yes, mock CBSE Class 4 Mathematics Question Paper Set R are available in multiple languages, including English, Hindi | 1,533 | 6,997 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.4375 | 4 | CC-MAIN-2024-38 | latest | en | 0.879947 |
https://reviewgamezone.com/preview.php?id=17883 | 1,519,126,168,000,000,000 | text/html | crawl-data/CC-MAIN-2018-09/segments/1518891812938.85/warc/CC-MAIN-20180220110011-20180220130011-00750.warc.gz | 767,908,980 | 4,856 | # Dividing Unit Fractions: Question Preview (ID: 17883)
### Below is a preview of the questions contained within the game titled DIVIDING UNIT FRACTIONS: Fractions .To play games using this data set, follow the directions below. Good luck and have fun. Enjoy! [print these questions]
Chase made 4 batches of chili and divides each batch into fourths. How many ¼ batches of chili does he have?
a) 4 batches
b) 1/16 of a batch
c) 8 batches
d) 16 batches
A painter has 5 gallons of paint. He uses ¼ gallon of paint on each wall. How many walls can be he paint?
a) 10 walls
b) 4 walls
c) 20 walls
d) 40 walls
Grace baked 4 apple pies. She cuts the pies into halves and will give them to her neighbors. How many neighbors will get ½ of an apple pie.
a) 8 neighbors
b) 4 neighbors
c) 2 neighbors
d) Just Grace
Kayla made 2 pepperoni pizza and cut them into eighths. How many 1/8 size slices does she have?
a) 16 slices
b) 1/16 of the pizza
c) 14 slices
d) 4 slices
In 1 second, sound travels 1/5 of a mile. How many seconds will it take to travel 4 miles?
a) 20 seconds
b) 3 seconds
c) 1/20 of a second
d) 8 seconds
Felix has \$20 to spend on holiday gifts and he plans on spending 1/5 of his money on each person he is buying for. How many people will he be buying for?
a) 8 people
b) 20 people
c) 16 people
d) 4 people
One-half of a pan of brownies will be shared equally by 4 people. What fraction of the pan will each person receive?
a) 1/16
b) 1/8
c) 8 brownies
d) 16.5 brownies
2 divided by 1/8
a) 16
b) 1/16
c) 4
d) 1/4
¼ divided by 3
a) 12
b) 4
c) 1/12
d) 1/14
If your equation starts with a fraction, your quotient will be a…
a) Whole number
b) Unit fraction
c) Mixed Number
d) Improper Fraction
Play Games with the Questions above at ReviewGameZone.com
To play games using the questions from the data set above, visit ReviewGameZone.com and enter game ID number: 17883 in the upper right hand corner at ReviewGameZone.com or simply click on the link above this text.
TEACHERS / EDUCATORS | 622 | 2,004 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.03125 | 4 | CC-MAIN-2018-09 | latest | en | 0.892428 |
https://paspolini.studio/en/will-i-lose-water-pressure-going-from-a-2-pipe-to-a-1-2-pipe-then-back-to-2/ | 1,717,104,804,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971684053.99/warc/CC-MAIN-20240530210614-20240531000614-00792.warc.gz | 381,400,718 | 14,983 | You say that the water appears to be “losing pressure” as it comes out of the 1/2″ pipe, but in reality that expansion doesn’t really cause any restriction. Much of the flow loss occurs where the water enters the 1/2″ section, and then the long length of small pipe causes additional flow losses.
## Does changing pipe size affect water pressure?
In water flowing pipeline, pipe size and water pressure are dependent on each other. Because if the diameter of a pipe decreased, then the pressure in the pipeline will increase. As per Bernoulli’s theorem, pressure can be reduced when the area of conveyance is reduced.
## Does pressure change with pipe diameter?
* Obviously as you force a given amount of water through a smaller size pipe, the velocity of the water must increase for it to get through the smaller pipe. According to Bernoulli’s Principle that will decrease the water pressure!
## What happens to pressure when pipe size decreases?
A smaller pipe would lessen the flow of water as well as reduce the pressure loss in the pipes. This in turn would cause more pressure but render a sprinkler system inoperative. This is because sprinklers need a certain level of flow working in tandem with pressure to function properly.
## Does splitting a water line reduce pressure?
Decreasing the diameter, increases the velocity. Increasing the velocity, decreases the pressure.
## How can I make my water pressure stronger?
Look on the main supply pipe near your water meter for a conical valve that has a bolt sticking out of the cone. To raise pressure, turn the bolt clockwise after loosening its locknut. Keep an eye on the gauge to make sure the pressure is within bounds, then retighten the locknut.
## Does a smaller diameter hose increase water pressure?
Smaller diameter hoses and shorter hosepipes can increase water pressure. They confuse flow rate with pressure. Smaller pipes would increase water pressure.
## Does a larger diameter pipe increase water pressure?
A larger pipe, and lower velocity, has less pressure loss. The fittings in a larger pipe also have less pressure loss. So, all things considered, if you want to lose less pressure through a series of pipes and fittings, you increase the size.
## Does pipe length affect pressure?
Pipe length (L).
Friction pushes against the flow, resulting in pressure loss as the pipe length increases. Pipe length also factors in fittings, which increase the “length” input for the equation. Velocity of fluid (v): As flow velocity increases, pressure loss also increases and efficiency decreases.
## Does flow rate split between two pipes?
Youtube quote:Remember that the amount of fluid flowing through the pipe RIA in a time delta v delta t is always going to be equal to the product of the cross-sectional area times of velocity.
## How do you split water pipes?
Youtube quote:Slide the stop inlet onto the copper tube until you feel resistance. Hand tighten the nut. Be sure to position the outlets of the stop to the desired directions for easy connection to the faucet.
## Can pressure be divided?
No, on increasing the Pressure. But, You can increase the Volume by Increasing the Pipe Size, which Can Be a Significant Improvement.
## Does fluid flow from high to low pressure?
particles in high pressure air always flow to lower pressure. In a pipe with a constriction, fluid flows from from low to high pressure after the constriction.
## Why head loss is same in parallel pipes?
Head loss in two identical parallel pipes ( same diameter and same length made of same material) is equal because the loss of head due to friction depends on the flow speed ( which depends on diameter ) and length and pipe material.
## Is pressure the same in parallel pipes?
Re: Pressure in a parallel pipe system
Pressure will not necessarily be equal in the two branches. However, pressure drop across the two paths will be equal.
## When pipes are connected in parallel what happens?
The main pipeline divides into two or more parallel pipes which again join together downstream. Discharge in Parallel Pipe: The rate of flow in the main pipe is the sum rate of flow in branched pipes.
## How do you calculate pressure drop in a parallel pipe?
Youtube quote:So HF a is related to the pressure drop one or two if I've got this pipe and the pressure drop in this pipe is 10 psi the pressure drop in pipe B. Has to be 10 psi has to be 10 psi.
## Is back pressure and pressure drop the same?
So the two are different. Pressure drop is caused by agents inside whereas back pressure is controlled by you from outside.
## What will be loss of head when pipes are connected in series?
Detailed Solution. Pipes in series or compound pipes are defined as the pipes of different dimensions connected end to end to form a pipe. For the series pipes the flow through all the pipes will be same and the total head loss will be sum of head losses for individual pipes. | 1,002 | 4,949 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.546875 | 4 | CC-MAIN-2024-22 | latest | en | 0.930609 |
http://www.briddles.com/2014/02/linear-thinking-riddle.html | 1,516,324,832,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084887692.13/warc/CC-MAIN-20180119010338-20180119030338-00066.warc.gz | 400,358,053 | 6,143 | There are four people in a house. A fireman, an athlete, an old woman and a drunk guy. The house catches fire and before the fact is known, it is too late. All they know is that the entire house is in flames and it will collapse exactly after twelve minutes. Now they can move out of the house but for that, they will have to pass the hallway which is entirely blazing with flames. Thus to move, one must carry a fire extinguisher to keep the flames away. Seeking the burnt wooden floor, only two person can run through that hallway at one time. But for others to go, one must return back with the fire extinguisher. The fireman, is trained for such tasks and can run through the hallway in a minute. The athlete can make it in a couple of minutes. The old woman can run slowly and will cover the hallway in four minutes. The drunk guy will take five minutes to run through it. If all of them can make it through the hallway in twelve minutes, all of them will be saved. When two move together, they will run with the speed of the slower one.
How will all four of them manage to run to safety?
1. Fireman and the fit guy will move first which will take 2 minutes.
2. Fireman will return back with fire extinguisher, taking 1 minutes.
3. Drunk guy will move with old woman taking 5 minutes.
4. The fit guy will return taking 2 minutes.
5. Both the fit guy and fireman will move taking 2 minutes.
2 + 1 + 5 + 2 + 2 = 12 minutes.
Discussion | 340 | 1,441 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.3125 | 3 | CC-MAIN-2018-05 | latest | en | 0.954373 |
http://www.edurite.com/kbase/at-what-temperature-does-water-become-a-gas | 1,464,104,552,000,000,000 | text/html | crawl-data/CC-MAIN-2016-22/segments/1464049272349.32/warc/CC-MAIN-20160524002112-00133-ip-10-185-217-139.ec2.internal.warc.gz | 491,668,435 | 16,773 | #### • Class 11 Physics Demo
Explore Related Concepts
# at what temperature does water become a gas
Question:AND WHY?
Answers:Ethanol is a liquid at RT Propane is a gas at RT NaCl is a solid at RT These properties are all due to intermoleccular attractions between the molecules in the samples. In gases, molecules are all floating around seperate from one an other. They hardly interact with each other at all. In liquids, molecules are quite close to each other and are held together in a smaller volume then gases. The molecules do not have as much energy as gas molecules. For a molecules to go from liquid to gas phase you have to provide it with enough energy to break away from all the other molecules. In solid phase molecules are very close to each other and hardly move at all. To go from solid to liquid you have to provide enough energy for the molecules to break their attractive forces with each other enough to be able to move around more freely. Propane is a hydrocarbon and is a non-polar compound. There are only very weak attractive forces between 1 propane molecule and another propane molecule. Because of this there are only very small forces holding the molecules in a sample together, so it does not take very much energy for the molecules to be in the gas phase. RT is enough. Ethanol has a polar group on it (OH). Polar groups have a slightly positive and a slightly negative end. The positive and of one molecule is attracted to the negative end of another molecule. So they are held close to each other by these "intermolecular" forces. These forces are moderate in stranght. Because ethanol molecules are attracted to each other like this it takes quite a bit of energy from an individual molecule to break away from these attractions. So ethanol is a liquid at RT. NaCl is an ionic compound. It consists of a cation Na+ and anion Cl- Cations and anions are extremely attracted to each other. In solid NaCl the cations and anions are arranged in a crystal lattice structure with alternating cations and anions that are all very strongly attracted to each other. It takes a large amount of energy to provide enough energy for these bonds to break. For this reason NaCl is solid at RT.
Question:Does water actually change from a liquid to a solid at exactly 0 degrees Celsius/32 degrees F? Does it change from a liquid to a gas at exactly 212 degrees F/ 100 degrees Celsius?
Answers:Great question! Also correct! Also open-air boiling point of water is typically considered to be 100 C or 212 F, although pressure and a change in composition of the liquid may alter the boiling or freezing point of the liquid. If you were to take something (anything at all) and break it up into almost the smallest things you could, you d get molecules. Everything there is is made up of lots of little molecules. If the molecules are stuck together really tightly, then they re called a solid. The solid form of water is ice. This actually makes a lot of sense, because it certainly does seem like all the little parts of a solid (like ice) are stuck together very tightly. When you heat something up, it makes the molecules move faster. If you heat up a solid, it melts and becomes a liquid. In a liquid (like water), the molecules are still stuck together, but they can move around some. What actually happens is that the molecules are still sort of sticking together, but they re constantly breaking apart and sticking to different molecules. This also makes sense when you think about water. Water sort of sticks together, but it breaks apart /really/ easily. If you heat a liquid like water up even more (like if you put it in a pot on the stove), then the molecules will move around so fast that they can t even hold on to each other at all. When this happens, all of the molecules go flying apart and become a gas (like when you boil water to make steam). The process of heating up a liquid like water to make a gas like steam is called "evaporation." When you do the opposite and cool a gas down to make a liquid, then it s called "condensation." Hope this answers your question! -Jerry
Question:i am looking for PSI
Answers:The critical point of air is 132 Kelvin and 61 atmospheres. This means that at normal temperature of 20C or 292 Kelvin it is impossible to compress air to a liquid form - it remains a gas at ALL pressures. You can only compress it to a liquid once it is already below 132 Kelvin, and this will take 61 atmospheres pressure at that point.
Question:
Answers:The Atmosphere How is the atmosphere structured? The atmosphere consists of four distinct layers, which are seperated by a boundaries that end in the suffix "pause". There is a lot of information in the Earth Science Reference Tables page 14. Troposphere-This layer of the atmosphere is closest to the earth and 99% of all weather takes place here because most of the water vapor is located here. Stratosphere-This layer is above the troposphere and contains ozone, which is an important gas, because it absorbs cancer-causing ultra violet rays from the Sun. Mesosphere-This layer is above the stratosphere. Thermosphere-This is the uppermost layer of the atmosphere. Earth Science Reference Table page 14: How do we describe the present condition of the atmosphere? The present condition of the atmosphere is called WEATHER and it is described using observations of temperature, air pressure, humidity, wind, cloudiness and precipitation. What is temperature? Temperature is a measure of the average kinetic energy of the particles of a substance. In other words, the more molecules vibrate within a substance, the higher the temperature. Measurement - There are three scales that measure temperature: CELSIUS, FAHRENHEIT AND KELVIN scales. There is a conversion scale on page 13 of your ESRT. Maps - Meteorologists find it useful to map temperatures and they use lines to connect equal temperatures called Isotherms. What is air pressure and how is it used to predict the weather? Air Pressure - Air pressure is also called barometric or atmospheric pressure. By definition, it is the weight of the atmosphere pushing down on the earth. Barometer - A barometer is the instrument that measures air pressure and it uses units of millibars (mb) or inches of mercury (inches). Factors that affect air pressure: Temperature - As the temperature of the air increases, the air pressure decreases, which is an indirect relationship. Humidity - As the humidity of the air increases, the air pressure decreases, which is an indirect relationship. You might ask "WHY?": Humid air is lighter than dry air. When water vapor enters dry air, it does not squeeze in between the molecules of air. Instead, the water vapor pushes out many of the air's heaviest molecules like Nitrogen and Oxygen. In other words, lighter molecules of water vapor replace heavier molecules and the result is the air weighs less and therefore has less air pressure. High air pressure - means that the atmosphere is heavy at that location and from our notes above, you can conclude that the air is probably cool and dry (if you are not sure, see notes above). So, if the barometer indicates high pressure or rising air pressure, you can conclude that the weather will be cool and dry. Low air pressure - means that the atmosphere is not heavy at that location and from our notes above, you can conclude that the air is probably warm and humid (if you are not sure, see notes above). So, if the barometer indicates low pressure or falling air pressure, you can conclude that the weather will be warmer and humid. Maps - Meteorologists find it useful to map air pressure and they use lines to connect equal equal air pressures called Isobars | 1,624 | 7,715 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.828125 | 3 | CC-MAIN-2016-22 | longest | en | 0.960654 |
https://brainmass.com/business/accounting/realized-gain-loss-362118 | 1,638,064,797,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964358443.87/warc/CC-MAIN-20211128013650-20211128043650-00631.warc.gz | 240,919,370 | 75,526 | Explore BrainMass
# Realized gain or loss
Not what you're looking for? Search our solutions OR ask your own Custom question.
This content was COPIED from BrainMass.com - View the original, and get the already-completed solution here!
46. Elaine exchanges a van that is used exclusively for business purposes for another van that also is used exclusively for business. The adjusted basis for the old van is \$18,000 and its FMV is \$14,500.
a. Calculate Elaines recognized gain or loss on the exchange.
b. Calculate Elaine's basis for the van she receives.
52. Patti's garage (used to store business property) is destroyed by a fire. She decides not to replace it and uses the insurance proceeds to invest in her business. The garage had an adjusted basis of \$50,000.
a. If the insurance proceeds total \$20,000, what is Patti's recognized gain or
loss?
b. If the insurance proceeds total \$60,000, what is Patti's recognized gain or
loss?
57. Virginia is an accountant for a global CPA firm. She is being temporarily transferred from the Raleigh, North Carolina, office to Tokyo. She will leave Raleigh on October 7, 2009, and will be out of the country for four years. She sells her personal residence on September 30, 2009, for \$250,000 (her adjusted basis is \$190,000). Upon her return to the United States in 2013, she purchases a new residence in Los Angeles for \$220,000, where she will continue working for the same firm.
a. What are Virginia's realized and recognized gain or loss?
b. What is Virginia's basis in the new residence?
61. On January 1, 2009, Myron sells stock that has a \$50,000 FMV on the date of the sale (basis \$75,000) to his son Vernon. On October 21,2009, Vernon sells the stock to an unrelated party. In each of the following, determine the tax consequences of these transactions to Myron and Vernon"
a. Vernon sells the stock for \$40,000.
b. Vernon sells the stock for \$80,000.
c. Vernon sells the stock for \$65,000.
#### Solution Preview
46.
Since the van is used exclusively for business purpose, this transaction qualifies as a like-kind exchange.
And there's no boots received, so there's no gain or loss recognized in a like-kind exchange and the basis of the van received equals the basis of the van given, which is \$18,000.
52.
Patti can't defer any gain or loss because she decides not to replace the garage, which ...
#### Solution Summary
Realized gain or losses are examined.
\$2.49 | 579 | 2,450 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.71875 | 3 | CC-MAIN-2021-49 | latest | en | 0.936073 |
http://mathhelpforum.com/advanced-statistics/166088-standard-normal-distrbution-question.html | 1,529,393,306,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267861980.33/warc/CC-MAIN-20180619060647-20180619080647-00267.warc.gz | 202,078,169 | 10,474 | # Thread: Standard Normal Distrbution question
1. ## Standard Normal Distrbution question
Hello All,
I've found this forum to be a lot of help. The answers provided to my last question were helpful and I'm hoping I can get similar help for this question.
Y has the standard normal distribution.
1. Determine the distribution function of |Y|, the absolute value of Y.
2. Determine the probability density function of
|Y|
3. Find P(
|Y|>1.70).
For part 1, I did this:
P(|Y|) ≤ x)= P(−x ≤ Y≤ x) = ϕ(x) ϕ(-x)
= ϕ(x) -(1 − ϕ(x) ) = 2ϕ(x) − 1 .
Does that look correct?
For 2, this is my work,
f(x)= d
P(|Y|) ≤ x)= 2ϕ'(x)
dx
Which is equal to 2*(1/
√2π)*e^(-x^2/2)
Is my work right for this?
I'm not sure about how I would go about doing 3, so all tips are appreciated. Thanks.
2. This is tough to type with an I(dot) pad, but the prob of abs y greater than 1.7 is twice that y is greater than 1.7
3. Originally Posted by hwill205
Hello All,
I've found this forum to be a lot of help. The answers provided to my last question were helpful and I'm hoping I can get similar help for this question.
Y has the standard normal distribution.
1. Determine the distribution function of [FONT=verdana, geneva, helvetica][SIZE=2][COLOR=#000000]|Y|, the absolute value of Y.
$\displaystyle F_{|Y|}(u)=Pr(|Y|<u)=Pr(-u<Y<u),\ \mbox{ when } u>0 \mbox{ and } 0 \mbox{ otherwise }$
Now write the right most probability in terms of the cumulative standard normal. Which is what you did, so yes.
CB
4. Originally Posted by hwill205
2. Determine the probability density function of [/COLOR][/SIZE][/FONT][FONT=verdana, geneva, helvetica][SIZE=2][COLOR=#000000]|Y|
Differentiate the cumulative distribution you get in part (1). Again this is what you did so yes.
CB | 527 | 1,762 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.671875 | 4 | CC-MAIN-2018-26 | latest | en | 0.865804 |
https://www.aqua-calc.com/calculate/weight-to-volume/substance/-op-e-cp--stilbene | 1,726,303,445,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651559.58/warc/CC-MAIN-20240914061427-20240914091427-00467.warc.gz | 610,933,173 | 13,054 | # Volume of (E)-Stilbene
## (e)-stilbene: convert weight to volume
### Volume of 100 grams of (E)-Stilbene
centimeter³ 103.02 milliliter 103.02 foot³ 0 oil barrel 0 Imperial gallon 0.02 US cup 0.44 inch³ 6.29 US fluid ounce 3.48 liter 0.1 US gallon 0.03 meter³ 0 US pint 0.22 metric cup 0.41 US quart 0.11 metric tablespoon 6.87 US tablespoon 6.97 metric teaspoon 20.6 US teaspoon 20.9
### The entered weight of (E)-Stilbene in various units of weight
carat 500 ounce 3.53 gram 100 pound 0.22 kilogram 0.1 tonne 0 milligram 100 000
#### How many moles in 100 grams of (E)-Stilbene?
There are 554.79 millimoles in 100 grams of (E)-Stilbene
#### Foods, Nutrients and Calories
SWEET CORN FREEZE DRIED VEGETABLES, UPC: 085239801918 weigh(s) 39 grams per metric cup or 1.3 ounces per US cup, and contain(s) 429 calories per 100 grams (≈3.53 ounces) [ weight to volume | volume to weight | price | density ]
Inulin in CHICORY ORGANIC HERBAL TEA, DANDELION COCONUT, UPC: 795239433407
#### Gravels, Substances and Oils
CaribSea, Freshwater, Super Naturals, Tahitian Moon weighs 1 473.7 kg/m³ (92.00009 lb/ft³) with specific gravity of 1.4737 relative to pure water. Calculate how much of this gravel is required to attain a specific depth in a cylindricalquarter cylindrical or in a rectangular shaped aquarium or pond [ weight to volume | volume to weight | price ]
Lead sulphate [PbSO4] weighs 6 200 kg/m³ (387.05336 lb/ft³) [ weight to volume | volume to weight | price | mole to volume and weight | mass and molar concentration | density ]
Volume to weightweight to volume and cost conversions for Refrigerant R-422B, liquid (R422B) with temperature in the range of -51.12°C (-60.016°F) to 60°C (140°F)
#### Weights and Measurements
A mile per hour or mph is a non-SI (non-System International) measurement unit of speed or velocity with which to measure how many miles traveled per one hour.
Luminous intensity is the luminous flux emitted by a light source in a particular direction within a solid angle.
troy/pm² to g/Ų conversion table, troy/pm² to g/Ų unit converter or convert between all units of surface density measurement.
#### Calculators
Weight to Volume conversions for sands, gravels and substrates | 662 | 2,237 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.78125 | 3 | CC-MAIN-2024-38 | latest | en | 0.692076 |
https://nz.education.com/common-core/CCSS.MATH.CONTENT.5.NF.A/ | 1,604,099,825,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107911792.65/warc/CC-MAIN-20201030212708-20201031002708-00634.warc.gz | 353,265,204 | 28,192 | # 5.NF.A Worksheets, Workbooks, Lesson Plans, and Games
#### CCSS.maths.CONTENT.5.NF.A
:
"Use equivalent fractions as a strategy to add and subtract fractions."
These worksheets, lesson plans, and exercises can help students practise this Common Core State Standards skill.
## Worksheets
Equivalent Fractions: Bar Models
Worksheet
Equivalent Fractions: Bar Models
With this worksheet, your students will use bar models to demonstrate their understanding of equivalent fractions.
Year 5
Maths
Worksheet
Scary Skeleton Word Problems
Worksheet
Scary Skeleton Word Problems
Don't let skeletons scare you from doing Halloween maths! practise addition, subtraction, multiplication, and even fractions with scary skeleton word problems.
Year 6
Maths
Worksheet
Make It Work! Adding Fractions with Unlike Denominators
Worksheet
Make It Work! Adding Fractions with Unlike Denominators
Help students add fractions easily with this step-by-step worksheet.
Year 6
Maths
Worksheet
Subtracting with Unlike Denominators Word Problem
Worksheet
Subtracting with Unlike Denominators Word Problem
Your child can conquer fraction word problems with this simple, step-by-step worksheet.
Year 6
Maths
Worksheet
Fraction Word Problems: Pie Time
Worksheet
Fraction Word Problems: Pie Time
What fraction of the pie do you want? This tasty maths worksheet will help your student familiarize with fractions.
Year 6
Maths
Worksheet
Fraction Word Problems: Strawberry Stand
Worksheet
Fraction Word Problems: Strawberry Stand
Amy needs help calculating her strawberry sales! This maths worksheet will be great for practical application of fractions in everyday life.
Year 6
Maths
Worksheet
## Lesson Plans
Organizing Word Problems
Lesson plan
Organizing Word Problems
Help students solve fraction word problems with a graphic organizer! This lesson addresses Common Core standards while teaching useful skills for standardized test taking.
Year 5
Maths
Lesson plan
Make It Work! Adding Fractions with Unlike Denominators
Lesson plan
Make It Work! Adding Fractions with Unlike Denominators
Adding fractions with unlike denominators just doesn't work like it does when the denominators are the same. Let's make it work! Students will add fractions with unlike denominators with sums between one and two.
Year 6
Maths
Lesson plan
When Having Like Denominators Isn’t Enough!
Lesson plan
When Having Like Denominators Isn’t Enough!
Does chaos ensue when your students subtract like mixed numbers and the larger fraction has a lesser numerator? Using this lesson plan, teach your class how to overcome this tricky obstacle: When having like denominators isn’t enough!
Year 6
Maths
Lesson plan
Solving Mixed Number Word Problems Using the Three-Reads Strategy
Lesson plan
Solving Mixed Number Word Problems Using the Three-Reads Strategy
There’s more than one way to add and subtract mixed numbers! Use this witty lesson plan to teach your students to illustrate mixed number sums and difference while giggling with delight.
Year 6
Maths
Lesson plan
Word Problems: Adding Mixed Number Fractions
Lesson plan
Word Problems: Adding Mixed Number Fractions
Children often wonder if algorithms are practical. Using this lesson plan, your students will see how the standard algorithm for adding mixed numbers brings life to real world real-world situations.
Year 6
Maths
Lesson plan
Fraction Simplest Form by Greatest Common Factor
Lesson plan
Fraction Simplest Form by Greatest Common Factor
When fractions have big numerators and denominators, it can be tough to find the simplest form. Fear not! Use this lesson plan to teach your students to find the simplest form using the greatest common factor.
Year 6
Maths
Lesson plan
## Workbooks
No workbooks found for this common core node.
## Games
No games found for this common core node.
## Exercises
Multiplying Fractions by Whole Numbers
Exercise
Multiplying Fractions by Whole Numbers
Multiplying Fractions by Whole Numbers will help students practise this key year six skill. Try our free exercises to build knowledge and confidence.
Year 6
Maths
Exercise
Fractions and Greatest Common Factors
Exercise
Fractions and Greatest Common Factors
Fractions and Greatest Common Factors will help students practise this key year six skill. Try our free exercises to build knowledge and confidence.
Year 6
Maths
Exercise
Multiplying Mixed Number Fractions and Arrays
Exercise
Multiplying Mixed Number Fractions and Arrays
Multiplying Mixed Number Fractions and Arrays will help students practise this key year six skill. Try our free exercises to build knowledge and confidence.
Year 6
Maths
Exercise
Exercise
Adding Mixed Number Fractions will help students practise this key year six skill. Try our free exercises to build knowledge and confidence.
Year 6
Maths
Exercise
Addition and Subtraction of Mixed Numbers
Exercise
Addition and Subtraction of Mixed Numbers
Teach students how to add and subtract mixed numbers by assigning this straightforward Education.com exercise.
Year 6
Maths
Exercise
Subtraction of Mixed Numbers
Exercise
Subtraction of Mixed Numbers
Subtraction of Mixed Numbers will help students practise this key year six skill. Try our free exercises to build knowledge and confidence.
Year 6
Maths
Exercise
Create new collection
0
### New Collection>
0Items
What could we do to improve Education.com? | 1,121 | 5,339 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.234375 | 3 | CC-MAIN-2020-45 | latest | en | 0.797002 |
https://gamedev.stackexchange.com/questions/203577/what-dynamic-range-of-numbers-do-games-typically-use/203589 | 1,708,932,879,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474653.81/warc/CC-MAIN-20240226062606-20240226092606-00624.warc.gz | 279,937,484 | 38,261 | # What dynamic range of numbers do games typically use?
Most of the numeric quantities used in games (aside from integers for counting or indexing objects) are single-precision floating point. This is IEEE format, which allows for numbers as big as 1e38 and as small as 1e-38.
Such a large dynamic range makes sense in the context in which the IEEE floating-point standard was defined: roughly speaking, for scientists, who routinely deal with very large and very small numbers.
Intuitively, games seem unlikely to have to deal with such a wide range. Say you measure distance in meters. Draw distance maybe a few kilometers? Smallest objects a few millimeters? Granted that multiplying numbers can temporarily take them out of the original range, it would seem sufficient to have a dynamic range like 1e8 to 1e-6, whereupon more of the bits could've been used for greater precision.
But maybe I am missing something about the data and calculations routinely used in games.
What sort of dynamic range do games actually use?
• I'm curious what actionable steps you're hoping to take with the answers you get to this question. Floating point may not be ideal, but it's what we have the hardware to work with. You could of course implement your own number format using integer math under the hood (or use an existing library that provides this), but that gives you flexibility to define a completely bespoke type with exactly the dynamic range you need for ONE game, or even just each single feature of one game. So knowing ranges used in "typical" games doesn't really help with that task. What's your goal we can help with? Dec 6, 2022 at 15:59
• @DMGregory The same as my goal in reading about the orbital dynamics of galaxies or Napoleon's campaign in Russia: to understand some aspect of how the world works. Dec 6, 2022 at 19:39
As games cover such a wide range of things, there is really no typical. However, for 3D worlds, greater scale is becoming more popular. Open worlds, for example, can cover any size you care to imagine.
Based on my experiments, if you want accuracy in the millimeters throughout, then with single precision, a world of about 70,000m is fine before you can start to notice some jitter.
Some math on this: single precision has a max mantissa value of 2^23-1, or 8,388,607. 70,000 / 8,388,607 is about 0.008m or 8mm. However, the resolution or error in 3D space is a 3D geometric formula, not the 1D calculation I just did.
If you factor in that the base error is a 3D geometric formula, the actual error is a factor of 3.4xdistance (ref geometric relative error) greater than what I calculated. I.E instead of 8mm accuracy the worst case accuracy much lower - probably in the centimetres to a meter.
Then if you do any calculations, the error is magnified by propagation. E.g. a multiply by 10 will make it 10x bigger.
The takeaway is that mm resolution is not achievable beyond 70km and most likely at a smaller range.
Practical proof: In the video: I demonstrate visible errors at the 70km range from distant relative jitter: Distant relative jitter (bookmark "Attempt to move to a point intersection". As I try to move to a point where two squares meet, the visible error would appear to be about 0.5 to a meter. The jitter at times appears to be less: in the cm range. The sliding motion demonstrates loss of resolution in at least one axis.
In contrast, at the bookmark: "At least mm accuracy when move to target" motion to target is smooth and accurate. This is because the design manages the resolution at that place by nesting a high-resolution virtual space within the World space.
• That is helpful, thanks! Sounds like confirmation of my conjecture: the limiting factor is precision, not the dynamic range of the exponent, so the practically achievable range would be better if there were one or two fewer bits of exponent, and one or two more bits of precision? Dec 6, 2022 at 13:02
• @rwallace Unfortunately current CPUs and GPUs don't have hardware support for custom floating point formats. Theoretically you could emulate them by using raw bytes and implementing your own mathematical operators, but then you lose a lot of performance... and sanity. Dec 6, 2022 at 14:03
• You may also be interested in What's the largest "relative" level I can make using float? Dec 6, 2022 at 18:39
• @rwallace Although "the limiting factor is precision" is the way I, and most people started, this is not actually true. You can travel the Solar system as one continuous space with just float, evidence: youtu.be/_04gv3CnjDU. The limiting factor is how we think about it and the resulting design we apply. Dec 6, 2022 at 23:15
• @MadMan, that is one of mine. I don't use separate local coordinates or any sectorized map of absolute coordinates - for the moving player. I use dynamic virtual spaces. The coordinate system around the stationary traveler never changes. However, the player may exist in multiple virtual spaces at any time. Travel is via relative movement (differentials) and the precision limits don't apply. Dec 8, 2022 at 4:00
I can give the game I'm working on as an example. It certainly does not fit "routinely used" being something of an outlier, but still might be illustrative as an example that the storage of the world doesn't need to be limited to what the graphics card can handle.
I'm working on a game set in a realistically scaled galaxy (10^18 km across, or even larger, and I also want to be able to fly outside it), with a unit scale of 10 km. For this, single precision floats or even double precision floats would have inadequate precision, and single precision floats are even problematic with dynamic range in intermediate calculations (I have overflowed them several times). My solution is that positions are stored as 64 bit integers, with 128 bit integers for some intermediate calculations.
The main target for this is Android. I can't assume the graphics card can handle anything bigger than single precision floats, so there is some juggling going on. One technique I use is to pass the graphics card offsets (as floats) from a particular point in space (the zero point), and occasionally re-center the zero point to be near the viewpoint to limit the distortion that would be caused by the zero point being far from the viewpoint. (This sort of distortion can be seen in Minecraft if you go very far from the start area).
I will also have to consider changing scales to something finer than a 10 km resolution for planets or moons when the viewpoint goes close to them. | 1,474 | 6,561 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.828125 | 3 | CC-MAIN-2024-10 | latest | en | 0.963096 |
http://www.scirp.org/journal/paperinformation.aspx?paperid=68634 | 1,571,614,732,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570986726836.64/warc/CC-MAIN-20191020210506-20191020234006-00278.warc.gz | 320,072,948 | 21,812 | Non-Standard Difference Method for Numerical Solution of Linear Fredholm Integro-Differential Type Two-Point Boundary Value Problems
In this article we have considered Fredholm integro-differential equation type second-order boundary value problems and proposed a rational difference method for numerical solution of the problems. The composite trapezoidal quadrature and non-standard difference method are used to convert Fredholm integro-differential equation into a system of equations. The numerical results in experiment on some model problems show the simplicity and efficiency of the method. Numerical results showed that the proposed method is convergent and at least second-order of accurate.
Conflicts of Interest
The authors declare no conflicts of interest.
Cite this paper
Pandey, P. (2015) Non-Standard Difference Method for Numerical Solution of Linear Fredholm Integro-Differential Type Two-Point Boundary Value Problems. Open Access Library Journal, 2, 1-10. doi: 10.4236/oalib.1101465.
[1] Delves, L.M. and Mohamed, J.L. (1985) Computational Methods for Integral Equations. Cambridge University Press, Cambridge. http://dx.doi.org/10.1017/CBO9780511569609 [2] Liz, E. and Nieto, J.J. (1996) Boundary Value Problems for Second Order Integro-Differential Equations of Fredholm Type. Journal of Computational and Applied Mathematics, 72, 215-225. http://dx.doi.org/10.1016/0377-0427(95)00273-1 [3] Zhao, J. and Corless, R.M. (2006) Compact Finite Difference Method Has Been Used for Integro-Differential Equations. Applied Mathematics and Computation, 177, 271-288. http://dx.doi.org/10.1016/j.amc.2005.11.007 [4] Ortiz, E.L. and Samara, L. (1981) An Operational Approach to the Tau Method for the Numerical Solution of Nonlinear Differential Equations. Computing, 27, 15-25. http://dx.doi.org/10.1007/BF02243435 [5] Chang, S.H. (1982) On Certain Extrapolation Methods for the Numerical Solution of Integro-Differential Equations. Mathematics of Computation, 39, 165-171. http://dx.doi.org/10.1090/S0025-5718-1982-0658220-4 [6] Yalcinbas, S. (2002) Taylor Polynomial Solutions of Nonlinear Volterra-Fredholm Integral Equations. Applied Mathematics and Computation, 127, 195-206. http://dx.doi.org/10.1016/S0096-3003(00)00165-X [7] Phillips, D.L. (1962) A Technique for the Numerical Solution of Certain Integral Equations of the First Kind. Journal of the ACM, 9, 84-96. http://dx.doi.org/10.1145/321105.321114 [8] Tikhonov, A.N. (1963) On the Solution of Incorrectly Posed Problem and the Method of Regularization. Soviet Mathematics, 4, 1035-1038. [9] He, J.H. (2000) Variational Iteration Method for Autonomous Ordinary Differential Systems. Applied Mathematics and Computation, 114, 115-123. http://dx.doi.org/10.1016/S0096-3003(99)00104-6 [10] Wazwaz, A.M. (1999) A Reliable Modification of the Adomian Decomposition Method. Applied Mathematics and Computation, 102, 77-86. http://dx.doi.org/10.1016/S0096-3003(98)10024-3 [11] Saadati, R., Raftari, B., Abibi, H., Vaezpour, S.M. and Shakeri, S. (2008) A Comparison between the Variational Iteration Method and Trapezoidal Rule for Solving Linear Integro-Differential Equations. World Applied Sciences Journal, 4, 321-325. [12] Hu, S., Wan, Z. and Khavanin, M. (1987) On the Existence and Uniqueness for Nonlinear Integro-Differential Equations. Journal of Mathematical and Physical Sciences, 21, 93-103. [13] Hairer, E., Nørsett, S.P. and Wanner, G. (1993) Solving Ordinary Differential Equations I Nonstiff Problems (Second Revised Edition). Springer-Verlag, New York. [14] Van Niekerk, F.D. (1988) Rational One Step Method for Initial Value Problem. Computers & Mathematics with Applications, 16, 1035-1039.http://dx.doi.org/10.1016/0898-1221(88)90259-3 [15] Pandey, P.K. (2013) Nonlinear Explicit Method for First Order Initial Value Problems. Acta Technica Jaurinensis, 6, 118-125. [16] Ramos, H. (2007) A Non-Standard Explicit Integration Scheme for Initial Value Problems. Applied Mathematics and Computation, 189, 710-718. http://dx.doi.org/10.1016/j.amc.2006.11.134 [17] Jain, M.K., Iyenger, S.R.K. and Jain, R.K. (1987) Numerical Methods for Scientific and Engineering Computation. Willey Eastern Limited, New Delhi. [18] Lambert, J.D. (1991) Numerical Methods for Ordinary Differential Systems. Wiley, England. [19] Pandey, P.K. (2013) A Non-Classical Finite Difference Method for Solving Two Point Boundary Value Problems. Pacific Journal of Science and Technology, 14, 147-152. [20] Varga, R.S. (2000) Matrix Iterative Analysis, Second Revised and Expanded Edition. Springer-Verlag, Heidelberg. http://dx.doi.org/10.1007/978-3-642-05156-2 [21] Henrici, P. (1982) Discrete Variable Methods in Ordinary Differential Equations. John Wiley and Sons, New York. [22] Volkov, Y.S. and Miroshnichenko, V.L. (2009) Norm Estimates for the Inverses of Matrices of Monotone Type and Totally Positive Matrics. Siberian Mathematical Journal, 50, 982-987. http://dx.doi.org/10.1007/s11202-009-0108-2 [23] Varah, J.M. (1975) A Lower Bound for the Smallest Singular Value of a Matrix. Linear Algebra and Its Applications, 11, 3-5. http://dx.doi.org/10.1016/0024-3795(75)90112-3 [24] Ahlberg, J.H. and Nilson, E.N. (1963) Convergence Properties of the Spline Fit. Journal of the Society for Industrial and Applied Mathematics, 11, 95-104. http://dx.doi.org/10.1137/0111007 [25] Horn, R.A. and Johnson, C.R. (1990) Matrix Analysis. Cambridge University Press, New York. [26] Shaw, R.E., Garey, L.E. and Lizotte, D.J. (2001) A Parrllel Numerical Algorithm for Fredholm Integro-Differential Two Point Boundary Value Problems. International Journal of Computer Mathematics, 77, 305-318. http://dx.doi.org/10.1080/00207160108805067 | 1,686 | 5,697 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.671875 | 3 | CC-MAIN-2019-43 | latest | en | 0.717455 |
https://wcipeg.com/problem/wc01Sp2#comment647 | 1,721,760,366,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763518059.67/warc/CC-MAIN-20240723163815-20240723193815-00636.warc.gz | 539,136,085 | 5,672 | ## Tuesday - Temptation Island
On Monday, the number of frosh were reduced in half. To further reduce the number of engineers to a manageable number, the following challenge was devised for the second day. Each of the students would have to take this challenge individually.
Each student would be placed at a vertex of perimeter fence of Waterloo (oh yeah, some background: to keep UofT’s engineering Lady Godiva band out of Waterloo, a fence was erected surrounding the university. The fence just happens to be an N-gon). At some other vertex along the fence would be located a temptation so seductive that no Waterloo student could resist – an extra-credit assignment. The challenge of each student is to go from his starting vertex to the vertex with the prize. There are, however, 3 rules:
1. The student can only travel from vertex to vertex (backwards or forwards) along the polygonal fence.
2. The student has to make contact with exactly K vertices (the vertex he starts at doesn’t count unless he returns to it). The K vertices need not be unique. The final vertex has to be the one with the prize.
3. If the student cannot reach the prize and make contact with exactly K vertices, he fails the test and is kicked out of the university.
Of course, no Waterloo student is satisfied with only 1 solution to any problem. Therefore, inevitably, each student determines all ways that he/she can win. Note that there may be no solution to the problem (the astute student has figured out that this will result in a class size of 0 – this is entirely allowable as the variable used to quantify enrollment was incorrectly defined as a whole number instead of a natural number).
### Input
N K (N, K ≤ 50)
A B (A = the starting vertex number, B = destination vertex number)
There will be multiple test cases, one after another: a case where N = -1, K = -1 terminates input.
### Output
The total number of ways of reaching the destination from the starting point by following the above rules. The total number of ways will be less than 2,147,483,647 (ie. a longint). Output 0 if there are no solutions.
### Sample Input
```8 5 1 4 -1 -1```
### Sample Output
`6`
Point Value: 10
Time Limit: 2.00s
Memory Limit: 16M
Added: Dec 05, 2008
Languages Allowed:
C++03, PAS, C, HASK, ASM, RUBY, PYTH2, JAVA, PHP, SCM, CAML, PERL, C#, C++11, PYTH3
## Comments (Search)
• (1/1)
Good answer.
No wonder you got -4 rating -.-
• (1/1)
-6*
and I didn't use a math formula.
Although it seems like there is one for some reason.
• (0/0)
There is a sort of math formula.
(It involves N choose K)
It probably isn't easier than the 'regular' way.
• (2/2)
need to generate some math formula?
• (0/0)
There are 4 numbers for every test case:
N and K are different every time.
• (0/2)
or is there a person on every vertex
• (0/0)
What do you mean?
There is only one student trying to get from the start to the end. The 'other students' are irrelevant. | 733 | 2,947 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.890625 | 3 | CC-MAIN-2024-30 | latest | en | 0.956717 |
https://nrich.maths.org/public/topic.php?code=-5&cl=3&cldcmpid=1966 | 1,571,230,525,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570986668569.22/warc/CC-MAIN-20191016113040-20191016140540-00240.warc.gz | 613,518,763 | 9,744 | # Search by Topic
#### Resources tagged with Divisibility similar to Obviously?:
Filter by: Content type:
Age range:
Challenge level:
### There are 54 results
Broad Topics > Numbers and the Number System > Divisibility
### Obviously?
##### Age 14 to 18 Challenge Level:
Find the values of n for which 1^n + 8^n - 3^n - 6^n is divisible by 6.
### Check Codes
##### Age 14 to 16 Challenge Level:
Details are given of how check codes are constructed (using modulus arithmetic for passports, bank accounts, credit cards, ISBN book numbers, and so on. A list of codes is given and you have to check. . . .
### Knapsack
##### Age 14 to 16 Challenge Level:
You have worked out a secret code with a friend. Every letter in the alphabet can be represented by a binary value.
### Transposition Fix
##### Age 14 to 16 Challenge Level:
Suppose an operator types a US Bank check code into a machine and transposes two adjacent digits will the machine pick up every error of this type? Does the same apply to ISBN numbers; will a machine. . . .
### Squaresearch
##### Age 14 to 16 Challenge Level:
Consider numbers of the form un = 1! + 2! + 3! +...+n!. How many such numbers are perfect squares?
### Novemberish
##### Age 14 to 16 Challenge Level:
a) A four digit number (in base 10) aabb is a perfect square. Discuss ways of systematically finding this number. (b) Prove that 11^{10}-1 is divisible by 100.
### Odd Stones
##### Age 14 to 16 Challenge Level:
On a "move" a stone is removed from two of the circles and placed in the third circle. Here are five of the ways that 27 stones could be distributed.
### The Chinese Remainder Theorem
##### Age 14 to 18
In this article we shall consider how to solve problems such as "Find all integers that leave a remainder of 1 when divided by 2, 3, and 5."
### Expenses
##### Age 14 to 16 Challenge Level:
What is the largest number which, when divided into 1905, 2587, 3951, 7020 and 8725 in turn, leaves the same remainder each time?
### Fac-finding
##### Age 14 to 16 Challenge Level:
Lyndon chose this as one of his favourite problems. It is accessible but needs some careful analysis of what is included and what is not. A systematic approach is really helpful.
### Multiplication Magic
##### Age 14 to 16 Challenge Level:
Given any 3 digit number you can use the given digits and name another number which is divisible by 37 (e.g. given 628 you say 628371 is divisible by 37 because you know that 6+3 = 2+7 = 8+1 = 9). . . .
### 396
##### Age 14 to 16 Challenge Level:
The four digits 5, 6, 7 and 8 are put at random in the spaces of the number : 3 _ 1 _ 4 _ 0 _ 9 2 Calculate the probability that the answer will be a multiple of 396.
### Big Powers
##### Age 11 to 16 Challenge Level:
Three people chose this as a favourite problem. It is the sort of problem that needs thinking time - but once the connection is made it gives access to many similar ideas.
### Mod 3
##### Age 14 to 16 Challenge Level:
Prove that if a^2+b^2 is a multiple of 3 then both a and b are multiples of 3.
### Sixational
##### Age 14 to 18 Challenge Level:
The nth term of a sequence is given by the formula n^3 + 11n . Find the first four terms of the sequence given by this formula and the first term of the sequence which is bigger than one million. . . .
### LCM Sudoku
##### Age 14 to 16 Challenge Level:
Here is a Sudoku with a difference! Use information about lowest common multiples to help you solve it.
### Differences
##### Age 11 to 14 Challenge Level:
Can you guarantee that, for any three numbers you choose, the product of their differences will always be an even number?
### Eminit
##### Age 11 to 14 Challenge Level:
The number 8888...88M9999...99 is divisible by 7 and it starts with the digit 8 repeated 50 times and ends with the digit 9 repeated 50 times. What is the value of the digit M?
### Skeleton
##### Age 11 to 14 Challenge Level:
Amazing as it may seem the three fives remaining in the following `skeleton' are sufficient to reconstruct the entire long division sum.
### Take Three from Five
##### Age 14 to 16 Challenge Level:
Caroline and James pick sets of five numbers. Charlie chooses three of them that add together to make a multiple of three. Can they stop him?
### Why 24?
##### Age 14 to 16 Challenge Level:
Take any prime number greater than 3 , square it and subtract one. Working on the building blocks will help you to explain what is special about your results.
### Oh! Hidden Inside?
##### Age 11 to 14 Challenge Level:
Find the number which has 8 divisors, such that the product of the divisors is 331776.
### What an Odd Fact(or)
##### Age 11 to 14 Challenge Level:
Can you show that 1^99 + 2^99 + 3^99 + 4^99 + 5^99 is divisible by 5?
### Elevenses
##### Age 11 to 14 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?
##### Age 11 to 14 Challenge Level:
Powers of numbers behave in surprising ways. Take a look at some of these and try to explain why they are true.
### Going Round in Circles
##### Age 11 to 14 Challenge Level:
Mathematicians are always looking for efficient methods for solving problems. How efficient can you be?
### What Numbers Can We Make?
##### Age 11 to 14 Challenge Level:
Imagine we have four bags containing a large number of 1s, 4s, 7s and 10s. What numbers can we make?
### Powerful Factorial
##### Age 11 to 14 Challenge Level:
6! = 6 x 5 x 4 x 3 x 2 x 1. The highest power of 2 that divides exactly into 6! is 4 since (6!) / (2^4 ) = 45. What is the highest power of two that divides exactly into 100!?
### Remainders
##### Age 7 to 14 Challenge Level:
I'm thinking of a number. My number is both a multiple of 5 and a multiple of 6. What could my number be?
### Just Repeat
##### Age 11 to 14 Challenge Level:
Think of any three-digit number. Repeat the digits. The 6-digit number that you end up with is divisible by 91. Is this a coincidence?
### Ewa's Eggs
##### Age 11 to 14 Challenge Level:
I put eggs into a basket in groups of 7 and noticed that I could easily have divided them into piles of 2, 3, 4, 5 or 6 and always have one left over. How many eggs were in the basket?
### Counting Factors
##### Age 11 to 14 Challenge Level:
Is there an efficient way to work out how many factors a large number has?
### Dozens
##### Age 7 to 14 Challenge Level:
Do you know a quick way to check if a number is a multiple of two? How about three, four or six?
### Legs Eleven
##### Age 11 to 14 Challenge Level:
Take any four digit number. Move the first digit to the end and move the rest along. Now add your two numbers. Did you get a multiple of 11?
### Square Routes
##### Age 11 to 14 Challenge Level:
How many four digit square numbers are composed of even numerals? What four digit square numbers can be reversed and become the square of another number?
### AB Search
##### Age 11 to 14 Challenge Level:
The five digit number A679B, in base ten, is divisible by 72. What are the values of A and B?
### Digat
##### Age 11 to 14 Challenge Level:
What is the value of the digit A in the sum below: [3(230 + A)]^2 = 49280A
### Flow Chart
##### Age 11 to 14 Challenge Level:
The flow chart requires two numbers, M and N. Select several values for M and try to establish what the flow chart does.
### What Numbers Can We Make Now?
##### Age 11 to 14 Challenge Level:
Imagine we have four bags containing numbers from a sequence. What numbers can we make now?
### Gaxinta
##### Age 11 to 14 Challenge Level:
A number N is divisible by 10, 90, 98 and 882 but it is NOT divisible by 50 or 270 or 686 or 1764. It is also known that N is a factor of 9261000. What is N?
### Digital Roots
##### Age 7 to 14
In this article for teachers, Bernard Bagnall describes how to find digital roots and suggests that they can be worth exploring when confronted by a sequence of numbers.
### Factoring Factorials
##### Age 11 to 14 Challenge Level:
Find the highest power of 11 that will divide into 1000! exactly.
### Divisively So
##### Age 11 to 14 Challenge Level:
How many numbers less than 1000 are NOT divisible by either: a) 2 or 5; or b) 2, 5 or 7?
### Remainder
##### Age 11 to 14 Challenge Level:
What is the remainder when 2^2002 is divided by 7? What happens with different powers of 2?
### Three Times Seven
##### Age 11 to 14 Challenge Level:
A three digit number abc is always divisible by 7 when 2a+3b+c is divisible by 7. Why?
##### Age 11 to 14 Challenge Level:
List any 3 numbers. It is always possible to find a subset of adjacent numbers that add up to a multiple of 3. Can you explain why and prove it?
### Peaches Today, Peaches Tomorrow...
##### Age 11 to 14 Challenge Level:
Whenever a monkey has peaches, he always keeps a fraction of them each day, gives the rest away, and then eats one. How long could he make his peaches last for?
### Repeaters
##### Age 11 to 14 Challenge Level:
Choose any 3 digits and make a 6 digit number by repeating the 3 digits in the same order (e.g. 594594). Explain why whatever digits you choose the number will always be divisible by 7, 11 and 13.
### There's Always One Isn't There
##### Age 14 to 16 Challenge Level:
Take any pair of numbers, say 9 and 14. Take the larger number, fourteen, and count up in 14s. Then divide each of those values by the 9, and look at the remainders. | 2,445 | 9,468 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.03125 | 4 | CC-MAIN-2019-43 | latest | en | 0.841462 |
https://www.millisecond.com/download/library/v6/nback/singlenback_locations/singlenbacktask_locations_german/singlenbacktask_locations_german.manual | 1,721,122,493,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514742.26/warc/CC-MAIN-20240716080920-20240716110920-00368.warc.gz | 774,310,313 | 8,470 | # User Manual: Inquisit NBackTask (locations)
___________________________________________________________________________________________________________________
(German version)
___________________________________________________________________________________________________________________
Script Author: Katja Borchert, Ph.D. (katjab@millisecond.com) for Millisecond Software LLC
Date: March, 2012
last updated: 11-15-2022 by K. Borchert (katjab@millisecond.com) for Millisecond Software, LLC
Translation provided by K. Borchert for Millisecond Software
___________________________________________________________________________________________________________________
BACKGROUND INFO
___________________________________________________________________________________________________________________
This script implements a basic shape n-back procedure. The n-back task is a go/nogo working-memory performance task
with increasing levels of difficulty.
The implemented procedure is based on:
Jaeggi, Susanne M.; Studer-Luethi, Barbara; Buschkuehl, Martin; Su, Yi-Fen; Jonides, John; Perrig, Walter J. (2010). The
relationship between n-back performance and matrix reasoning - implications for training and transfer. Intelligence, 38, 625–635.
Millisecond Software thanks Dr. Susanne Jaeggi for kindly providing original instruction materials, stimuli, as well as assistance!
Adjustments to z-scores as recommended by:
Gregg, A. & Sedikides, C. (2010). Narcissistic Fragility:
Rethinking Its Links to Explicit and Implicit Self-esteem, Self and Identity, 9:2, 142-161 (p.148)
___________________________________________________________________________________________________________________
___________________________________________________________________________________________________________________
In the 1-key single n-back task, participants are shown a sequence of blue squares in varying screen locations
(3x3 matrix with center being used for fixation)
and are asked to indicate whether the screen location of the current square fulfills the following criteria:
for N=0 trials:
- is the location of the square directly above the fixation cross? If so (it's a target), and press "A". If not, don't respond.
for N=1 trials:
- is the location of the square the same as the one that preceded it? If so (it's a target), and press "A". If not, don't respond.
for N=2 trials:
- is the location of the square the same as the one presented two trials before? If so (it's a target), and press "A". If not, don't respond.
for N=3 trials:
- is the location of the square the same as the one presented three trials before? If so (it's a target), and press "A". If not, don't respond.
etc. for increasing levels of N
___________________________________________________________________________________________________________________
DURATION
___________________________________________________________________________________________________________________
the default set-up of the script takes appr. 15 minutes to complete
___________________________________________________________________________________________________________________
DATA FILE INFORMATION
___________________________________________________________________________________________________________________
The default data stored in the data files are:
(1) Raw data file: 'singlenbacktask_locations_german_raw.iqdat' (a separate file for each participant)
build: The specific Inquisit version used (the 'build') that was run
computer.platform: the platform the script was run on (win/mac/ios/android)
date, time: date and time script was run
subject, group, with the current subject/groupnumber
session: with the current session id
blockcode, blocknum: the name and number of the current block (built-in Inquisit variable)
trialcode, trialnum: the name and number of the currently recorded trial (built-in Inquisit variable)
Note: trialnum is a built-in Inquisit variable; it counts all trials run; even those
that do not store data to the data file such as feedback trials. Thus, trialnum
may not reflect the number of main trials run per block.
TotalBlocks: the total number of experimental blocks run
N: the current level N tested
starttrialcounter: keeps track of how many start trials have been run
currenttarget: position number of the current target
currentposition: position number of the current square
xpos: the horizontal coordinate of the current square (in canvas pct)
ypos: the vertical coordinate of the current square (in canvas pct)
response: the Response of the participant (scancode of response button):
0 = no response
31 = 'A' key (positive response)
responseCategory: "Hit" vs. "Omission Error" vs. "CorrReject" vs. "Commission Error"
correct: the correctness of the response (1 = correct; 0 = otherwise)
latency: how fast a participant responded within the given timeframe, if at all (in ms)
no responses = the entire trialduration
list.blockACC.mean: proportion correct during the current block (start trials are excluded from block ACC assessment by default, editable)
(2) Summary data file: 'singlenbacktask_locations_german_summary*.iqdat' (a separate file for each participant)
inquisit.version: Inquisit version run
computer.platform: the platform the script was run on (win/mac/ios/android)
startDate: date script was run
startTime: time script was started
subjectid: assigned subject id number
groupid: assigned group id number
sessionid: assigned session id number
elapsedTime: time it took to run script (in ms); measured from onset to offset of script
completed: 0 = script was not completed (prematurely aborted);
1 = script was completed (all conditions run)
Round: the total number of experimental blocks run
the following summary variables are only based on performance on trial.target/trial.nontarget
(excludes trial.start even if parameters.excludeStartTrialfromPerformanceMeasure = false)
propCorrect: overall proportion correct (across all test blocks)
meanHit_RT: overall mean correct response time (in ms); is equal to the overall mean hit response time
Note: z-score calculations: adjustments (see Gregg & Sedikides, 2010, p.148)
If the hit rate FA rate is 0 => 0.005 is used instead (aka 0.005 is added to the hitFA rate)
IF the hit rate FA rate is 1.0 => 0.995 is used instead (aka 0.005 is subtracted from the hitFA rate)
hitRate_overall: overall hit rate (across all test blocks) = hitting 'A' for targets
list.hits.itemCount: the absolute number of target trials (across all test blocks)
meanHit_RT: overall mean hit response time (in ms; across all test blocks)
FArate_overall: overall False Alarm/Commission Error Rate (across all test blocks) = hitting 'A' for nontargets
list.commissions.itemCount: the absolute number of nontarget trials (across all test blocks)
meanFA_RT: overall mean False Alarm response time (in ms; across all test blocks)
zhitrate_overall: the z-score of the overall hit rate
zFArate_overall: the z-score of the overall False Alarm rate
dprime_overall: Computes d' (parametric measure of sensitivity):
The higher the value, the better targets were overall correctly distinguished
from nontargets (d' = 0: chance performance; negative values: nontargets were treated as targets
and targets were treated as nontargets)
DV: dependent variable suggested by Jaeggi et al (2010):
(TotalHits - TotalFA)/number of experimental blocks
The following variables look at the different levels of N. If a level of N was not run,
the variables will store "not run"
//Level N = 0
hitRate_N0: hit rate (hitting 'A' for targets) for level N = 0
meanHit_RT_N0: mean hit response time (in ms) for level N = 0
FArate_N0: False Alarm/Commission Error Rate (hitting 'A' for nontargets) for level N = 0
meanFA_RT_N0: mean False Alarm response time (in ms) for level N = 0
zhitrate_N0: the z-score of the hit rate for level N = 0
zFArate_N0: the z-score of the False Alarm rate for level N = 0
dprime_N0: Computes d' (parametric measure of sensitivity):
The higher the value, the better targets were overall correctly distinguished
from nontargets (d' = 0: chance performance) for level N = 0
(same for Level N=1-6)
Note: if additional levels of N are run, go to section DATA and follow further instructions
___________________________________________________________________________________________________________________
EXPERIMENTAL SET-UP
___________________________________________________________________________________________________________________
- After instructions, participants receive 10 trials of practice per level N tested (here: N = 2 to N = 4),
Note: edit list.nextN_practice to change levels of N practiced
- once practice is done, participants get the option to repeat practice (see parameters.allowPracticeRepeat)
- After practice, participants receive 3 blocks per level of N tested. By default, the level of N increases from
1 to 3.
Note: edit list.Nlevel to change a) N levels tested b) number of testblocks c)selection of N-levels
Block Information:
- Each single n-task block consists of 20+ trials using 8 different locations (3x3 matrix with center being used for fixation).
=> + trials are the N trials that cannot display target locations yet (=start trials). Their numbers vary depending on N.
The data collected during these start trials are NOT included in performance counts for Correct Rejections/False Alarms.
=> Of the actual 20 experimental trials, 6 present a target and 14 do not (3:7 ratio).
To change the ratio, you need to change the relevant code under block.Test
- The computer selects randomly
a) whether it is a target trial or not and
b) what shape to show if it is not a target trial
Trial Information:
- Each trial presents the square for 500ms and waits another 2500ms before presenting the next square in the sequence.
The trialduration can be edited by experimenter.
- Participants have the entire 3000ms to respond by pressing "A" if they detect a target.
___________________________________________________________________________________________________________________
STIMULI
___________________________________________________________________________________________________________________
1 blue square, see section Editable Stimuli
Note: if the image is changed, the instruction pages need to be updated accordingly
___________________________________________________________________________________________________________________
INSTRUCTIONS
___________________________________________________________________________________________________________________
provided by Millisecond Software - can be edited under section Editable Instructions
Instructions are not original to the task. They are provided by Millisecond Software
as htm pages and simple page elements and can be edited either by changing
the provided html files or directly under Editable Instructions.
Note: in order for the instructions to reflect alternative parameter settings,
some of the conditionalized code used in the htm files can be found in the script itself.
___________________________________________________________________________________________________________________
EDITABLE CODE
___________________________________________________________________________________________________________________
check below for (relatively) easily editable parameters, stimuli, instructions etc.
Keep in mind that you can use this script as a template and therefore always "mess" with the entire code
The parameters you can change are:
/allowPracticeRepeat: true: practice can be repeated as often as indicated (default)
false: practice cannot be repeated
/excludeStartTrialfromPerformanceMeasure: true (1): performance during start trials that cannot be targets yet is not considered
for Block level performance analyses, e.g. for feedback purposes (default)
false (0): performance during start trials that cannot be targets yet IS considered for block performance analyses (default)
/SOA: Stimulus Onset Asynchrony (default: 3000ms)
/stimulusPresentationtime: the presentation time (in ms) of the stimuli (default: 500ms)
/stimHeight_inpct: the height of the stim (here: blue square) relative to canvas height
(Note: should not exceed 25%)
debugmode = 0 debugmode = 1: targetalerts are shown on screen,
debugmode = 0, no targetalerts are shown (default)
(in general: if levels of N are changed, adjustments need to be made to instructions and list.Nlevel) | 2,674 | 12,695 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2024-30 | latest | en | 0.659119 |
http://www.mathworks.com/matlabcentral/fileexchange/41147-environmental-modeling-using-matlab/content/viscosity_dyn.m | 1,438,162,872,000,000,000 | text/html | crawl-data/CC-MAIN-2015-32/segments/1438042986357.49/warc/CC-MAIN-20150728002306-00239-ip-10-236-191-2.ec2.internal.warc.gz | 579,566,118 | 8,186 | Code covered by the BSD License
# Environmental Modeling - using MATLAB
### Ekkehard Holzbecher (view profile)
Springer Book, 2nd Edition
viscosity_dyn.m
```% Viscosity Intercomparison
% Dynamic viscosity of water in dependency of temperature
%
% Ref:
% \$Ekkehard Holzbecher \$Date: 2006/03/10 \$
T_1 = [0:1:16];
mu_1 = 1.8e-3-06.55e-5*T_1+1.44e-6*T_1.*T_1; % Hagen 1839 acc. to Prandtl/Tientjes
T_2 = [0:1:75];
mu_2 = 0.001779./(1+0.03368*T_2+0.00022099*T_2.*T_2); % Poiseuille 1840 acc. to Lamb
T_3 = [100:1:300];
mu_3 = 241.4e-7*10.^(247.8./(T_3+132.15)); % JSME 1968
T_4 = [5:1:25];
mu_4 = 1.31*1.e-3./(0.7 + 0.03*T_4); % Gavich 1985
T_5 = [0:1:100];
mu_5 = 0.001*(1.+0.636*(T_5-20)/41).^(-1/0.636); % Pawlowski 1991
T_6 = [15:1:35];
mu_6 = 1.98404e-6*exp(1825.85./(273+T_6)); % Lin e.a. 2003
plot (T_1,mu_1,'r',T_2,mu_2,'g',T_4,mu_4,'m',T_5,mu_5,'c',T_6,mu_6,'b')
legend ('Hagen','Poiseuille','Gavich e.a.','Pawlowski','Lin e.a.');
xlabel ('Temperature [C]'); ylabel ('dynamic viscosity [Pa s]');
grid;``` | 494 | 1,115 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.640625 | 3 | CC-MAIN-2015-32 | longest | en | 0.356164 |
http://www.chestnut.com/en/eg/4/mathematics/5/6/examples/5/ | 1,477,192,583,000,000,000 | text/html | crawl-data/CC-MAIN-2016-44/segments/1476988719139.8/warc/CC-MAIN-20161020183839-00470-ip-10-171-6-4.ec2.internal.warc.gz | 364,194,946 | 8,905 | 4.5.6. Approximation
Approximate 84 121 638.772 to the nearest hundred thousand.
• A
84 100 000
• B
84 120 000
• C
84 122 000
• D
84 121 600
Example
Approximate 84 121 638.772 to the nearest hundred thousand.
Solution
The digit in the ten thousands place (2) is less than 5, so we replace every digit to the right of the hundred thousands place by zero and keep the other digits as they are. Remove the decimal part of the number.
0
correct
0
incorrect
0
skipped | 142 | 469 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.6875 | 4 | CC-MAIN-2016-44 | latest | en | 0.792304 |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.