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 |
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
https://www.doubtnut.com/question-answer/which-of-the-following-represents-direction-cosines-of-the-line--96593332 | 1,627,768,164,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046154126.73/warc/CC-MAIN-20210731203400-20210731233400-00243.warc.gz | 744,702,887 | 69,032 | Home
>
English
>
Class 12
>
Maths
>
Chapter
>
Three Dimensional Geometry
>
Which of the following represe...
# Which of the following represents direction cosines of the line ?
Step by step solution by experts to help you in doubt clearance & scoring excellent marks in exams.
Updated On: 19-2-2020
Apne doubts clear karein ab Whatsapp par bhi. Try it now.
Watch 1000+ concepts & tricky questions explained!
39.5 K+
2.0 K+
Text Solution
0,(1)/(sqrt(2)),(1)/(2)0,-(sqrt(3))/(2),(1)/(sqrt(2))0,(sqrt(3))/(2),(1)/(2)(1)/(2),(1)/(2),(1)/(2)
Solution :
0, (sqrt(3))/(2), (1)/(2)
Image Solution
75728347
1.7 K+
34.1 K+
27:22
10867
2.7 K+
54.7 K+
2:54
100279
338.6 K+
413.8 K+
4:21
1488093
4.5 K+
90.1 K+
3:58
2599
231.1 K+
294.2 K+
1:34
8495759
62.6 K+
78.9 K+
2:10
8431
2.5 K+
50.7 K+
2:31
3868338
2.2 K+
43.5 K+
2:12
8495760
1.7 K+
33.2 K+
2:52
51237251
24.8 K+
35.0 K+
2:45
1488128
1.1 K+
22.1 K+
2:17
96593301
1.3 K+
26.1 K+
1:45
98739534
700+
15.2 K+
2:10
53806338
1.9 K+
37.7 K+
1:54
11004
4.4 K+
88.7 K+
3:04
## Latest Questions
Class 12th
Three Dimensional Geometry
Class 12th
Three Dimensional Geometry
Class 12th
Three Dimensional Geometry
Class 12th
Three Dimensional Geometry
Class 12th
Three Dimensional Geometry
Class 12th
Three Dimensional Geometry
Class 12th
Three Dimensional Geometry
Class 12th
Three Dimensional Geometry
Class 12th
Three Dimensional Geometry
Class 12th
Three Dimensional Geometry | 553 | 1,499 | {"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} | 2.78125 | 3 | CC-MAIN-2021-31 | latest | en | 0.653734 |
https://kidsworksheetfun.com/law-of-sines-worksheet-answers/ | 1,721,322,304,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514848.78/warc/CC-MAIN-20240718161144-20240718191144-00749.warc.gz | 296,842,710 | 27,055 | # Law Of Sines Worksheet Answers
The law of sines worksheets apply the law of sines to establish a relationship between the sides and angles of a triangle. Free printable worksheet plus answer key on the ambiguous case of the law of sines includes visual aides model problems an online component and challenge problems.
Law Of Sines And Cosines Picture Law Of Sines Worksheets For Kids Worksheets
### Worksheet by kuta software llc kuta software infinite precalculus the law of sines name date period 1 state the number of possible triangles that can be formed using the given measurements.
Law of sines worksheet answers. Worksheet answers law of sines worksheet answers as recognized adventure as competently as experience about lesson amusement as skillfully as pact can be gotten by just checking out a book law of sines worksheet answers afterward it is not directly done you could receive even more on the subject. A 150 c 20 a 200 4. Round your answers to the nearest tenth.
And it says that. 1 m a 31 c mi a mi two triangles. Students will practice applying the law of sines to calculate side lengths and angle measurements.
Law of sines practice worksheet answers worksheet january 09 2020 09 29 many law firms have been known to use the law of sines and the law of opposites to help find the best candidates for their law practice jobs. A sin a b sin b c sin c. It is a long standing idea that helps them to find potential candidates that can be hired to go into their law practice.
A b and c are angles. Side a faces angle a side b faces angle b and. 1 find bc 8 ba c 61 30 7.
It works for any triangle. This worksheet includes word problems as well as challenging bonus problems. A 33 b 29 b 41 c 79 a 9 4 b 13 6 c 118 a 46 1 c 74 7 3.
A 30 b 45 a 10 b 10 b 69 5 c 105 b 14 1 c 19 3 c 136 8 find the area of each triangle. Side c faces angle c. Round your answers to the nearest tenth.
Worksheet by kuta software llc trigonometry law of sines and cosines review worksheet name date period p h2 0i1p6 eknu tsaw uslotfktiwya rjec hlolxcj o ra lklh nrxi gphwtase irrewswe rgv e d 1 find each measurement indicated. A 38 b 63 c 15 2. Round to the nearest tenth.
A b and c are sides. Round to the nearest tenth. 1 find ac 24 a c b 118 22 14 2 find ab 7 c a b 53 44 8 3 find bc 27 c b a 51 39 17 4 find ab 9 b c a 101 63 29 1 5 find bc 16 a b c 93 58 33 6 find m c 21 26 16 1 a c b 88 53 8 7 find m c 24 20 c 29.
This set of trigonometry worksheets covers a multitude of topics on applying the law of sines like finding the missing side or unknown angle missing sides and angles find the area of sas triangle and so on. The law of sines or sine rule is very useful for solving triangles. The law of sines date period find each measurement indicated.
C 4 a 37 b 69. The law of sines solve each triangle.
Love Using Mazes In The Geometry Classroom This Law Of Sines Law Of Cosines Maze Would Be A Great Review Law Of Sines Law Of Cosines Teach Math High School
Sin And Cosine Worksheets Law Of Cosines Math Pages Worksheets
Sin And Cosine Worksheets Law Of Cosines Law Of Sines Triangle Worksheet
Sin And Cosine Worksheets Law Of Cosines Trigonometry Worksheets Worksheets
Sin And Cosine Worksheets Law Of Cosines Trigonometry Worksheets Worksheets
Trigonometry Law Of Sines Worksheet Activity Law Of Sines Trigonometry Law Of Cosines
Laws Of Sines And Cosines Solve And Match Law Of Sines Trigonometry Law Of Cosines
Trigonometry Laws Of Sines And Cosines Solve And Match Law Of Sines Trigonometry Law Of Cosines
Law Of Sines And Law Of Cosines Geometry Puzzle Worksheet Trigonometry Activity For High High School Math High School Math Activities Trigonometry Worksheets
Law Of Sines And Law Of Cosines Maze Worksheet Pythagorean Theorem Law Of Sines Teaching Geometry
Law Of Sines Ambiguous Case Fruit Method Law Of Sines Math Precalculus
50 Law Of Sines Worksheet Answers In 2020 Law Of Sines
Law Of Sines Worksheet Answers What Is The Law Of Sines Simply Explained With 4 Examples In 2020 Law Of Sines Math Methods Math Formulas
Right Triangles Unit The Laws Of Cosines Sines Quiz Fr Law Of Cosines Right Triangle Law Of Sines
Sin And Cosine Worksheets Law Of Cosines Worksheets Trigonometry Worksheets
Law Of Sines Worksheet Answers Awesome Law Of Sines And Cosines How To Know Which Formula You In 2020 Law Of Sines Nouns Verbs Adjectives Nouns And Verbs Worksheets
Missing Sides Law Of Sines Law Of Sines Trigonometry Worksheets Triangle Worksheet
This Self Checking Maze Has 11 Problems Involving The Law Of Sines And The Law Of Cosines Students Will Be Law Of Cosines Law Of Sines Word Problem Worksheets
Law Of Sines And Law Of Cosines Maze Law Of Cosines Law Of Sines Distance Formula | 1,209 | 4,731 | {"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} | 3.265625 | 3 | CC-MAIN-2024-30 | latest | en | 0.893758 |
https://se.mathworks.com/matlabcentral/cody/problems/43026-determinant-of-a-3x3-matrix/solutions/1468614 | 1,603,969,032,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107904039.84/warc/CC-MAIN-20201029095029-20201029125029-00512.warc.gz | 518,488,348 | 16,997 | Cody
# Problem 43026. Determinant of a 3x3 Matrix
Solution 1468614
Submitted on 22 Mar 2018 by J. S. Kowontan
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 Pass
filetext = fileread('dtrmnt_3x3.m'); assert(isempty(strfind(filetext, 'det')),'det() forbidden')
2 Pass
x = [3 0 2; 2 1 -2; 0 1 1]; y_correct = 13; assert(sum(sum(abs(dtrmnt_3x3(x)-y_correct)))<1e-3)
3 Pass
x = [1 40 2; 7 -9 10; 1 0 1]; y_correct = 129; assert(sum(sum(abs(dtrmnt_3x3(x)-y_correct)))<1e-3)
### Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting! | 247 | 720 | {"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-2020-45 | latest | en | 0.590785 |
http://www.java-forums.org/new-java/54430-menu-driven-program-print.html | 1,438,259,102,000,000,000 | text/html | crawl-data/CC-MAIN-2015-32/segments/1438042987174.71/warc/CC-MAIN-20150728002307-00209-ip-10-236-191-2.ec2.internal.warc.gz | 512,198,607 | 3,554 | • 01-24-2012, 11:35 AM
Hi, I'm supposed to write a loan calculator menu-driven program.
The user inputs the payment amount, annual interest rate (which needs to be converted to months), and the loan period in months.
Below is what I've written so far but my program outputs all the wrong numbers.
When I try to calculate the PAYMENT it always spits out -\$0.00, the INTEREST just returns what I input as the interest rate, and the TOTAL is always \$0.00.
Code:
```import java.util.*; import java.text.DecimalFormat; public class Menu { public static void main(String[] args) { // initialize money format DecimalFormat money = new DecimalFormat("\$0.00"); // variables String input; char inputCode = ' ' ; double PRINCIPLE = 0; double RATE = 0; int TERM = 0; // equations double PAYMENT = PRINCIPLE * ((RATE / 1200) / 1 - Math.pow(1 + (RATE / 1200), -TERM)); double INTEREST = TERM * PAYMENT - PRINCIPLE; double TOTAL = PRINCIPLE + INTEREST; /* menu */ System.out.println("Enter [p] for payment amount"); System.out.println("Enter [i] for interest rate"); System.out.println("Enter [t] for total"); System.out.println("Enter [q] to terminate program"); System.out.print("\nPlease enter your choice: "); // create Scanner object for keyboard input Scanner sc = new Scanner(System.in); input = sc.nextLine(); inputCode = input.charAt(0); while ((inputCode != 'p' && inputCode != 'i' && inputCode != 't' && inputCode !='q') && (inputCode != 'P' && inputCode != 'I' && inputCode != 'T' && inputCode !='Q')) { System.err.println("Invalid choice! Please select p, i, t, or q: "); input = sc.nextLine(); inputCode = input.charAt(0); } /* display results */ switch (inputCode) { case 'p': case 'P': System.out.print("\nEnter principle (loan amount), e.g. 120000.95: "); PRINCIPLE = sc.nextDouble(); System.out.print("Enter annual interest rate (between 1-100), e.g. 5.25: "); RATE = sc.nextDouble(); if (RATE < 1 || RATE > 100) { System.err.println("Invalid input! Please re-enter annual interest rate: "); RATE = sc.nextDouble(); } System.out.print("Enter term (loan period) in months as an integer, e.g. 8: "); TERM = sc.nextInt(); System.out.println("\nThe payment amount is " + money.format(PAYMENT)); break; case 'i': case 'I': System.out.print("\nEnter (principle) loan amount, e.g. 125000.00: "); PRINCIPLE = sc.nextDouble(); System.out.print("Enter annual interest rate (between 1-100), e.g. 5.25: "); RATE = sc.nextDouble(); if (RATE < 1 || RATE > 100) { System.err.println("Invalid input! Please re-enter annual interest rate: "); RATE = sc.nextDouble(); } System.out.print("Enter term (loan period) in months as an integer , e.g. 8: "); TERM = sc.nextInt(); System.out.println("\nThe interest rate is " + RATE); break; case 't': case 'T': System.out.print("\nEnter (principle) loan amount, e.g. 125000.00: "); PRINCIPLE = sc.nextDouble(); System.out.print("Enter annual interest rate (between 1-100), e.g. 5.25: "); RATE = sc.nextDouble(); if (RATE < 1 || RATE > 100) { System.err.println("Invalid input! Please re-enter annual interest rate: "); RATE = sc.nextDouble(); } System.out.print("Enter term (loan period) in months as an integer, e.g. 8: "); TERM = sc.nextInt(); System.out.println("\nThe total amount is " + money.format(TOTAL)); break; case 'q': case 'Q': System.out.println("Program terminated"); } } }```
Can anyone help me find where I went wrong?
I know it's terrible that I'm struggling with basic concepts but I've never taken a programming course before and this class also has a prereq that I didn't take. I best learn through example so any help or advice to make my code more efficient will be appreciated!
• 01-24-2012, 11:40 AM
Tolls | 1,103 | 5,433 | {"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-2015-32 | longest | en | 0.280512 |
https://www.ruby-forum.com/t/loser-question/118336 | 1,709,352,655,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947475727.3/warc/CC-MAIN-20240302020802-20240302050802-00525.warc.gz | 963,991,019 | 6,509 | # Loser question
Ok. I’m a loser. I’ve been told and warned. But I still got bit.
If I do:
irb(main):001:0> GOLF = [ 1, 2, 3]
=> [1, 2, 3]
irb(main):002:0> g = GOLF
=> [1, 2, 3]
irb(main):003:0> g << 4
=> [1, 2, 3, 4]
irb(main):004:0> g
=> [1, 2, 3, 4]
irb(main):005:0> h = GOLF
=> [1, 2, 3, 4]
Notice that h is [ 1, 2, 3, 4 ] and GOLF has changed its value.
I understand why. Thats not my question.
My question is what or how do most ruby people do this?
Do you simply do this:
irb(main):001:0> GOLF = [ 1, 2, 3 ]
=> [1, 2, 3]
irb(main):002:0> g = GOLF.dup
=> [1, 2, 3]
irb(main):003:0> g << 4
=> [1, 2, 3, 4]
irb(main):004:0> h = GOLF
=> [1, 2, 3]
Or, do you usually hide GOLF inside a method – which does the dup. How
is this handled most of the time?
Thank you very much,
pedz
On 10/25/07, Perry S. [email protected] wrote:
Ok. I’m a loser. I’ve been told and warned. But I still got bit.
Man, you’re too hard on yourself!
=> [1, 2, 3, 4]
is this handled most of the time?
If you want GOLF to be immutable, you can freeze it:
\$ irb --simple-prompt
GOLF = [ 1, 2, 3 ]
=> [1, 2, 3]
GOLF.freeze
=> [1, 2, 3]
h = GOLF
=> [1, 2, 3]
h << 4
TypeError: can’t modify frozen array
from (irb):4:in `<<’
from (irb):4
from :0
I would say that it depends on the context. If this is something you
will do often, make a function out of it to keep things DRY, but if it’s
just once or twice , do it inline. Since we are not talking about a
large number of lines of code and the intent is fairly obvious, it
really doesn’t matter too much.
my \$.02
-Bill
Michael G. wrote:
–Michael
Sincerely,
William P.
I have a lot of functions that do things like this:
I have a function that takes an array, and sets various additional
"default’ values in it. I first clone what is passed in:
newvals = args.clone
and then modify newvals.
I assume .dup and .clone are about the same?
–Michael
On 10/25/07, Perry S. [email protected] wrote:
=> [1, 2, 3, 4]
irb(main):004:0> g
=> [1, 2, 3, 4]
irb(main):005:0> h = GOLF
=> [1, 2, 3, 4]
Notice that h is [ 1, 2, 3, 4 ] and GOLF has changed its value.
Not really, GOLF still points to the same array.
The array is mutable though, and you may want to freeze it as others
have pointed out.
Sincerely,
Isak
I’m not quite sure what you’re driving at here. Were you expecting
different behavior than what you showed in your original post? I mean,
if h and g are references to the same object, of course modifying the
object will have a same result no matter which reference you use to
“view” the object.
If your end goal is to create references to two different objects (by
object identity) that begin life with the same contents, then it’s
very common in object oriented programming languages to clone (dup)
the object. Any other behavior than this would be “odd” at best, and
would completely confuse every object oriented programmer out there.
Here is an example of how a language other than Ruby would support
this. The following is a “factory” method used to construct a new
array using the contents of another array written in Objective C:
• (id)arrayWithArray:(NSArray *)anArray
Here the + indicates that this is a class method returning an array
constructed using a reference to a different array. The array returned
by this method references an new array and does not simply reference
the original array.
Ruby doesn’t actually have this type of factory method, but if you do
this a lot it could be easily added to the Array class. But, is that
really worth it when all you need is:
g = h.dup
I started doing this:
GOLF = [ 1, 2, 3].freeze
Now GOLF can’t change and what it points to can’t change and it is
reasonably compact and somewhat obvious.
irb(main):032:0> S = [ 1, 2, 3, 4 ].freeze
=> [1, 2, 3, 4]
irb(main):033:0> h = S
=> [1, 2, 3, 4]
irb(main):034:0> h << 8
TypeError: can’t modify frozen array
from (irb):34:in `<<’
from (irb):34
from :0
irb(main):035:0> S = 8
(irb):35: warning: already initialized constant S
=> 8 | 1,299 | 3,996 | {"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} | 3.359375 | 3 | CC-MAIN-2024-10 | latest | en | 0.884106 |
http://www.mywordsolution.com/question/solve-the-travelling-salesman-unruly-using-the/923397 | 1,524,317,361,000,000,000 | text/html | crawl-data/CC-MAIN-2018-17/segments/1524125945222.55/warc/CC-MAIN-20180421125711-20180421145711-00466.warc.gz | 491,045,231 | 7,757 | +1-415-315-9853
info@mywordsolution.com
## Statistics
Solve the travelling salesman unruly using the brute force algorithm.
1) describe why would a city street department want its snow plow operator's path to follow an Euler circuit if possible?
2) Use the brute force algorithm to resolve the traveling salesman problem for the graph of the four cities shown below.
Statistics and Probability, Statistics
• Category:- Statistics and Probability
• Reference No.:- M923397
Have any Question?
## Related Questions in Statistics and Probability
### A laboratory in california is interested in finding the
A laboratory in California is interested in finding the mean chloride level for a healthy resident in the state. A random sample of 50 healthy residents has a mean chloride level of 100 mEq/L. If it is known that the chl ...
### At a local college 316 of the male students are smokers and
At a local college, 316 of the male students are smokers and 474 are non-smokers. Of the female students, 70 are smokers and 630 are non-smokers. A male student and a female student from the college are randomly selected ...
### Your assignment this week is to describe structural
Your assignment this week is to describe structural equation modeling in which you include: (a) a comparison of the first and second generation models of structural equation modeling and their related major assumptions, ...
### The percentage of people who recycle their old phone books
The percentage of people who recycle their old phone books is 15.8% on average. If 240 people are randomly selected, what is the probability that more than 20% of them recycle their phone books?
### 1 what is the difference between simple also known as
1. What is the difference between simple (also known as univariate and bivariate) regression and multivariate regression? A) Simple regression includes one independent variable, and multivariate regression includes more ...
### You are at a magic show the magician has a coin and you as
You are at a magic show. The magician has a coin and you, as a member of the audience, do not know whether the coin has heads on both sides or heads on one side and tails on the other. The null hypothesis is that the coi ...
### The borda and condorcet methods- the following example was
The Borda and Condorcet Methods :- The following example was presented by Condorcet, in a critique of the Borda Method. A committee composed of 81 members is to choose a winner from among three candidates, A, B, and C. T ...
### A women in the sample on average were as healthy as people
(a) Women in the sample, on average were as healthy as people nationally from both a physical and mental health perspective as indicated by the SD deviation of 10.0 and a mean of 50.0 based on the national average. (b) T ...
### Suppose we obtain a random sample of 80 champaign voters of
Suppose we obtain a random sample of 80 Champaign voters, of which 55 support marijuana legalization. We also obtain a random sample of 100 Urbana voters, of which 75 support marijuana legalization. Let pC be the true pr ...
### Repeat given exercise using the following aumann model of
Repeat given Exercise, using the following Aumann model of incomplete information with beliefs: N = {I, II}, Y = {1, 2, 3, 4, 5}, F I = {{1, 2}, {3, 4}, {5}}, F II = {{1, 3, 5}, {2}, {4}} P(ω) = 1/5, ∀ ω ∈ Y. for A = {1, ...
• 13,132 Experts
## Looking for Assignment Help?
Start excelling in your Courses, Get help with Assignment
Write us your full requirement for evaluation and you will receive response within 20 minutes turnaround time.
### WalMart Identification of theory and critical discussion
Drawing on the prescribed text and/or relevant academic literature, produce a paper which discusses the nature of group
### Section onea in an atwood machine suppose two objects of
SECTION ONE (a) In an Atwood Machine, suppose two objects of unequal mass are hung vertically over a frictionless
### Part 1you work in hr for a company that operates a factory
Part 1: You work in HR for a company that operates a factory manufacturing fiberglass. There are several hundred empl
### Details on advanced accounting paperthis paper is intended
DETAILS ON ADVANCED ACCOUNTING PAPER This paper is intended for students to apply the theoretical knowledge around ac
### Create a provider database and related reports and queries
Create a provider database and related reports and queries to capture contact information for potential PC component pro | 998 | 4,538 | {"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-2018-17 | latest | en | 0.946281 |
http://www.bartre.com/what-is-tension-force-in-physics-reviews-amp-guide/ | 1,571,770,848,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570987823061.83/warc/CC-MAIN-20191022182744-20191022210244-00485.warc.gz | 237,208,901 | 10,444 | ## What Is Tension Force in Physics Reviews & Guide
Uncategorized Comments Off on What Is Tension Force in Physics Reviews & Guide
Besides your own personal taste, there are several things to take into account while on the watch for motorcycle chest protector. Hope starts with the ball within her mitt. These are essential to make lift.
After the cart collided on the opposite end, it stuck to the clay. A metallic bar functions as a framework to keep support to the forehead and chin. Our skins are especially elastic, particularly for the young.
You must make a bid to solve as many problems as possible without the help paraphrasing online of notes, solutions, teachers, and other students. As soon as you place your purchase, we’ll pick a writer who you may work on your project the very best way possible. Specifically, ask yourself the next question.
As soon as you have solved a issue, click the button to confirm your answers. This will guarantee that the brackets are in the exact same central line. The button beside mini-map was renewed.
At that moment, the individual’s seat is in fact pushing back on the individual, that is the standard force. The Set Creator helps you put together single components of distinct Outfits to make your own individual appearance. Whenever someone sits on a bicycle that’s not moving at all, it can be unbelievably get more difficult for the man to keep their balance.
As soon as an object has a lot of forces acting on it, the result is the exact same as one force in a particular direction. When it is stored, the object is provided potential energy that may be released to do work. 1 force is connected to the turbulence resulting from the object.
It’s the force necessary to attain maximum shielding effectiveness. Elastic energy is a kind of potential energy in truth, it’s often called ‘elastic possible energy’. Accordingly, in an inelastic collision, the whole kinetic energy isn’t conserved.
Therefore, the flexibility of any object or structure is contingent on its elastic modulus and geometric form. The concept was not new, however. Think of what keeps an object up after it’s thrown.
## Using What Is Tension Force in Physics
It is going to also be uncomfortable for everyone in the plane since they will tend to get thrown against the upwind” side of the fuselage. Proportional shift in the shape of a body, is called strain physically. These greater shear forces increase the chance of back injury through ruptured http://www.physics.umd.edu/courses/Phys122/Streets/14_Oscillations.pdf discs.
The first kind of material parameter is known as a modulus, which measures the sum of force per unit area necessary to attain a given amount of deformation. Furthermore, the outcomes of any hardness test are contingent on the elastomer thickness. The constant proportion of tension and strain is known as coefficient or modulus of elasticity of the fabric of the body.
## What Is Tension Force in Physics Explained
Tensile tests provide better info about the test material and so help continue the material quality. You will find that you’re more buoyant once you inhale, than when you exhale. Boundary conditions besides simply-supported will lead to different critical loads and mode shapes.
## Characteristics of What Is Tension Force in Physics
Otherwise, ball-string friction produces about exactly the same spin for each and every string. Cervical fillings are made from foam in addition to fiber. It is possible to generate maximum power and maximum speed with a light stone and strong bands.
Elastic force is a idea of physics that not all students may be able to comprehend. Thus the most significant point to remember regarding uniform circular motion is the fact that it is just a subset of the bigger topic of dynamics. Therefore, whichever concept of physics you’re facing a issue with, we will be able to help you deal with almost everything.
The corresponding acceleration is known as tangential acceleration. The object is believed to have gotten to a terminal velocity. In nonuniform circular motion, there’s also the perpendicular part of acceleration vector.
The yellow path in the video and in the above mentioned photo demonstrates that the ball must undergo centripetal acceleration as a portion of its overall acceleration. This is also called the law of acceleration. Increased speeds lead to a heightened amount of air resistance.
» Uncategorized » What Is Tension Force in...
On September 23, 2019
By | 898 | 4,508 | {"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} | 3.09375 | 3 | CC-MAIN-2019-43 | longest | en | 0.951976 |
https://graphsearch.epfl.ch/en/concept/1651906 | 1,722,977,129,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640508059.30/warc/CC-MAIN-20240806192936-20240806222936-00330.warc.gz | 235,286,344 | 17,673 | Concept
# Ordinary least squares
Summary
In statistics, ordinary least squares (OLS) is a type of linear least squares method for choosing the unknown parameters in a linear regression model (with fixed level-one effects of a linear function of a set of explanatory variables) by the principle of least squares: minimizing the sum of the squares of the differences between the observed dependent variable (values of the variable being observed) in the input dataset and the output of the (linear) function of the independent variable. Geometrically, this is seen as the sum of the squared distances, parallel to the axis of the dependent variable, between each data point in the set and the corresponding point on the regression surface—the smaller the differences, the better the model fits the data. The resulting estimator can be expressed by a simple formula, especially in the case of a simple linear regression, in which there is a single regressor on the right side of the regression equation. The OLS estimator is consistent for the level-one fixed effects when the regressors are exogenous and forms perfect colinearity (rank condition), consistent for the variance estimate of the residuals when regressors have finite fourth moments and—by the Gauss–Markov theorem—optimal in the class of linear unbiased estimators when the errors are homoscedastic and serially uncorrelated. Under these conditions, the method of OLS provides minimum-variance mean-unbiased estimation when the errors have finite variances. Under the additional assumption that the errors are normally distributed with zero mean, OLS is the maximum likelihood estimator that outperforms any non-linear unbiased estimator. Linear regression model Suppose the data consists of observations . Each observation includes a scalar response and a column vector of parameters (regressors), i.e., . In a linear regression model, the response variable, , is a linear function of the regressors: or in vector form, where , as introduced previously, is a column vector of the -th observation of all the explanatory variables; is a vector of unknown parameters; and the scalar represents unobserved random variables (errors) of the -th observation. | 428 | 2,216 | {"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-2024-33 | latest | en | 0.876924 |
https://electronics.stackexchange.com/questions/53214/what-is-the-right-way-to-understand-the-behaviour-of-shocking-voltage | 1,701,645,704,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100518.73/warc/CC-MAIN-20231203225036-20231204015036-00567.warc.gz | 275,180,520 | 45,368 | # What is the right way to understand the behaviour of shocking voltage?
I am just a starter and I am reading some beginner material about electricity.
The writer explained why the bird is not shocked while the man got shocked:
This might lend one to believe that its impossible to be shocked by electricity by only touching a single wire. Like the birds, if we're sure to touch only one wire at a time, we'll be safe, right? Unfortunately, this is not correct. Unlike birds, people are usually standing on the ground when they contact a "live" wire. Many times, one side of a power system will be intentionally connected to earth ground, and so the person touching a single wire is actually making contact between two points in the circuit (the wire and earth ground):
While I know that current flows from the negative end to the positive end, I had a concept ( which is probably wrong ) that after passing through the load, the energy would be consumed and therefore it is safe for anybody to tough the positive side of the wire.
On the other hand, the writer said the bird is not shocked because the bird didn't complete the circuit if the bird only put 1 foot on the wire. And even though if the bird put both legs on the wire, a circuit for the current was completed, the 2 legs were electrically common so there is no voltage across them. I think the man in the picture are in similar situation but he was shocked. How come? Does the present of the load make a difference? What if the load is removed, and the man grabs both the negative end and positive end of the powered wire?
EDIT
I read the chapter again. According to it:
If connecting wire resistance is very little or none, we can regard the connected points in a circuit as being electrically common. That is, points 1 and 2 in the above circuits may be physically joined close together or far apart, and it doesn't matter for any voltage or resistance measurements relative to those points. The same goes for points 3 and 4. It is as if the ends of the resistor were attached directly across the terminals of the battery, so far as our Ohm's Law calculations and voltage measurements are concerned. This is useful to know, because it means you can re-draw a circuit diagram or re-wire a circuit, shortening or lengthening the wires as desired without appreciably impacting the circuit's function. All that matters is that the components attach to each other in the same sequence.
The whole theory of electrically common is based on the fact that the wire got very little or none Resistance so that it require almost no voltage to push the current from point 2 to point 1. However, if we consider the bird is on its 2 legs, it would form another path for the current and the resistance is somehow much larger than the wire. I think it is the Resistance of the bird itself matters. Am I right?
• This reminds me a few days ago I used metal scissors to touch both ends of a broken light bulb in series. The other light bulbs started to glow and I felt nothing dangerous. Because I effectively touched only one point in the circuit. I stupidly thought that there is just 5 volts in those wires (45 light bulbs in series on 230 V i.e. 45x5=225V) and then I stupidly touched both wires around broken light bulb by hand. I completed the circuit and felt 230 V, not 5 V. That's how it works. ;-) Jan 7, 2013 at 11:43
The man gets shocked because his hand is touching the positive terminal, and his legs are touching the ground, which is connected to the negative terminal. Since the positive terminal is at a high potential relative to the ground, he gets a shock. If the load is removed, he still gets a shock.
The bird doesn't because it's only touching the positive terminal, so there is no potential difference across it's body.
Remember voltage is always between two points. You can't just say "this point is at 5V", you have to say "this point is at 5V relative to this point" (although people often say the former because a common reference point, like circuit ground is assumed)
Regarding the energy being consumed part, as long as the voltage source is strong enough (i.e. has a low internal resistance) an assuming it has "infinite" capacity to keep supplying power (e.g. a mains source as opposed to a battery or capacitor) then it will supply current to many loads connected in parallel to it, without it's voltage dropping appreciably.
If it is a weak source, then it's voltage will drop if a low resistance load is connected across it - in this scenario it would be safe(r) for the man to touch it (depending on how much the voltage drops - this sort of stuff is all covered by Ohm's law, Norton and Thevenin)
Regarding edit
In an ideal situation where we assume the wires have zero resistance, then the resistance of the bird does not matter. You are right that the resistance of the bird could matter in the real world though.
We can take a look at this with a SPICE simulation. We will use your circuit above, and assume the resistance of the wire between the birds legs is 1mΩ. By varying the resistance of the bird we can see what would be needed to cause an appreciable current to flow through it:
Circuit:
Simulation, seeping the birds resistance from 1Ω to 1kΩ:
You can see the current through the load R1 (green trace in the bottom graph) is around 2A, and hardly changes at all.
For the bird, even with a resistance of only 1Ω, it only has 2mA flowing through it. At 1kΩ, it has 2μA flowing through it (note the logarithmic graph used on the X axis due to the large variation)
In reality, unless the bird is soaking wet, and the cable it's standing on is very thin, then the above values are a long way from typical. Typical bird resistance will probably be > 1MΩ, and the resistance between it's legs probably less than a few μΩ.
Hopefully this makes some sense - it all agrees with Ohm's law, and once you get the hang of it it will seem quite natural. It might be good idea to grab something like LTSpice (the simulator used in the above examples and have a play around with it to see what happens when you change various things. Or build a circuit on a breadboard and measure with a multimeter. Also, have a good book that covers this stuff handy (see "Art of Electronics", "Practical Electronics for Inventors", etc)
• Why does the voltage across the terminal has to be the voltage of the power source? Does it mean the current must find a way to dissipate the energy before returning? Jan 8, 2013 at 4:12
• @lamwaiman1988 - I added a bit extra on the edit. Jan 8, 2013 at 8:38
You only get shocked when current passes through your body. The bird doesn't get shocked since the bird is only touching one wire and it isn’t touching the ground either. Because of this, the current doesn't have anywhere to go since there is no potential difference (voltage) between 2 points since the bird is only touching one "point" (1 wire), and like I mentioned earlier, isn’t grounded. The human gets shocked because they are touching 2 points that have a potential difference (voltage) between them, which results in current flow. The live wire has a higher potential than the ground.
For example, in the US, live wires are 120 volts, and the ground (wire) is always 0 volts. If you touched both the ground and live at the same time, it would result in a potential difference of 120, allowing for current to flow. Voltage is the cause and current flow is the effect. You need voltage in order to have current flow.
• Welllll...if the bird puts both feet on the same wire there will be a voltage difference between the two feet and some current will flow through the bird as a result. But, the voltage difference is very small and the resistance of the bird is very high compared to the resistance of the short section of wire, so a very, very small current will flow through the bird. Not enough to get the bird's attention, but not strictly zero either. Feb 27, 2021 at 19:55
• @ElliotAlderson Yeah, good catch. I just didn't mention that because I wanted to keep it simple, but you are totally right. Feb 28, 2021 at 19:15 | 1,831 | 8,122 | {"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} | 3.65625 | 4 | CC-MAIN-2023-50 | longest | en | 0.969102 |
http://math.stackexchange.com/questions/273297/rearrange-y-frac-tan-left-fracn-x2-rightn-to-give-n | 1,443,973,037,000,000,000 | text/html | crawl-data/CC-MAIN-2015-40/segments/1443736675218.1/warc/CC-MAIN-20151001215755-00063-ip-10-137-6-227.ec2.internal.warc.gz | 202,044,310 | 15,917 | # Rearrange $y = \frac{\tan\left(\frac{N x}{2}\right)}{N}$ to give N
Is it possible to rearrange
$$y = \frac{ \tan \left(\frac{N x}{2}\right)}{N}$$
where $x \lt \pi$ as a function of x and y that gives N?
-
What do you mean here by "rearrange"? I don't think there is a way to write the right-hand side that's any simpler than what you already have. – Greg Martin Jan 9 '13 at 6:11
I mean to give N as a function of x and y. – geometrikal Jan 11 '13 at 3:43
The program "maple" could not solve for N as a function of x,y. – coffeemath Jan 11 '13 at 6:41
Agreed, even the special case $\tan(N)/N = 1$ doesn't admit a closed-form solution, I believe. – Greg Martin Jan 11 '13 at 21:21
Thankyou all ... i will have to come up with an approximation. – geometrikal Jan 12 '13 at 0:05 | 267 | 787 | {"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} | 2.875 | 3 | CC-MAIN-2015-40 | longest | en | 0.876127 |
https://www.unitconverters.net/power/kilocalorie-th-hour-to-kilocalorie-it-second.htm | 1,659,944,806,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882570767.11/warc/CC-MAIN-20220808061828-20220808091828-00403.warc.gz | 923,284,752 | 3,660 | Home / Power Conversion / Convert Kilocalorie (th)/hour to Kilocalorie (IT)/second
# Convert Kilocalorie (th)/hour to Kilocalorie (IT)/second
Please provide values below to convert kilocalorie (th)/hour to kilocalorie (IT)/second, or vice versa.
From: kilocalorie (th)/hour To: kilocalorie (IT)/second
### Kilocalorie (th)/hour to Kilocalorie (IT)/second Conversion Table
Kilocalorie (th)/hourKilocalorie (IT)/second
0.01 kilocalorie (th)/hour2.7759200874707E-6 kilocalorie (IT)/second
0.1 kilocalorie (th)/hour2.77592E-5 kilocalorie (IT)/second
1 kilocalorie (th)/hour0.000277592 kilocalorie (IT)/second
2 kilocalorie (th)/hour0.000555184 kilocalorie (IT)/second
3 kilocalorie (th)/hour0.000832776 kilocalorie (IT)/second
5 kilocalorie (th)/hour0.00138796 kilocalorie (IT)/second
10 kilocalorie (th)/hour0.0027759201 kilocalorie (IT)/second
20 kilocalorie (th)/hour0.0055518402 kilocalorie (IT)/second
50 kilocalorie (th)/hour0.0138796004 kilocalorie (IT)/second
100 kilocalorie (th)/hour0.0277592009 kilocalorie (IT)/second
1000 kilocalorie (th)/hour0.2775920087 kilocalorie (IT)/second
### How to Convert Kilocalorie (th)/hour to Kilocalorie (IT)/second
1 kilocalorie (th)/hour = 0.000277592 kilocalorie (IT)/second
1 kilocalorie (IT)/second = 3602.4091778203 kilocalorie (th)/hour
Example: convert 15 kilocalorie (th)/hour to kilocalorie (IT)/second:
15 kilocalorie (th)/hour = 15 × 0.000277592 kilocalorie (IT)/second = 0.0041638801 kilocalorie (IT)/second | 499 | 1,470 | {"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-2022-33 | longest | en | 0.465865 |
https://docest.com/chapter-4-continuous-distribution | 1,713,818,015,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296818337.62/warc/CC-MAIN-20240422175900-20240422205900-00251.warc.gz | 168,836,324 | 9,092 | # Chapter 4 Continuous Distribution ( การแจกแจงความน่าจะเป็นแบบต่อเนื่อง)
Chapter 4
Continuous Distribution
## Definition Continuous Random Variable
A random variable is continuous if it can assume any value in some interval or intervals of real numbers and the probability that it assume any specific value is 0.
Note 1.Owing to the limits of our ability to measure, experimental data never seem to come from a continuous scale.
Ex. T=74.8C is not a point on acontinuous scale. Actually it could be any value between74.75-74.85
2. Sincethe probability that a continuous random variable will take on a point on a scale is 0, therefore there is no difference in writing intervals.
P (a x b ) = P (a x < b)
= P (a < x b)
= P(a < x <b)
## Probability Distribution Function
Definition Let x be a continuous random variable. A function f is called probability density function if:
(1) f(x) 0 for x real
(2)
(3) P (a x b) = for a and b real
Ex.1 If a random variable has the probability density
12.5x -1.25 0.1 x 0.5
f(x) =
0 ,elsewhere
Find P (0.2 x 0.3)
Def. Cumulative Distribution Function
Let x be continuous with density f, the cumulative distribution function for x,
denoted by F, is defined by:
F(x) = P( X x ) for x real
Computationally ,
P( X x ) = F (x)
=
Determining probability from F(x)
1. P (X x) = 1 - F(x)
2. P ( x1 X x2 ) = F (x2 ) – F ( x1)
Properties of F (x)
1. F(x) is nondecreasing as x increases, meaning that
if x1 < x2 then F (x1) F(x2)
2. = 0 and = 1
Ex.2 Suppose the pdf of the magnitude X of a dynamic load on a bridge (in newtons) is
f (x) = , 0 x 2
0 otherwise
(a) Determine F(x)
(b) Find the prob. that the load is between 1 and 1.5 newtons
(c) Find the prob. that the load exceeds 1 newton
## Expectation
The expected value (mean value) of a continuous random variable with pdf f(x) is
= E (X) =
If X is a continuous random variable with pdf f(x) and h(x) is any function of X, then
E [h(X)] = =
Ex. The distribution of the amount of gravel (in tons) sold by a particular construction supply company in a given week is a continuous random variable X with pdf
f(x) = , 0 x 1
0otherwise
( a) Find F(x) (b) Find E (x)
Variance
Definition The varianace of a continuous random variable X with pdf f(x) and mean value is
= V(x) =E =
The standard deviation of X is
=
Shortcut formula V(x) = E (x2) - [E(x)]2
Ex. For X = weekly gravel sales, we computed E(X) = 3/8. Find V(X) and (x)
Rules for Variancesame as discrete case
------midterm------
Normal Distribution N (,2) or f (x; ,2)
underlies many of the statistical methods
is the limiting form of binomial density as the number of trials become infinite
describes the behavior of errors inmeasurements
Definition A random variable x has a normal distribution with parameters and if its density is given by :
f (x) =
, - x
E (X) =
Var X = 2
Graph of N (,2)
## Moment Generating Function
Definition Let x be a random variable with density f . The moment generating function for x (m.g.f) is denoted by mx(t) and is given by:
= E (etx) for t real
Theorem Let be the moment generating functionfor x, then
= E [x k]
Standard Normal Distribution Table
Standard Normal Distribution หมายถึงnormal distribution ที่มี =0, = 1
F (Z) =
Ex. Find the probabilities that a random variable having the standard normal distribution will take on a value
(a)between 0.87 and 1.28
(b) between -0.34 and 0.62
(c) greater than 0.85
(d) greater than –0.65
In finding theprobability of random variable x withmeanand variance 2, we can use the standard normal table after changing x to z (normal standard normal)
Z =
Thus
P (a< x <b) = -
Ex. Let x = number of grams of hydrocarbons emitted by an automobile per mile. Assuming that x is normal with =1 g and = 0.25g, find the probability that a randomly selected automobile will emit between 0.9 and 1.54 g of HC per mile
Ex. Let x = the amount of radiation that can be absorbed by an individual before death ensues. Assume thatx is normal with a mean of 500roentgensand of 150 roentgens. Above what dosage level will only 5 % of those exposed survive?
## The Normal Approximations to the Bionomial Distribution
Theorem Let x be binomial with parameters n and p. For large n, x is approximately normal with mean np and variance np(1-p).
Acceptable if
p 0.5 and np > 5 Or
p > 0.5 and n(1-p) > 5
Ex. If 20 % of the diodes made in a certain plant are defective, what are the prob. that in a lot of 100 randomly chosen for inspection.
(a)at most 15 will be defective.
(b) exactly 15 will be defective ?
## Linear Combinations of Normal Distributed Variable
Theorem1 If the random variables X1 ,…, Xk are independent and if Xihas a normal distribution with mean i and variance (i = 1, …, k), then the sum X1 + … + Xk
has a normal distribution with mean 1 + … + k and variance +…+
Theorem 2 If x has a normal distribution with mean and variance 2 and if y = ax + b, where a and b are constants and a 0, then Y has a normal distribution with mean a + b and variance
Ex. An assembly consists of 3 linkage components as shown here. The properties X1, X2, X3 are given below with means in cm, variance in cm2.
X1 N (12, 0.02)
X2 N (24, 0.03)
X3 N (18, 0.04)
If the links are produced by different machines and operators, determine P (53.8 Y 54.2)
Gamma Distribution
- very important because it allows us to define the exponential and chi-square
distribution, that are used extensively in applied statistics.
The theoretical basis for the gamma distribution = gamma function:
Definition The gamma function is defined by
=
where > 0
Properties of the Gamma Function
1. If > 1, then
2.
= 1
3. For any positive integer n,
= (n-1)!
Gamma Distribution
Definition A random variable X has a Gamma Distribution with parameters and if its density is :
f (x) =
where x > 0
> 0
> 0
E(x) =
Var x = 2
Shape of Gamma Distribution
, = shape parameters
Exponential Distribution
= Gamma Distribution with = 1
Recall: Gamma Density
f (x) =
Thus, exponential density
f (x) = where x > 0
and > 0
##### E (x) =
Var x = 2
###### Application of Exponential Distribution
Usually occur in common with Poisson process
If = mean arrival rate
then t = the time between successive arrivals will have an exponential distribution with = 1/
Since x is the time, we can write f(x) as f(t)
where t > 0 and > 0
t = the time that an arrival occurs
f (t) = probability of an arrival at time t
Ex. The time T between the arrival of orders at a regional warehouse is recorded for 24 orders: 17 19 25 34 35 35 37 39 40 40 41 41 42 42 44 46 51 52 52 56 71 72 80 93
(a) Estimate and interpret the parameter
(b) Assume arrivals follow an exponential distribution. Plot and on the graph.
Ex. Engineers have collected data from 100 compressors on natural gas pipelines and found that the average life is 5.75 years and that failures follow the exponential distribution.
(a) Compute the probability of failure during the first year after installation, and during the first 3 months.
(b) Compute the probability of failure prior to the average life.
© Compute the probability of operating at least 10 years.
Chi-squared distribution
Use in making inference about variance
Def. Let X be a gamma random variable with = 2 and for being a positive integer. X is said to have a chi-squared distribution with degrees of freedom. We denote this variable by
, x 0
0 , x 0
E (X) =
Var X = 2
Weibull Distribution
Def. A random variable X is said to have a Weibull distribution with parameters and if its density is given by:
where x > 0, > 0, > 0
= E (x) =
###### Shape
- varies depending on the values of and
-General shape resembles that of gamma density with the curve becoming more symmetic as the value of increases.
Uniform Distribution
A uniform variable is a variable whose values have equal probability of occurrence throughout a specified interval.
, x
f(x) =
0 , otherwise
E (X) =
V(X) =
Proof E (X) =
=
=
V(X) =
=
Graph of U (x; , )
0 ; x
F (X) = = ; x
1 ; x
\
Ex. If a value is randomly chosen from a uniform random variable in the interval [0, 10], determine the probability that the value is between 3/2 and 7/2
Ex. The noise level (N) in a workshop is a uniform random variable between 80-95 db. If the safe limit of noise exposure must not exceed 90 db, find the pdf of the random variable N and determine the probability that the noise level exceeds the safe limit.
1 | 2,582 | 8,609 | {"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} | 4.53125 | 5 | CC-MAIN-2024-18 | latest | en | 0.874429 |
https://www.futureschool.com/international-baccalaureate-curriculum/mathematics-pyp-grade-3/ | 1,585,484,201,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585370494331.42/warc/CC-MAIN-20200329105248-20200329135248-00202.warc.gz | 953,137,255 | 10,359 | # TOPIC TITLE
1 Self Assessment Self Assessment – PYP Grade 3
Objective: Assessment
2 Calculation-larger numbers The numbers 100 to 999
Objective: On completion of the lesson the student will be able to count to 999, skip count by 10s and 100s to 999 and use pictures or objects to represent the numbers 100 to 999.
Objective: On completion of the lesson the student will be able to use place value to add two 2-digit numbers together up to a total of 99.
4 Place value The numbers 1000 to 9999
Objective: On completion of the lesson the student will be able to count to 999, skip count by 10s and 100s to 999 and use pictures or objects to represent the numbers 100 to 999.
Objective: On completion of the lesson the student will be able to use place value to solve and record addition problems with three digit numbers.
6 Subtraction Subtraction up to the number 99
Objective: On completion of the lesson the student will know how to find the answer to take away number sentences with bigger numbers and will also know how to write these number sentences.
7 Subtraction Subtraction with borrowing
Objective: On completion of the lesson the student will be able to record subtraction number sentences and will understand the need to trade between place value columns using the renaming method
8 Subtraction Subtraction of two-digit numbers Involving comparison.
Objective: On completion of the lesson the student will know how to solve and record the answer to word problems and will understand how estimation helps to find out if an answer is correct.
9 Subtraction Subtraction up to the number 999 using the renaming method
Objective: On completion of the lesson the student will know another way to trade between columns in subtraction using the renaming method. The student will also learn how to solve number sentences with missing numbers.
10 Data Pictograms
Objective: On completion of the lesson the student will be able to organise, read and summarise information in picture graphs.
11 Data Bar Charts
Objective: On completion of the lesson the student will be able to organise, read and summarise information in column graphs.
12 Data Line graphs.
Objective: On completion of the lesson the student will be able to organise, read and summarise information in line graphs.
13 Calculation-grouping Multiplication using equal groups
Objective: On completion of the lesson the student will be able to write about equal groups and rows and will understand how to count them.
14 Calculation-grouping Multiplication using repeated addition
Objective: On completion of the lesson the student will be able to write about equal groups and rows using another method and also learn different ways to count them.
15 Calculation-multiplication The multiplication sign
Objective: on completion of the lesson the student will be able to multiply single numbers to solve multiplication problems.
16 Calculation-multiples Multiples of 10 up to 100
Objective: On completion of the lesson the student will be able to read, write, rename and add multiples of ten.
17 Multiplication Multiplication – important facts.
Objective: On completion of the lesson the student will know the connection between multiplication and division and recognise the strategies to help solve multiplication number sentences.
18 Multiplication Multiplying 2-digit numbers by multiple of 10
Objective: On completion of the lesson the student will be able to multiply any 2 digit number by any multiple of 10 using the process of long multiplication.
19 Calculation sharing/division Strategies for division
Objective: On completion of the lesson the student will be able to share objects equally and will also learn how to write about them.
20 Problems Solve and record division using known facts and sharing
Objective: On completion of the lesson the student will be able to use sharing concepts to solve and record division problems.
21 Angles Measure and classify angles
Objective: On completion of the lesson the student will be able to recognise, measure and classify angles and measure the angles in a triangle.
22 2-D shapes Using the prefix to determine polygons
Objective: On completion of the lesson the student will be able to recognise and name two-dimensional shapes such as pentagons and hexagons, using the prefix of the shapes name to determine number of angles and sides.
23 2-D shapes Spatial properties of quadrilaterals
Objective: On completion of the lesson the student will know how to analyse and explain the spatial properties of quadrilaterals.
24 Tessellating 2-D shapes Use grids to enlarge/reduce 2D shapes
Objective: On completion of the lesson the student will be able to use grids to enlarge or reduce two dimensional shapes and also to recognise shapes that will and won’t tessellate.
25 3-D shapes Recognise and name prisms according to spatial properties
Objective: On completion of the lesson the student will be able to recognise and name various prisms according to their spatial properties.
26 3-D shapes Viewing 3-D shapes.
Objective: On completion of the lesson the student will be able to use conventional representations of three-dimensional shapes to show depth etc when drawing or viewing shapes from various angles.
27 Lines and angles Describing position.
Objective: On completion of the lesson the student will be able to use and understand conventional location language to describe position.
28 Lines and angles Mapping and grid references
Objective: On completion of the lesson the student will be able to identify specific places on a map and use regions on a grid to locate objects or places.
29 Lines and angles Informal coordinate system
Objective: On completion of the lesson the student will be able to use an informal coordinate system to specify location, and locate coordinate points on grid paper.
30 Fractions Using fractions 1/2, 1/4, 1/8 to describe part of a whole
Objective: On completion of the lesson the student will be able to: name fractions, and use fractions to describe equal parts of a whole.
31 Fractions Using fractions 1/2, 1/4, 1/8 to describe parts of a group or collection
Objective: On completion of the lesson the student will be able to use the fractions to describe equal parts of a collection of objects.
32 Fractions Fractions 1/5, 1/10, 1/100
Objective: On completion of the lesson the student will be able to: compare fractions with the denominators 5, 10, 100, and represent fractions with the denominator 5, 10, 100.
33 Fractions Comparing and ordering fractions
Objective: On completion of the lesson the student will be able to compare and order fractions with the same number of equal parts, and compare and order fractions with a different number of equal parts.
34 Fractions mixed numbers (mixed numerals)
Objective: On completion of the lesson the student will be able to: name and recognise mixed numbers (mixed numerals), count by halves and quarters, and use a number line to represent halves and quarters beyond one.
35 Fractions Improper fractions
Objective: On completion of the lesson the student will be able to: use diagrams and number lines to recognise and represent mixed numbers and improper fractions, and develop strategies for changing improper fractions to mixed numbers and vice versa, mentally.
36 Fractions Finding equivalent fractions
Objective: On completion of the lesson the student will be able to name and find fractions that represent equal amounts between halves, quarters and eighths – using diagrams and number lines.
37 Fractions Subtracting fractions from whole numbers
Objective: On completion of the lesson the student will be able to: use a diagram to subtract fractions from a whole number, develop mental strategies for subtracting fractions from whole numbers, and recognise and use the written form for subtracting fractions from
38 Fractions Adding and subtracting fractions with the same denominator
Objective: On completion of the lesson the student will be able to add and subtract fractions with the same denominator.
39 Length Compare length by using informal units of measurement
Objective: On completion of the lesson the student will be able to measure objects around the student’s home and compare their length to each other. The student will also be able to compare the different ways the student measured the same object.
40 Length Using the metre as a formal unit to measure perimeter
Objective: On completion of the lesson the student will be able to calculate the perimeter of different shapes in metres.
41 Length Using the formal unit of the centimetre to measure length and perimeter
Objective: On completion of the lesson the student will be able to measure length and perimeter in centimetres.
42 Area Comparing and ordering areas.
Objective: On completion of the lesson the student will be able to estimate and compare the area of shapes using a standard unit and order shapes according to their area.
43 Area Introduction to the square centimetre.
Objective: On completion of the lesson the student will be able to calculate the area in square centimetres of surfaces or objects and record their results correctly.
44 Weight/mass Introducing the concept of mass
Objective: On completion of the lesson the student will understand the concepts of mass.
45 Weight/mass The kilogram
Objective: On completion of the lesson the student will know: how to use the kilogram as a measure of mass., and how to weigh items accurately using scales.
46 Weight/mass The gram and net mass
Objective: On completion of the lesson the student will understand: why there’s a zero button on digital scales, how to zero some other types of scales, and also how to measure mass in grams.
47 Capacity Using the cubic cm and displacement to measure volume and capacity
Objective: On completion of the lesson the student will know a way to find volume and capacity.
48 Capacity Using the cubic cm as a standard unit of measurement for volume and capacity
Objective: On completion of the lesson the student will understand what a cubic centimetre is and how it can be used to find out the volume and capacity of a three dimensional shape.
49 Capacity The relationship between the common units of capacity, the litre and the millilitre
Objective: On completion of the lesson the student will understand the relationship between the two common units of capacity, the litre and millilitre.
50 Capacity Estimate, measure and compare the capacity of containers
Objective: On completion of the lesson the student will know why and when we might need to estimate and a way to go about it.
51 Time, digital O’clock and half past using digital time
Objective: On completion of the lesson the student will be able to use the terms ‘o’clock’ and ‘half past’, and read hour and half-hour time on a digital clock.
52 Time, analogue O’clock and half past on the analogue clock
Objective: On completion of the lesson the student will be able to use the terms ‘o’clock’ and ‘half past’, and read hour and half-hour time on an analogue clock.
53 Time, minutes Analogue – Telling time – minutes in the hour
Objective: On completion of the lesson the student will be able to recognize the coordinated movements of the hands on an analogue clock.
54 Time, quarter to, past Quarter past and quarter to
Objective: On completion of the lesson the student will be able to associate the numerals 3, 6 and 9 with 15, 30 and 45 minutes, and use the terms ‘quarter to’ and ‘quarter past’.
55 Time, units Units of time
Objective: On completion of the lesson the student will be able to convert units of time and read and interpret simple timetables, timelines and calendars.
56 Time, minutes past the hour Minutes past
Objective: On completion of the lesson the student will be able to understand and read the minutes past the hour on an analogue clock.
57 Time, minutes to the hour Minutes to
Objective: On completion of the lesson the student will be able to understand and read the minutes to the hour on an analogue clock.
58 Time, digital, analogue Comparing analogue and digital time
Objective: On completion of the lesson the student will be able to: read analogue and digital clocks to the minute, be familiar with the recording of digital time, and tell the same time using both analogue and digital clocks.
59 Decimals Introduction to decimals
Objective: On completion of the lesson the student will be able to represent decimals to two decimal places.
60 Decimals Comparing and ordering decimals to two decimal places
Objective: On completion of the lesson the student will be able to compare and order decimals to two decimal places and understand decimal notation to two places.
61 Decimals Decimals with whole numbers 10th and 100th
Objective: On completion of the lesson the student will be able to use and understand place value to show whole numbers, tenths and hundredths as decimals.
62 Multiplication Multiplying 3 and 4-digit numbers by multiples of 100
Objective: On completion of the lesson the student will be able to multiply any 3 or 4 digit numbers by any multiple of 100 using long multiplication.
63 Multiplication Multiplying 2-digit numbers by 2-digit numbers
Objective: On completion of the lesson the student will be able to multiply any 2 digit number by any other 2 digit number.
64 Multiplication Multiples and factors of whole numbers
Objective: On completion of the lesson the student will be able to specify multiples and factors of whole numbers, and calculate the product of squared numbers.
65 Division Division with and without a remainder.
Objective: On completion of the lesson the student will understand division with and without a Remainder.
66 Division Dividing two and three digit numbers by a single digit number.
Objective: On completion of the lesson the student will understand dividing two and three digit numbers by a single digit number.
67 Exam Exam – PYP Grade 3
Objective: Exam | 2,824 | 13,904 | {"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-2020-16 | latest | en | 0.893597 |
https://www.mrexcel.com/board/threads/calculated-column-circular-error-need-alternative-solution.1103495/?s=ddc2922d2a66368fcb48895505529078 | 1,597,266,433,000,000,000 | text/html | crawl-data/CC-MAIN-2020-34/segments/1596439738944.95/warc/CC-MAIN-20200812200445-20200812230445-00526.warc.gz | 764,978,838 | 17,022 | # Calculated Column Circular Error need alternative solution
#### ollieotis
##### New Member
Hello,
I unfortunately/fortunately learned the hard way yesterday that using the calculate function in calculated columns doesn't always work out so well and have since spent entirely too much time spinning my wheels trying to figure out an alternative solution to my report requirement. Hoping someone can help point me in the right direction from here.
Simplified version of what I'm doing here (4 tables):
Two data tables as follows (accounts receivable balance current week and prior week)
Company Customer Parent Customer A/R Balance
<tbody>
</tbody>
Fact table connecting the two A/R tables derived using power query using concatenated [Company]&[Customer]&[Parent Customer] fields, and eliminating duplicates, and adding same field to end of A/R tables. Called Master_CoCP (company+Customer+Parent Cust)
Last table is a Customer Table
Customer Unit
<tbody>
</tbody>
Now here's where I'm getting stuck. Ultimately, what I'm trying to do is generate a pivot table that looks like the following, that groups customers based on whether they are currently in a top ten list, or whether they were and fell off that list in the subsequent report:
Parent Customer Current Rank Current Balance Prior Rank Prior Balance Top Ten Parent Customers Customer A Customer B... Formerly Top Ten Parent Customers Customer X
<tbody>
</tbody>
To achieve this, I used lookup values to bring in the Customer[Unit] to my master fact table Master_CoCP, then wrote the following calculated columns into that same table:
Customer Balance Current = Calculate(sum(Aging_Current[Balance]),allexcept(Master_CoCP,Master_CoCP[Parent Customer],Master_CoCP[Unit]))
Customer Rank Current = rankx(CALCULATETABLE(Master_CoCP,allexcept(Master_CoCP,Master_CoCP[Unit])),[Customer Balance Current],,,DENSE)
Repeated for prior aging report then wrote a simple if statement in another column to say if <10 current, > 10 prior.
This worked beautifully (maybe not elegant) until such time as I loaded new Aging Reports and tried refreshing all queries and connections.... learn something new every day I guess.
Sorry if this is too long and doesn't make sense. I thought the length of my post should follow suit with how long I've been beating my head on the desk on this one.
Any ideas would be much appreciated. Thanks in advance!
### Excel Facts
Round to nearest half hour?
Use =MROUND(A2,"0:30") to round to nearest half hour. Use =CEILING(A2,"0:30") to round to next half hour.
#### ollieotis
##### New Member
Hoping to bump this one, as I'm still struggling to find a solution. Maybe it will help if I put my problem a different way:
My desired end pivot table is a list of customers with their current balance and prior balance (from two receivables aging reports). I have created two measures to rank my customers: Top Ten for current period and Top Ten for prior period.
What I'm trying to do now is categorize the customers on the rows in the following three buckets:
Current Top Ten Balances
No Longer in Top Ten (if statement saying if they were top ten previous week, but are no longer)
All Other
Is there a way to create this type of categorization derived from measures? I've done a little reading on disconnected tables, but am struggling a little. Any help or reference to other posts that might help would be very much appreciated!
Thanks!
Last edited:
#### Louis_Guionneau
##### New Member
I haven't found a solution to this problem either. You're not alone.
#### nikio8
##### Board Regular
Hi, I did something like this and it was the most difficult task that I have ever done. lots of these Filter(ALL(....
Pivot tables already have top 10.
You can create two pivot tables and say filter by week.
Previous top 10, filter out paid date - last day or last week,
Today, no filters.
You could create another "Calendar table" and use timeline to control which data you see.
Kind regards
#### Louis_Guionneau
##### New Member
@ nikio - nice one. did you have to create a bunch of calculated columns to do that? maybe it's possible to get around it using a combination of top n / if?
1,102,643
Messages
5,488,074
Members
407,622
Latest member
### This Week's Hot Topics
• Timer in VBA - Stop, Start, Pause and Reset
[CODE=vba][/CODE] Option Explicit Dim CmdStop As Boolean Dim Paused As Boolean Dim Start Dim TimerValue As Date Dim pausedTime As Date Sub...
• how to updates multiple rows in muliselect listbox
Hello everyone. I need help with below code. code is only chaning 1st row in mulitiselect list box. i know issue with code...
• Delete Row from Table
I am trying to delete a row from a table using VBA using a named range to find what I need to delete. My Range is finding the right cell. In the...
• Assigning to a variable
I have a for each block where I want to assign the value in column 5 of the found row to the variable Serv. [CODE=vba] For Each ws In...
• Way to verify information
Hi All, I don't know what to call this formula, and therefore can't search. I have a spreadsheet with information I want to reference...
• Active Cell Address – Inactive Sheet
How to use VBA to get the cell address of the active cell in an inactive worksheet and then place that cell address in a location on the current... | 1,205 | 5,336 | {"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-2020-34 | latest | en | 0.889138 |
https://answers.everydaycalculation.com/divide-fractions/30-56-divided-by-7-24 | 1,725,920,051,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651157.15/warc/CC-MAIN-20240909201932-20240909231932-00643.warc.gz | 85,125,471 | 2,839 | Solutions by everydaycalculation.com
## Divide 30/56 with 7/24
30/56 ÷ 7/24 is 90/49.
#### Steps for dividing fractions
1. Find the reciprocal of the divisor
2. Now, multiply it with the dividend
So, 30/56 ÷ 7/24 = 30/56 × 24/7
3. = 30 × 24/56 × 7 = 720/392
4. After reducing the fraction, the answer is 90/49
5. In mixed form: 141/49
MathStep (Works offline) | 135 | 364 | {"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} | 4.125 | 4 | CC-MAIN-2024-38 | latest | en | 0.801695 |
http://strengthandstiffness.com/2_statics/page_2a.htm | 1,505,867,363,000,000,000 | text/html | crawl-data/CC-MAIN-2017-39/segments/1505818686077.22/warc/CC-MAIN-20170919235817-20170920015817-00185.warc.gz | 324,977,705 | 3,336 | 2a. Statics Fundamentals (SSES Chapt. 2.0) • Vectors and Forces • Free-Body Diagrams • Equilibrium Back | Index | Next
Vectors and Forces
A solid grasp of vectors is vital to solve problems in Statics and Strength of Materials. Vectors describe the magnitude and direction of forces and torques/moments. It is often convenient to reduce vectors into their scalar components; e.g., 2D-force F can be broken into its x- and y-components:
Fx = Fcosq and Fy = Fsinq
where F is the magnitude of force F, and q is its direction measured from the x-axis.
Free-Body Diagrams
The construction of free-body diagrams (FBDs) is the first and most important step in solving a Strength of Materials problem. The following steps are involved in constructing a FBD:
• Isolate from its surroundings the part(s) of the system that you are interested in analyzing; visualize or actually draw a boundary around the body of interest.
• Identify and represent ALL external forces (and moments) acting on the body; i.e., forces that act across the boundary. Include weight when comparable to the applied forces.
• Include a coordinate system and necessary dimensions on the diagram for convenience in applying equilibrium equations and communicating geometry.
• The diagram should be free of clutter and extensive information. The forces, moments and key dimensions are the primary information required.
To help illustrate the construction of a FBD, consider the rear suspension of the mountain bike below:
Full Suspension Bicycle Free-body diagram of rear suspension
Equilibrium
Once a FBD system is constructed, the next step is to apply the conditions of equilibrium. A body in static equilibrium is not accelerating, so the forces and moments acting on it must sum to zero in any direction. Equilibrium requires that:
• the sum of the forces in any direction must be zero: S F = 0.
• the sum of the moments about any non-accelerating point must be zero: S MO= 0.
In 3-dimensions, six equations must be satisfied:
S Fx = 0 ; S Fy = 0 ; S Fz = 0 S Mx = 0 ; S My = 0 ; S Mz = 0
In 2-dimensions, only three equilibrium equations are required. When the object is in the x-y plane, one of the following three sets of equations may be used:
S Fx = 0 S Fy = 0 S Mz,A = 0 S Fx = 0 S Mz,A = 0 S Mz,B = 0 S Mz,A = 0 S Mz,B = 0 S Mz,C = 0 The sum of forces in x and in y, and the sum of moments about any Point A (about a z-axis through Pt. A), each equal zero Points A and B are two different points. A, B and C are three points, not all on the same line. The points define a plane.
Many 3D-problems can be reduced to 2-D.
Top Back | Index | Next
Updated: 05/24/2009 DJD | 675 | 2,679 | {"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} | 4.375 | 4 | CC-MAIN-2017-39 | latest | en | 0.882622 |
https://etastudyph.com/math/1-learning-task-1-write-the-prime-f-524840124 | 1,686,440,126,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224646652.16/warc/CC-MAIN-20230610233020-20230611023020-00540.warc.gz | 277,675,248 | 14,139 | , 13.11.2022 21:55 elaineeee
# 1 Learning Task 1: Write the prime factorization of the following numbers using the factor tree. Write your answer in your notebook. 1) 18 3 X 6 i 45 1) 32 4 x8 1) 12ā¬ 6 x 20 1) 72 8 x9
### Another question on Math
Math, 28.10.2019 14:46
Match the verbal sentences in column a with the corresponding mathematical statements in columnb.4) twice a number x is less than 28. a. 2x < 28 b. x + 2 > 28 c. x + 2 > 28 d. x < 28 e. x ā 2 < 28
Math, 28.10.2019 21:29
Wvaries directly as x and y and inversely as vĀ² and w=1200 when x =4, y=9 and v=6. find w when x=3, y=12 and v=9.
Math, 28.10.2019 22:28
After a cash discount of 50,000, mr. perez paid 1,200 for the goods he purchased. what was the rete of discount? | 294 | 746 | {"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-2023-23 | latest | en | 0.796483 |
https://www.physicsforums.com/threads/confusion-with-superposition-of-states.975171/ | 1,726,782,801,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700652067.20/warc/CC-MAIN-20240919194038-20240919224038-00028.warc.gz | 862,601,156 | 19,826 | Confusion with superposition of states
• I
• ntt
In summary, the wavefunction can be written as a linear combination of eigenfunctions due to completeness property. If an electron is excited, when it is relaxing back to ground state, it can be in a superposition of two discrete energy eigenstates. During relaxation, there is a dynamical law that describes the change of coefficient of each eigenstate with time.
ntt
TL;DR Summary
Can an electron be in a superposition of energy eigenstates during relaxation after excited?
Since my major is not physics. My QM knowledge is not pretty good (Mostly self study). I am sorry if this question was asked multiple times in the forum.
I've learned that wavefunction can be written as a linear combination of eigenfunctions due to completeness property.
If an electron is excited. When it is relaxing back to ground state. Can the state of the electron during relaxation a superposition of two discrete energy eigenstates?
If so, during relaxation, is there dynamical law that describes the change of coefficient of each eigenstate with time? (I've looked up the spontaneous emission but still don't have a clear idea what governs the time of relaxation like the design of laser which utilizes a metastable level)
From my understanding, when we observed photon emitted from relaxation, this means we already observe the system which the wave function then has already collapsed to the ground state and the relaxation time is very short. Is my understanding correct?
Any interpretation and correction of my concept and additional reading resources would be greatly appreciated.
Thank everyone in advance for the replies.
You have to carefully distinguish about which eigenstates you talk.
Take a hydrogen atom as a concrete example. You first treat it just as an electron in the Coulomb potential of the nucleus. Then you can evaluate the energy eigenstates of this Hamiltonian. If this were the full story, and your atom is in some energy eigenstate at ##t=0##, then it would stay in this state forever.
Now, however, there's also the electromagnetic field itself, and it's a quantum field, i.e., there are quantum fluctuations, leading to a coupling of the electron to the electromagnetic field. Taking this additional part of the Hamiltonian into account, which can be done perturbatively, you'll get a finite probability for a spontaneous transition from an excited atomic state to a lower atomic state and the emission of a photon.
aaroman
vanhees71 said:
You have to carefully distinguish about which eigenstates you talk.
Take a hydrogen atom as a concrete example. You first treat it just as an electron in the Coulomb potential of the nucleus. Then you can evaluate the energy eigenstates of this Hamiltonian. If this were the full story, and your atom is in some energy eigenstate at ##t=0##, then it would stay in this state forever.
Now, however, there's also the electromagnetic field itself, and it's a quantum field, i.e., there are quantum fluctuations, leading to a coupling of the electron to the electromagnetic field. Taking this additional part of the Hamiltonian into account, which can be done perturbatively, you'll get a finite probability for a spontaneous transition from an excited atomic state to a lower atomic state and the emission of a photon.
Thank you very much for your reply.
Your reply lead my reading to Fermi's Golden rule which is what I was looking for.
vanhees71
1. What is superposition of states?
Superposition of states is a principle in quantum mechanics that states that a quantum system can exist in multiple states at the same time. This means that the system is in a combination of all possible states until it is observed or measured, at which point it will collapse into a single state.
2. How does superposition of states differ from classical physics?
In classical physics, a system can only exist in one state at a time. However, in quantum mechanics, a system can exist in multiple states at the same time, known as superposition. This is one of the key differences between classical and quantum physics.
3. What is the significance of superposition of states in quantum computing?
In quantum computing, the ability to exist in multiple states at the same time allows for the creation of quantum bits, or qubits. These qubits can represent and process much more information than classical bits, making quantum computers potentially much more powerful than classical computers.
4. Can superposition of states be observed in everyday life?
No, superposition of states is a phenomenon that occurs at the quantum level and cannot be observed in everyday life. This is because the act of observing or measuring a system causes it to collapse into a single state, making it impossible to observe superposition in larger, macroscopic objects.
5. How is superposition of states related to the concept of entanglement?
Entanglement is a phenomenon in which two or more particles become connected in such a way that the state of one particle is dependent on the state of the other, even when they are physically separated. Superposition of states is often used to create entangled particles, as the particles must exist in multiple states at the same time in order to become entangled.
Replies
124
Views
6K
Replies
2
Views
1K
Replies
1
Views
1K
Replies
13
Views
2K
Replies
29
Views
3K
Replies
5
Views
3K
Replies
10
Views
3K
Replies
3
Views
6K
Replies
1
Views
1K
Replies
15
Views
3K | 1,162 | 5,496 | {"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} | 3.1875 | 3 | CC-MAIN-2024-38 | latest | en | 0.955557 |
https://gmatclub.com/forum/milankovitch-proposed-in-the-early-twentieth-century-that-153493.html?sort_by_oldest=true | 1,531,979,656,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676590559.95/warc/CC-MAIN-20180719051224-20180719071224-00524.warc.gz | 668,239,310 | 70,749 | GMAT Question of the Day - Daily to your Mailbox; hard ones only
It is currently 18 Jul 2018, 22:54
### GMAT Club Daily Prep
#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.
Customized
for You
we will pick new questions that match your level based on your Timer History
Track
every week, we’ll send you an estimated GMAT score based on your performance
Practice
Pays
we will pick new questions that match your level based on your Timer History
# Milankovitch proposed in the early twentieth century that
Author Message
TAGS:
### Hide Tags
Intern
Joined: 14 May 2013
Posts: 3
Location: Singapore
Concentration: Finance, Social Entrepreneurship
Schools: NUS '16
GMAT 1: 700 Q51 V34
GMAT 2: 700 Q50 V35
GPA: 2.9
WE: Other (Other)
Milankovitch proposed in the early twentieth century that [#permalink]
### Show Tags
29 May 2013, 02:54
2
11
Question 1
00:00
Question Stats:
88% (04:21) correct 12% (16:48) wrong based on 797
### HideShow timer Statistics
Question 2
00:00
Question Stats:
81% (00:31) correct 19% (00:46) wrong based on 777
### HideShow timer Statistics
Question 3
00:00
Question Stats:
55% (00:47) correct 45% (00:58) wrong based on 775
### HideShow timer Statistics
Question 4
00:00
Question Stats:
85% (00:43) correct 15% (00:45) wrong based on 747
### HideShow timer Statistics
Question 5
00:00
Question Stats:
65% (00:48) correct 35% (00:50) wrong based on 756
### HideShow timer Statistics
Question 6
00:00
Question Stats:
93% (00:39) correct 7% (00:18) wrong based on 693
### HideShow timer Statistics
Question 7
00:00
Question Stats:
83% (00:40) correct 17% (00:32) wrong based on 712
### HideShow timer Statistics
Question 8
00:00
Question Stats:
90% (00:34) correct 10% (00:22) wrong based on 673
### HideShow timer Statistics
Question 9
00:00
Question Stats:
82% (00:55) correct 18% (01:41) wrong based on 684
### HideShow timer Statistics
Milankovitch proposed in the early twentieth century that the ice ages were caused by variations in the Earth’s orbit around the Sun. For sometime this theory was considered untestable, largely because there was no sufficiently precise chronology of the ice ages with which the orbital variations could be matched.
To establish such a chronology it is necessary to determine the relative amounts of land ice that existed at various times in the Earth’s past. A recent discovery makes such a determination possible: relative land-ice volume for a given period can be deduced from the ratio of two oxygen isotopes, 16 and 18, found in ocean sediments. Almost all the oxygen in water is oxygen 16, but a few molecules out of every thousand incorporate the heavier isotope 18. When an ice age begins, the continental ice sheets grow, steadily reducing the amount of water evaporated from the ocean that will eventually return to it. Because heavier isotopes tend to be left behind when water evaporates from the ocean surfaces, the remaining ocean water becomes progressively enriched in oxygen 18. The degree of enrichment can be determined by analyzing ocean sediments of the period, because these sediments are composed of calcium carbonate shells of marine organisms, shells that were constructed with oxygen atoms drawn from the surrounding ocean. The higher the ratio of oxygen 18 to oxygen 16 in a sedimentary specimen, the more land ice there was when the sediment was laid down.
As an indicator of shifts in the Earth’s climate, the isotope record has two advantages. First, it is a global record: there is remarkably little variation in isotope ratios in sedimentary specimens taken from different continental locations. Second, it is a more continuous record than that taken from rocks on land. Because of these advantages, sedimentary evidence can be dated with sufficient accuracy by radiometric methods to establish a precise chronology of the ice ages. The dated isotope record shows that the fluctuations in global ice volume over the past several hundred thousand years have a pattern: an ice age occurs roughly once every 100,000 years. These data have established a strong connection between variations in the Earth’s orbit and the periodicity of the ice ages.
However, it is important to note that other factors, such as volcanic particulates or variations in the amount of sunlight received by the Earth, could potentially have affected the climate. The advantage of the Milankovitch theory is that it is testable: changes in the Earth’s orbit can be calculated and dated by applying Newton’s laws of gravity to progressively earlier configurations of the bodies in the solar system. Yet the lack of information about other possible factors affecting global climate does not make them unimportant.
17. In the passage, the author is primarily interested in
(A) suggesting an alternative to an outdated research method.
(B) introducing a new research method that calls an accepted theory into question.
(C) emphasizing the instability of data gathered from the application of a new scientific method.
(D) presenting a theory and describing a new method to test that theory.
(E) initiating a debate about a widely accepted theory.
18. The author of the passage would be most likely to agree with which of the following statements about the Milankovitch theory?
(A) It is the only possible explanation for the ice ages.
(B) It is too limited to provide a plausible explanation for the ice ages, despite recent research findings.
(C) It cannot be tested and confirmed until further research on volcanic activity is done.
(D) It is one plausible explanation, though not the only one, for the ice ages.
(E) It is not a plausible explanation for the ice ages, although it has opened up promising possibilities for future research.
19. It can be inferred from the passage that the isotope record taken from ocean sediments would be less useful to researchers if which of the following were true?
(A) It indicated that lighter isotopes of oxygen predominated at certain times.
(B) It had far more gaps in its sequence than the record taken from rocks on land.
(C) It indicated that climate shifts did not occur every100,000 years.
(D) It indicated that the ratios of oxygen 16 and oxygen18 in ocean water were not consistent with those found in fresh water.
(E) It stretched back for only a million years.
20. According to the passage, which of the following is true of the ratios of oxygen isotopes in ocean sediments?
(A) They indicate that sediments found during an ice age contain more calcium carbonate than sediments formed at other times.
(B) They are less reliable than the evidence from rocks on land in determining the volume of land ice.
(C) They can be used to deduce the relative volume of land ice that was present when the sediment was laid down.
(D) They are more unpredictable during an ice age than in other climatic conditions.
(E) They can be used to determine atmospheric conditions at various times in the past.
21. It can be inferred from the passage that precipitation formed from evaporated ocean water has
(A) the same isotopic ratio as ocean water
(B) less oxygen 18 than does ocean water
(C) less oxygen 18 than has the ice contained in continental ice sheets
(D) a different isotopic composition than has precipitation formed from water on land
(E) more oxygen 16 than has precipitation formed from fresh water
22. According to the passage, which of the following is (are) true of the ice ages?
Ⅰ. The last ice age occurred about 25,000 years ago.
Ⅱ. Ice ages have lasted about 10,000 years for at least the last several hundred thousand years.
Ⅲ. Ice ages have occurred about every 100,000 years for at least the last several hundred thousand years.
(A) Ⅰ only
(B) Ⅱ only
(C) Ⅲ only
(D) Ⅰand II only
(E) Ⅰ,Ⅱ and Ⅲ
23. It can be inferred from the passage that calcium carbonate shells
(A) are not as susceptible to deterioration as rocks.
(B) are less common in sediments formed during an ice age.
(C) are found only in areas that were once covered by land ice.
(D) contain radioactive material that can be used to determine a sediment's isotopic composition.
(E) reflect the isotopic composition of the water at the time the shells were formed.
24. According to the passage, one advantage of studying the isotope record of ocean sediments is that it
(A) corresponds with the record of ice volume taken from rocks on land.
(B)shows little variation in isotope ratios when samples are taken from different continental locations.
(C) corresponds with predictions already made by climatologists and experts in other fields.
(D) confirms the record of ice volume initially established by analyzing variations in volcanic emissions.
(E) provides data that can be used to substantiate records concerning variations in the amount of sunlight received by the Earth.
25. the purpose of the last paragraph of the passage is to
(a) offer a note of caution
(b) introduce new evidence
(c) present two recent discoveries
(d) summarize material in the preceding paragraphs
(e) offer two explanations for a phenomenon
Same passage with just 6 questions from OG: LINK
Manhattan Prep Instructor
Joined: 22 Mar 2011
Posts: 1289
Re: Milankovitch proposed in the early twentieth century that [#permalink]
### Show Tags
30 May 2013, 22:34
It's important to compare the content in the final paragraph to the rest of the passage. It introduces new considerations (volcanic ash, variations in sunlight) that might also explain ice ages. True, it reiterates that Milankovitch's theory is testable, but in no way does it summarize the passage. The paragraph is saying "Here are some other possible explanations. Milankovitch's explanation (orbital variations) is being tested, but that doesn't mean these other factors may not turn out to be important." Therefore, the author is sounding a note of caution. Despite the promising results so far, we can't be sure that this theory is the answer.
_________________
Dmitry Farber | Manhattan GMAT Instructor | New York
Manhattan GMAT Discount | Manhattan GMAT Course Reviews | View Instructor Profile |
Manhattan GMAT Reviews
Senior Manager
Joined: 19 Oct 2012
Posts: 341
Location: India
Concentration: General Management, Operations
GMAT 1: 660 Q47 V35
GMAT 2: 710 Q50 V38
GPA: 3.81
WE: Information Technology (Computer Software)
Re: Milankovitch proposed in the early twentieth century that [#permalink]
### Show Tags
06 Aug 2013, 10:35
9mins; 8/9 correct!! I guess I am ok with my speed!
_________________
Citius, Altius, Fortius
Manager
Joined: 21 Aug 2012
Posts: 170
Concentration: General Management, Operations
Schools: HBS '19 (S)
GMAT 1: 740 Q49 V42
Re: Milankovitch proposed in the early twentieth century that [#permalink]
### Show Tags
06 Aug 2013, 11:19
I chose B for q18 ... can anyone explain why D is the right choice?
Senior Manager
Joined: 19 Oct 2012
Posts: 341
Location: India
Concentration: General Management, Operations
GMAT 1: 660 Q47 V35
GMAT 2: 710 Q50 V38
GPA: 3.81
WE: Information Technology (Computer Software)
Re: Milankovitch proposed in the early twentieth century that [#permalink]
### Show Tags
06 Aug 2013, 11:32
roopika2990 wrote:
I chose B for q18 ... can anyone explain why D is the right choice?
I guess I can help.
The choice B has several issues:
Re-Stating the question and dissecting choices bit by bit.
18. The author of the passage would be most likely to agree with which of the following statements about the Milankovitch theory?
(A) It is the only possible explanation for the ice ages.
No where it is claimed in the passage that theory presented is the only claim. In context of the last para it is clearly implied that it is one of the plausible theory explaining the ice-ages.
(B) It is too limited to provide a plausible explanation for the ice ages, despite recent research findings.
phrase "too limited" makes this option incorrect. Referring back to passage, in particular last line, " Yet the lack of information about other possible factors affecting global climate does not make them unimportant." - implies that author is kinda CONCERN about neglecting these factors even while author is identifying advantages of Milankovitch theory. This choice is too negative and changes the author stand point altogether.
(C) It cannot be tested and confirmed until further research on volcanic activity is done.
Directly contradicting author's view in the last para.
(D) It is one plausible explanation, though not the only one, for the ice ages.
Bingo!!! This basically implies that the presented theory by Milankovitch is one of the possible explanation but not the only one since there are other factors or theories(implied) which might have contributed to the ice age.
(E) It is not a plausible explanation for the ice ages, although it has opened up promising possibilities for future research.
Author concludes the passage indicating skepticism of ignoring factors like volcanic eruptions and duration of sunlight. He never recommends,explicitly, that an extended research is needed to prove something. Hence out!
Hope it helps
_________________
Citius, Altius, Fortius
Current Student
Joined: 25 Sep 2012
Posts: 271
Location: India
Concentration: Strategy, Marketing
GMAT 1: 660 Q49 V31
GMAT 2: 680 Q48 V34
Re: Milankovitch proposed in the early twentieth century that [#permalink]
### Show Tags
09 Aug 2013, 02:14
My first RC in which I got all questions right!
Manager
Joined: 22 Mar 2014
Posts: 144
Location: United States
Concentration: Finance, Operations
GMAT 1: 530 Q45 V20
GPA: 3.91
WE: Information Technology (Computer Software)
Re: Milankovitch proposed in the early twentieth century that [#permalink]
### Show Tags
21 Feb 2017, 12:51
18 mins... all correct.... Not okay with my speed..... I wasted 5 mins on 1 question......
Manager
Joined: 23 Jul 2015
Posts: 162
Re: Milankovitch proposed in the early twentieth century that [#permalink]
### Show Tags
18 Mar 2017, 05:38
Relatively easy passage to understand. took too much time to read the passage
12 mins all correct. Guess I need to improve reading speed
Manager
Joined: 17 Apr 2016
Posts: 99
Re: Milankovitch proposed in the early twentieth century that [#permalink]
### Show Tags
22 Mar 2017, 19:20
Manhattan Prep Instructor
Joined: 22 Mar 2011
Posts: 1289
Re: Milankovitch proposed in the early twentieth century that [#permalink]
### Show Tags
Updated on: 28 Mar 2017, 12:50
1
19 and 21? I'm on it!
19. It can be inferred from the passage that the isotope record taken from ocean sediments would be less useful to researchers if which of the following were true?
(A) It indicated that lighter isotopes of oxygen predominated at certain times.
(B) It had far more gaps in its sequence than the record taken from rocks on land.
(C) It indicated that climate shifts did not occur every 100,000 years.
(D) It indicated that the ratios of oxygen 16 and oxygen18 in ocean water were not consistent with those found in fresh water.
(E) It stretched back for only a million years.
We're looking for something that would make this method less useful to researchers. B is a problem because we're taking away one of the stated advantages of the method. (Specifically, this knocks out the second advantage described in paragraph 3.)
A) The lighter isotope (16) always predominates, so this can't be a problem! (The passage states that only a few molecules out of every thousand have the heavier 18.)
C) This might be a problem for Milankovitch's theory, but not for researchers. The method is still useful, whether it supports the theory or casts doubt on it.
D) If fresh water samples gave us totally different indications of when the ice ages occurred, that would be a problem. However, just seeing variation between fresh and salt water doesn't tell us anything. Maybe the ratios in fresh water are supposed to be different, so we can't really interpret this without more information.
E) A million years is a long time! How do we know it's long enough for what we want? Ice ages appear to occur about once every hundred thousand years, and the author feel comfortable announcing this pattern after looking back a few hundred thousand years, or a few ice ages. By this standard, going back a million years would seem to be more than enough.
_________________
Dmitry Farber | Manhattan GMAT Instructor | New York
Manhattan GMAT Discount | Manhattan GMAT Course Reviews | View Instructor Profile |
Manhattan GMAT Reviews
Originally posted by DmitryFarber on 28 Mar 2017, 12:46.
Last edited by DmitryFarber on 28 Mar 2017, 12:50, edited 1 time in total.
Manhattan Prep Instructor
Joined: 22 Mar 2011
Posts: 1289
Re: Milankovitch proposed in the early twentieth century that [#permalink]
### Show Tags
28 Mar 2017, 12:49
1
21. It can be inferred from the passage that precipitation formed from evaporated ocean water has
(A) the same isotopic ratio as ocean water
(B) less oxygen 18 than does ocean water
(C) less oxygen 18 than has the ice contained in continental ice sheets
(D) a different isotopic composition than has precipitation formed from water on land
(E) more oxygen 16 than has precipitation formed from fresh water
The support for B is right here: "When an ice age begins, the continental ice sheets grow, steadily reducing the amount of water evaporated from the ocean that will eventually return to it. Because heavier isotopes tend to be left behind when water evaporates from the ocean surfaces, the remaining ocean water becomes progressively enriched in oxygen 18."
Translation? When water evaporates, the heavier 18 gets left behind. When that evaporated water gets trapped in ice and doesn't return, the ocean gets steadily higher in oxygen 18. This can only happen if the missing water has less 16 than normal ocean water. (For a real-world connection, compare this to what happens with salt. When ocean water evaporates, the salt is left behind.)
_________________
Dmitry Farber | Manhattan GMAT Instructor | New York
Manhattan GMAT Discount | Manhattan GMAT Course Reviews | View Instructor Profile |
Manhattan GMAT Reviews
Intern
Joined: 23 Mar 2017
Posts: 2
Re: Milankovitch proposed in the early twentieth century that [#permalink]
### Show Tags
27 Apr 2017, 10:30
Took 15 mins, and got only 2 out of the 6 questions right!
I did the question from the OG'17 book. I know that the the online prep at wiley [dot] com also shows the difficulty level (Easy, Medium, Hard).
Does anyone know the difficulty level for these questions? I also shared my answer choices.
Q1: B
Q2: B
Q3: A
Q4: C (yay!)
Q5: C
Q6: E (yay!)
RC takes away much of my time! I get dizzy looking at the passage and the answer choices. I have my GMAT in 4 weeks (+a few days). How do you suggest I get quicker and better in that timeline?
Senior Manager
Joined: 05 Dec 2016
Posts: 259
Concentration: Strategy, Finance
GMAT 1: 620 Q46 V29
Re: Milankovitch proposed in the early twentieth century that [#permalink]
### Show Tags
28 Jun 2017, 05:55
16 minutes, all correct:
4 minutes to read and take notes, 12 minutes to tackle 9 questions.
Lengthy passage but easy questions, wish to face any such on the real test
Intern
Joined: 03 Sep 2017
Posts: 2
Concentration: Finance, General Management
WE: Sales (Manufacturing)
Re: Milankovitch proposed in the early twentieth century that [#permalink]
### Show Tags
06 Sep 2017, 05:40
16 mins all correct. The passage was easy to comprehend. But i think i need to improve my reading speed.
Ideally how much time one should take for a passage of this type?
Manager
Status: Aiming MBA!!
Joined: 19 Aug 2017
Posts: 140
Location: India
GMAT 1: 620 Q49 V25
GPA: 3.75
WE: Web Development (Consulting)
Re: Milankovitch proposed in the early twentieth century that [#permalink]
### Show Tags
31 Oct 2017, 05:47
13:24 secs 7/9
Intern
Joined: 14 Aug 2012
Posts: 42
Re: Milankovitch proposed in the early twentieth century that [#permalink]
### Show Tags
14 Apr 2018, 12:50
1
DmitryFarber wrote:
21. It can be inferred from the passage that precipitation formed from evaporated ocean water has
(A) the same isotopic ratio as ocean water
(B) less oxygen 18 than does ocean water
(C) less oxygen 18 than has the ice contained in continental ice sheets
(D) a different isotopic composition than has precipitation formed from water on land
(E) more oxygen 16 than has precipitation formed from fresh water
The support for B is right here: "When an ice age begins, the continental ice sheets grow, steadily reducing the amount of water evaporated from the ocean that will eventually return to it. Because heavier isotopes tend to be left behind when water evaporates from the ocean surfaces, the remaining ocean water becomes progressively enriched in oxygen 18."
Translation? When water evaporates, the heavier 18 gets left behind. When that evaporated water gets trapped in ice and doesn't return, the ocean gets steadily higher in oxygen 18. This can only happen if the missing water has less 16 than normal ocean water. (For a real-world connection, compare this to what happens with salt. When ocean water evaporates, the salt is left behind.)
Why is C incorrect here? Doesn't the oxygen 18 enriched water eventually turn into ice?
Manhattan Prep Instructor
Joined: 22 Mar 2011
Posts: 1289
Re: Milankovitch proposed in the early twentieth century that [#permalink]
### Show Tags
01 May 2018, 00:00
bpdulog There's no indication that the 18-enriched water turns to ice. It's in the ocean, and the passage doesn't say that the ocean freezes.
However, that doesn't matter anyway. The question is asking us to compare rain from evaporated ocean water with continental ice sheets. That rain is exactly what forms the ice sheets, so they should have roughly the same concentration of oxygen 18 (a very low one, if any at all!).
_________________
Dmitry Farber | Manhattan GMAT Instructor | New York
Manhattan GMAT Discount | Manhattan GMAT Course Reviews | View Instructor Profile |
Manhattan GMAT Reviews
Director
Joined: 23 Sep 2015
Posts: 590
Re: Milankovitch proposed in the early twentieth century that [#permalink]
### Show Tags
11 Jul 2018, 17:16
Is it really a 700 level Passage ??? 13:40 - all correct. but i call it easy cause most of the time i don't need to read it back.
_________________
Thanks!
Do give some kudos.
Simple strategy:
“Once you’ve eliminated the impossible, whatever remains, however improbable, must be the truth.”
Best Gmat Resource:
GmatPrep CR|GmatPrep SC|GmatPrep RC
Re: Milankovitch proposed in the early twentieth century that [#permalink] 11 Jul 2018, 17:16
Display posts from previous: Sort by
# Events & Promotions
Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®. | 5,485 | 23,011 | {"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} | 3.140625 | 3 | CC-MAIN-2018-30 | latest | en | 0.809756 |
http://gmatclub.com/forum/water-is-leaking-out-from-a-cylinder-at-the-rate-of-0-31m-28830.html | 1,427,580,896,000,000,000 | text/html | crawl-data/CC-MAIN-2015-14/segments/1427131297831.4/warc/CC-MAIN-20150323172137-00232-ip-10-168-14-71.ec2.internal.warc.gz | 125,187,702 | 43,655 | Find all School-related info fast with the new School-Specific MBA Forum
It is currently 28 Mar 2015, 14:14
### GMAT Club Daily Prep
#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.
Customized
for You
we will pick new questions that match your level based on your Timer History
Track
every week, we’ll send you an estimated GMAT score based on your performance
Practice
Pays
we will pick new questions that match your level based on your Timer History
# Events & Promotions
###### Events & Promotions in June
Open Detailed Calendar
# Water is leaking out from a cylinder at the rate of 0.31m^3
Author Message
TAGS:
Senior Manager
Joined: 08 Aug 2005
Posts: 251
Followers: 1
Kudos [?]: 17 [0], given: 0
Water is leaking out from a cylinder at the rate of 0.31m^3 [#permalink] 26 Apr 2006, 02:47
Water is leaking out from a cylinder at the rate of 0.31m^3 per minute. 10 minutes later, the water level decreases 0.25 meter, what is the value of the radius of the cylinder?
0.5
1.0
1.5
2.0
2.5
Intern
Joined: 02 Mar 2006
Posts: 21
Followers: 0
Kudos [?]: 1 [0], given: 0
volume emptied per minute = 0.31
for 10 minutes = 3.1
volume of cylinder = 22/7 * r^2 * l
here l = 0.25
22/7 * r^2 * 0.25 = 3.1
solving the above equation we get
r = 2
GMAT Club Legend
Joined: 07 Jul 2004
Posts: 5079
Location: Singapore
Followers: 21
Kudos [?]: 180 [0], given: 0
Water leaking out at 0.31 cubic meter/min
10 mintues --> will have leaked 3.1 cubic meters
Volume of cylinder = pi * r^2 * h
3.1 = pi * r^2 * 0.25
pi*r^2 = 12.4
r = 1.9 (approximately 2)
Intern
Joined: 12 Mar 2006
Posts: 26
Followers: 0
Kudos [?]: 0 [0], given: 0
Yeah, Ans: 2.
Good Q though.
Manager
Joined: 27 Mar 2006
Posts: 136
Followers: 1
Kudos [?]: 1 [0], given: 0
Agree with 2
Manager
Joined: 21 Mar 2006
Posts: 89
Followers: 1
Kudos [?]: 1 [0], given: 0
Guys, this is an excellent question. I got this one on the real thing a couple weeks ago. Be sure to familiarize yourself with it.
VP
Joined: 06 Jun 2004
Posts: 1059
Location: CA
Followers: 2
Kudos [?]: 42 [0], given: 0
ywilfred wrote:
Water leaking out at 0.31 cubic meter/min
10 mintues --> will have leaked 3.1 cubic meters
Volume of cylinder = pi * r^2 * h
3.1 = pi * r^2 * 0.25
pi*r^2 = 12.4
r = 1.9 (approximately 2)
Agree with ywildred's working.
_________________
Don't be afraid to take a flying leap of faith.. If you risk nothing, than you gain nothing...
Manager
Joined: 14 Mar 2006
Posts: 209
Followers: 1
Kudos [?]: 4 [0], given: 0
Charlie45 wrote:
Guys, this is an excellent question. I got this one on the real thing a couple weeks ago. Be sure to familiarize yourself with it.
Exact same question? It is a very good question though, I have having trouble finding L but know I understand. Reminds me of HS math questions. Is there are formula for these type of questions, I mean we have rate of change and time.
Display posts from previous: Sort by | 986 | 3,056 | {"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} | 3.890625 | 4 | CC-MAIN-2015-14 | longest | en | 0.854257 |
https://descheepsjongensvanbontekoe.nl/01235-candlestick-in-binary-trading-strategies/ | 1,586,474,053,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585371880945.85/warc/CC-MAIN-20200409220932-20200410011432-00263.warc.gz | 407,017,141 | 10,065 | ### Candlestick in binary trading strategies
These are shown in the graphic below: This is a strong pattern to trade Binaries. With an 80 payout a binary option trade of 100 risks 100 and wins. Pretty simple, isnt it? When these patterns are seen, traders can enter into PUT options based on these expectations. Automated, trading using, candlestick, charts, candlesticks are not only useful for viewing the markets and getting a quick understanding of price action, they also are easy to incorporate into automated trading systems.
#### Binary Strategy The 5 Most Powerful Candlestick Patterns
The charts make a clear contrast between the real body (between the open and close) and wicks (between the high and low). The shooting star strategy The first strategy is called the shooting star strategy. Video Describing the Trading Strategy and how it can be Backtested Using Excel to Backtest the Binary Option Strategy Microsoft Excel is a very useful tool for backtesting trading strategies. Such as, for example, if the real body is very short and both the upper and the lower shadow are of equal length and are very large, then in this case, the value of an asset is expected to remain constant. They are therefore much easier to program compared to systems that rely on data from many bars. The initial candlestick consists of a tiny white body. Options allow traders to take advantage of margin to make bigger profits and losses they would do by trading the underlying instrument. Your capital may be at risk. Refer to chapter 8 for detailed guidance about how to perform this task. Predicting the Movement of Assets with Candlesticks. By short term we mean anything between 5 minutes to 1 hour. The length between the highest and closing values of a candlestick is referred to as its wick.
The real body, the real body or body of the candlestick is the rectangle in the middle of the candlestick. However, after a few rounds of practice youre set. Formulas: In the video I showed how the rules for this simple candlestick strategy can be programmed into Excel. This position was closed at expiration following the appearance of a bearish engulfing formation and hammer as demonstrated towards the top-right candlestick in binary trading strategies of the next diagram. In this case, if you notice a very short red or green real body and a very long upper shadow you can guess that the value of an asset will begin to increase shortly.
#### Trading Candlestick Formations - Binary Options Strategy 2017
Imagine that the value of an asset is continually candlestick in binary trading strategies increasing. The logic behind this approach comes from the fact that prices are already at extreme lows but markets have snapped back (evidenced by the long lower Hammer wick). You will be able to use the hanging man binary options strategy during a session of trading when the value of an asset is continually decreasing. There are also two considerations you will have to remember. Seeing this, you would have most likely bet on the outcome that the value of the asset would continue to decrease even further. First we must understand the anatomy of the candle. After further 30 minutes, the value of the asset becomes 110 because of the high number of buys. Conversely, longer bodies suggest stronger buying and selling pressure. The ensuing figure presents such a setup.
But one under-utilized aspect of these charts can be seen in the candle formations, which can give strong indications of how prices are likely to move in the future. These strategies will not be accurate for long positions. Using Candle Stick Patterns to Spot Price Reversals From the examples above, we can see that chart candlestick patterns can provide a way to determine potential reversals in prices. Candlesticks are comprised of information explaining the High, Low, Open and Close for the given time period. When prices are showing a strong downtrend, traders can look for bullish trading opportunities once a Hammer formation becomes apparent. Ut wisi enim ad minim veniam, quis nostrud exerci tation ullamcorper suscipit lobortis nisl ut aliquip ex ea commodo consequat. Binary options are comparatively simple way of trading and are ideal to be backtested using Excel.
Next, we look at the candlestick chart as a whole to see how these candles fit into the larger picture: A closer look how candlesticks can help you as a trader. Trade Entry Conditions: Entry conditions are defined as follows. Automatic trading relies on the designer being able to replicate what is happening on the screen into a series of logical steps. For instance, if you have chosen the 1 hour time period, then every candlestick will present the price action that occurred during each successive hourly period. This strategy is used to predict the sudden upward change in the movement of assets. In normal trading, a winning percentage of more than.5 would be easily attainable, however, for binary options the problem is that the trade will expire at a fixed time.
#### Closed captioning work at home work from home singapore forum
For investors, options act as a form of portfolio insurance. The logic behind this approach comes from the fact that prices are already at extreme highs (too expensive) but markets have failed after reaching these heights (evidenced by volatility of the long upper wick). A very important part of learning to use this strategy is to remember the designs of the candlesticks mentioned above. Candlesticks are now the default view in most trading software and glancing at a chart shows why. This can be seen in the graphics below: Trading Binary Options with Candlesticks can be easy. A long red real body means that the value of the asset has decreased a lot in a very short time frame. The initial candlestick consists of a tiny black body.
However, you are welcome to check out our additional guides and articles in order to learn more binary options winning tips and tricks. Hanging man or shooting star patterns may be more profitable. 7.5.3 Bearish Engulfing Pattern This formation comprises of two candlesticks and is a serious sign that a bullish candlestick in binary trading strategies trend could be ending. . You also notice that the real body has a very (!) small or even non-existing upper shadow but a very long lower shadow. When prices are showing a strong uptrend, a bearish reversal pattern can be a good indication that the rally is over and that traders should consider PUT options. When prices are showing a strong downtrend, a bullish reversal candle can help to create solid opportunities for call options. Conversely, when prices are showing a strong uptrend, traders can look for bearish trading opportunities once. Doji body to be a minimum size that can be varied. A chart can consequently exhibit a large number of candlesticks, as illustrated by the following figure: Many prosperous investors utilize candlesticks because they offer the next advantages: Numerous candlestick formations have been defined and have been comprehensively analyzed over extensive time periods. The use of colours to distinguish bull and bear bars makes them easy to identify. . You would have been able to purchase the right binary options contract in this case.
#### Cryptocurrency, brokers the Complete Trading Guide Finance Magnates
If this would be possible, then you would have been able to see that a huge number of people decided candlestick in binary trading strategies to buy after the price dropped, meaning that the price was expected to increase in the future. To win the trader must correctly guess whether the market will be higher or lower than the current price at a set time. What are Candlesticks in, binary, options? 4th candle must be a Doji with a small body. The most popular type of binary option trade is the Higher-Lower trade.
The expiry time selected is the Daily. Use well-proven risk and money management concepts to determine your position size so that you do not risk more than 2 of your entire equity per position. However, its most efficient in binary options trading. For example, a very long line on top means that a very large number of traders have decided to buy the given asset. The most common alternative is white and black where white represents an increase while black represents a decrease. I have set the number of preceding candles. The logic behind this approach comes from the fact that the previously bullish sentiment is now being overshadowed by bearish momentum, and prices are likely to continue lower. Now that you know what candlesticks actually are in binary options and how to read them, we will reveal you how you can use them in order to predict the future movement of an asset. The trading strategy is a reversal strategy. The logic behind this approach comes from the fact that the previously bearish sentiment is overextended and is being overcome by bullish momentum. If you are a newcomer, then you should initially focus on the shooting star strategy only. The profitability of the pattern may be affected by the preceding momentum.
#### 75 Legitimate Work from Home Jobs for 2019 - ivetriedthat
This will lead to the following: The value of the asset will highly likely increase. This type of bet often has a payout around 80 and so the trader must be correct more than.5 of the time in order to be profitable. Open a new call binary option after you identify a bullish engulfing pattern and EMA9 is higher than EMA50. Purchasers of an option have the right to buy or sell the underlying instrument at a certain price before a certain time. I did this using an IF statement The long trades were opened using the following: S67 Short trades were opened using the following: X67 Results Wining Trades 296 Losing Trades 190 Win.9 How to Improve the. Either or both of these could be tweaked. In order to understand how this works, first you will have to know what a candlestick is made. A candlestick can exhibit numerous formations with each one possessing a unique interpretation. In other words, its also likely than in the case you see a pattern like the ones mentioned above, the opposite of the expected outcome will happen. The asset chosen is the USD/CHF. Wed say they are accurate around 75-90 of the time. General Tips and Considerations As you could have noticed above, using this binary options winning strategy is not that complicated after all. A series of famous patterns are now presented.
And its this is how these binary candlestick in binary trading strategies options candlestick strategies work. This information can be critical when looking to establish a trading bias using binary options. There are basically two main strategies that work best. Since prices are likely to continue to move higher, traders can look to establish call options when these patterns become apparent. A red color means that the value of the asset was decreasing.
#### Hoeveel is 1 bitcoin in euro / Forex Trading
Candlesticks are comparatively simple to study and interpret. Candlestick, trading for, binary. The primary indicators are the Bullish and Bearish Engulfing Patterns. A number of famous candlestick patterns were introduced in section. Excel can handle quite a lot of data, in the video above I am testing 100,000 15 minute periods. Short candle bodies indicate restricted price movement and consolidation. But how can we interpret the information given by these charts? Now that we understand how to interpret these charts, we will now look at ways to spot potential reversals in price (which is key for constructing binary options trade ideas ). Different shapes of wicks give the pattern a different look. We could test whether the pattern is more effective in a downtrend or an uptrend.
#### What is the Best Forex Scalping Indicator?
Mehr erklärt, wie die Raten bei verschiedenen Krediten entstehen und auf was es dabei ankommt. Schlagzeilen der letzten Wochen bestätigen diese Vermutung. Another strategy that involves candlesticks is the doji candlestick binary options strategy. Schließlich gibt es auch weniger fröhliche Zeitgenossen, die sich auf den Schlips getreten fühlen könnten. In diesem Zusammenhang entsteht häufig die Frage: Darf der Vermieter die Gebäudeversicherung, auch Wohngebäudeversicherung genannt. Mehr Inflationsgeschützter Vermögensaufbau lässt sich mit Immobilien besonders gut bewerkstelligen. Mehr Dual Sim ermöglicht die Nutzung von zwei Rufnummern mit einem Smartphone. Reichen sie dann alle Bescheinigungen mit ihrer Steuererklärung samt Anlage KAP ein, bekommen sie die bei der anderen Bank abgeführte Abgeltungsteuer zumindest anteilig erstattet. Mehr In der Regel besteht erst nach einem Jahr Ehe ein Anspruch auf Hinterbliebenenrente. Mehr Die Wohngebäudeversicherung tritt in Kraft bei Schäden durch Feuer, Leitungswasser oder Sturm.
#### Steuererklärung für 2014 machen - Tipps zum
Mehr Billigstrom hat trotz des geringeren Preises die gleiche Qualität wie Normalstrom. Es gibt jedoch Ausnahmen. Binary, option, candlestick Trading Strategy, posted on April 25, 2014 August 25, 2016 by Tradinformed This article discusses why candlestick trading is an ideal way to trade binary options. Ein in den letzten Jahren entstandenes, neues Konzept, ist die Investition in Pflegeimmobilien. Mehr Wer seine Rechtsschutzversicherung kündigen möchte, hat je nach Umstand verschiedene Möglichkeiten die Kündigung der Rechtsschutzversicherung zu vollziehen. Mehr Unser Rechtsexperte hat sich mal die Mühe gemacht und alle Rechtsänderungen bezüglich Finanzfragen zum Beginn des neuen Jahres 2015 aufgelistet. Unter anderem basierend auf der gesetzlichen Regelung, dass in Deutschland niemand mehr ohne Versicherungsschutz sein darf. Dies ist grundsätzlich nicht nur für den Kauf von Indexfonds notwendig, sondern Sie benötigen ein derartiges Depotkonto immer dann, wenn Sie Aktien, Anleihen oder Fonds erwerben möchten. Falls nicht, geben auch die zuständigen Gerichte für den Firmensitz oder die entsprechende Industrie- und Handelskammer Auskunft. Wir erklären, was dabei zu beachten ist.
Mehr Einmal im Jahr erhalten Sie von Ihrem Gasanbieter eine Jahresabrechnung für Ihren Gasverbrauch. Wer oben ohne Speed gibt, zahlt die Hälfte selbst mehr Radfahren ohne Helm kann auch bei einem unverschuldeten Unfall zur Mitschuld führen mehr In der Regel entstehen durch Festgeldkonten keine Kosten. Verluste aus Aktienverkäufen können steuerlich nur mit Gewinnen aus verkauften Aktien ausgeglichen werden. Unser Rechtsexperte beantwortet einige häufige Fragen an dieser Stelle. Mehr Ein DSL Wechsel kann einmalige Kosten verursachen. Their huge popularity has lowered reliability because. Mehr Günstige Onlinekredite finden mit Hilfe von.
If a near percent. 01 ISK-ing Once you get into trading, you how aetna jobs work at home pharmacists to remove a directorymand line…..
Elliott wave analysis can help you to understand market cycles and sometimes predict whats going to happen next. The primary trend is bearish and…..
Forex pro usd inr
Indian Malik 1 hour ago. United States Dollar vs Indian R Chart - usdinr advfn. ForexTrader UsdInr 1 hour ago. Today ( Tuesday…
Bitcoin eos bitcointalk
Clean in ether in technology form of monthly Channels. A how to calculate risk in forex n?met thread a t?maspecifikus child board-okkal. EOS is latest family's garden…
Mit forex reich werden
Preisliste: Foto im Passepartout, preisliste: Holz, preisliste: Alu-Dibond. Fotopuzzle Mousepad Handyh?lle Preisliste: Foto im Rahmen. Classic Tasse, premium Tasse, magic Mug, bunte Tasse. Quadratisch, digital, klassisch, panorama, preisliste…
Bitfinex cryptocurrency exchange was founded in forex asli rusia 2012 and has its headquarters in Hong Kong. All cryptocurrencies follow the same general system. Since…
Non phone work from home jobs part time
Convergys hires virtual workers for sales and service, customer service, and tech support. Must have at least 1 year of sales experience and six months customer… | 3,407 | 16,029 | {"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-2020-16 | latest | en | 0.928053 |
https://management.ind.in/forum/union-bank-india-clerk-exam-previous-year-question-papers-pdf-format-47155.html | 1,628,196,033,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046157039.99/warc/CC-MAIN-20210805193327-20210805223327-00258.warc.gz | 390,474,328 | 21,194 | 2021 2022 Student Forum Union Bank of India clerk exam previous year question papers in PDF format
#1
31st July 2014, 01:24 PM
Unregistered Guest
Union Bank of India clerk exam previous year question papers in PDF format
Will you please share with me the Union Bank of India clerk exam previous year question papers in PDF format?
#2
1st August 2014, 11:08 AM
Super Moderator Join Date: Apr 2013
Re: Union Bank of India clerk exam previous year question papers in PDF format
As you want to get the Union Bank of India clerk exam previous year question papers in PDF format so here it is for you:
1. How many such pairs of letter are there in the word FREQUENT, each of which has as many letters between them in the word (in both forward and backward directions) as they have between in the English alphabetical series?
(1) None
(2) One
(3) Two
(4) Three
(5) More than three
2. In a class of 35 children, Ameya’s rank sixth from the top. Annie is seven ranks below Amerya. What is Annie’s rank for the bottom?
(1) 22
(2) 20
(3) 19
(4) 23
(5) Cannot be determined
3. Four of the following five are alike in a certain way and so form a group. Which is one that does not belong to that group?
(1) Lens
(2) Shutter
(3) Film
(4) Camera
(5) Zoom
4. What will come in place of the question mark (?) in the following series? AB BE DH ? KQ
(1) HL
(2) GL
(3) GK
(4) EI
(5) IM
5. If two is subtracted from each odd digit and if two is added to each even digit in the number 9275436, what will be the difference between the digits which are third from the right and second from the left of the new number thus formed?
(1) 6
(2) 8
(3) 2
(4) 1
(5) 5
6. The position of how many alphabets will remain unchanged if each of the alphabets in the word FORGET is arranged in alphabetical order from left to right?
(1) None
(2) One
(3) Two
(4) Three
(5) More than three
7. Which of the following groups of alphabets should replace the blank space so that the group of alphabets, given in bold, follow a logical pattern from the preceding and the following group of alphabets?
a b _ y
a b c _ x w
a b c d e v _
(1) z, d, u
(2) d, x, u
(3) c, d, u
(4) z, y, w
(5) c, d, w
8. In which of the following expressions will the expression ‘P < F’ be definitely false?
(1) F = B > P M
(2) P > B M = F
(3) P B < F M
(4) B < P M < F
(5) None of these
If ‘A × B’ means A is the son of B.
If ‘A + B’ means A is the father of B.
If ‘A > B’ means A is the daughter of B.
If ‘A < B’ means A is the wife of B.
9. Which of the following pairs of people represent first cousins with regard to the relations given in the expressions, if it is provided that A is the sister of J: ‘L > V < J + P’ and ‘S × A < D + F < E + K’
(1) LP
(2) SP
(3) SK
(4) SF
(5) cannot be determined
10. What will come in the place of the question mark, if it is provided that M is the grandmother of F in the expression: ‘F × R < S ? M’
(1) >
(2) <
(3) +
(4) ×
(5) cannot be determined
Attached Files
Union Bank of India clerk exam previous year question papers in PDF format.zip (498.2 KB, 48 views)
#3
24th March 2015, 11:01 AM
Unregistered Guest
Re: Union Bank of India clerk exam previous year question papers in PDF format
Please provide me Previous Year Question Papers of Union Bank of India Clerk Exam for preparation of Exam?
#4
24th March 2015, 11:07 AM
Super Moderator Join Date: Apr 2013
Re: Union Bank of India clerk exam previous year question papers in PDF format
Here I am providing you Previous Year Question Papers of Union Bank of India Clerk Exam for preparation of your Exam.
1. You can keep your personal files/folders in—
(1) My folder
(2) My Document
(3) My Files
(4) My Text
(5) None of these
2. The primary purpose of software is to turn data into—
(1) Wed sites
(2) information
(3) programs
(4) objects
(5) None of these
3. A directory within a directory is called—
(1) Mini Directory
(2) Junior Directory
(3) Part Directory
(4) Sub Directory
(5) None of these
4. A complier translates a program written in a high-level language into—
(1) Machine language
(2) An algorithm
(3) A debugged program
(4) Java
(5) None o these
5. When you turn on the computer, the boot routine will perform this test—
(1) RAM test
(2) disk drive test
(3) memory test
(4) power-on self-test
(5) None of these
6. A _______ is a unique name that you give to a file information—
(1) device letter
(2) folder
(3) filename
(4) filename extension
(5) None of these
7. Hardware includes ________.
(1) all devices used to input data into a computer
(2) sets of instructions that a computer runs or executes
(3) the computer and all the devices connected to it that used to input and output data
(4) all devices involved in processing information including the central processing unit, memory, and storage
(5) None of these
8. A _______ contains specific rules and words that express the logical steps of an algorithm.
(1) programming language
(2) syntax
(3) programming structure
(4) logic chart
(5) None of these
9. All the deleted files go to—
(1) Recycle Bin
(3) Tool Bar
(4) My Computer
(5) None of these
10. The simultaneous processing of two or more programs by multiple processors is—
(1) multiprogramming
(3) time-sharing
(4) multiprocessing
(5) None of these
1. In a certain code MAIN is written as '9364' and DEAR is written as '8532'. How is MEND written in that code ?
(1)9548 (2)9458
(3) 9538 (4) 9528
(5) None of these
2. In a certain code DREAMING is written as BFSEFMHL. How is TREATISE written in that code ?
(l)USFBDRHS
(2) BFSUDRHS
(3) BFSUSHRD
(4) BDQSDRHS
(5) None of these
3. The positions of how many digits in the number 5314697 will remain unchanged if the digits are rearranged in ascending order within the number ?
(l)None (2) One
(3) Two (4) Three (5) More than three
4. Among A, B, C, D and E each having different amount of money, C has more money than only E. B has more money than D out less than A. Who among them has the highest amount of money?
(1)B (2)A (3)D (4) Data inadequate (5) None of these
5. Prakash walked 30 metres towards West, took a left turn and walked 20 metres. He again took a left tum and walked 30 metres. He then took a right turn and stopped. Towards which direction was he facing when he stopped ?
(1) South
(2) North
(3) East
(5) None of these
6. How many meaningful English words can be made with the letters RTOU using each letter only once in each word ?
(l)None (2) One
(3) Two (4) Three
(5) More than three
7. If 'P' denotes '-': 'Q' denotes '÷', 'R' denotes 'x' and 'W denotes'+' then —
48 Q 12 R 10 P 8 W 4 = ?
(1)56 (2)40
(3) 52 (4) 44
(5) None of these
8. How many such pairs of letters are there in the word ENGULFED each of which has as many letters between them in the word as in the English aiphabet ? (l)None (2) One
(3) Two (4) Three
(5) More than three
9. Which of the following is the middle digit of the second highest number among the five three digit numbers given below ? 512 739 428 843 654
(1) 1 (2) 3
(3) 2 (4) 4
(5) 5
10. In a certain code language 'green grass everywhere' is written as 'dik pa sok' and 'cow eats grass' is written as 'nok ta pa'. How is 'cow' written in that code language ?
(l)nok (2)ta
(3) nok or ta
(5) None of these
Directions (11- 16): Study the following arrangement carefully and answer the questions given below :
11. How many such consonants are there in the above arrangement, each of which is immediately followed by a number and not immediately preceded by a number ?
(l)None (2) One
(3) Two (4) Three
(5) More than three
12. How many such numbers ar there in the above arrangemen each of which is immediately pn ceded by a symbol and immed ately followed by a consonant ? (l)None (2) One
(3) Two (4) Three
(5) More than three
13. How many such vowels are thei in the above arrangement, eac of which is immediately precede by a number but not immediate followed by a number ? (l)None (2) One
(3) Two (4) Three
(5) Four
14. If all the Symbols are droppe from the above arrangemen which of the following will be Ü eleventh from the left end ?
(1)P (2)1
(3) 2 (4) B
(5) None of these
15. Four of the following five are alil in a certain way based on the positions in the above arrang ment and so form a group. Whic is the one that does not belor to that group ?
(l)KP© (2)UQ★ (3) 9 E F (4) I B 7 (5) R 1 %
16. Which of the following is tf eighth to the right of the twent eth from the right end of tl above arrangement ?
(1)6 (2)%
(3) 8 (4) A
(5) None of these
Directions (17 - 22): In each i the questions below are given thre Statements followed by two conclusior numbered 1 and II. You have to tal< the given Statements to be true even they seem to be at variance from con monly known facts. Read all the coi clusions and then decide which of the
given conclusions logically follows from the glven Statements disregarding commonly known facts.
Give ans wer (1) if only Conclusion I follows.
Give answer (2) if only Conclusion II follows.
Give answer (3) if either Conclusion I or II follows.
Give answer (4) if neither Conclusion I nor II follows.
Give answer (5) if both Conclusions I and II follow.
17. Statements:
Some trees are forests.
Some forests are houses.
Some houses are tents.
Conclusions:
I. Some tents are forests.
II. Some houses are trees.
18. Statements:
All cards are boxes.
No box is slate.
Some slates are tiles.
Conclusions:
I. No slate is card.
II. Some tiles are boxes.
19. Statements :
Some papers are arrows.
All arrows are sticks.
Some sticks are boards.
Conclusions:
I. Some boards are papers.
II. No board is paper.
20. Statements:
All ropes are tiles.
Some tiles are bangles.
All bangles are nails.
Conclusions:
I. Some nails are ropes.
II. Some nails are tiles.
21. Statements :
Some days are nights.
All nights are stars.
All stars are clouds.
Conclusions:
I. Some clouds are days.
II. Some stars are days.
22. Statements:
All bells are hammers.
All hammers are dogs.
All dogs are packets.
Conclusions :
I. Some packets are hammers.
II. Some dogs are bells.
Directions (23 - 28) : Study the following Information carefully and answer the questions giyen below :
P, Q. R. S, T, V. W and Z are sitting around a circk facing the Center. R is second to the left of Z, who is third to the left of P. T is third to the right of W who is not an immediate neighbour of eithsr^or Z- S is fourth to the right of Z^Q is fourth to the right of T.
23. In which of the following pairs is the first person sitting to the immediate left of the second person ?
(1)RV (2) ZV
(3) WQ (4) SP
(5) None of these
24. In which of the following combinations is the first person sitting in between the second and the third persons ?
(l)TRV (2) PST
(3) WPQ (4)QZV (5) Data inadequate
25. Who is second to the right of S ? (1)T (2)V
(3)W
(5) None of these
26. What is S's position with respect toW?
(1) Third to the right
(2) Third to the left
(3) Fourth to the right
(4) Fourth to the left
(5) Second to the right
27. Who is to the immediate left of T?
(1)S i2)P
(3) R
(5) None of these
28. Who is to the immediate right of 0?
(DP (2)W
(3)Z (4)V
(5) None of these
Directions (29-34) : In the following questions, the Symbols \$, %, ★,δ and # are used with the following meaning as illustrated below:
P % Q' means 'P is not smaller than Q '.
'P # Q' means 'P is not greater than Q'.
'P \$ Q' means 'P is neither smaller than nor equal to Q'.
'P Q' means 'P is neither greater than nor equal to Q'.
★'P δ means ‘P is neilher greater than nor smaller than Q'.
Now in each following questions assuming the given Statements to be true, find which of the two conclusions I and II given below them is/ are definitely true ?
Give answer (1) if only Conclusion I is true.
Give answer (2) if only Conclusion II is true.
Give answer (3) if either Conclusion I or II is true.
Give answer (4) if neither Conclusion I nor II is true. •
Give answer (5) if both Conclusions I and II are true.
29. Statements:
H % R,R # K, K \$ B Conclusions: I. B ★ R
II. K\$H
30. Statements:
N % F, F# H, H ★ U
Conclusions: I. U\$F
II. I1%N
31. Statemen
ts:K\$R, R δ W, W # T
Conclusions: I. T % R
II. KW
32. Statements:
\$K★T. T \$ M, M δ J
Conclusions : I. J★K
II. J ★T
33. Statements:
D # R, R 5 M, M \$ B
Conclusions : I. M 5 D
II. M \$ D
34. Statements:
B 6 V, V%M. M\$J
Conclusions: I. J ★ V
II. M # B
Detailed Question Papers of Union Bank of India Clerk Exam:
Attached Files
Union Bank of India Clerk Exam Answer.pdf (1.84 MB, 84 views) Union Bank of India Clerk Exam Paper.pdf (1.05 MB, 148 views)
Message:
Options | 3,726 | 12,471 | {"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-2021-31 | latest | en | 0.919098 |
http://makezine.com/images/13/poisonpills.html | 1,369,310,149,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368703298047/warc/CC-MAIN-20130516112138-00068-ip-10-60-113-184.ec2.internal.warc.gz | 161,324,064 | 1,824 | Choose any eight of the pills and put four of them on each side of the balance.
There are two possibilities:
1. One side of the balance comes out lighter. In this case, you know that the abnormal (safe) pill is one of the pills already on the balance. Label the pills on the lighter side A B C and D, and the pills on the heavier side E F G and H. Label the pills not on the balance NORM (you know they're normal pills).
2. The balance is even. In this case, you know that the abnormal (safe) pill is one of the pills not on the balance. Label the pills already on the balance NORM, and label the four pills not on the balance I J K and L.
Let's proceed with possibility 1.
Consider why the side ABCD came out higher than the side EFGH. this could be because:
• A is the abnormal pill, and it's lighter than the other pills.
• B is the abnormal pill, and it's lighter than the other pills.
• C is the abnormal pill, and it's lighter than the other pills.
• D is the abnormal pill, and it's lighter than the other pills.
• E is the abnormal pill, and it's heavier than the other pills.
• F is the abnormal pill, and it's heavier than the other pills.
• G is the abnormal pill, and it's heavier than the other pills.
• H is the abnormal pill, and it's heavier than the other pills.
Now let's make another weighing, with two of the ABCD pills on either side, and one of the EFGH pills on either side. For example, let's weigh ABE versus CDF. How would this weighing come out given each of those 8 possibilities we just listed?
• If A is the light pill, the ABE/CDF weighing will come out with ABE high.
• If B is the light pill, the ABE/CDF weighing will come out with ABE high.
• If C is the light pill, the ABE/CDF weighing will come out with ABE low.
• If D is the light pill, the ABE/CDF weighing will come out with ABE low.
• If E is the heavy pill, the ABE/CDF weighing will come out with ABE low.
• If F is the heavy pill, the ABE/CDF weighing will come out with ABE high.
• If G is the heavy pill, the ABE/CDF weighing will come out even.
• If H is the heavy pill, the ABE/CDF weighing will come out even.
OK, so we observe how the ABE versus CDF weighing actually comes out.
1. If it comes out even, then we know that the abnormal pill is either G or H. For our third weighing, we can weigh G against one of the pills we already know to be normal (one of the pills we labelled NORM). If it comes out even, then G is normal and H must be the abnormal pill. If it comes out uneven, then G is the abnormal pill.
2. As we can see from the data above, if the ABE/CDF weighing comes out with ABE high, then the situation is either: A is the light pill, B is the light pill, or F is the heavy pill.
3. As we can see from the data above, if the ABE/CDF weighing comes out with ABE low, then the situation is either: C is the light pill, D is the light pill, or E is heavy pill.
So in either situation (b) or (c), we have two possible light pills and one possible heavy pill. What we do in that case is we put one of the possible light pills and the possible heavy pill on one side of the scale, and two NORM pills on the other side of the scale. This is our third weighing. If it comes out even, then we know that the other possible light pill is the abnormal pill. If it comes out with the two NORM pills high, then we know that one of the pills on the other side is abnormally heavy, so we know that the possible heavy pill is the culprit. If it comes out with the two NORM pills low, then we know that one of the pills on the other side is abnormally light, so we know that the possible light pill on the scale is the culprit.
That takes care of case (1), where the first weighing came out uneven.
What about case (2), where the first weighing comes out even?
Then we know the abnormal pill is one of I J K or L, and we have two weighings to find the abnormal pill in.
For our second weighing, we put I and J on one side of the scale, and two NORM pills on the other.
1. If this comes out uneven, we know the abnormal pill is I or J; we weigh I against one NORM pill to see if I is abnormal and if it isn't, we can conclude that J is the abnormal pill.
2. If the IJ versus 2 NORM weighing comes out even, we know the abnormal pill is K or L; we weight K against one NORM pill to see if K is abnormal and if it isn't, we can conclude that L is the abnormal pill. | 1,102 | 4,377 | {"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-2013-20 | latest | en | 0.946819 |
https://www.physicsforums.com/threads/total-differential.749245/ | 1,508,318,947,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187822851.65/warc/CC-MAIN-20171018085500-20171018105500-00586.warc.gz | 978,839,259 | 18,262 | # Total differential
1. Apr 17, 2014
### chemphys1
1. The problem statement, all variables and given/known data
If z(x,y) = f(x/y)
show that
x(∂z/∂x)y + y(∂z/∂y)x = 0
2. Relevant equations
so I understand z(x,y)
means I can write
dz = (∂z/∂x)ydx + (∂z/∂y)x dy
I do not understand the = f(x/y) bit though?
does that mean this?
df= (∂f/∂x)y dx+ (∂f/∂y)x dy
and (∂f/∂x)y = -y/x2 (∂f/∂y)x = 1/x
although that seems wrong can't manipulate to get the answer
any help on the method or explaining f(x/y) equalling z(x,y) would be appreciated. maths is not my strongest so if possible go as basic as it comes
2. Apr 17, 2014
### LCKurtz
That isn't true. Try it for z = f(x/y) = x/y. Generally speaking, if everything is continuous, you would have $z_{xy} = z_{yx}$ and you wouldn't expect to multiply one by x and the other by y and have them be equal with opposite signs. Did you copy the problem correctly, parentheses and all?
3. Apr 17, 2014
### chemphys1
complete question is attached, but the information in the original post is correctly copied as far as I can see
#### Attached Files:
• ###### question.png
File size:
14.6 KB
Views:
66
4. Apr 17, 2014
### LCKurtz
Like I said, it is false without more context. Check my example yourself.
5. Apr 17, 2014
### chemphys1
Sorry, I really do not follow
I find it hard to understand mathematical notation, so re: z = f(x/y) = x/y
I can't see how to check the example
6. Apr 17, 2014
### CAF123
You are familiar with the notation f(x) as describing a function of x. Similarly, f(x/y) just means you have some function in the variable x/y. You can instead consider u=x/y and then you are back to the more familiar f(u).
But with LCKurtz's example of f(x/y)=x/y, I get the equation to be satisfied. I also get it to be satisfied in general. I think LCKurtz simply misread the notation, that's all.
7. Apr 17, 2014
### LCKurtz
Are you saying that $\left(\frac {\partial z}{\partial x}\right)_y$ means something other than $\frac \partial {\partial y}\left (\frac{\partial z}{\partial x}\right )$ or, as I wrote $z_{xy}$?
8. Apr 18, 2014
### CAF123
Yes, I took $\left(\frac{\partial z}{\partial x}\right)_y$ to mean, say, differentiate z wrt x, keeping y held fixed. Similarly for the other case. I actually thought that was a standard notation, although I have seen cases where they simply suppress the variable being held constant because in a sense it is obvious from the problem.
9. Apr 18, 2014
### LCKurtz
Well, that's a new one on me. In over 40 years of teaching calculus using many different texts, I never encountered that notation. | 789 | 2,616 | {"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} | 3.59375 | 4 | CC-MAIN-2017-43 | longest | en | 0.919864 |
http://toccata.lri.fr/gallery/mergesort_list.en.html | 1,553,409,874,000,000,000 | text/html | crawl-data/CC-MAIN-2019-13/segments/1552912203378.92/warc/CC-MAIN-20190324063449-20190324085449-00354.warc.gz | 206,094,380 | 5,167 | ## Mergesort (lists)
Sorting lists using mergesort.
Authors: Jean-Christophe Filliâtre
Tools: Why3
see also the index (by topic, by tool, by reference, by year)
# Sorting lists using mergesort
Author: Jean-Christophe FilliĆ¢tre (CNRS)
```module Elt
use export int.Int
use export list.List
use export list.Length
use export list.Append
use export list.Permut
type elt
val predicate le elt elt
clone relations.TotalPreOrder
with type t = elt, predicate rel = le, axiom .
clone export list.Sorted with
type t = elt, predicate le = le, goal Transitive.Trans
end
```
recursive (and naive) merging of two sorted lists
```module Merge (* : MergeSpec *)
clone export Elt with axiom .
let rec merge (l1 l2: list elt) : list elt
requires { sorted l1 /\ sorted l2 }
ensures { sorted result }
ensures { permut result (l1 ++ l2) }
variant { length l1 + length l2 }
= match l1, l2 with
| Nil, _ -> l2
| _, Nil -> l1
| Cons x1 r1, Cons x2 r2 ->
if le x1 x2 then Cons x1 (merge r1 l2) else Cons x2 (merge l1 r2)
end
end
```
tail recursive implementation
```module EfficientMerge (* : MergeSpec *)
clone export Elt with axiom .
use list.Mem
use list.Reverse
use list.RevAppend
lemma sorted_reverse_cons:
forall acc x1. sorted (reverse acc) ->
(forall x. mem x acc -> le x x1) -> sorted (reverse (Cons x1 acc))
let rec merge_aux (acc l1 l2: list elt) : list elt
requires { sorted (reverse acc) /\ sorted l1 /\ sorted l2 }
requires { forall x y: elt. mem x acc -> mem y l1 -> le x y }
requires { forall x y: elt. mem x acc -> mem y l2 -> le x y }
ensures { sorted result }
ensures { permut result (acc ++ l1 ++ l2) }
variant { length l1 + length l2 }
= match l1, l2 with
| Nil, _ -> rev_append acc l2
| _, Nil -> rev_append acc l1
| Cons x1 r1, Cons x2 r2 ->
if le x1 x2 then merge_aux (Cons x1 acc) r1 l2
else merge_aux (Cons x2 acc) l1 r2
end
let merge (l1 l2: list elt) : list elt
requires { sorted l1 /\ sorted l2 }
ensures { sorted result /\ permut result (l1 ++ l2) }
=
merge_aux Nil l1 l2
end
```
Mergesort.
This implementation splits the input list in two according to even- and odd-order elements (see function `split` below). Thus it is not stable. For a stable implementation, see below module `OCamlMergesort`.
```module Mergesort
clone Merge (* or EfficientMerge *) with axiom .
let split (l0: list 'a) : (list 'a, list 'a)
requires { length l0 >= 2 }
ensures { let l1, l2 = result in
1 <= length l1 /\ 1 <= length l2 /\ permut l0 (l1 ++ l2) }
= let rec split_aux (l1 l2 l: list 'a) : (list 'a, list 'a)
requires { length l2 = length l1 \/ length l2 = length l1 + 1 }
ensures { let r1, r2 = result in
(length r2 = length r1 \/ length r2 = length r1 + 1) /\
permut (r1 ++ r2) (l1 ++ (l2 ++ l)) }
variant { length l }
= match l with
| Nil -> l1, l2
| Cons x r -> split_aux l2 (Cons x l1) r
end
in
split_aux Nil Nil l0
let rec mergesort (l: list elt) : list elt
ensures { sorted result /\ permut result l }
variant { length l }
= match l with
| Nil | Cons _ Nil -> l
| _ -> let l1, l2 = split l in merge (mergesort l1) (mergesort l2)
end
end
```
## OCaml's List.sort
There are several ideas here:
- instead of splitting the list in two, sort takes the length of the prefix to be sorted; hence there is nothing to do to get the first half and the second half is obtained with chop (which does not allocate at all)
- all functions are tail recursive. To avoid the extra cost of List.rev, sort is duplicated in two versions that respectively sort in order and in reverse order (`sort` and `sort_rev`) and merge is duplicated as well (`rev_merge` and `rev_merge_rev`).
Note: this is a stable sort, but stability is not proved here.
```module OCamlMergesort
clone export Elt with axiom .
use list.Mem
use list.Reverse
use list.RevAppend
lemma sorted_reverse_cons:
forall acc x1. sorted (reverse acc) ->
(forall x. mem x acc -> le x x1) -> sorted (reverse (Cons x1 acc))
lemma sorted_rev_append: forall acc l: list elt.
sorted (reverse acc) ->
sorted l ->
(forall x y. mem x acc -> mem y l -> le x y) ->
sorted (reverse (rev_append l acc))
let rec rev_merge (l1 l2 accu: list elt) : list elt
requires { sorted (reverse accu) /\ sorted l1 /\ sorted l2 }
requires { forall x y: elt. mem x accu -> mem y l1 -> le x y }
requires { forall x y: elt. mem x accu -> mem y l2 -> le x y }
ensures { sorted (reverse result) }
ensures { permut result (accu ++ l1 ++ l2) }
variant { length l1 + length l2 }
= match l1, l2 with
| Nil, _ -> rev_append l2 accu
| _, Nil -> rev_append l1 accu
| Cons h1 t1, Cons h2 t2 ->
if le h1 h2 then rev_merge t1 l2 (Cons h1 accu)
else rev_merge l1 t2 (Cons h2 accu)
end
lemma sorted_reverse_mem:
forall x l. sorted (reverse (Cons x l)) -> forall y. mem y l -> le y x
lemma sorted_reverse_cons2:
forall x l. sorted (reverse (Cons x l)) -> sorted (reverse l)
let rec rev_merge_rev (l1 l2 accu: list elt) : list elt
requires { sorted accu /\ sorted (reverse l1) /\ sorted (reverse l2) }
requires { forall x y: elt. mem x accu -> mem y l1 -> le y x }
requires { forall x y: elt. mem x accu -> mem y l2 -> le y x }
ensures { sorted result }
ensures { permut result (accu ++ l1 ++ l2) }
variant { length l1 + length l2 }
= match l1, l2 with
| Nil, _ -> rev_append l2 accu
| _, Nil -> rev_append l1 accu
| Cons h1 t1, Cons h2 t2 ->
if not (le h1 h2) then rev_merge_rev t1 l2 (Cons h1 accu)
else rev_merge_rev l1 t2 (Cons h2 accu)
end
function prefix int (list 'a) : list 'a
axiom prefix_def1:
forall l: list 'a. prefix 0 l = Nil
axiom prefix_def2:
forall n: int, x: 'a, l: list 'a. n > 0 ->
prefix n (Cons x l) = Cons x (prefix (n-1) l)
let rec lemma prefix_length (n: int) (l: list 'a)
requires { 0 <= n <= length l } ensures { length (prefix n l) = n }
variant { n } =
if n > 0 then match l with Nil -> () | Cons _ r -> prefix_length (n-1) r end
let rec lemma prefix_append (n: int) (l1 l2: list 'a)
requires { length l1 <= n <= length l1 + length l2 }
ensures { prefix n (l1 ++ l2) =
prefix (length l1) l1 ++ prefix (n - length l1) l2 }
variant { l1 }
= match l1 with Nil -> () | Cons _ t -> prefix_append (n-1) t l2 end
let rec chop (n: int) (l: list 'a) : list 'a
requires { 0 <= n <= length l }
ensures { l = prefix n l ++ result }
variant { n }
=
if n = 0 then l else
match l with
| Cons _ t -> chop (n-1) t
| Nil -> absurd
end
```
`sort n l` sorts `prefix n l` and `rev_sort n l` sorts `prefix n l` in reverse order.
``` use mach.int.Int
let rec sort (n: int) (l: list elt) : list elt
requires { 2 <= n <= length l }
ensures { sorted result }
ensures { permut result (prefix n l) }
variant { n }
=
if n = 2 then match l with
| Cons x1 (Cons x2 _) ->
if le x1 x2 then Cons x1 (Cons x2 Nil) else Cons x2 (Cons x1 Nil)
| _ -> absurd
end else if n = 3 then match l with
| Cons x1 (Cons x2 (Cons x3 _)) ->
if le x1 x2 then
if le x2 x3 then Cons x1 (Cons x2 (Cons x3 Nil))
else if le x1 x3 then Cons x1 (Cons x3 (Cons x2 Nil))
else Cons x3 (Cons x1 (Cons x2 Nil))
else
if le x1 x3 then Cons x2 (Cons x1 (Cons x3 Nil))
else if le x2 x3 then Cons x2 (Cons x3 (Cons x1 Nil))
else Cons x3 (Cons x2 (Cons x1 Nil))
| _ -> absurd
end else begin
let n1 = n / 2 in
let n2 = n - n1 in
let l2 = chop n1 l in
assert { prefix n1 l ++ prefix n2 l2 = prefix n l };
let s1 = rev_sort n1 l in
let s2 = rev_sort n2 l2 in
rev_merge_rev s1 s2 Nil
end
with rev_sort (n: int) (l: list elt) : list elt
requires { 2 <= n <= length l }
ensures { sorted (reverse result) }
ensures { permut result (prefix n l) }
variant { n }
=
if n = 2 then match l with
| Cons x1 (Cons x2 _) ->
if not (le x1 x2) then Cons x1 (Cons x2 Nil)
else Cons x2 (Cons x1 Nil)
| _ -> absurd
end else if n = 3 then match l with
| Cons x1 (Cons x2 (Cons x3 _)) ->
if not (le x1 x2) then
if not (le x2 x3) then Cons x1 (Cons x2 (Cons x3 Nil))
else if not (le x1 x3) then Cons x1 (Cons x3 (Cons x2 Nil))
else Cons x3 (Cons x1 (Cons x2 Nil))
else
if not (le x1 x3) then Cons x2 (Cons x1 (Cons x3 Nil))
else if not (le x2 x3) then Cons x2 (Cons x3 (Cons x1 Nil))
else Cons x3 (Cons x2 (Cons x1 Nil))
| _ -> absurd
end else begin
let n1 = n / 2 in
let n2 = n - n1 in
let l2 = chop n1 l in
assert { prefix n1 l ++ prefix n2 l2 = prefix n l };
let s1 = sort n1 l in
let s2 = sort n2 l2 in
rev_merge s1 s2 Nil
end
lemma permut_prefix: forall l: list elt. permut (prefix (length l) l) l
let mergesort (l: list elt) : list elt
ensures { sorted result /\ permut result l }
=
let n = length l in
if n < 2 then begin
assert { sorted l by match l with
Nil | Cons _ Nil -> sorted l | _ -> false end }; l end
else sort n l
end
``` | 2,872 | 8,549 | {"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} | 2.84375 | 3 | CC-MAIN-2019-13 | latest | en | 0.437548 |
https://www.yaclass.in/p/mathematics-state-board/class-8/rational-numbers-2788/four-basic-operations-on-rational-numbers-2659 | 1,642,952,636,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320304287.0/warc/CC-MAIN-20220123141754-20220123171754-00123.warc.gz | 1,128,923,777 | 13,025 | Get A+ with YaClass!
Register now to understand any school subject with ease and get great results on your exams!
### Theory
1 Addition of various digits 2 Subtraction of Rational numbers 3 Multiplication of Rational numbers 4 Division of rational numbers
### Exercises
1 Addition with same denominator Difficulty: easy 2 2 The sum of the negative fraction if the denominators are the same Difficulty: easy 3 3 Divide a fraction by integer Difficulty: easy 2 4 Multiplication of a rational number Difficulty: medium 4 5 The sum of negative mixed numbers Difficulty: medium 4 6 Find the required number Difficulty: medium 4 7 Multiplication of a mixed number Difficulty: medium 2.5 8 Divide and multiply Difficulty: medium 3 9 Multiplication of two fractions Difficulty: medium 2 10 Calculate the required number Difficulty: medium 4 11 Negative value added to the real rational number Difficulty: medium 4 12 Multiplication of three rational numbers Difficulty: medium 3 13 Divide a mixed number by a positive or negative integer Difficulty: medium 3 14 Subtract the negative part from the fraction Difficulty: medium 2 15 Divide the mixed number Difficulty: medium 3 16 Division and multiplication of rational numbers(text format) Difficulty: medium 2 17 Divide an integer by a real fraction Difficulty: medium 2 18 Road, speed, time (rational numbers) Difficulty: hard 5 19 Calculate the unknown additive from the sum Difficulty: hard 5 20 Mixed number and integer (negative numbers) Difficulty: hard 4 21 Rational expression (difference) Difficulty: hard 6 22 Rational expression substitution Difficulty: hard 7
### Tests
1 Training Difficulty: easy 11
### Teacher manual
1 Methodical recommendation | 376 | 1,714 | {"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} | 3.25 | 3 | CC-MAIN-2022-05 | longest | en | 0.803923 |
http://uva.onlinejudge.org/external/121/12190.html | 1,369,473,261,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368705884968/warc/CC-MAIN-20130516120444-00089-ip-10-60-113-184.ec2.internal.warc.gz | 278,571,793 | 2,335 | It's year 2100. Electricity has become very expensive. Recently, your electricity company raised the power rates once more. The table below shows the new rates (consumption is always a positive integer):
Range Price (Crazy-Watt-hour) (Americus) 1 100 2 101 10000 3 10001 1000000 5 > 1000000 7
This means that, when calculating the amount to pay, the first 100 CWh have a price of 2 Americus each; the next 9900 CWh (between 101 and 10000) have a price of 3 Americus each and so on.
For instance, if you consume 10123 CWh you will have to pay 2 x 100 + 3 x 9900 + 5 x 123 = 30515 Americus.
The evil mathematicians from the company have found a way to gain even more money. Instead of telling you how much energy you have consumed and how much you have to pay, they will show you two numbers related to yourself and to a random neighbor:
A:
the total amount to pay if your consumptions were billed together; and
B:
the absolute value of the difference between the amounts of your bills.
If you can't figure out how much you have to pay, you must pay another 100 Americus for such a service". You are very economical, and therefore you are sure you cannot possibly consume more than any of your neighbors. So, being smart, you know you can compute how much you have to pay. For example, suppose the company informed you the following two numbers: A = 1100 and B = 300. Then you and your neighbor's consumptions had to be 150 CWh and 250 CWh respectively. The total consumption is 400 CWh and then A is 2 x 100 + 3 x 300 = 1100. You have to pay 2 x 100 + 3 x 50 = 350 Americus, while your neighbor must pay 2 x 100 + 3 x 150 = 650 Americus, so B is | 350 - 650| = 300.
Not willing to pay the additional fee, you decided to write a computer program to find out how much you have to pay.
## Input
The input contains several test cases. Each test case is composed of a single line, containing two integers A and B, separated by a single space, representing the numbers shown to you (1A, B109). You may assume there is always a unique solution, that is, there exists exactly one pair of consumptions that produces those numbers.
The last test case is followed by a line containing two zeros separated by a single space.
## Output
For each test case in the input, your program must print a single line containing one integer, representing the amount you have to pay.
1100 300
35515 27615
0 0
350
2900 | 623 | 2,406 | {"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} | 3.46875 | 3 | CC-MAIN-2013-20 | latest | en | 0.954893 |
https://weisslab.cs.ucl.ac.uk/WEISS/SoftwareRepositories/SNAPPY/scikit-surgerycore/commit/c3390b068f09ad64d4cfed7de1cb162e646be766?view=parallel&w=1 | 1,582,345,150,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875145648.56/warc/CC-MAIN-20200222023815-20200222053815-00245.warc.gz | 624,921,796 | 16,213 | Commit c3390b06 by Mian Asbat Ahmad
### Issue #14: Test case added, imports corrected
`Test case added to check function with empty data.`
parent ab5215db
... @@ -3,7 +3,7 @@ ... @@ -3,7 +3,7 @@ """Functions for point based registration using Orthogonal Procrustes.""" """Functions for point based registration using Orthogonal Procrustes.""" import numpy as np import numpy as np from sksurgerycore.algorithms.procrustes import validate_procrustes_inputs import sksurgerycore.algorithms.procrustes as p def compute_fre(fixed, moving, rotation, translation): def compute_fre(fixed, moving, rotation, translation): ... @@ -19,7 +19,7 @@ def compute_fre(fixed, moving, rotation, translation): ... @@ -19,7 +19,7 @@ def compute_fre(fixed, moving, rotation, translation): """ """ # pylint: disable=assignment-from-no-return # pylint: disable=assignment-from-no-return validate_procrustes_inputs(fixed, moving) p.validate_procrustes_inputs(fixed, moving) transformed_moving = np.matmul(rotation, moving.transpose()) + translation transformed_moving = np.matmul(rotation, moving.transpose()) + translation squared_error_elementwise = np.square(fixed squared_error_elementwise = np.square(fixed ... ...
... @@ -3,7 +3,7 @@ ... @@ -3,7 +3,7 @@ """Functions for point based registration using Orthogonal Procrustes.""" """Functions for point based registration using Orthogonal Procrustes.""" import numpy as np import numpy as np from sksurgerycore.algorithms.errors import compute_fre import sksurgerycore.algorithms.errors as e def validate_procrustes_inputs(fixed, moving): def validate_procrustes_inputs(fixed, moving): ... @@ -116,6 +116,6 @@ def orthogonal_procrustes(fixed, moving): ... @@ -116,6 +116,6 @@ def orthogonal_procrustes(fixed, moving): T[0][0] = tmp[0] T[0][0] = tmp[0] T[1][0] = tmp[1] T[1][0] = tmp[1] T[2][0] = tmp[2] T[2][0] = tmp[2] fre = compute_fre(fixed, moving, R, T) fre = e.compute_fre(fixed, moving, R, T) return R, T, fre return R, T, fre
... @@ -3,6 +3,12 @@ ... @@ -3,6 +3,12 @@ import six import six import numpy as np import numpy as np import pytest import pytest import sksurgerycore.algorithms.procrustes as p import sksurgerycore.algorithms.errors as e def test_empty_fixed(): with pytest.raises(TypeError): e.compute_fre(None, None, np.ones(1, 3), np.ones(3, 3))
Markdown is supported
0% or
You are about to add 0 people to the discussion. Proceed with caution.
Finish editing this message first! | 670 | 2,436 | {"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-2020-10 | latest | en | 0.40496 |
https://kr.mathworks.com/matlabcentral/profile/authors/5570092?detail=answers | 1,712,972,236,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296816465.91/warc/CC-MAIN-20240412225756-20240413015756-00513.warc.gz | 317,134,575 | 24,909 | Community Profile
Sam Chak
Last seen: Today 2020년부터 활동
배지 보기
Content Feed
보기 기준
답변 있음
integrate for long equation
Hi @GUILU You can DIVIDE the integrand and then CONQUER it. Once you've done that, you can validate the result by comparing it ...
2일 전 | 0
답변 있음
How do I solve the following Algebra equation?
Algebraically speaking, since there is no equal sign '', it is not an equation and there is technically nothing to solve in the ...
4일 전 | 0
답변 있음
Particle swarm algorithm in Fuzzy Logic Designer is breaking unity sum property of membership functions, how can I stop this?
Hi @Ben Hatrick I have rarely used PSO to tune fuzzy systems, and I hadn't realized this phenomenon until you brought it up. ...
4일 전 | 0
| 수락됨
답변 있음
hi how can i evaluate the function below from x=1,x=2 into a step of 0.1 y=x/(x2+1)
syms x y = eval('x/(x^2 + 1)') % <-- check if the function is correct xa = 1; % start point of [xa ≤ x ≤...
5일 전 | 0
답변 있음
How to expand this formula in Matlab/simulink
If the symbol is related to the complex variable as defined in the Laplace transform, then you can use the Transport Delay bloc...
5일 전 | 1
| 수락됨
답변 있음
Plotting a 3-value graph
Refer to the examples in yyaxis documentation. x = linspace(0,10); y = sin(3*x); yyaxis left plot(x,y) z = sin(3*x).*exp(...
5일 전 | 0
답변 있음
How can I define a block ( Transfer Fcn) in simulink?
Hi @M1A Linear time-invariant models with more zeros than poles are not supported in Simulink. However, you can create such a m...
6일 전 | 1
| 수락됨
답변 있음
how to add subscript into transfer function and constant block parameter
Hi @Olarewaju Update: In order to create a model block with a visually appealing display of mathematical notation, I personally...
7일 전 | 0
| 수락됨
답변 있음
I am running into trouble answering the question asked in the description.
Hi @Jackson, You have been provided with the MATLAB script 'Project2_Q1.m' to plot the solution curves for question Q1(a). To ...
7일 전 | 0
답변 있음
problem with matlab function in simulink (inpunt problem)
Hi @Alya Bostani Is this the expected result for ? Try fixing the code in the matlabFunction block. t = 0:0.01:10; P = as...
7일 전 | 0
답변 있음
a nonlinear system and its control input in simulink
Hi @controlEE Great job on your progress shown in the comment. Now, I'd like to share my piecewise function formula with you: ...
7일 전 | 1
답변 있음
How to get graph properties on Live script? e.g. Rise time, setting time etc.
Can use the stepinfo() command. G = tf(1, [1, 10, 20]) % the transfer function stepinfo(G) step(G), grid on
8일 전 | 2
| 수락됨
답변 있음
Hello all, I have a little problem with my simulink model. So, I can't connect this block each other like I had in past model from old version Matlab. Do u mind, how solve ?
Hi @Bogdan The blocks you're looking for are likely still present in the MATLAB directory, although they may be hidden within s...
8일 전 | 0
답변 있음
I can't see the cosine graph. How do I generate the cosine graph using 2^10 samples?
deltat = 0.001; nsamples = 2000; time =[0 : deltat : deltat*(nsamples-1)]; size(time) % for i = 1:nsamples ...
8일 전 | 0
답변 있음
Solving a system of Non Linear Differential Equations
Hi @kdv0 Apart from the incorrect initial value for , which should be , the rest of the information in the code is correct. The...
9일 전 | 0
답변 있음
Plotting an Inverse Laplace Function
Not exactly an answer, but rather an attempt to recreate the 'answer' that was displayed in your Command Window. It appears th...
9일 전 | 0
답변 있음
How to fix my linear fit model?
Hi @Jenni Upon initial inspection, it appears that the data follows the trend of the ReLU function, which is a piecewise linear...
9일 전 | 1
답변 있음
Simulation for a standalone wind turbine with a DFIG
There is an example for a wind turbine with a Doubly Fed Induction Generator (DFIG). Type "DFIG" at the Help Center to search ...
10일 전 | 0
| 수락됨
답변 있음
Couple ODE System not enough Input arguments, Why?
Hi @Thanh Hoang I have added these four lines to your original code. The modifications made are minimal. %% ----- added these ...
10일 전 | 0
답변 있음
Hi @Gowtham Use this block.
10일 전 | 0
답변 있음
I want to find two missing parameters in an ODE system of equations using regression/optimization in MATLAB
Hi @naiva saeedia The system identification task at hand seems to be quite challenging, as you only have two parameters to mani...
10일 전 | 0
답변 있음
How do I optimize solutions of ODE?
Hi @Amito Usami-san Considering your objective of varying the force input 'Fc', it becomes necessary to create a Gain Schedule ...
10일 전 | 1
| 수락됨
답변 있음
solve differential equations use bvp4c
Hi @lvlv sun Add these two lines to code as shown below to get the desired number of points for the solution. numpts = 5; ...
10일 전 | 0
| 수락됨
답변 있음
hi.I have problem in spacecraft dynamic control simulation.
Hi @ali.heydari I renamed the ODE function to 'odefcn()' because 'ode' is a built-in function introduced in the R2023b release....
10일 전 | 0
| 수락됨
답변 있음
How can i pass S-curve through the center of a circle ??
Hi @Dr Sohaib Khan Check if the following is the desired 3D line plot? t1 = (0:pi/100:pi); % Top curve 0 to 180 deg t2 = (-p...
10일 전 | 0
| 수락됨
답변 있음
a nonlinear system and its control input in simulink
Hi @controlEE, If you define solely as a continuous-time function, it would appear as displayed in Figures 1 and 2. h = 2; ...
10일 전 | 1
답변 있음
Minimize the sum of squared errors between the experimental and predicted data in order to estimate two optimum parameters
Hi @Vikas Meena Taking into account the information provided, this solution is likely the most suitable option I can propose. ...
11일 전 | 0
답변 있음
Need Help with Position, Velocity, Acceleration model
Hi @Sinan When modeling a differential equation in Simulink, it is generally necessary to formulate the problem in integral for...
11일 전 | 0
답변 있음
i want to get the overall transfer function of the parallel conncected transfer functions blocks
@deekshith, Oh I see. Because you left the spaces. Either you ensure no spaces, or place commas (,) between the elements.
12일 전 | 0
답변 있음
i want to get the overall transfer function of the parallel conncected transfer functions blocks
Hi @deekshith, Check if this works for you. %% Transfer function 1 num1=[576 288]; den1=[4.8 34.992 73.392 1504.8]; G1 = tf...
12일 전 | 1
| 수락됨 | 1,810 | 6,358 | {"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} | 3.125 | 3 | CC-MAIN-2024-18 | latest | en | 0.763769 |
https://lists.osgeo.org/pipermail/grass-user/2011-November/062445.html | 1,606,160,853,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141164142.1/warc/CC-MAIN-20201123182720-20201123212720-00542.warc.gz | 387,753,123 | 2,608 | # [GRASS-user] Re: r.in.gdal oddity
Brian Oney zenlines at gmail.com
Thu Nov 3 06:28:25 EDT 2011
```Hmmm, maybe just a suggestion, although it won't fix the current problem
but to make the code better (I hope)... because it apperas the if one
has an imprecise 0-360 map that thing could go awry, which may justify
another flag (if I may) or something similar to the few changes below.
/* constrain to geographic coords */
if (flag_l->answer&& G_projection() == PROJECTION_LL) {
if (cellhd.north> 90.) cellhd.north = 90.;
if (cellhd.south< -90.) cellhd.south = -90.;
if (cellhd.east> 360.) cellhd.east = 360.;
/* for the case of imprecise coordinates 0-360: if the difference is greater than 360 AND it is the "other" ll projection AND the cellhd.west is strange THEN set cellhd.west to 0*/
if (cellhd.east - cellhd.west> 360.&& cellhd.east>= 181.&& cellhd.west< 0.) cellhd.west = 0.;
/* maybe just one degree of imprecision allowed? */
if (cellhd.east> 180.&& cellhd.east< 181.) cellhd.east = 180.;
if (cellhd.west< -180.) cellhd.west = -180.;
cellhd.ns_res = (cellhd.north - cellhd.south) / cellhd.rows;
cellhd.ew_res = (cellhd.east - cellhd.west) / cellhd.cols;
cellhd.ew_res3 = cellhd.ew_res;
cellhd.ns_res3 = cellhd.ns_res;
G_warning(_("Map bounds have been constrained to geographic "
"coordinates. You will almost certainly want to check "
"map bounds and resolution with r.info and reset them "
"with r.region before going any further."));
}
I am not sure how to perform math operations in if-expressions in C, but
above is the general idea.
Again, I doubt this will fix the r.in.gdal oddity, but it may improve
the -l flag.
Notice that the resolution in the mentioned post is the same as my
case... prob just a coincidence.
cheers,
Brian
On 11/03/2011 07:21 AM, Markus Neteler wrote:
> On Thu, Nov 3, 2011 at 5:17 AM, Hamish<hamish_b at yahoo.com> wrote:
>> Hi, coming to this thread a bit late, sorry if I miss the point;
>> I'll study the rest of the thread/situation better asap, but some
>>
>> Markus wrote:
>>>> Just force it to be 180W and 180E... I thought (the -l
>>>> flag) it is just a recent addition to GRASS 6.4, so maybe
>>>> it needs a some t.l.c...
>> AFAIK it is working as intended and without bugs. | 692 | 2,251 | {"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-2020-50 | latest | en | 0.809509 |
https://www.physicsforums.com/threads/product-of-inertia.390904/ | 1,660,777,701,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882573118.26/warc/CC-MAIN-20220817213446-20220818003446-00192.warc.gz | 785,142,699 | 13,915 | # Product of inertia
ian2012
I am confused about the concept of product of inertia.
Wikipedia says: "Here Ixx denotes the moment of inertia around the x-axis when the objects are rotated around the x-axis, Ixy denotes the moment of inertia around the y-axis when the objects are rotated around the x-axis, and so on."
So, when do you get a product of inertia(what's the rule)? I am guessing only when the rotating (about the center of mass) object has mass in regions x, y, z > 0. If one of the coordinates are zero and the object is stuck rotating in a plane, then I am assuming it has stable rotation?
## Answers and Replies
Gold Member
I would say that non-zero products of inertia means that rotation about the axis in question cannot be pure or, similarly, that the inertial body in question is not inertial symmetric around that axis.
If you look at the equation that really defines the meaning of the inertia tensor (in inertial space), namely $L = I \omega$ you can see that if any of the product of inertia (the off-diagonal elements of I) is non-zero, the angular momentum vector will in general not point in the same direction as the rotation axis and any rotation will be a non-pure rotation (for pure rotation the angular momentum vector is parallel to the rotation axis).
For a body with fixed direction of rotation axis this means that there must be a resulting torque that will make the "tip" of the angular momentum vector move in circles around the the rotation axis, like what happens for dynamically unbalanced wheels, for instance. As the inertia tensor is usually defined in body coordinates, which rotates around the rotation axis, the inertia tensor in inertial space will thus always have off-diagonal elements that are non-zero.
On the other hand, for a torque free body (where the resulting torque is zero) the angular momentum vector stays fixed in inertial space and the rotation vector must then move around the angular momentum in some way, usually giving rise to body tumbling.
Also, as you probably know, it is possible to select a (principal) coordinate system for any rigid body such that the rotation around each of these principal axis is pure and in this coordinate system all the products of inertia is zero. | 473 | 2,256 | {"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} | 3.90625 | 4 | CC-MAIN-2022-33 | latest | en | 0.912631 |
https://billwadge.com/2022/01/25/topology-and-computability/ | 1,680,422,730,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296950422.77/warc/CC-MAIN-20230402074255-20230402104255-00051.warc.gz | 167,234,189 | 28,085 | ## Topology and Computability[3400 views]
Readers of this blog are familiar with notions of computability – basically, the question is, what can machines do without human assistance? And you are familiar with machines. Electronic ones of course, but I always like to think of machines as composed of gears, levers and pulleys.
Topology? That’s another story. Rubber doughnuts being continuously stretched but always preserving that hole. Or calculus and differential equations.
So what’s the connection? You’d be surprised
Topology
The formal definition of topology doesn’t shed much light. A topology is a nonempty space (set) X together with a collection of ‘open’ sets that includes X and ∅ and is closed under finite intersection and arbitrary union. A set its closed iff its complement is open. A function from X to X is continuous iff the inverse image of an open set is open. Clear? Hardly. (But remember these concepts.)
What topology is really about is the idea of closeness. A set is open iff all points sufficiently close to a member of the set are also in the set. A set is closed iff any point that has members of the set arbitrarily close to it, is also in the set. A function is continuous iff f(x) is arbitrarily close to f(x’) as long as x’ is sufficiently close to x.
So what has topology to do with computability? A lot, as it turns out. (Besides network topologies, which is not what we’re talking about.) (This is based on lecture given by my PhD supervisor J. W. Addison jr way way back in 1967 (!) in the graduate logic course at UC Berkeley. I remember it vividly.)
The Baire Space
The study of the topological notions of open, closed, and continuous began in analysis with the study of the real numbers and functions over the real numbers. As the discipline progressed, mathematicians realized that the the reals were somewhat inconvenient. For example, the product of the reals with itself (the set of pairs of reals) is not isomorphic to the reals.
These minor difficulties could be cleared up by taking the irrationals as the space studied. Every irrational has a unique decimal expansion but not all expansions (e.g. those that repeat after a certain point) denote irrationals. However every positive irrational has a unique infinite continued fraction expansion
a0+1/(a1+1/(a2+1/(a3+…
Soon topologists dealt directly with the sequence
a0, a1, a2, a3, …
and thus was born the Baire space of infinite sequences of natural numbers (isomorphic to the irrationals).
What is the topology of the Baire space? If you trace back the definitions it turns out that a set G of infinite sequences is open iff whenever a (= <a0,a1,a2,…>) is in G then there is an n such that every sequence with initial segment <a0,a1,a2,…,a(n)> is also in G.
Open sets in the Baire topology
In other words, if a is in G then there is a finite amount of information about a that implies that a is in G. And this is where computability comes in.
Suppose that we have a Turing machine M with a two-way infinite tape. We can write a on one half of the tape and start the machine. Let’s say that M accepts a iff it eventually halts and prints “1”. Then the set accepted by M is open.
For example the sets {a: a0 = 5}, {a: a(a0) = 8}, and {a: a(i) = 7 for some i} are all open. Whereas {a: a is increasing} is not.
The converse is not true for computability-over-the-naturals reasons. Let S be any non-re set, e.g. codes for true statements (with quantifiers) of arithmetic. Then {a: a0 is in S} is open but is not the set accepted by any M. We can restore the symmetry by allowing M to have an infinite data base in the form of an element d of the Baire space written on a second tape. The tape d could enumerate all the initial segments that guarantee membership in G.
Now it is true that a set is open iff it is the set accepted by a machine augmented as described.
We know from computability theory that a set defined by being those numbers accepted by a machine is recursively enumerable (re), and vice versa. So in a sense the open subsets of the Baire space are the re subsets. This is the basis for the analogy between computability over the naturals and topology over the Baire space (which is really the irrational numbers).
Closed sets
For the closed sets, let us say that an (augmented) machine rejects a iff at some point it halts and prints “0”. A set F is closed iff there is a machine that rejects exactly the non elements of F. The first two example sets given above are (also) closed, as is the set of increasing sequences. Notice that (unlike over the reals) a set can be clopen – both closed and open.
If a set D is both open (witnessed by augmented machine M1) and closed (witnessed by augmented machine M2) then we can run M1 and M2 in parallel until one of them halts, as eventually must happen. It’s easy to combine M1 and M2 and their augments into an augmented machine M which accepts elements of D and rejects elements of -D. In this way we get a machine that decides membership in D.
In computability theory a set whose membership is decidable by Turing machine is recursive (decidable, effective). Thus the clopen subsets correspond to decidable subsets of the naturals.
Functions
What about functions? What functions from the Baire space to itself deserve to be called computable (by augmented Turing machines)? No prize for guessing the continuous functions, and here’s why.
If you trace through the topology definitions it turns out that b = f(a) is continuous iff every initial segment of b is determined by at least one initial segment of a. In other words, b begins
<b0, b1, b2, … b(m)>
if a begins with some
<a0, a1, a2, …, a(n)>
Thus, any finite amount of information about f(a) is determined by a finite amount of information about a.
Continuous functions as filters
We can code up the (countable) set {(s,t): a has s as an initial segment implies b has t as an initial segment} in an element d of the Baire space. It is then easy to define a machine M augmented with d that computes f incrementally. M starts with a on its main tape then starts reading in a0, a1, a2, … in turn and eventually prints b0 on the other half tape, then eventually b1, then eventually b2, etc.
(Latex images by John Plaice)
Modern computer scientists will recognize M as a filter that transforms the stream a into the stream b:
Dataflow was practically an unknown concept back in 1967 but Addison anticipated it by decades!
Of course that is not the end of the analogy. The G-delta sets (countable intersection of open sets) correspond to the ∀∃ level of the arithmetic hierarchy. The Borel sets are those closed under complement and intersection and correspond to the hyperarithmetic sets (more precisely, the hyperarithmetic-in-some-d sets).
For me personally, the analogy lead me directly to the game characterization of the ‘Wadge’ reducibility, which formed the basis of my PhD research.
Not only that, my next major research project was dataflow and Lucid – once again, infinite sequences and filters.
Addison’s 1967 lecture was for me the most important of my whole educational career !
I am a retired Professor in Computer Science at UVic.
This entry was posted in Uncategorized. Bookmark the permalink.
### 1 Response to Topology and Computability[3400 views]
1. dm00 says:
Interesting!
You may enjoy Maurice Herlihy’s exploration of the light combinatorial topology has to shed on distributed algorithms in *Distributed Computing through Combinatorial Topology*.
As I understand it, one can model the state of a distributed computation as a simplicial complex (a higher-dimensional generalization of a graph). The steps in a communications protocol can then be modeled as a continuous transformation of that simplicial complex (i.e., mapping one complex to another).
Many classic problems in distributed computing then become: can one simplicial complex representing the start state of the problem be continuously transformed into another representing the desired end-state of the problem? In many cases (e.g. a connected complex transformed into a disconnected one, or a complex with one “hole” transformed into a complex with a different number of “holes”), the answer is a pretty trivial (from the point of view of a topologist) “no”.
This site uses Akismet to reduce spam. Learn how your comment data is processed. | 1,920 | 8,396 | {"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-2023-14 | latest | en | 0.969505 |
https://xypi.wordpress.com/vfc/ | 1,558,277,666,000,000,000 | text/html | crawl-data/CC-MAIN-2019-22/segments/1558232254889.43/warc/CC-MAIN-20190519141556-20190519163556-00035.warc.gz | 1,004,667,633 | 12,175 | # Virtual File Cabinet
The MTBoS search engine and Problem Based Learning Search are great places to start.
Practice Structures
War (Evaluating Functions) – Sarah Carter
Risk & Four in a Row – Sarah Carter
Two Truths and a Lie (Classifying Function Types) – Math = Love
I Have, Who Has? Compilation – via Math = Love
Vocab Review Cards – Math = Love
Printable Flash Card Maker
Partner Problems (Similar to Player Coach) – Simplifying Radicals
Question Stack (Rational Expressions) (Factoring Trinomials) (Similar to Around the Room) – Math = Love
Other Thoughts– Do Nows, super quick stuff
If A is the answer, what is the question? – Pam Wilson
Word Problem Choice – Foussinating
Which One Doesn’t Belong?
Algebra
Number Line Intro Task – Restructuring Algebra
Integer Temperature Number Lines – Divisible by 3
Integer Ideas: Use army men in two colors to practice zero pairs
Square Dance (Clothesline Roots) – Divisible by 3
Combining Like Terms – Dan Meyer
Cut and Paste Combine Like Terms – Math = Love
Distributive Property – Math with Bad Drawings
Simplify – Math with Bad Drawings
Evaluating Expressions Sorting Cards – Sarah Carter
Draw “monsters” to represent the variable for solving equations
Absolute Value Equations – Simplifying Radicals
Function Auction – Sarah Carter
Stacking Cups Linear Equations – Restructuring Algebra writeup | Dan Meyer
Graphles to Graphles (Domain & Range) – Kate Nowak
Build a Function (for Absolute Value, but works on any transformations) – Misscalcul8
Line Game Linear Graph Review – Simplifying Radicals
Graphing Pictures Project (Desmos + a rubric) – Restructuring Algebra
Graph an Inequality (Student created) – Pam Wilson
Systems of Equations Elimination – Misscalcul8
Quadratic Function Blanks – Math = Love
Candy Catapult for Quadratics – I Speak Math
Parabolas: Will It Hit the Hoop? – Dan Meyer
Unit Circle Table – Misscalcul8
Algebra 2 Remix – Infinite Sums
Geometry
3 Acts Nissan Girl Scout Cookie Volume – Dan Meyer
Statistics
Phone Usage Data – I Speak Math
Penny/Balcony Crash Exploration – Radical Rational
Survey Project – Math Equals Love
Fun Bonus Stuff
Bulletin Board Ideas & Files, Reference Pages
Growth Mindset – Math = Love
Mathematicians Toolbox Foldable Reference – Math = Love
Problem Solving Strategies – Math = Love
Six P’s Class Rules – Math = Love | 563 | 2,370 | {"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} | 3.40625 | 3 | CC-MAIN-2019-22 | latest | en | 0.716734 |
https://elsmar.com/elsmarqualityforum/threads/setting-aql-acceptable-quality-level-on-c-zero-inspections.5417/ | 1,603,653,850,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107889651.52/warc/CC-MAIN-20201025183844-20201025213844-00673.warc.gz | 293,821,212 | 32,259 | # Setting AQL (Acceptable Quality Level) on C=Zero inspections
N
#### Norman V
Here's a general question:
We use a C=Zero sampling system for our incoming inspection, using "associated AQLs". What is the best way to determine which AQL to select for a particular part being inspected under this type of system? I have a Quality Engineer that wants to develop a set of "standard AQL levels" for each type of product being inspected.
For example: Rubber seals would always have an AQL of 1.0, nuts and bolts, 10.0, and so on.
Does anybody within the sound of my voice have any input into how this should be handled? We're trying hard not to be arbitrary.
Thanks.
Norman V.
B
#### Bill Ryan - 2007
Sorry, I've got a major disconnect going right now but I'm responding anyway (they say the first thing that goes.......).
I don't see how an AQL has anything to do with a C=0 "sampling plan". By definition, doesn't AQL imply defective/nonconforming parts are OK?
Bill
N
#### Norman V
Ok, let's clarify
In the traditional sense of AQL, (MIL-105-E), you are correct, AQL determines (among other things) how many rejects you can accept in a lot. In a C=Zero system, the "associated AQL" merely is used in a table to help you determine how many samples to pull based on lot size.
The AQL levels are "associated", because the Operating Characterisitic (OC) Curves would look about the same in terms of a percent defective for each AQL level, even though the acceptance number is Zero.
My main questions was, what is the best way to decide how to set the AQL level during the Quality Planning process, aside from just picking an AQL arbitrarily?
Thanks,
Norman V.
#### Mike S.
##### Happy to be Alive
Trusted Information Resource
Since it is incoming inspection, I'd sit down with your internal customer(s) who will use each part and discuss it with them. Discuss what impact a defect will have on them - i.e. if there is 1% defective how would that affect you and downstream operations, etc. Also, what has been the incoming quality level in the past, and was it "acceptable" to them in the past. Also consider time of inspection, etc. Some will rant and rave and say no defect is ever "acceptable", but in the real world, IMO, zero defects is a pipe dream. JHO. Hope this helps.
C
#### Craig H.
Hi
Dr Ken Stephens (Applied Acceptance Sampling, available through ASQ) gives the following steps:
1. Determine and list each quality characteristic
2. Determine the product unit.
3. Develop and specify the test method.
4. Determine the criteria for conformity.
5. Determine and list the calssification of nonconforming units or nonconformities and/or groups of nonconformities.
6. Establish an Index Quality Level (AOQL, etc).
7. Determine if sampling/acceptability should be based on lot inspection, on-line (continuous, etc.
I have excerpted here, and he goes on to more specifics. We used this textg for one of my classes at SPSU, and I can say that if you want to find out about acceptance sampling, this is the text.
The class was mind boggling.
Hope this helps.
N
#### Norman V
Mind Boggling?
Craig,
When you say "mind-boggling", do you mean that in a good way or a bad way?
Anyhow, I saw that book in the ASQ catalog, and it does look good. The CD ROM that accompanies it seem interesting as well.
Maybe I'll put it on my Christmas list.
Thanks,
Norman V.
C
#### Craig H.
Norman:
Mind boggling in both a good and bad way, really.
Before this class, I assumed that Dr. Deming ment that inspection was bad. Not so. What he said, and ment, was that DEPENDANCE on inspection for quality is bad. We still inspect, but our process is such that we don't make bad stuff.
In a bad way, we actually designed acceptance sampling plans - those are some of the major files on the CD. Not for those with less-than-complete statistical backgrounds, IMHO. The class was the most difficult in the SPSU Master's program. We used the draft for Dr. Stephen's book, and he taught the class (online). He is retired now, but his list of publications is impressive, and this particular book is highly recommended.
#### Douglas E. Purdy
##### Quite Involved in Discussions
But the question still has not been answered, unless it is in The BOOK? I believe the act of deciding is a step (#6), but are there any generally accepted parameters in making that decision, OR do we have to read the book?
C
#### Craig H.
Douglas:
In short, if you want an optimum answer, read the (a) book, and use it to decide what is best for your situation.
Acceptance sampling is like a lot of other tools - it can be powerful, but only if it is used in a manner that fits the situation. I could give you a "canned" answer, but without knowing the situation, it would likely be worse than the "arbitrary" alternative.
Sorry I can't be of more help, but to give an answer here might do more harm than good.
##### One of THE Original Covers!
Staff member
Craig,
Good information.
Kevin
Setting Acceptance Quality Limits (AQL) for New Medical Devices ISO 13485:2016 - Medical Device Quality Management Systems 4
Use of sampling plans like MIL-STD-105 E or ANSI Z1.4 for setting AQL's? AQL - Acceptable Quality Level 13
Setting action completion dates Nonconformance and Corrective Action 3
Heat Sealers Upper and lower limit setting during OQ for Heat Sealer Qualification and Validation (including 21 CFR Part 11) 0
GDPR in Urgent Healthcare Setting Other ISO and International Standards and European Regulations 1
Setting deadlines (ex. 45 days) for Document Registration & Review Cycle Document Control Systems, Procedures, Forms and Templates 3
What are the criteria for setting a target KPI value against a quality objective? Benchmarking 7
Setting Quality Objectives (for the first time) AS9100, IAQG, NADCAP and Aerospace related Standards and Requirements 17
ISO 11137 Gamma sterilization validation, dose setting, dose audit Other Medical Device Related Standards 1
Best Practice for setting tolerances on a Drawing Inspection, Prints (Drawings), Testing, Sampling and Related Topics 8
Setting up a quality department in an Engineering Firm Quality Management System (QMS) Manuals 11
Software for setting up New Warehouse Manufacturing and Related Processes 0
Setting up the process - First piece inspections Statistical Analysis Tools, Techniques and SPC 3
Add-On software for VMC to calculate setting time, Idle Time, OEE etc.. Manufacturing and Related Processes 2
Setting Up a Basic Quality Lab for a small manufacturing company General Measurement Device and Calibration Topics 2
Setting up an LLC as a Consultant Paid Consulting, Training and Services 27
Setting initial Expiry Date of an IVD per BS EN 23640:2015 Other Medical Device Related Standards 12
Q Setting Quality Objectives with a Timeframe ISO 9000, ISO 9001, and ISO 9004 Quality Management Systems Standards 4
Setting up Prolink's QC-Gage to a Keyence IM-6225 Vision System Statistical Analysis Tools, Techniques and SPC 2
Cpk Setting Tolerance - Cart before the Horse? (Wifi Routers) Capability, Accuracy and Stability - Processes, Machines, etc. 3
Bioburden 0 cfu for Dose Setting using the VDmax25 Other Medical Device Related Standards 9
A New company, no employees, Setting up a QMS ISO 9000, ISO 9001, and ISO 9004 Quality Management Systems Standards 5
F Setting up a new sterilization unit in a new facility 21 CFR Part 820 - US FDA Quality System Regulations (QSR) 6
C ISO 14001 Recertification without setting Targets ISO 14001:2015 Specific Discussions 3
T Guidance - Setting up an integrated quality & environmental management system Career and Occupation Discussions 12
M Setting up a M1003 QA Program for the AAR Various Other Specifications, Standards, and related Requirements 2
A Setting up a Medical Device Distributor in Malaysia Other Medical Device Regulations World-Wide 16
Irradiation Sterilization Dose Setting and Dose Audit Flow Charts ISO 13485:2016 - Medical Device Quality Management Systems 0
J Setting Up New QMS - New US Location ISO 9000, ISO 9001, and ISO 9004 Quality Management Systems Standards 2
A Setting air gauging outside of a part tolerance Capability, Accuracy and Stability - Processes, Machines, etc. 2
S Target Setting - Internal metric(s) to drive Internal Processes Six Sigma 1
S MSA Help or guidelines - Plastics - Setting up an MSA on a multi cavity fixture Gage R&R (GR&R) and MSA (Measurement Systems Analysis) 2
G Setting up a corporate umbrella QMS ISO 9000, ISO 9001, and ISO 9004 Quality Management Systems Standards 4
T Setting Quality Objectives ISO 9000, ISO 9001, and ISO 9004 Quality Management Systems Standards 10
W Setting up from scratch - Looking for help (ISO9001 / ISO13485) ISO 9000, ISO 9001, and ISO 9004 Quality Management Systems Standards 4
Setting up a New QMS for Certification when ISO 9001 is about to Change ISO 9000, ISO 9001, and ISO 9004 Quality Management Systems Standards 23
P New to AS9100 Rev. C - Setting KPIs and FAI Reports AS9100, IAQG, NADCAP and Aerospace related Standards and Requirements 14
Setting Lower Control Limit on Attributes Charts Statistical Analysis Tools, Techniques and SPC 9
J Lean 5S - Setting Min/Max Inventory Levels Lean in Manufacturing and Service Industries 3
Setting Specification Limits for UOD (Uniformity) for Live Bacteria or Spores Pharmaceuticals (21 CFR Part 210, 21 CFR Part 211 and related Regulations) 4
D How do I make the Corporate QMS work in the manufacturing setting? ISO 9000, ISO 9001, and ISO 9004 Quality Management Systems Standards 9
J Setting Quality Objectives for Service, Repair and Sales ISO 9000, ISO 9001, and ISO 9004 Quality Management Systems Standards 4
Setting Specification Limits Quality Tools, Improvement and Analysis 11
C Mitutoyo SJ210 Settings - Is there an Industry Standard Surface Measurement Setting? General Measurement Device and Calibration Topics 4
W Setting up a Zmike for Pin Gage Calibration General Measurement Device and Calibration Topics 3
Zero setting of correlation graph when plotting best fit linear regression Statistical Analysis Tools, Techniques and SPC 7
G Setting up a 3 Factor, 2 Level,2 response DOE in Minitab Using Minitab Software 7
B Quality Goals and Improvements in a Bioanalytical Laboratory Setting Quality Tools, Improvement and Analysis 1
N Setting up MSA system in a machine shop environment. Gage R&R (GR&R) and MSA (Measurement Systems Analysis) 5
N Setting up a new Calibration Room in a Machine Shop Calibration and Metrology Software and Hardware 3 | 2,487 | 10,531 | {"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-2020-45 | longest | en | 0.946468 |
https://mathhelpforum.com/threads/integrate.147924/ | 1,582,060,260,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875143815.23/warc/CC-MAIN-20200218210853-20200219000853-00161.warc.gz | 478,938,857 | 15,923 | Integrate?
Warrenx
(t^2 + e^(2t) + e^(-2t))^1/2
anyone got any ideas, I suck at integration so its hard for me to see what to do, if someone can start me in the right track I am sure I can finish.
thanks,
Warren.
p0oint
Hello!
Is the original integral that you want to solve:
$$\displaystyle \int \sqrt{t^2 + e^{2t} + e^{-2t}} dt$$
Warrenx
Hello!
Is the original integral that you want to solve:
$$\displaystyle \int \sqrt{t^2 + e^{2t} + e^{-2t}} dt$$
Yes that is it. I am sorry I did not use Latex, have headache right now and couldn't think
Warrenx
Dear Warrenx,
This integration could not be evaluated using elementary functions. You can get the answer using, Function calculator
hmm... I think I wrote the problem write:
It is asking for the arc length:
so s(t) =$$\displaystyle \int_0^1 |r'(t)|$$
r(t) = $$\displaystyle \sqrt2t , e^t , e^{-t}$$
so I get r'(t) = $$\displaystyle t , e^t , e^{-t}$$
then lr'(t)l = $$\displaystyle \sqrt{(t)^2 + (e^t)^2 + (-e^{-t})^2}$$
(sorry it was -e for the third term)
and then I am supposed to integrate it, but I suck at integratin' so I are stuck
simplependulum
MHF Hall of Honor
hmm... I think I wrote the problem write:
It is asking for the arc length:
so s(t) =$$\displaystyle \int_0^1 |r'(t)|$$
r(t) = $$\displaystyle \sqrt2t , e^t , e^{-t}$$
so I get r'(t) = $$\displaystyle t , e^t , e^{-t}$$
then lr'(t)l = $$\displaystyle \sqrt{(t)^2 + (e^t)^2 + (-e^{-t})^2}$$
(sorry it was -e for the third term)
and then I am supposed to integrate it, but I suck at integratin' so I are stuck
Oh , the derivative of $$\displaystyle \sqrt{2} t$$ is $$\displaystyle \sqrt{2}$$
so what the integral you are going to evaluate should be :
$$\displaystyle \int \sqrt{ \sqrt{2}^2 + e^{2t} + e^{-2t} }~dt$$
Trick is also found
$$\displaystyle 2 + e^{2t} + e^{-2t} = (e^t)^2 + 2 (e^t)(e^{-t}) + (e^{-t})^2 = (e^t + e^{-t} )^2$$ (Happy)
the heaven is just in front of you now .
Warrenx
Warrenx
Oh , the derivative of $$\displaystyle \sqrt{2} t$$ is $$\displaystyle \sqrt{2}$$
so what the integral you are going to evaluate should be :
$$\displaystyle \int \sqrt{ \sqrt{2}^2 + e^{2t} + e^{-2t} }~dt$$
Trick is also found
$$\displaystyle 2 + e^{2t} + e^{-2t} = (e^t)^2 + 2 (e^t)(e^{-t}) + (e^{-t})^2 = (e^t + e^{-t} )^2$$ (Happy)
the heaven is just in front of you now .
If you don't tell anyone about this, I wont tell anyone...
but thank you! | 856 | 2,416 | {"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} | 3.609375 | 4 | CC-MAIN-2020-10 | longest | en | 0.861231 |
http://www.cram.com/flashcards/logic-phl-321-269881 | 1,488,178,884,000,000,000 | text/html | crawl-data/CC-MAIN-2017-09/segments/1487501172649.58/warc/CC-MAIN-20170219104612-00178-ip-10-171-10-108.ec2.internal.warc.gz | 360,551,159 | 23,383 | • Shuffle
Toggle On
Toggle Off
• Alphabetize
Toggle On
Toggle Off
• Front First
Toggle On
Toggle Off
• Both Sides
Toggle On
Toggle Off
Toggle On
Toggle Off
Front
### How to study your flashcards.
Right/Left arrow keys: Navigate between flashcards.right arrow keyleft arrow key
Up/Down arrow keys: Flip the card between the front and back.down keyup key
H key: Show hint (3rd side).h key
A key: Read text to speech.a key
Play button
Play button
Progress
1/44
Click to flip
### 44 Cards in this Set
• Front
• Back
If A,B A-->B A only if B A --> B A is a sufficient condition for B A --> B A is sufficient for B A --> B Provided that A,B A --> B B provided that A A --> B B on the condition that A A --> B B is a necessary condition for A A-->B B is necessary for A A-->B Whenever A,B A-->B B if A A-->B Given that A, B A--> B In case A,B A-->B A only on the condition that B A-->B If B,A B-->A B only if A B-->A B is a sufficient condition for A B-->A B is sufficient for A B-->A B is sufficient for A B-->A Provided that B,A B-->A A provided that B B-->A A on the condition that B B-->A A is a necessary condition for B B-->A A is necessary for B B-->A Whenever B,A B-->A A if B B-->A Given that B,A B-->A In case B,A B-->A B only on the condition that A B-->A A and B A & B Both A and B A & B A, but B A & B A, although B A & B A as well as B A & B Though A, B A & B A, also B A & B A or B AvB Either A or B AvB A unless B AvB = (~B --> A) A if and only if B A <--> B A is equivalent to B A <--> B A is necessary and sufficient for B A <--> B A just in case B A <--> B Neither A nor B ~(AvB) = (~A &~B) | 496 | 1,615 | {"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} | 3.25 | 3 | CC-MAIN-2017-09 | longest | en | 0.714749 |
http://ecoursesonline.iasri.res.in/mod/page/view.php?id=1393 | 1,695,947,501,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510462.75/warc/CC-MAIN-20230928230810-20230929020810-00472.warc.gz | 13,766,140 | 8,880 | ## LESSON 5. PASCAL’S LAW, FLOW, ENERGY, WORK AND POWER
5.1 INTRODUCTION
Hydraulic devices are concerned with flow of liquid in enclosed paths such as pipes, tubes, valves, actuators etc. Flow can be quantified in terms of volumetric flow rate and mass flow rates. Flow can be classified as steady flow and unsteady flow based upon state of conditions at a point with respect to time. Further flow may be classified as laminar flow and turbulent flow based upon flow pattern.
5.2 Volumetric Flow
Volumetric flow rate is the volume of fluid moving past a given point in unit time. It can be calculated as the product of average flow velocity and area of cross-section of the path. It is usually measured in litre per second (l/s) or cubic centimeter per second (cusec).
Volumetric flow rate, V = v.A, cm3/s (5.1)
Where, ‘v’ is the velocity in cm/s and ‘A’ is area of pipe cross-section in cm2.
Example: Calculate the volumetric flow rate of a liquid flowing with an average velocity of 15 cm/s in a pipe of diameter 5 cm.
Solution:
5.3 Mass Flow
Mass flow rate can be defined as the measure of the mass of fluid passing through a point in the system per unit time. It is equal to the product of density of the fluid and its volumetric flow rate.
Mass flow rate, M = rV gm3/s (5.2)
Where, ‘ρ’ is the density in gm/cm3 and ‘V’ is volumetric flow rate in cm3/s.
Example: Calculate the mass flow rate of a liquid flowing with an average velocity of 15 cm/s in a pipe of diameter 5 cm. Density of the liquid is 60 gm/cc.
Solution: Volumetric flow rate,
Mass flow rate = ρV
= 60 X 294.52 = 17,671.2 gm/cc.
In steady flow, fluid properties like pressure, temperature and velocity at point in the system do not change with time. In the steady state flow rate there is no mass accumulation in the system or component.
Unsteady flow is a transient phenomenon which may become a steady flow or zero flow with time. For example, when a valve is opened suddenly, liquid flow into the pipe is unsteady which become steady with time. Similarly, when a valve is suddenly closed at the discharge end of a pipe, there will be fluctuations in both velocity and pressure in the pipe before the flow becomes zero in the pipe.
5.5 Laminar and Turbulent Flow
When a fluid is flowing through a pipe or pipe like structure ,there may be two types of flow that generally occur i.e. laminar flow or turbulent flow depending on the velocity of the fluid.
At low flow velocities, the flow pattern is smooth and linear. This is known as laminar or streamline flow. In this type of flow, velocity is minimum at the pipe walls and maximum at the centre of the pipe. As flow velocity increases, eddies start to form until at high flow velocities complete turbulence results. At this stage, flow velocty is virtually uniform across the pipe cross-section. Laminar flow and turbulent flow are depicted in Fig. 5.1
Laminar flow occurs in parallel layers, with no disruption between the layers. At low velocities the fluid tends to flow without lateral mixing, and adjacent layers move over one another. There are no cross currents perpendicular to the direction of flow. In laminar flow the motion of the particles of fluid is such that the particles are moving in straight lines parallel to the pipe walls.
In turbulent flow the fluid undergoes irregular fluctuations, or mixing. It is in contrast to laminar flow, in which the fluid moves in smooth paths or layers. In turbulent flow the velocity of the fluid at a point is continuously changing in both magnitude and direction. Common examples of turbulent flow are river flow, blood flow in arteries, oil transport in pipelines, ocean currents and flow through pumps and turbines.
5.6 Reynolds Number
The nature of the flow i.e. whether the flow is laminar or turbulent is determined by Reynolds Number, Re, given as:
Where ‘v’ is flow velocity, ‘d’ is pipe diameter, ‘ρ’ is the fluid density and ‘$\eta$’ is the fluid viscosity. For a given liquid flowing in a pipe the fluid density, fluid viscosity and the pipe diameter are constant. Thus, the value of Reynolds Number depends upon flow velocity.
Reynolds Number is a ratio and hence dimensionless. For a fully developed flow in a pipe flow is laminar for Re < 2000 and flow is turbulent for Re > 4000.
5.7 Bernoulli’s Principle
Recall that the total energy ‘E’ of a liquid flowing in a pipe is given by
E = Kinetic Energy + Potential Energy + Pressure Energy
And Kinetic Energy, $K.E.\; = \;\frac{1}{2}m{v^2}$
Potential Energy, PE = mgh
Pressure Energy = PV
Where ‘m’ is the mass, ‘v’ is the velocity, ‘h’ is the height, ‘P’ is the pressure and ‘V’ is the volume of liquid.
KE, PE and pressure energy when expressed as per unit weight of liquid are written as
$\;\frac{{{v^2}}}{{2g}}$ , h and $\frac{P}{{\rho g}}$ respectively.
According to Bernoulli’s principle, the total energy of a liquid is constant neglecting any losses
$E = \;\frac{{{v^2}}}{{2g}}\; + \;h\; + \;\frac{P}{{\rho g}}\quad=$ = constant (5.4)
Therefore, total energy of a flowing liquid at point 1 will be equal to total energy at point 2, i.e.
mgz1 + ½ mV12 + p1v1 = mgz2 + ½ mV22 + p2v2 (5.6)
where,
m = mass
g = acceleration due to gravity
V1 = velocity at 1
v1 = volume at 1
p1 = pressure at 1
V2 = velocity at 2
v2 = volume at 2
p2 = pressure at 2
Therefore, Bernoulli's principle states that for a fluid flow, an increase in the speed of the fluid occurs simultaneously with a decrease in pressure or a decrease in the fluids potential energy. Bernoulli's principle can be applied to various types of fluid flow, and equation (5.4) is called Bernoulli's equation. Bernoulli’s equation is best applicable under the following assumptions:
• Fluid is under steady state flow, hence flow rate at all the positions in the pipe is not changing.
• Fluid is having streamline flow, hence flow at all the points is exhibiting streamline or laminar flow.
• Fluid is assumed to be of negligible viscosity, hence the fluid does not have tendency to stick on to the pipe. | 1,536 | 6,354 | {"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} | 4.375 | 4 | CC-MAIN-2023-40 | latest | en | 0.899603 |
https://mathematica.stackexchange.com/questions/194075/parsing-output-of-a-reduce-operation?noredirect=1 | 1,585,698,440,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585370504930.16/warc/CC-MAIN-20200331212647-20200401002647-00077.warc.gz | 595,683,329 | 32,495 | # Parsing output of a reduce operation
This question is a broader question than this. The output of Reduce can be of different forms. The solutions proposed for that question works well when the output of Reduce is as provided in that question. The output of Reduce can also be of the forms:
$$n_2\in \mathbb{Z}\land n_1=1\land 0\leq n_2\leq 1993.$$
or
$$\left(n_1|n_2\right)\in \mathbb{Z}\land \left(\left(n_1=0\land 1.\leq n_2\leq 19979.\right)\lor \left(1.\leq n_1\leq 2222.\land 0\leq n_2\leq \text{3.5534074004528205\grave{ }*{}^{\wedge}-79} \sqrt{-4.55384\times 10^{157} n_1^2+1.50383\times 10^{162} n_1+7.9031\times 10^{164}}+\text{1.7152185859604652\grave{ }*{}^{\wedge}-56} \left(3.20659\times 10^{56} n_1+5.82404\times 10^{59}\right)\right)\lor \left(2223.\leq n_1\leq 33540.\land \text{1.7152185859604652\grave{ }*{}^{\wedge}-56} \left(3.20659\times 10^{56} n_1+5.82404\times 10^{59}\right)-\text{3.5534074004528205\grave{ }*{}^{\wedge}-79} \sqrt{-4.55384\times 10^{157} n_1^2+1.50383\times 10^{162} n_1+7.9031\times 10^{164}}\leq n_2\leq \text{3.5534074004528205\grave{ }*{}^{\wedge}-79} \sqrt{-4.55384\times 10^{157} n_1^2+1.50383\times 10^{162} n_1+7.9031\times 10^{164}}+\text{1.7152185859604652\grave{ }*{}^{\wedge}-56} \left(3.20659\times 10^{56} n_1+5.82404\times 10^{59}\right)\right)\right)$$
In such cases, how can I find the maximum value of $$n_1$$ and $$n_2$$ from the Reduce output?
Edit: Please find below the function and the reduce operations that produce the two kind of outputs:
driftParamSet = (-0.72
\!$$\*SubsuperscriptBox[\(n$$, $$1$$, $$2$$]\) -
Subscript[n,
1] (0.35 (0.8 - 0.39 Subscript[n, 2]) +
0.8 (-2.35 - 0.1 Subscript[n, 2])) -
0.19 Subscript[n,
2] (0.39 - 0.1 Subscript[n, 2] +
0.1 (-3 + 2 Subscript[n, 2])))/(0.8 Subscript[n, 1] +
0.19 Subscript[n, 2])
Reduce[driftParamSet > -5 && Subscript[n, 1] >= 0 &&
Subscript[n, 2] >= 0 , {Subscript[n, 1], Subscript[n,
2]}, Integers]
Reduce[driftParamSet > -1000 && Subscript[n, 1] >= 0 &&
Subscript[n, 2] >= 0 , {Subscript[n, 1], Subscript[n, 2]}, Integers]
• Copyable code would enhance your chances of getting a nice answer. – Carl Woll Mar 28 '19 at 2:43
• @CarlWoll, I have added the function and Reduce instances that produce the two kinds of outputs. I apologize for the delay. – gaganso Mar 28 '19 at 16:52
You can use Maximize on the first example:
r1 = Quiet @ Reduce[
driftParamSet > -5 && Subscript[n, 1] >= 0 && Subscript[n, 2] >= 0,
{Subscript[n, 1], Subscript[n, 2]},
Integers
];
Maximize[{Subscript[n, 2], r1}, {Subscript[n, 1], Subscript[n, 2]}]
{657., {Subscript[n, 1] -> 102., Subscript[n, 2] -> 657}}
For the second example, Maximize is unable to find a result, and then uses NMaximize:
r2 = Quiet @ Reduce[
driftParamSet > -1000 && Subscript[n, 1] >= 0 && Subscript[n, 2] >= 0,
{Subscript[n, 1], Subscript[n, 2]},
Integers
];
Maximize[{Subscript[n, 2], r2}, {Subscript[n, 1], Subscript[n, 2]}]
NMaximize::cvmit: Failed to converge to the requested accuracy or precision within 100 iterations.
{22712., {Subscript[n, 1] -> 1118, Subscript[n, 2] -> 22712}}
As the error message says, 100 iterations were not sufficient. So, switch to using NMaximize, and raise the iteration maximum:
NMaximize[
{Subscript[n, 2], r2},
{Subscript[n, 1], Subscript[n, 2]},
MaxIterations -> 2000
]
{114663., {Subscript[n, 1] -> 17793, Subscript[n, 2] -> 114663}} | 1,330 | 3,372 | {"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": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 4, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.15625 | 3 | CC-MAIN-2020-16 | latest | en | 0.440434 |
https://leanprover-community.github.io/archive/stream/116395-maths/topic/pow_lt_pow_of_lt.20golf.html | 1,621,024,827,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243991207.44/warc/CC-MAIN-20210514183414-20210514213414-00158.warc.gz | 375,768,936 | 8,454 | ## Stream: maths
### Topic: pow_lt_pow_of_lt golf
#### Kevin Buzzard (Nov 14 2018 at 14:36):
variable (x : ℝ) -- or [linear_ordered_field]
theorem pow_lt_pow_of_lt {i j : ℕ} : x > 1 → i < j → x^i < x^j := sorry
Is this in mathlib?
#### Mario Carneiro (Nov 14 2018 at 14:44):
all the other versions of this are in group_power.lean, but it looks like this one was missed
#### Kevin Buzzard (Nov 14 2018 at 14:44):
nat.pow_lt_pow_of_lt_right : ∀ {x : ℕ}, x > 1 → ∀ {i j : ℕ}, i < j → x ^ i < x ^ j
This should be a theorem about partially ordered semirings or something, right?
#### Mario Carneiro (Nov 14 2018 at 14:45):
linear_ordered_semiring
#### Mario Carneiro (Nov 14 2018 at 14:45):
because we don't have partially ordered semirings
#### Kevin Buzzard (Nov 14 2018 at 14:47):
I spotted this hole this time last year, when I didn't understand the purpose of the mathlib library. At the time, I just figured this was the sort of thing you had to prove yourself, because I had no concept of what "should be there already" (so I proved it in the case I needed it). I understand this concept much better now.
#### Mario Carneiro (Nov 14 2018 at 14:51):
here's my proof:
theorem pow_lt_pow {a : α} {n m : ℕ} (ha : 1 < a) (h : n < m) : a ^ n < a ^ m :=
lt_of_lt_of_le
((lt_mul_iff_one_lt_left (pow_pos (lt_trans zero_lt_one ha) _)).2 ha)
(pow_le_pow (le_of_lt ha) h)
#### Kevin Buzzard (Nov 14 2018 at 14:53):
I'm doing all my own example sheet questions again after last year's attempts. Some of the code I wrote a year ago was absolutely terrible.
Here's the library I wrote this year, to do a question on my problem sheet about n'th roots (n a positive integer)
https://github.com/ImperialCollegeLondon/M1F_example_sheets/blob/master/src/xenalib/real_nth_root.lean
Any stylistic comments or anything would be welcome. I only care about the reals but really that's for stylistic reasons -- I am trying to write a library with a lot of tactic mode proofs so maths students can follow them more easily, and I wanted to make it as simple as possible. Maybe some of this stuff is in mathlib but I understand my own proofs better -- I find them much more readable.
#### Mario Carneiro (Nov 14 2018 at 14:54):
My proof differs from yours in the proof strategy, which is most of why it is shorter
#### Mario Carneiro (Nov 14 2018 at 14:54):
lt_of_pow_lt also has a very short proof using pow_le_pow
#### Mario Carneiro (Nov 14 2018 at 14:55):
The lesson is "use lemmas"
#### Mario Carneiro (Nov 14 2018 at 14:56):
And I don't just mean use theorems that have already been proven, I mean arrange the proofs of similar facts to make the best use of commonality
#### Mario Carneiro (Nov 14 2018 at 14:59):
your assumptions to lt_of_pow_lt are also stronger than they need to be - it's nice when you can learn this by attempting the proof itself
#### Mario Carneiro (Nov 14 2018 at 15:00):
nth_root_unique is reducible in the sense that it has an equality hypothesis - I would prove a lemma which doesn't have that hypothesis first
#### Mario Carneiro (Nov 14 2018 at 15:02):
it also factors into x ^ n = y ^ n -> x = y, which should also be in group_power in some generality
#### Kevin Buzzard (Nov 14 2018 at 15:23):
here's my proof:
theorem pow_lt_pow {a : α} {n m : ℕ} (ha : 1 < a) (h : n < m) : a ^ n < a ^ m :=
lt_of_lt_of_le
((lt_mul_iff_one_lt_left (pow_pos (lt_trans zero_lt_one ha) _)).2 ha)
(pow_le_pow (le_of_lt ha) h)
pow_le_pow (le_of_lt ha) h is a dirty trick, isn't it?
#### Kevin Buzzard (Nov 14 2018 at 15:26):
So the reason I have noticed that pow_le_pow trick is because I manually completely unfolded your proof:
theorem pow_lt_pow' {a : α} {n m : ℕ} (ha : 1 < a) (h : n < m) : a ^ n < a ^ m :=
begin
apply lt_of_lt_of_le,
{ exact ((lt_mul_iff_one_lt_left (pow_pos (lt_trans zero_lt_one ha) _)).2 ha)},
exact (pow_le_pow (le_of_lt ha) h)
end
theorem pow_lt_pow'' {a : α} {n m : ℕ} (ha : 1 < a) (h : n < m) : a ^ n < a ^ m :=
begin
apply lt_of_lt_of_le,
{ refine (lt_mul_iff_one_lt_left _).2 ha,
refine pow_pos _ _,
-- got it
exact lt_trans zero_lt_one ha
},
{ refine pow_le_pow _ h, -- dirty trick?
exact le_of_lt ha
}
end
into a form which I can actually read. Could there be some code which helps me do this unravelling? It is so much easier for me to inspect nodes of the tree when in tactic mode. @Simon Hudon can code do this? Break down some simple class of term mode functions into a tactic proof?
#### Mario Carneiro (Nov 14 2018 at 15:29):
explode does this
#### Kevin Buzzard (Nov 14 2018 at 15:29):
Even after this breaking-down I still lose information. For example after that first lt_of_lt_of_le -- when I do it in tactic mode I get an extra metavariable goal which Lean has solved in the term mode proof but has not solved in the tactic mode proof. I just want to inspect the metavariable-free goal which is actually proved at each function application I think. How does one do this?
#### Kevin Buzzard (Nov 14 2018 at 15:30):
How do I run explode?
found it
#### Mario Carneiro (Nov 14 2018 at 15:34):
ooh, explode actually works pretty well on that proof
#### Kevin Buzzard (Nov 14 2018 at 15:34):
Quick, we need an emoji
#### Kevin Buzzard (Nov 14 2018 at 15:35):
hey #explode is exactly the answer to my question!
:grinning:
#### Kevin Buzzard (Nov 14 2018 at 15:38):
import tactic.explode
import algebra.group_power
variables {α : Type*} [linear_ordered_semiring α]
theorem pow_lt_pow {a : α} {n m : ℕ} (ha : 1 < a) (h : n < m) : a ^ n < a ^ m :=
lt_of_lt_of_le
((lt_mul_iff_one_lt_left (pow_pos (lt_trans zero_lt_one ha) _)).2 ha)
(pow_le_pow (le_of_lt ha) h)
#explode pow_lt_pow
MWE. Say kids! Understand Mario's proofs in seconds with #explode! Cool name, cool tactic.
#### Kevin Buzzard (Nov 14 2018 at 15:52):
variables {α : Type*} [linear_ordered_semiring α]
theorem pow_lt_pow {a : α} {n m : ℕ} (ha : 1 < a) (h : n < m) : a ^ n < a ^ m :=
lt_of_lt_of_le
( iff.mpr
( lt_mul_iff_one_lt_left $pow_pos ( lt_trans zero_lt_one ha ) _ ) ha ) ( pow_le_pow ( le_of_lt$
ha
)
h
)
Here is another way of taking your proof apart Mario. I have tried to have some sort of a system when unravelling. Is there some sort of name for a form like this? Again I feel like I applied an algorithm. I used $for functions of one variable and indented for two or more. How do I tell which term fills the underscore in that proof, by the way? What's the easiest way? #### Reid Barton (Nov 14 2018 at 15:54): I discovered recently if you replace a _ by a hole {! !} then Lean will give you an error saying there's only one way to fill the hole and tell you what it should be #### Bryan Gin-ge Chen (Nov 14 2018 at 15:54): I was just about to say that ^ #### Bryan Gin-ge Chen (Nov 14 2018 at 15:55): In this case it's n #### Kevin Buzzard (Nov 14 2018 at 15:59): theorem pow_lt_pow' {a : α} {n m : ℕ} (ha : 1 < a) (h : n < m) : a ^ n < a ^ m := lt_of_lt_of_le -- 14 ( iff.mpr -- 11 ( lt_mul_iff_one_lt_left$ -- 10
pow_pos -- 9 -- takes two inputs
( lt_trans -- 8
zero_lt_one -- 7
ha -- 5
)
_ -- gaargh explode doesn't tell me
)
ha -- 5
)
( pow_le_pow -- 13
( le_of_lt \$ -- 12
ha -- 5
)
h -- 6
)
I can't immediately see how to fill in that hole using the output of #explode though
#### Kevin Buzzard (Nov 14 2018 at 15:59):
Funky numbering by the way.
#### Bryan Gin-ge Chen (Nov 14 2018 at 16:07):
This is post-hoc and probably not immediate enough, but first note that the underscore is the second argument to pow_pos (easy to see right away with the bracket colorizer extension), and then compare with the corresponding line of #explode, which says 0 < a ^ n.
#### Kevin Buzzard (Nov 14 2018 at 16:07):
Maybe this indentation is better:
import tactic.explode
import algebra.group_power
variables {α : Type*} [linear_ordered_semiring α]
theorem pow_lt_pow' {a : α} {n m : ℕ} (ha : 1 < a) (h : n < m) :
a ^ n < a ^ m :=
lt_of_lt_of_le -- 14
( iff.mpr -- 11
( lt_mul_iff_one_lt_left -- 10
( pow_pos -- 9 -- takes two inputs
( lt_trans -- 8
zero_lt_one -- 7
ha -- 5
)
_ -- gaargh explode doesn't tell me
)
)
ha -- 5
)
( pow_le_pow -- 13
( le_of_lt -- 12
ha -- 5
)
h -- 6
)
#explode pow_lt_pow'
#### Kevin Buzzard (Nov 14 2018 at 16:08):
import tactic.explode
import algebra.group_power
theorem pow_lt_pow'
{α : Type*} -- 0
[linear_ordered_semiring α] -- 1
{a : α} -- 2
{n m : ℕ} -- 3,4
(ha : 1 < a) -- 5
(h : n < m) : -- 6
a ^ n < a ^ m := -- proof starts
lt_of_lt_of_le -- 14
( iff.mpr -- 11
( lt_mul_iff_one_lt_left -- 10
( pow_pos -- 9
( lt_trans -- 8
zero_lt_one -- 7
ha) -- 5
_)) -- gaargh explode doesn't tell me
ha) -- 5
( pow_le_pow -- 13
( le_of_lt -- 12
ha) -- 5
h) -- 6
#explode pow_lt_pow'
explode covers basically every other line of code
#### Kevin Buzzard (Nov 14 2018 at 16:11):
#explode zmodp.quadratic_reciprocity
-- (deterministic) timeout
boo
#### Bryan Gin-ge Chen (Nov 14 2018 at 16:12):
Hmm, so the line in the output of #explode corresponding to pow_pos is this:
9 │8 │ pow_pos │ 0 < a ^ n
Why doesn't the second column read 8,3 (3 is the line corresponding to n : nat)? It's the same whether I put in n or _.
#### Mario Carneiro (Nov 14 2018 at 16:13):
Non-propositional arguments are automatically suppressed, because they would otherwise dominate the output and they are inferrable from the (fully elaborated) types in the right column
#### Jeremy Avigad (Nov 14 2018 at 16:14):
Wow, explode is really nice. It is great for teaching. I just tried
theorem foo (A B C : Prop) : (A → B → C) → (A ∧ B → C) :=
λ h hab, and.elim hab (λ ha hb, h ha hb)
#explode foo
It made me realize that the proof can be shortened:
theorem foo' (A B C : Prop) : (A → B → C) → (A ∧ B → C) :=
λ h hab, and.elim hab h
The tactic doesn't behave well with have, though.
theorem bar (A B C : Prop) : A ∧ (B ∨ C) → (A ∧ B) ∨ (A ∧ C) :=
assume h : A ∧ (B ∨ C),
have h₁ : A, from and.left h,
have h₂ : B ∨ C, from and.right h,
or.elim h₂
(assume h₃ : B,
or.inl (and.intro h₁ h₃))
(assume h₃ : C,
or.inr (and.intro h₁ h₃))
#explore bar
@Mario Carneiro Would it be hard to handle have nicely in the tactic?
#### Mario Carneiro (Nov 14 2018 at 16:18):
hm, it's explode not explore but that is also an interesting name
#### Jeremy Avigad (Nov 14 2018 at 16:19):
Oops, sorry, I forgot to cut-and-paste that part and added it manually. You are right, it is also a good name.
#### Mario Carneiro (Nov 14 2018 at 16:20):
have should be handled well since it will translate to a proof line that is referred to twice
#### Mario Carneiro (Nov 14 2018 at 16:20):
what does it look like now? (on my phone)
#### Mario Carneiro (Nov 14 2018 at 16:22):
indeed it is best suited to the basic dtt proofs used in intro logic
#### Kevin Buzzard (Nov 14 2018 at 16:22):
In general, I think I'd rather read the output from the bottom up when trying to figure out what you're doing.
yes absolutely
#### Kevin Buzzard (Nov 14 2018 at 16:22):
so...it's upside-down?
#### Mario Carneiro (Nov 14 2018 at 16:23):
well no, it is meant to be read bottom up
#### Mario Carneiro (Nov 14 2018 at 16:23):
but it follows the usual proof order
#### Mario Carneiro (Nov 14 2018 at 16:24):
it is a fitch style proof display
#### Kevin Buzzard (Nov 14 2018 at 16:25):
Thanks a lot for alerting me to it.
#### Bryan Gin-ge Chen (Nov 14 2018 at 16:25):
#explode bar looks like this now:
bar : ∀ (A B C : Prop), A ∧ (B ∨ C) → A ∧ B ∨ A ∧ C
0│ │ A ├ Prop
1│ │ B ├ Prop
2│ │ C ├ Prop
3│ │ h ├ A ∧ (B ∨ C)
4│ │ λ (h₁ : A),
have h₂ : B ∨ C, from h.right,
or.elim h₂ (λ (h₃ : B), or.inl ⟨h₁, h₃⟩) (λ (h₃ : C), or.inr ⟨h₁, h₃⟩) │ A → A ∧ B ∨ A ∧ C
5│3 │ and.left │ A
6│4,5│ ∀E │ A ∧ B ∨ A ∧ C
#### Mario Carneiro (Nov 14 2018 at 16:26):
also pp.beta helps sometimes
Last updated: May 14 2021 at 20:13 UTC | 3,778 | 12,525 | {"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": 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} | 3.171875 | 3 | CC-MAIN-2021-21 | latest | en | 0.882525 |
http://www.mathworks.ch/matlabcentral/answers/40085 | 1,369,440,478,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368705300740/warc/CC-MAIN-20130516115500-00082-ip-10-60-113-184.ec2.internal.warc.gz | 587,375,525 | 6,337 | # Matrix Simple Index Question
Asked by Steve on 1 Jun 2012
Latest activity Commented on by Steve on 2 Jun 2012
Hello Experts,
1) Given matrix A with m rows and n columns, I want to check if there is an entry A(i,j)>alpha and if yes to make it A(i,j) = beta. How to make it without for and if? 2) How to make it in a vector of size (1,m) or (1,n)?
## Products
Answer by Oleg Komarov on 1 Jun 2012
Sample inputs:
```m = 10;
n = 13;
A = rand(m,n);
alpha = .38;
```
Use logical indexing
```idx = A > alpha;
A(idx) = 10;
```
Same applies to a vector (just play around with m and n to see that).
Oleg Komarov on 1 Jun 2012
Check out the getting started guide and especially this section: http://www.mathworks.co.uk/help/techdoc/learn_matlab/f2-8955.html
Oleg Komarov on 1 Jun 2012
Demos are also useful: http://www.mathworks.co.uk/products/matlab/demos.html
Steve on 2 Jun 2012
Dear Oleg, again thank you very very much!!!
It was my great pleasure to get your answer and experts guide!
Steve | 298 | 1,003 | {"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} | 2.84375 | 3 | CC-MAIN-2013-20 | latest | en | 0.88233 |
https://www.fresherslive.com/placement-papers/apps-associates-interview-questions-answers/2 | 1,597,015,019,000,000,000 | text/html | crawl-data/CC-MAIN-2020-34/segments/1596439738595.30/warc/CC-MAIN-20200809222112-20200810012112-00496.warc.gz | 667,535,881 | 46,680 | # Apps Associates Interview Questions
Posted on :20-04-2016
Q1. Find the next term in series? 25 16 9 4 1 0
A.-1
B.1
C.0
D.none
ANS: B
Explanation:
5 ^ 2, 4 ^ 2, 3 ^ 2, 2 ^ 2, 1^ 2, 0 ^ 2, (-1) ^2 = 1
Q2. 123456147, 12345614, 2345614, 234561, (.....)
A.3456
B.2345
C.23456
D.34561
ANS: D
Explanation:
The digits are removed one by one from the end as well as from the beginning in order so as to obtain the next term of the series.
Q3. 45 grinders brought at Rs. 2215/-.transpot expense Rs 2190/-.2760/- on octroi. Find selling price/piece to make profit of 20%?
A.2585
B.2225
C.2790
D.3325
ANS: C
Explanation:
Grider pricepiece = 2215 Transport exppiece= (2190+2760)/45 = 110 Total acquiring price of grinderpiece = 2215+110 = 2325
20% profit on C = (2325 * 20)/100 = 465 Selling price C+D = 2790.
Q4. The last digit of (2004) ^5 is
A.4
B.8
C.6
D.2
ANS: A
Explanation:
As 2004 = 2000 + 4, the last digits of (2004) ^5 and 4^5 are equal. i.e. 4^5 = 1024 so unit digit is 4.
Q5. In a 2 digit number unit's place is halved and ten's place is doubled. Difference between the numbers is 37.Digit in unit's place is 2 more than ten's place. Find the number?
A.24
B.46
C.42
D.NONE
ANS: B
Explanation:
Let the ten's digit be x then unit digit= x+2 Number = 10x+(x+2) = 11x+2 -------- (1) The number obtained by doubling digit in ten's place
& by reducing unit's place digit to half the original value = 2(10x) + (x+2)/2 -------- (2) Difference between the numbers (1) - (2) = 37.
By solving equation X= 4 Therefore number 11* 4 + 2= 46.
Q6. If x - y + z = 19, y + z =20, x - z = 3, find the value of x + 4y - 5z ?
A.22
B.38
C.17
D.none
ANS: D
Explanation:
X - y + z = 19.......Eqn 1 y + z = 20............Eqn 2 x - z = 3...............Eqn 3 Adding Eqn 1 and Eqn 2 gives
x - y + z + y + z = x + 2z = 19 + 20 = 39 Subtracting Eqn 3 from the above gives x + 2z -(x - z) = 3z = 39 - 3 = 36............z = 12 From Eqn 2,
y + 12 = 20 ......................y = 8 From Eqn 1, x - 8 + 12 = x + 4 = 19....x = 15 Then, x + 4y - 5z = 15 + (4x8) - (5x12) = 15 + 32 - 60 = -13
Q7. A is 4 yrs old and B is thrice A. when A is 12 yrs, how old will B be ?
A.16
B.20
C.24
D.28
ANS: B
Explanation:
A= 4 years B = 3times A= 12 years A will be 12 after 8 years when A will be 12 years B will be 12 + 8 = 20 years
Q8. A motorboat whose speed is 15 kmph in still water goes 30 kmph downstream and comes back in a total of 4hrs 30min.Find the speed of the stream in kmph?
A.4
B.5
C.24
D.10
ANS: B
Explanation:
Let the speed of the stream be x km/hr. Then, Speed downstream = (15 + x) km/hr, Speed upstream = (15 - x) km/hr.
Therefore 30 / (15 + x) + 30 / (15 - x) = 4(1/2) 900/ (225 - x^ 2) = 9/2 9x2 = 225 x2 = 25 x = 5 km/hr.
Q9. A garrison had provision for a certain number of days. After 10 days, 1/5 of the men desert and it is found that the provisions will now last just as long as before. How long was that ?
A.15
B.25
C.35
D.50
ANS: D
Explanation:
Let initially there be X men having food for Y days. After 10 days X men had food for Y-10 days. Also X - X/5 = 4/5X had food for Y days.
Apply chain rule X(Y-10) = (4/5) X *Y (X*Y) - 10 X=[4(X*Y)/5] 5(X*Y) - 4(X*Y)=50X X*Y = 50XY = 50
Q10. Find the equation whose roots are 9 and 5 ?
A.x2 + 14x + 45 = 0
B.x2 + 14x + 49 = 0
C.x2 + 10x + 45 = 0
D.NONE
ANS: A
Explanation:
Finding the roots of given equations x2 - 14x + 45 = 0 x2 - 9 x - 5 x + 45 = 0 (x - 5) (x - 9) = 0
Q11. Find the next term in series - Y W U S Q O M ?
A.P
B.J
C.L
D.NONE
ANS: D
Explanation:
Correct ans should be K.
Q12. In a triangle ABC, the lengths of the sides AB and AC equal 17.5 cm and 9 cm respectively. Let D be a point on the line segment BC such that AD is perpendicular to BC. If AD = 3 cm, then what is the radius (in cm) of the circle circumscribing the triangle ABC?
A.17.05
B.27.85
C.22.45
D.26.25
ANS: D
Q13. Train A starts from Meerut at 12:00 pm & reach at Delhi on 2:30 pm & train B starts from Delhi at 12:15pm and reach at Meerut on 2:15pm. So, when both trains crosses each other ?
A.1.00PM
B.1.15PM
C.1.30PM
D.NONE
ANS: B
Explanation:
Let the distance between Meerut & Delhi be x kms & let trains meet y hours after 12.15pm Speed of A= x/150 kms/min.
speed of B = x/120 kms / min. Distance covered by A in (y+15) min+distance covered by B in y hours=x (x/150)* (y+15) + (x/120) * y = x Equating
the equations Y=60 Hence trains meet at 1.15 pm.
Q14. In election a candidate who gets 42% of votes, is lost by 112 votes. What is the total number of votes polled?
A.650
B.680
C.710
D.700
ANS: D
Explanation:
Let the total number of votes polled be x, then votes polled by other candidate = (100-42)% of x = 58% of x Therefore 58% of x- 42% of x = 112 X= 700.
Q15. A boat covers 2 kms distance downstream in 1/2 hour. while it comes back in 1 hour. Then the velocity of the current is
A.1 km/hr
B.2 km/hr
C.1.15 km/hr
D.3 km/hr
ANS: A
Explanation:
Rate down stream = (2/0.5) kmph Rate upstream = (2/1) kmph Therefore velocity of current = ½ (4-2) = 1kmph.
Q16. A man took loan from a bank at the rate of 12% p.a. simple interest. After 3 years he had to pay Rs. 5400 interest only for the period. The principal amount borrowed by him was:
A.2000
B.10000
C.15000
D.20000
ANS: C
Explanation:
Principal = Rs. (100 x 5400) / (12 x 3)= Rs. 15000.
Q17. A man traveled a certain distance at the rate of 15 miles/hour and came back at the rate of 10 miles/hour. What is his average speed ?
A.10 miles/hour
B.15 miles/hour
C.17 miles/hour
D.12 miles/hour
ANS: D
Explanation:
Average speed = (2* 15* 10) / 25 = 300/25 12 miles/ hour. Aver
Q18. The addition of 2 number difference of 2 number is a perfect square & the difference of both perfect square also a perfect square. Then find out this number ?
A.2,4
B.4,4
C.6,9
D.6,2
ANS: D
Explanation:
Checking the given options =>6, 2 =>6 + 2 = 8, 6 - 2 = 4 and 8 - 4 = 4 By c
Q19. The sum of 3rd and 15th elements of an arithmetic progression is equal to the sum of 6th, 11th and 13th elements of the same progression. Then which element of the series should necessarily be equal to zero ?
A.1st
B.9th
C.12th
D.NONE
ANS: C
Explanation:
lets the first term of ap is a and difference is d then third term will be =a+2d 15th will be=a+14d 6th will be =a+5d 11th will be =a+10d 13th will be=a+12d then
the equation will be as a+2d+a+14d = a+5d+a+10d+a+12d 2a+16d = 3a+27d a+11d=0 simply we can understand a+11d will be the 12th term of arithmetic progression so the correct answer is 12
Q20. X + Y = 6, then XY = ?
A.36
B.8
C.3
D.30
ANS: B
Explanation:
Lets us consider x = 4, y = 2 x + y = 4 + 2 = 6 xy = 4*2 = 8
Q21. There are three coins of Re1, 50 ps, 25 ps having ratio of 13:11:3. The total sums of money are 77, and then find out how much Re. 1 coin is there ?
A.22
B.52
C.27
D.NONE
ANS: B
Explanation:
Let number if coins of each denomination be x. Then as they are in ratio of 13:11:3 Then (13/27)x + (11/27)(1/2)x + (3/27)(1/4)x = 77 Taking L.C.M.
& equating LHS & RHS gives x = 108 Number of 1 rupee coins = (13/27)*108 = 52.
Q22. Ten years ago, the ages of the members of a joint family of eight people added up to 231 years. Three years later, one member died at the age of 60 years and a child was born during the same year. After another three years, one more member died, again at 60, and a child was born during the same year. The current average age of this eight member joint family is nearest to:
A.23years
B.22years
C.21years
D.24years
ANS: D
Explanation:
The sum of the ages of the members of the family ten years ago = 231. The sum of the ages of the members of the family seven years ago = 231 + (3 × 8) - 60 =195.
The sum of the ages of the members of the family four years ago = 195 + (3 × 8) - 60 = 159 ∴ The sum of the ages of the members of the family
now = 159 + (4 × 8) = 191 ∴ Required average = 191/8 = 23.875 ≈ 24.
Q23. A test has 50 questions. A student scores 1 mark for a correct answer, -1/3 for a wrong answer, and -1/6 for not attempting a question. If the net score of a student is 32, the number of questions answered wrongly by that student cannot be less than ?
A.3
B.6
C.12
D.9
ANS: A
Explanation:
Let the number of correct answers be 'x', number of wrong answers be 'y' and number of questions not attempted be 'z'.
Thus, x + y + z = 50 ... (i) And x - y/3 - z/6 = 32 The second equation can be written as, 6x - 2y - z = 192 ... (ii) Adding the two equations we get,
7x - y = 242 or x = (242 + y)/7 Since, x and y are both integers, y cannot be 1 or 2. The minimum value that y can have is 3.
Q24. How many even integers n, where 100
A.40
B.37
C.38
D.39
ANS: D
Explanation:
There are 101 integers in all, of which 51 are even. From 100 to 200, there are 14 multiples of 7, of which 7 are even.
There are 11 multiples of 9, of which 6 are even. But there is one integer (i.e. 126) that is a multiple of both 7 and 9 and also even.
Hence the answer is (51 - 7 - 6 + 1) = 39
Q25. DIRECTIONS for Questions 25 and 26:
Answer the questions on the basis of the information given below. The Head of a newly formed governmentdesires to appoint five of the six elected members A, B, C, D, E and F to portfolios of Home, Power, Defense, Telecom and Finance.
F does not want any portfolio if D gets one of the five. C wants either Home or Finance or no portfolio. B says that if D gets either Power or Telecom then she must get the other one. E insists on a portfolio if A gets one. Which is a valid assignment ?
A.A-Home, B-Power, C-Defense, D-Telecom, E-Finance.
B.C-Home, D-Power, A-Defense, B-Telecom, E-Finance
C.A-Home, B-Power, E-Defense, D-Telecom, F-Finance.
D.B-Home, F-Power, E-Defense, C-Telecom, A -Finance
ANS: B
Explanation:
From the above information we can note that => If D gets any then F does not want any .So, option C invalid. => C wants in either
Q26. If A gets Home and C gets Finance, then which is NOT a valid assignment for Defense and Telecom ?
A.D-Defense, B-Telecom.
B.F-Defense, B-Telecom
C.B-Defense, E-Telecom.
D.B-Defense, D-Telecom.
ANS: D
Explanation:
As B says that if D gets either Power or Telecom then she must get the other one option D is invalid.
in Home or Finance. Therefore option A and D invalid => So, option B answer.
Q27. DIRECTIONS for Questions 27 to 28:
Answer the questions on the basis of the information given below. Rang Barsey Paint Company (RBPC) is in the business of manufacturing paints. RBPC buys RED, YELLOW, WHITE, ORANGE, and PINK paints. ORANGE paint can be also produced by mixing RED and YELLOW paints in equal proportions. Similarly, PINK paint can also be produced by mixing equal amounts of RED and WHITE paints. Among other paints, RBPC sells CREAM paint, (formed by mixing WHITE and YELLOW in the ratio 70:30) AVOCADO paint (formed by mixing equal amounts of ORANGE and PINK paint) and WASHEDORANGE paint (formed by mixing equal amounts of ORANGE and WHITE paint). The following table provides the price at which RBPC buys paints. Color Rs. /liter RED 20 YELLOW 25 WHITE 15 ORANGE 22 PINK18 WASHEDORANGE can be manufactured by mixing .......
A.CREAM and RED in the ratio 14:10
B.CREAM and RED in the ratio 3:1
C.YELLOW and PINK in the ratio 1:1
D.RED, YELLOW, and WHITE in the ratio 1:1:2
ANS: D
Explanation:
Mixing equal amounts of ORANGE and WHITE can make WASHED ORANGE; ORANGE can be made by mixing equal amounts of RED and YELLOW.
So the ratio of RED, YELLOW and WHITE is 1:1:2
Q28. Assume that AVOCADO, CREAM, and WASHEDORANGE each sell for the same.....
B.CREAM.
C.WASHEDORANGE
D.Sufficient data is not available.
ANS: B
Explanation:
If cost of AVOCADO paint is Rs.19.75 The cost of the CREAM is [(7 x 15) + (3 x 75)]/10 = Rs. 18 And cost of WASHEDORANGE is Rs.18.50 So CREAM is the most profitable.
Q29. DIRECTIONS for Questions 29 to 30:
Answer the questions on the basis of the information given below. A, B, C, D, E, and F are a group of friends. There are two housewives, one professor, one engineer, one accountant and one lawyer in the group. There are only two married couples in the group. The lawyer is married to D, who is a housewife. No woman in the group is either an engineer or an accountant. C the accountant is married to F, who is a professor. A is married to a housewife. E is not a housewife. Which of the following is one of the married couples ?
A.A&B
B.B&E
C.D&E
D.A&D
ANS: D
Explanation:
One couple is C and F, here C is male and F is female. Now there can be only one more couple. A is married to a housewife, and D is housewife. So the other couple is A (lawyer) & D.
Q30. What is E's profession ?
A.Engineer
B.Lawyer
C.Professor
D.Accountant
ANS: A
Explanation:
A is Lawyer, D is house wife, C is Accountant, and F is Professor. B is housewife (since E is not a housewife.) then E is Engineer.
## Apps Associates Placement Papers
➤ Apps Associates Technical Paper
FreshersLive - No.1 Job site in India. Here you can find latest 2020 government as well as private job recruitment notifications for different posts vacancies in India. Get top company jobs for both fresher and experienced. Job Seekers can get useful interview tips, resume services & interview Question and answer. Practice online test free which is helpful for interview preparation. Register with us to get latest employment news/rojgar samachar notifications. Also get latest free govt and other sarkari naukri job alerts daily through E-mail.
✖ | 4,511 | 13,476 | {"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} | 3.921875 | 4 | CC-MAIN-2020-34 | latest | en | 0.751046 |
https://oeis.org/A087265 | 1,723,288,321,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640805409.58/warc/CC-MAIN-20240810093040-20240810123040-00730.warc.gz | 341,731,663 | 5,854 | The OEIS is supported by the many generous donors to the OEIS Foundation.
Hints (Greetings from The On-Line Encyclopedia of Integer Sequences!)
A087265 Lucas numbers L(8*n). 10
2, 47, 2207, 103682, 4870847, 228826127, 10749957122, 505019158607, 23725150497407, 1114577054219522, 52361396397820127, 2459871053643326447, 115561578124838522882, 5428934300813767249007, 255044350560122222180447 (list; graph; refs; listen; history; text; internal format)
OFFSET 0,1 COMMENTS a(n+1)/a(n) converges to (47+sqrt(2205))/2 = 46.9787137... a(0)/a(1)=2/47; a(1)/a(2)=47/2207; a(2)/a(3)=2207/103682; a(3)/a(4)=103682/4870847; etc. Lim_{n->infinity} a(n)/a(n+1) = 0.02128623625... = 2/(47+sqrt(2205)) = (47-sqrt(2205))/2. a(n) = a(-n). - Alois P. Heinz, Aug 07 2008 From Peter Bala, Oct 14 2019: (Start) Let F(x) = Product_{n >= 0} (1 + x^(4*n+1))/(1 + x^(4*n+3)). Let Phi = 1/2*(sqrt(5) - 1). This sequence gives the partial denominators in the simple continued fraction expansion of the number F(Phi^8) = 1.0212763906... = 1 + 1/(47 + 1/(2207 + 1/(103682 + ...))). Also F(-Phi^8) = 0.9787231991... has the continued fraction representation 1 - 1/(47 - 1/(2207 - 1/(103682 - ...))) and the simple continued fraction expansion 1/(1 + 1/((47 - 2) + 1/(1 + 1/((2207 - 2) + 1/(1 + 1/((103682 - 2) + 1/(1 + ...))))))). F(Phi^8)*F(-Phi^8) = 0.9995468962... has the simple continued fraction expansion 1/(1 + 1/((47^2 - 4) + 1/(1 + 1/((2207^2 - 4) + 1/(1 + 1/((103682^2 - 4) + 1/(1 + ...))))))). 1/2 + 1/2*F(Phi^8)/F(-Phi^8) = 1.0217391349... has the simple continued fraction expansion 1 + 1/((47 - 2) + 1/(1 + 1/((103682 - 2) + 1/(1 + 1/(228826127 - 2) + 1/(1 + ...))))). (End) REFERENCES R. P. Stanley. Enumerative combinatorics. Vol. 2, volume 62 of Cambridge Studies in Advanced Mathematics. Cambridge University Press, Cambridge, 1999. LINKS Indranil Ghosh, Table of n, a(n) for n = 0..596 Tanya Khovanova, Recursive Sequences A. V. Zarelua, On Matrix Analogs of Fermat's Little Theorem, Mathematical Notes, vol. 79, no. 6, 2006, pp. 783-796. Translated from Matematicheskie Zametki, vol. 79, no. 6, 2006, pp. 840-855. Index entries for recurrences a(n) = k*a(n - 1) +/- a(n - 2) Index entries for linear recurrences with constant coefficients, signature (47,-1). FORMULA a(n) = 47*a(n-1) - a(n-2), starting with a(0) = 2 and a(1) = 47. a(n) = ((47+sqrt(2205))/2)^n + ((47-sqrt(2205))/2)^n (a(n))^2 = a(2n)+2. G.f.: (2-47*x)/(1-47*x+x^2). - Alois P. Heinz, Aug 07 2008 From Peter Bala, Oct 14 2019: (Start) a(n) = F(8*n+8)/F(8) - F(8*n-8)/F(8) = A049668(n+1) - A049668(n-1). a(n) = trace(M^n), where M is the 2 X 2 matrix [0, 1; 1, 1]^8 = [13, 21; 21, 34]. Consequently the Gauss congruences hold: a(n*p^k) = a(n*p^(k-1)) ( mod p^k ) for all prime p and positive integers n and k. See Zarelua and also Stanley (Ch. 5, Ex. 5.2(a) and its solution). 45*Sum_{n >= 1} 1/(a(n) - 49/a(n)) = 1: (49 = Lucas(8) + 2 and 45 = Lucas(8) - 2) 49*Sum_{n >= 1} (-1)^(n+1)/(a(n) + 45/a(n)) = 1. x*exp(Sum_{n >= 1} a(n)*x^/n) = x + 47*x^2 + 2208*x^3 + ... is the o.g.f. for A049668. (End) E.g.f.: 2*exp(47*x/2)*cosh(21*sqrt(5)*x/2). - Stefano Spezia, Oct 18 2019 EXAMPLE a(4) = 4870847 = 47*a(3) - a(2) = 47*103682 - 2207=((47+sqrt(2205))/2)^4 + ( (47-sqrt(2205))/2)^4 =4870846.999999794696 + 0.000000205303 = 4870847. MAPLE a:= n-> (Matrix([[2, 47]]). Matrix([[47, 1], [ -1, 0]])^(n))[1, 1]: seq(a(n), n=0..14); # Alois P. Heinz, Aug 07 2008 MATHEMATICA LucasL[8*Range[0, 20]] (* or *) LinearRecurrence[{47, -1}, {2, 47}, 20] (* Harvey P. Dale, Oct 23 2017 *) PROG (Magma) [ Lucas(8*n) : n in [0..100]]; // Vincenzo Librandi, Apr 14 2011 CROSSREFS Cf. A000032. Cf. Lucas(k*n): A005248 (k = 2), A014448 (k = 3), A056854 (k = 4), A001946 (k = 5), A087215 (k = 6), A087281 (k = 7), A087287 (k = 9), A065705 (k = 10), A089772 (k = 11), A089775 (k = 12). a(n) = A000032(8n). Sequence in context: A246543 A119776 A373920 * A079307 A368193 A005814 Adjacent sequences: A087262 A087263 A087264 * A087266 A087267 A087268 KEYWORD easy,nonn AUTHOR Nikolay V. Kosinov (kosinov(AT)unitron.com.ua), Oct 19 2003 EXTENSIONS Terms a(22)-a(27) from John W. Layman, Jun 14 2004 STATUS approved
Lookup | Welcome | Wiki | Register | Music | Plot 2 | Demos | Index | Browse | More | WebCam
Contribute new seq. or comment | Format | Style Sheet | Transforms | Superseeker | Recents
The OEIS Community | Maintained by The OEIS Foundation Inc.
Last modified August 10 05:56 EDT 2024. Contains 375044 sequences. (Running on oeis4.) | 1,878 | 4,482 | {"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} | 3.53125 | 4 | CC-MAIN-2024-33 | latest | en | 0.526954 |
http://www.infobarrel.com/How_To_Multiply_And_Divide_Mixed_Fractions | 1,540,317,199,000,000,000 | text/html | crawl-data/CC-MAIN-2018-43/segments/1539583516892.84/warc/CC-MAIN-20181023174507-20181023200007-00389.warc.gz | 475,667,469 | 14,007 | Mixed fractions can be multiplied and divided just like standard fractions once they are rewritten into the form of improper fractions. This article will explain how this process works.
Fractions include two parts: a numerator and a denominator. Improper fractions also include the same two parts except that the numerator is greater than the denominator. Mixed fractions on the other hand include an integer term along with the numerator and denominator. Mixed fractions are essentially the same thing as improper fractions, they are just written differently. An improper fraction can be transformed into the mixed fraction form by dividing the denominator into the numerator. Since the numerator is greater than the denominator, the denominator will divide evenly into the numerator some number of times, this is the integer term. The remainder that is left over after the division will result in a fraction with a new numerator and the original denominator. An example shows this quite clear. The fraction, 17/4 is an improper fraction and to rewrite this into a mixed fraction, we will divide the denominator of 4 into the numerator of 17. 4 can be divided into 17, 4 times, so 4 is our integer term. Since 4 times 4 is 16, we are left with a remainder of 1 which is the numerator of our new fraction that also includes the old denominator, so we have 4 1/4.
To rewrite mixed fractions into improper fractions, we must simply do the opposite of the procedure above. Recognizing that in a mixed number, the integer term is the number of times the denominator goes evenly into a large numerator, we realize that multiplying the integer term by the denominator and adding the remainder in the numerator will give us the numerator of the improper fraction, whereas the denominator remains unchanged. An example will further explain this. Consider we want to rewrite the mixed number of 4 2/9 into the form of an improper fraction. We see that 4 (the integer term) x 9 (the denominator) gives 36 and we add the remainder of 2 which yields 38. So the improper fraction is 38/9. If we rewrite this back into the mixed fraction form, we get that 9 divides into38, 4 times and leaves a remainder of 2 and so as expected we have 4 2/9.
Now once we have rewritten our mixed numbers, we can multiply and divide them just like we would with any fraction.
Some examples will clarify these points.
1.) 2 5/7 x 6 3/7 = [((2 x 7) + 5)/7][((6 x 7) + 3)/7] = (19/7)(45/7)
(19 x 45)/(7 x 7) = 855/49
We will further reduce this answer into the mixed fraction form.
855/49 = 17 22/49
2.) 10 2/9 divided by 4 5/8 = ?
10 2/9 = ((10 x 9) + 2)/9 = 92/9
4 5/8 = ((4 x 8) + 5)/8 = 37/8
So we have (92/9) / (37/8) = (92/9) (8/37) = (92 x 8)/(9 x 37) = 736/333 = 2 70/333
To perform this division, we found the reciprocal of 37/8 and multiplied it by (92/9) and then reduced the product to a mixed fraction. | 735 | 2,895 | {"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} | 4.90625 | 5 | CC-MAIN-2018-43 | longest | en | 0.927905 |
https://www.lmfdb.org/ModularForm/GL2/Q/holomorphic/1815/2/c/k/ | 1,611,186,365,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703522133.33/warc/CC-MAIN-20210120213234-20210121003234-00060.warc.gz | 867,969,481 | 55,720 | # Properties
Label 1815.2.c.k Level $1815$ Weight $2$ Character orbit 1815.c Analytic conductor $14.493$ Analytic rank $0$ Dimension $24$ CM no Inner twists $2$
# Related objects
## Newspace parameters
Level: $$N$$ $$=$$ $$1815 = 3 \cdot 5 \cdot 11^{2}$$ Weight: $$k$$ $$=$$ $$2$$ Character orbit: $$[\chi]$$ $$=$$ 1815.c (of order $$2$$, degree $$1$$, minimal)
## Newform invariants
Self dual: no Analytic conductor: $$14.4928479669$$ Analytic rank: $$0$$ Dimension: $$24$$ Twist minimal: no (minimal twist has level 165) Sato-Tate group: $\mathrm{SU}(2)[C_{2}]$
## $q$-expansion
The dimension is sufficiently large that we do not compute an algebraic $$q$$-expansion, but we have computed the trace expansion.
$$\operatorname{Tr}(f)(q) =$$ $$24q - 24q^{4} - 2q^{5} + 8q^{6} - 24q^{9} + O(q^{10})$$ $$\operatorname{Tr}(f)(q) =$$ $$24q - 24q^{4} - 2q^{5} + 8q^{6} - 24q^{9} + 6q^{10} - 12q^{14} + 48q^{16} - 32q^{19} - 2q^{20} + 16q^{21} - 24q^{24} + 2q^{25} + 32q^{26} + 8q^{30} - 12q^{34} - 10q^{35} + 24q^{36} - 36q^{39} - 34q^{40} + 2q^{45} + 56q^{46} - 24q^{49} - 46q^{50} + 36q^{51} - 8q^{54} + 12q^{56} - 40q^{59} - 26q^{60} + 40q^{61} + 12q^{64} - 10q^{65} - 2q^{70} + 64q^{71} + 136q^{74} + 20q^{75} + 68q^{76} - 64q^{79} + 76q^{80} + 24q^{81} - 60q^{84} - 72q^{86} + 20q^{89} - 6q^{90} + 4q^{94} - 64q^{95} + 56q^{96} + O(q^{100})$$
## Embeddings
For each embedding $$\iota_m$$ of the coefficient field, the values $$\iota_m(a_n)$$ are shown below.
For more information on an embedded modular form you can click on its label.
Label $$a_{2}$$ $$a_{3}$$ $$a_{4}$$ $$a_{5}$$ $$a_{6}$$ $$a_{7}$$ $$a_{8}$$ $$a_{9}$$ $$a_{10}$$
364.1 2.72592i 1.00000i −5.43063 1.85250 1.25229i 2.72592 0.486331i 9.35163i −1.00000 −3.41365 5.04977i
364.2 2.59201i 1.00000i −4.71851 −0.0953833 + 2.23403i 2.59201 3.41835i 7.04639i −1.00000 5.79063 + 0.247234i
364.3 2.36377i 1.00000i −3.58741 0.979185 + 2.01027i −2.36377 1.71423i 3.75227i −1.00000 4.75182 2.31457i
364.4 2.21395i 1.00000i −2.90157 −2.14116 + 0.644543i −2.21395 1.59339i 1.99602i −1.00000 1.42699 + 4.74042i
364.5 2.10651i 1.00000i −2.43738 −0.823234 2.07901i 2.10651 3.31633i 0.921348i −1.00000 −4.37946 + 1.73415i
364.6 1.90743i 1.00000i −1.63829 −2.10308 + 0.759628i 1.90743 4.45734i 0.689943i −1.00000 1.44894 + 4.01149i
364.7 1.27093i 1.00000i 0.384732 1.01340 1.99324i −1.27093 0.483291i 3.03083i −1.00000 −2.53328 1.28796i
364.8 0.803366i 1.00000i 1.35460 2.13623 0.660705i 0.803366 0.508505i 2.69497i −1.00000 −0.530788 1.71617i
364.9 0.784131i 1.00000i 1.38514 −1.83964 + 1.27111i −0.784131 1.51801i 2.65439i −1.00000 0.996719 + 1.44252i
364.10 0.488299i 1.00000i 1.76156 2.10928 0.742259i 0.488299 5.10895i 1.83677i −1.00000 −0.362444 1.02996i
364.11 0.298064i 1.00000i 1.91116 −1.68093 + 1.47460i 0.298064 2.32107i 1.16578i −1.00000 0.439527 + 0.501026i
364.12 0.288813i 1.00000i 1.91659 −0.407156 2.19869i −0.288813 3.66740i 1.13116i −1.00000 −0.635009 + 0.117592i
364.13 0.288813i 1.00000i 1.91659 −0.407156 + 2.19869i −0.288813 3.66740i 1.13116i −1.00000 −0.635009 0.117592i
364.14 0.298064i 1.00000i 1.91116 −1.68093 1.47460i 0.298064 2.32107i 1.16578i −1.00000 0.439527 0.501026i
364.15 0.488299i 1.00000i 1.76156 2.10928 + 0.742259i 0.488299 5.10895i 1.83677i −1.00000 −0.362444 + 1.02996i
364.16 0.784131i 1.00000i 1.38514 −1.83964 1.27111i −0.784131 1.51801i 2.65439i −1.00000 0.996719 1.44252i
364.17 0.803366i 1.00000i 1.35460 2.13623 + 0.660705i 0.803366 0.508505i 2.69497i −1.00000 −0.530788 + 1.71617i
364.18 1.27093i 1.00000i 0.384732 1.01340 + 1.99324i −1.27093 0.483291i 3.03083i −1.00000 −2.53328 + 1.28796i
364.19 1.90743i 1.00000i −1.63829 −2.10308 0.759628i 1.90743 4.45734i 0.689943i −1.00000 1.44894 4.01149i
364.20 2.10651i 1.00000i −2.43738 −0.823234 + 2.07901i 2.10651 3.31633i 0.921348i −1.00000 −4.37946 1.73415i
See all 24 embeddings
$$n$$: e.g. 2-40 or 990-1000 Embeddings: e.g. 1-3 or 364.24 Significant digits: Format: Complex embeddings Normalized embeddings Satake parameters Satake angles
## Inner twists
Char Parity Ord Mult Type
1.a even 1 1 trivial
5.b even 2 1 inner
## Twists
By twisting character orbit
Char Parity Ord Mult Type Twist Min Dim
1.a even 1 1 trivial 1815.2.c.k 24
5.b even 2 1 inner 1815.2.c.k 24
5.c odd 4 1 9075.2.a.dx 12
5.c odd 4 1 9075.2.a.ea 12
11.b odd 2 1 1815.2.c.j 24
11.d odd 10 2 165.2.s.a 48
33.f even 10 2 495.2.ba.c 48
55.d odd 2 1 1815.2.c.j 24
55.e even 4 1 9075.2.a.dy 12
55.e even 4 1 9075.2.a.dz 12
55.h odd 10 2 165.2.s.a 48
55.l even 20 2 825.2.n.o 24
55.l even 20 2 825.2.n.p 24
165.r even 10 2 495.2.ba.c 48
By twisted newform orbit
Twist Min Dim Char Parity Ord Mult Type
165.2.s.a 48 11.d odd 10 2
165.2.s.a 48 55.h odd 10 2
495.2.ba.c 48 33.f even 10 2
495.2.ba.c 48 165.r even 10 2
825.2.n.o 24 55.l even 20 2
825.2.n.p 24 55.l even 20 2
1815.2.c.j 24 11.b odd 2 1
1815.2.c.j 24 55.d odd 2 1
1815.2.c.k 24 1.a even 1 1 trivial
1815.2.c.k 24 5.b even 2 1 inner
9075.2.a.dx 12 5.c odd 4 1
9075.2.a.dy 12 55.e even 4 1
9075.2.a.dz 12 55.e even 4 1
9075.2.a.ea 12 5.c odd 4 1
## Hecke kernels
This newform subspace can be constructed as the intersection of the kernels of the following linear operators acting on $$S_{2}^{\mathrm{new}}(1815, [\chi])$$:
$$T_{2}^{24} + \cdots$$ $$T_{19}^{12} + \cdots$$ | 2,716 | 5,271 | {"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 | 3 | CC-MAIN-2021-04 | latest | en | 0.36751 |
http://couplescounselingofelpaso.com/pvc-fence/7270.html | 1,550,745,384,000,000,000 | text/html | crawl-data/CC-MAIN-2019-09/segments/1550247503844.68/warc/CC-MAIN-20190221091728-20190221113728-00213.warc.gz | 65,864,648 | 5,079 | # figuring price per linear foot
### How to Calculate Linear Feet | U-Pack
12 Feb 2018 . Find out what is a linear foot, how to measure linear feet and determine pricing based on the footage.
### 4 Easy Ways to Calculate Linear Feet - wikiHow
You will need to calculate the amount of material in each category. . Find the price of each type of material (per foot) and multiply by the total linear foot value obtained for the type of .
### LTL Freight Linear Footage Calculator - FreightorGator
Have you ever experienced extra fees and charges due to the linear foot rule? Freight carriers will add unexpected fees if your shipment doesn't fit pre-determined parameters for size, shape and weight. Avoid these costly fees and expenses .
### FAQ - What is linear feet? How do I convert it to feet??
linear feet (often called Lineal feet) are the same as regular feet. No conversion is necessary. If something is 6 linear feet tall, it is 6 feet tall. It should be noted, that the correct term is linear, since Lineal refers to a line of ancestry, not to length.
### Calculator Tools | Transcendia
Calculator Tools. Billing Unit . Billing Unit Conversion Formulas. price per POUND TO price per linear foot: ÷. x. x 12 = \$ per linear foot . MSI price TO price per SQUARE foot: 144 ÷ 1000 x. = \$ per Square foot. MSI price .
### How to Calculate Linear Feet the Right Way - Unpakt
The cost of move typically depends on how much linear feet you use up in the moving truck. Learn how to calculate linear feet the right way and save money.
### How to Calculate Board Feet: 13 Steps (with Pictures) - wikiHow
Example: Calculate the cost of three separate pieces of hardwood available at a price of \$5.45 per board foot. The first piece is 4.3 board feet, the second is 6.8 board feet, and the third is .
### How to Calculate Linear Feet - Inch Calculator
12 Mar 2015 . Calculating the cost of lumber that is priced by the linear foot requires finding the total feet needed and then multiplying by the price per foot. When ordering material this way it's a good idea to consider the lengths that lumber .
### Lumber Calculators - Timber Creek
Now Mobile Friendly, take this lumber calculator with you when you're in the lumber yard or on the road with your smartphone! . Calculate total cost of board footage, linear footage, or pieces from any known unit of measure. Convert any .
### What Does Per Linear Foot for Cabinets Mean? | Home Guides | SF .
An older method of pricing cabinets is by the linear foot. You can calculate the linear feet of your cabinets by measuring their length. Measure each section of cabinetry from wall to wall, and be sure to include both the upper and lower cabinets.
### Price Per Linear Foot - Midland Paper
price per linear foot This tool calculates the price per linear foot of paper coming off a roll. . price per linear foot of paper coming off a roll. M-Weight: (width of a roll x 12"). If you don't know the m-weight, click here to calculate it. M-Weight (lbs).
### What Is the Difference Between a Linear and a Square Foot? | Hunker
16 Aug 2017 . When material in a roll is sold by the linear foot, divide the length of the roll into the area you're covering to get the . When you're buying lumber for a project or material for a floor, you may see the price quoted based on linear feet or square feet. . For example, when finding the length of a bedroom wall, you would measure from one corner to another and express the length in feet.
### How to Calculate a Linear Foot - YouTube
4 Sep 2014 . http://blog.upack.com/posts/how-to-calculate-linear-feet We'll show you how to measure a linear foot to calculate the cost of your long-distance move. A line.
### Sales Per Linear Foot Calculator - CSGNetwork.Com
This script calculates the sales per linear foot of selling rack or shelf space (not including storage or invetory space). This is done . This is an important figure to most retail stores as linear selling footage is one of the largest overhead costs.
### Linear Feet to Square Feet Conversion Calculator - Trestlewood
Trestlewood makes no representations or warranties whatsoever relative to the accuracy of this calculator (or any of its other calculators) and accepts no liability or responsibility for results obtained from same. Any user of this or any other .
### How to Calculate Linear Feet of Wall Space | Hunker
When installing molding, wainscoting, wallpaper or other decorative elements, it is important to know a wall's measurements in linear feet. When hanging art in galleries or displaying merchandise for sale in a retail space, it is important to know .
### How to Calculate a Linear Foot for Carpeting | Hunker
Although the area of your floor is measured in square feet, the carpet to cover that floor may be sold by the linear foot. Carpet manufacturers and retailers often don't provide any information about what a linear foot really is. In fact, "linear foot" is .
### How to Calculate Weight Per Linear Foot | Sciencing
14 Mar 2018 . By determining the weight per linear foot of a material, you know how much any length of the substance weighs. The weight per foot is also known as linear weight density. This equals the weight in pounds of the object, such .
### Lumber Pricing Calculator | Gibsons Building Supplies
Lumber Pricing Calculator. per. PC (Pieces), LNFT (linear foot), BDFT (Board foot), MBFT (Thousand Board foot). to: PC (Pieces), LNFT (linear foot), BDFT (Board foot), MBFT (Thousand Board foot). Quantity / Amount of Lumber to be .
### Cost Calculator for Length - Cleave Books
Comparative Costs per Unit. per inch, per millimetre. per foot, per centimetre. per yard, per decimetre. per furlong . the Total cost. and this can be done in the top part of this calculator. It can also be used to work out what the 'Cost per unit' was . | 1,332 | 5,917 | {"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} | 3.75 | 4 | CC-MAIN-2019-09 | latest | en | 0.867083 |
https://metanumbers.com/210724 | 1,638,141,490,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964358673.74/warc/CC-MAIN-20211128224316-20211129014316-00591.warc.gz | 459,931,796 | 7,399 | # 210724 (number)
210,724 (two hundred ten thousand seven hundred twenty-four) is an even six-digits composite number following 210723 and preceding 210725. In scientific notation, it is written as 2.10724 × 105. The sum of its digits is 16. It has a total of 4 prime factors and 12 positive divisors. There are 104,328 positive integers (up to 210724) that are relatively prime to 210724.
## Basic properties
• Is Prime? No
• Number parity Even
• Number length 6
• Sum of Digits 16
• Digital Root 7
## Name
Short name 210 thousand 724 two hundred ten thousand seven hundred twenty-four
## Notation
Scientific notation 2.10724 × 105 210.724 × 103
## Prime Factorization of 210724
Prime Factorization 22 × 139 × 379
Composite number
Distinct Factors Total Factors Radical ω(n) 3 Total number of distinct prime factors Ω(n) 4 Total number of prime factors rad(n) 105362 Product of the distinct prime numbers λ(n) 1 Returns the parity of Ω(n), such that λ(n) = (-1)Ω(n) μ(n) 0 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 210,724 is 22 × 139 × 379. Since it has a total of 4 prime factors, 210,724 is a composite number.
## Divisors of 210724
12 divisors
Even divisors 8 4 2 2
Total Divisors Sum of Divisors Aliquot Sum τ(n) 12 Total number of the positive divisors of n σ(n) 372400 Sum of all the positive divisors of n s(n) 161676 Sum of the proper positive divisors of n A(n) 31033.3 Returns the sum of divisors (σ(n)) divided by the total number of divisors (τ(n)) G(n) 459.047 Returns the nth root of the product of n divisors H(n) 6.79025 Returns the total number of divisors (τ(n)) divided by the sum of the reciprocal of each divisors
The number 210,724 can be divided by 12 positive divisors (out of which 8 are even, and 4 are odd). The sum of these divisors (counting 210,724) is 372,400, the average is 310,33.,333.
## Other Arithmetic Functions (n = 210724)
1 φ(n) n
Euler Totient Carmichael Lambda Prime Pi φ(n) 104328 Total number of positive integers not greater than n that are coprime to n λ(n) 8694 Smallest positive number such that aλ(n) ≡ 1 (mod n) for all a coprime to n π(n) ≈ 18809 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 104,328 positive integers (less than 210,724) that are coprime with 210,724. And there are approximately 18,809 prime numbers less than or equal to 210,724.
## Divisibility of 210724
m n mod m 2 3 4 5 6 7 8 9 0 1 0 4 4 3 4 7
The number 210,724 is divisible by 2 and 4.
## Classification of 210724
• Deficient
### Expressible via specific sums
• Polite
• Non-hypotenuse
## Base conversion (210724)
Base System Value
2 Binary 110011011100100100
3 Ternary 101201001121
4 Quaternary 303130210
5 Quinary 23220344
6 Senary 4303324
8 Octal 633444
10 Decimal 210724
12 Duodecimal a1b44
20 Vigesimal 166g4
36 Base36 4ilg
## Basic calculations (n = 210724)
### Multiplication
n×y
n×2 421448 632172 842896 1053620
### Division
n÷y
n÷2 105362 70241.3 52681 42144.8
### Exponentiation
ny
n2 44404604176 9357115810383424 1971768872027236638976 415499023789067413511578624
### Nth Root
y√n
2√n 459.047 59.5074 21.4254 11.6076
## 210724 as geometric shapes
### Circle
Diameter 421448 1.32402e+06 1.39501e+11
### Sphere
Volume 3.9195e+16 5.58005e+11 1.32402e+06
### Square
Length = n
Perimeter 842896 4.44046e+10 298009
### Cube
Length = n
Surface area 2.66428e+11 9.35712e+15 364985
### Equilateral Triangle
Length = n
Perimeter 632172 1.92278e+10 182492
### Triangular Pyramid
Length = n
Surface area 7.6911e+10 1.10275e+15 172055
## Cryptographic Hash Functions
md5 658c4d861c60be2216aa36664e9cb577 c9d8418fb41aba58bf128e54ae411bc6e99c9677 1d115db7d35ce90ffefb61b453d5787a04cbbc9c8368303f3e3cd398530064f0 db3448515fb338a9e479097a3ca16c276b512ae01ada8cf1e68e07f657f40144415306a9a088a9e0b66529ebd1002413ba28d39b90c063b6b12ff8bf66064971 ee26e28fb61b62a046ed960cfcda8bd29470cece | 1,443 | 4,201 | {"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} | 3.484375 | 3 | CC-MAIN-2021-49 | latest | en | 0.813016 |
http://mathhelpforum.com/geometry/130736-two-circles-intersecting-each-other-print.html | 1,503,464,055,000,000,000 | text/html | crawl-data/CC-MAIN-2017-34/segments/1502886117519.92/warc/CC-MAIN-20170823035753-20170823055753-00303.warc.gz | 269,163,361 | 2,963 | # Two circles intersecting each other...
• Feb 25th 2010, 07:30 AM
snigdha
Two circles intersecting each other...
Two circles intersect each other at A and B as shown in the figure. The common chord AB is produced to meet a common tangent PQ to the circles at D. Prove that DP=DQ.
• Feb 25th 2010, 07:58 AM
It follows from the alternate segment theorem that the triangles PDB and ADP are similar. Deduce that $PD^2 = AD*BD$. (In fact, this is a well-known theorem: from a point D, draw a tangent DP to a circle, and a chord DBA cutting the circle; then the square of PD is equal to the product of DA and DB.)
Now do the same for the other circle, to see that $QD^2$ is also equal to $AD*BD$. Conclusion: PD = QD. | 203 | 713 | {"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": 0, "img_math": 0, "codecogs_latex": 3, "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} | 3.453125 | 3 | CC-MAIN-2017-34 | longest | en | 0.937829 |
https://mathspace.co/textbooks/syllabuses/Syllabus-98/topics/Topic-1473/subtopics/Subtopic-17367/?activeTab=interactive | 1,721,054,202,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514707.52/warc/CC-MAIN-20240715132531-20240715162531-00891.warc.gz | 331,333,705 | 56,385 | Hong Kong
Stage 1 - Stage 3
# Deferred Payment Plans
## Interactive practice questions
Han purchased a fridge, valued at $\$702$$702, on terms. He paid a 6%6% deposit, followed by monthly instalments of \21$$21 over $3$3 years.
a
How much did Han pay as the deposit?
b
What was the balance owing after he paid the deposit?
c
How much in total did he pay in instalments?
d
How much did he pay for the fridge altogether?
e
How much interest did he pay?
f
What was the interest charged as a percentage of the cost of the fridge? Write your answer as a percentage to two decimal places.
g
What was the annual rate of simple interest charged for buying on terms?
Easy
11min
Han purchased a couch, valued at $\$726$$726, on terms. He paid a 11%11% deposit, followed by fortnightly instalments of \10$$10 over $3$3 years.
Easy
9min
James purchased a car, valued at $\$36653$$36653, on a deferred payment plan. He paid a \1714$$1714 deposit, followed by nothing for the first $6$6 months and then $31$31 monthly instalments of $\$1266$$1266. Easy 9min A store offered interest free terms for 2020 months on all purchases. Valentina purchased a \220$$220 treadmill by paying an initial $\$37$$37 deposit followed by 55 monthly instalments. If she was also charged an account keeping fee of \5$$5 per month, what is:
Easy
4min | 352 | 1,338 | {"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} | 2.828125 | 3 | CC-MAIN-2024-30 | latest | en | 0.967778 |
https://forum.azimuthproject.org/plugin/ViewComment/18349 | 1,575,695,099,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575540495263.57/warc/CC-MAIN-20191207032404-20191207060404-00319.warc.gz | 382,455,522 | 2,701 | Nice work in [comment #5](https://forum.azimuthproject.org/discussion/comment/18341/#Comment_18341), Jonathan! I'm glad to see someone tackle these puzzles.
Sophie's point is a good one, though:
> Does this mean that you are defining \$$\leq\$$ according to the amount of resources in each complex? I thought that we were using \$$\leq\$$ to represent possible reactions. Remember, \$$y \leq y'\$$ means that we can get \$$y\$$ from \$$y'\$$ not that there are fewer resources in \$$y\$$ than in \$$y'\$$.
Nothing in [comment #5](https://forum.azimuthproject.org/discussion/comment/18341/#Comment_18341) mentions how the ordering is defined in terms of the reactions we have available, so the argument must be incomplete, at least... though I think it's on the right track, and maybe Sophie fixed it in [comment #10](https://forum.azimuthproject.org/discussion/comment/18346/#Comment_18346). I'll think about that tomorrow. For now, something simpler:
I think your argument would be exactly right if we were dealing with a collection of reactions that can destroy one item of each time. Given a set \$$S = \\{s_1, \dots, s_n\\} \$$ and reactions
$s_i \to 0$
the preorder we get on \$$\mathbb{N}(S)\$$ is the one where
$a_1 s_1 + \cdots a_n s_n \le b_1 s_2 + \cdots + b_n s_n \; \textrm{ iff } \; a_i \le b_i \textrm{ for all } 1 \le i \le n .$
In other words, we can get from one complex to another only by destroying things.
Let's consider this case. Suppose we have any function \$$\phi : S \to T \$$. We can use it to define a homomorphism
$f: \mathbb{N}[S] \to \mathbb{N}[T]$
by
$f( a_1 s_1 + \cdots + a_n s_n) = a_1 s_{\phi(1)} + \cdots + a_n s_{\phi(n)}$
The \$$f\$$ in my puzzles is an example of this idea. But let's not solve those puzzles now: instead, give \$$\mathbb{N}[S] \$$ and \$$\mathbb{N}[T] \$$ the preorders described above.. Then \$$f\$$ is a strict monoidal monotone. I believe Jonathan's argument shows \$$f\$$ has no right adjoint unless \$$\phi\$$ is onto. | 608 | 1,994 | {"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} | 3.625 | 4 | CC-MAIN-2019-51 | longest | en | 0.879452 |
https://why-android.com/ | 1,501,245,662,000,000,000 | text/html | crawl-data/CC-MAIN-2017-30/segments/1500550967030.98/warc/CC-MAIN-20170728123647-20170728143647-00514.warc.gz | 742,790,772 | 38,168 | ## A simple QuickSort Program in Java
Introduction to Quick Sort
QuickSort algorithm is based on the Divide and Conquer Approach.
For a given sequence S there are 4 basic steps involved:
• Pick a number X from S. This program picks the last element.
• Create two Lists. One list will store elements less than X and the other will store greater than X.
• Sort the two lists recursively.
• Combine the left list elements with X and the right list elements.
The program below sorts a sequence of Integers. After which another version has been provided which can be used to sort a sequence of characters.These are my own versions of the QuickSort Algorithm.
Both the programs use a utility file which is also provided. The Source Code is also available here
```package impl;
import java.util.ArrayList;
/*
* Ankur Srivastava
* QuickSort Implementation using ArrayList for Sorting a sequence of Integers
* */
public class ArrayListQuickSort {
public ArrayListQuickSort() { }
public static void main(String[] args) {
Utility.printList(sort(Utility.createMockList()));
}
public static ArrayList<Integer> sort(ArrayList<Integer> input){
int size = input.size();
ArrayList<Integer> result = new ArrayList<Integer>();
if(size > 1){
int last = input.get(size-1);
ArrayList<Integer> left = new ArrayList<Integer>();
ArrayList<Integer> right = new ArrayList<Integer>();
for(int i=0;i<size;i++){
if(input.get(i) < last){
}else if(input.get(i) > last){
}
}
//Utility.printList("left", left);
//Utility.printList("right", right);
if(left != null && left.size() > 1){
left = sort(left);
}
if(right != null && right.size() > 1){
right = sort(right);
}
result = combine(left, last, right);
}
//Utility.printArray(result);
return result;
}
public static ArrayList<Integer> combine(ArrayList<Integer> left, int center, ArrayList<Integer> right){
ArrayList<Integer> result = new ArrayList<Integer>();
if(!left.isEmpty()){
for(int i =0; i<left.size(); i++){
}
}
if(!right.isEmpty()){
for(int i = 0; i<right.size(); i++){
}
}
//Utility.printList("merge", result);
return result;
}
}```
Next one demonstrates how to sort a character sequence
```package impl;
import java.util.ArrayList;
public class ArrayListCharQS {
public ArrayListCharQS() {
// TODO Auto-generated constructor stub
}
public static void main(String[] args) {
Utility.printCharList(sort(Utility.createMockCharList()));
}
public static ArrayList<Character> sort(ArrayList<Character> input){
int size = input.size();
ArrayList<Character> result = new ArrayList<Character>();
if(size > 1){
char last = input.get(size-1);
ArrayList<Character> left = new ArrayList<Character>();
ArrayList<Character> right = new ArrayList<Character>();
for(int i=0;i<size;i++){
if(input.get(i) < last){
}else if(input.get(i) > last){
}
}
//Utility.printList("left", left);
//Utility.printList("right", right);
if(left != null && left.size() > 1){
left = sort(left);
}
if(right != null && right.size() > 1){
right = sort(right);
}
result = combine(left, last, right);
}
//Utility.printArray(result);
return result;
}
public static ArrayList<Character> combine(ArrayList<Character> left, char center, ArrayList<Character> right){
ArrayList<Character> result = new ArrayList<Character>();
if(!left.isEmpty()){
for(int i =0; i<left.size(); i++){
}
}
if(!right.isEmpty()){
for(int i = 0; i<right.size(); i++){
}
}
//Utility.printList("merge", result);
return result;
}
}```
The Utility file used by both the programs above
```package impl;
import java.util.ArrayList;
public class Utility {
public Utility() {
// TODO Auto-generated constructor stub
}
public static void show(String str){
System.out.println(str);
}
public static void printArray(int[] A){
System.out.println("");
for(int i=0; i<A.length; i++){
System.out.print(A[i]+" ");
}
System.out.println("");
}
//Check if Array has 0's
public static boolean emptyArray(int[] input){
if(input[0] == 0 && input[input.length-1] == 0){
return true;
}
return false;
}
public static ArrayList<Integer> createMockList(){
ArrayList<Integer> inputArray = new ArrayList<Integer>();
/*
*/
printList(inputArray);
return inputArray;
}
public static ArrayList<Character> createMockCharList(){
ArrayList<Character> inputArray = new ArrayList<Character>();
/*
*/
printCharList(inputArray);
return inputArray;
}
public static void printList(ArrayList<Integer> list){
printList("", list);
}
public static void printList(String message, ArrayList<Integer> list){
System.out.println("");
System.out.println(message);
for(int i:list){
System.out.print(i+" ");
}
System.out.println("");
}
public static void printCharList(ArrayList<Character> list){
System.out.println("");
for(char i:list){
System.out.print(i+" ");
}
System.out.println("");
}
}```
## Learn how to create different versions of your App like Free and Paid
Android framework makes it quite easy for you to generate different flavors and variants of your App. Like you can create a free version or a paid version of your App.
In order to do this you need to follow these steps:
You need to specify the variants or flavors in your App’s build.gradle file.
Like
```productFlavors {
demo {
applicationId "com.edocent.demo"
versionName "1.0-demo"
}
full {
applicationId "com.edocent.full"
versionName "1.0-full"
}
}```
In all likelihood your Free App version will need a different Activity compared to the Paid one. So you need to create different folders for the two flavors.
So go to the Project->src folder and create a new Java Folder – free.
Create another one for paid similarly.
Add appropriate Activity and other files to this folder.
And you are all set. Execute Generate Signed APK task from the Build option. This will ask you which flavor you need the APK for.
Cool isn’t it.
Source code is available here
If you are like me then you might have a tendency to bite your nails, pull your hairs or keep doing something which you want to get rid off. Problem in such cases is awareness.
Soon you might be able to tame these habits using a wearable device. HabitAware has created a device called Liv.
A smart bracelet that creates awareness of subconscious behaviors like hair pulling, nail biting, skin picking and thumb-sucking in older children.
Willingness to change is the first step. Awareness is the second. Liv helps with this second step.
## Learn how to generate a signed APK file
All Android Apps must be digitally signed before they are installed on a device.
This is required to verify identity of developer who published it. And also to verify if it has been tampered.
Follow the video to generate the signed APK file and also to learn how to automate the process.
## Learn how to create and use a Android Module in your Android Project
You saw why using modules is important and how to create a Java module in an earlier Post. In a similar fashion you can add an Android module to your project.
Android Studio makes it easy to add a new module. Follow the video to understand the steps.
Source code is available here
## What is Android Debug Bridge or ADB ?
An ADB is a command line tool which can be used to communicate with Emulator or Android Devices.
When your development machine needs to communicate with an Android Device it does so using ADB. It’s a process that is controlled by a command also known as adb.
The adb command works by talking to an adb server which runs in the background at port 5037. The server is also known as adb daemon or adbd.
Android Studio also talks to this server when it needs to run an app via an Android Device.
To work with ADB you need Android SDK.
## Learn how to create and use a Java Module in Android Project
Creating modules is a good practice since it aids in reusability. In your Android project you can easily add a Java or Android module.
Follow the video to create a Java module and add it as a dependency in your Android Project.
Project source code is available in GitHub | 1,742 | 7,939 | {"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-2017-30 | longest | en | 0.48573 |
http://www.justanswer.com/multiple-problems/942ne-1-solve-following-linear-programming-problem.html | 1,498,686,737,000,000,000 | text/html | crawl-data/CC-MAIN-2017-26/segments/1498128323801.5/warc/CC-MAIN-20170628204133-20170628224133-00409.warc.gz | 556,018,751 | 19,478 | • 100% Satisfaction Guarantee
linda_us, Master's Degree
Category: Multiple Problems
Satisfied Customers: 1402
Experience: A tutor for Business, Finance, Accounts and other related topics.
19873544
Type Your Multiple Problems Question Here...
linda_us is online now
1 Solve the following linear programming problem
Customer Question
B.1 Solve the following linear programming problem graphically: Maximize profit = 4X + 6Y Subject to: X + 2Y < 8 5X + 4Y < 20 X, Y > 0 B.5 Solve the following LP problem graphically: Minimize cost = 24X + 15Y Subject to: 7X + 11Y > 77 16X + 4Y > 80 X,
Y > 0 B.7 The Attaran Corporation manufactures two electrical products: portable air conditioners and portable heaters. The assembly process for each is similar in that both require a certain amount of wiring and drilling. Each air conditioner takes 3 hours
of wiring and 2 hours of drilling. Each heater must go through 2 hours of wiring and 1 hour of drilling. During the next production period, 240 hours of wiring time are available and up to 140 hours of drilling time may be used. Each air conditioner sold yields
a profit of \$25. Each heater assembled may be sold for a \$15 profit. Formulate and solve this LP production-mix situation, and find the best combination of air conditioners and heaters that yields the highest profit. B.11 The Sweet Smell Fertilizer Company
markets bags of manure labeled “not less than 60 lb dry weight.” The packaged manure is a combination of compost and sewage wastes. To provide good-quality fertilizer, each bag should contain at least 30 lb of compost but no more than 40 lb of sewage. Each
pound of compost costs Sweet Smell 5 cents and each pound of sewage costs 4 cents. Use a graphical LP method to determine the least-cost blend of compost and sewage in each bag. B.21. Par, Inc., produces a standard golf bag and a deluxe golf bag on a weekly
basis. Each golf bag requires time for cutting and dyeing and time for sewing and finishing, as shown in the following table: Par, Inc., will sell whatever quantities it produces of these two products. a) Find the mix of standard and deluxe golf bags to produce
per week that maximizes weekly profit from these activities. b) What is the value of the profit?
Submitted: 2 years ago.
Category: Multiple Problems
Expert: F. Naz replied 2 years ago.
Customer: replied 2 years ago.
Sunday. Thanks
Expert: F. Naz replied 2 years ago.
Sorry unable to do this, therefore opting out so other experts may help you, thanks.
Expert: PDtax replied 2 years ago.
Hi from Just Answer. I'm PDtax, and can assist.
I will post the answers this afternoon.
Customer: replied 2 years ago.
Thank you
Expert: PDtax replied 2 years ago.
I see that I did a project for you the other day, and the service shows that you opted me out.
Please confirm that there were no issues with that project, and that you want me to finish this one.
PDtax | 703 | 2,900 | {"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-2017-26 | latest | en | 0.872861 |
https://www.enotes.com/homework-help/annual-income-b-ratio-3-4-their-annual-343326?en_action=hh-question_click&en_label=hh_carousel&en_category=internal_campaign | 1,611,333,846,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703530835.37/warc/CC-MAIN-20210122144404-20210122174404-00509.warc.gz | 766,260,694 | 19,327 | # The annual income of A and B are in the ratio 3: 4 and their annual expenditure are in the ratio 5:7. If each saves 15000 per month find their annual income.
The annual income of A and B are in the ratio 3 : 4. If the income of A is 3x, the income of B is 4x. Their annual expenditure is in the ratio 5:7, if A's expenditure is 5y, B's expenditure is 7y. Each of them saves 15000 per month.
=> 3x - 5y = 15000 and 4x - 7y = 15000
3x - 5y = 15000
=> y = (3x - 15000)/5
Substituting in 4x - 7y = 15000
=> 4x - 7*(3x - 15000)/5 = 15000
=> 20x - 21x + 105000 = 75000
=> x = 30000
This gives the income of A as 90000 and of B as 120000
Approved by eNotes Editorial Team
Start your 48-hour free trial and unlock all the summaries, Q&A, and analyses you need to get better grades now.
• 30,000+ book summaries
• 20% study tools discount | 296 | 843 | {"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-2021-04 | latest | en | 0.932666 |
https://prep4gmat.com/qod-gmat-in-consecutive-order/ | 1,660,361,649,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882571869.23/warc/CC-MAIN-20220813021048-20220813051048-00779.warc.gz | 437,692,234 | 9,464 | # GMAT in Consecutive Order
Explanation
The missing word here is "average." For a life of simplified computations, when you hear the word "sum" you want to think "average" (and vice versa).
We can use the cross-multiplied average formula: sum of items = (average of items)(number of items).
Let's apply it to the fact that the sum of the first 4 integers is 656:
656 = (average of first 4 numbers)(4) average of first 4 numbers = 656/4 = 164
If this is the average of consecutive odd integers, then the integers must be 161, 163, 165, and 167, which balance in pairs around 164.
Therefore, the last four numbers in the sequence must be 169, 171, 173, 175. We could add these up to find the answer... or, we could use the average formula to find their sum!
They are balanced around 172, so that is their average, and there are four of them, so sum of second 4 numbers = (172)(4) = 400 + 280 + 8 = 688. | 244 | 906 | {"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} | 4.53125 | 5 | CC-MAIN-2022-33 | latest | en | 0.920786 |
https://lavelle.chem.ucla.edu/forum/viewtopic.php?f=9&t=45893&view=print | 1,611,465,342,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703547333.68/warc/CC-MAIN-20210124044618-20210124074618-00016.warc.gz | 404,488,711 | 2,094 | Page 1 of 1
### E21b. Sig Figs
Posted: Wed Oct 02, 2019 12:10 am
If the given number is 25.92 mg of HF why is the amount of significant figures for the answer 3?
### Re: E21b. Sig Figs
Posted: Wed Oct 02, 2019 1:09 am
I'm also confused by this, but it may have to do with the work that's done for the problem? so when you divide the 25.92 x 10^-3 by 20.01 mol and get 1.30 x 10^-3 the answer for this part (1.30 x 10^-3) has 3 sig figs, which is the least amount of sig figs in part, so this might be why the final answer for b is 3 sig figs.
### Re: E21b. Sig Figs
Posted: Wed Oct 02, 2019 7:54 am
I can't seem to find the problem, but the amount of sig figs is determined by the sig figs of the amount given (by the given amount with the least sig figs), so although you may get a number with less sig figs when you do your calculations you should refer back to the given amounts. So maybe the book was mistaken if it didn't give the amount of sig figs as the given amount with the least amount of sig figs. I hope this helps!
### Re: E21b. Sig Figs
Posted: Wed Oct 02, 2019 4:20 pm
This was also confusing, but it may be possible that when calculating the molar mass of HF, you use 1.01 as the molar mass for hydrogen which is given in 3 significant figures so the book would like to keep using 3 significant figures.
### Re: E21b. Sig Figs
Posted: Wed Oct 02, 2019 4:22 pm
it depends on the number of sig figs used for when you calculate molar mass.
### Re: E21b. Sig Figs
Posted: Wed Oct 02, 2019 4:26 pm
I agree with the above answers; it must have something to do with the molar mass calculations. | 490 | 1,616 | {"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} | 3.53125 | 4 | CC-MAIN-2021-04 | latest | en | 0.924088 |
http://www.globalspec.com/reference/29086/203279/10-5-constrained-deconvolution | 1,368,876,373,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368696382360/warc/CC-MAIN-20130516092622-00080-ip-10-60-113-184.ec2.internal.warc.gz | 478,356,160 | 8,504 | Sites:
## 10.5 Constrained Deconvolution
Another concept of restoration is represented by constrained deconvolution. This method does not impose the requirement of minimising estimation errors but only requests that the original noise power in the restored signal be preserved. It comes from the general tendency of deconvolution methods to lead to a substantial degradation in signal-to-noise ratio and therefore it seeks an approximate approach which would correct the convolution distortion without increasing the noise level.
The method uses the notion of a discrete model of deterioration consisting in distortion caused by a discrete linear system with a finite impulse response { h n}, n = 0, 1, , J - 1, and additive noise { ? n},
(10.47)
For a finite section { x n} of the input signal, M samples long, obviously n ÃŽ ?0, M + J - 2 ?, i ÃŽ ?0, M - 1 ?. The model equation can then be rewritten into vector form
(10.48)
where the nonzero elements of the matrix are H n,i = h n - i, ( n - i) ÃŽ ?0, J - 1 ? so that it is a circulant matrix. The noise energy of the observed signal, that is before restoration, is
(10.49)
If the noise is centred, its energy is given by the variance,
We shall define the residuum r as being the difference vector dependent...
##### Products & Services
Thermocouple Temperature Transmitters
Thermocouple temperature transmitters convert the small millivolt output of a thermocouple to a current signal (typically 4-20 mADC) that is immune to noise and voltage drops over a long distance.
Audio Analyzers
Audio analyzers measure the noise and audio spectrum of output from an amplifier or a stereo.
Noise Figure Meters
RF noise figure meters measure the noise contribution of an amplifier relative to a noise-free amplifier at a reference temperature. Usually expressed in dB for Ku-band amplifiers.
Frame Grabbers
Frame grabbers are image processing computer boards that capture and store image data for industrial applications such as quality control.
Audio Processor ICs
Audio processors IC are semiconductor devices used to detect, decode and process analog or digital audio.
##### Topics of Interest
10.6 Deconvolution via Impulse-Response Optimisation Another approach to deconvolution is represented by the method of optimising the shape of a specially defined impulse response. It is based on...
10.2 Plain Deconvolution and Pseudo-Inversion As far as the influence of the linear system (distortion) prevails in the deterioration according to models (10.1) or (10.2), the purpose of restoration...
10.7 Generalised Discrete Mean-Square-Error Minimisation The method that we shall introduce in this section is, in a sense a generalisation of Wiener filtering. The concept assumes that both the...
15.1 Useful Matrix Types A Toeplitz matrix A from a sequence a = [ a ? N +1, , a M ?1] of length M + N ? 1 is an M N matrix where A i, j = a i ? j, i = 0, , M ? 1, j = 0, , N ? 1, in...
10.8 Other Approaches to Signal Restoration We shall briefly mention two other approaches to restoration. The first of these is the method of maximum a posteriori probability (MAP). The concept of...
Product Announcements
Acromag, Inc.
Pyromation, Inc.
Conax Technologies | 739 | 3,219 | {"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} | 2.515625 | 3 | CC-MAIN-2013-20 | latest | en | 0.874711 |
https://www.wyzant.com/resources/answers/16429/show_that_2_x_2_3x_4_is_equal_to_x_5_when_x_1 | 1,524,548,436,000,000,000 | text/html | crawl-data/CC-MAIN-2018-17/segments/1524125946564.73/warc/CC-MAIN-20180424041828-20180424061828-00517.warc.gz | 931,212,562 | 15,664 | 0
# Show that 2[x-(2-3x)-4] is equal to x-5 when x=1.
please solve if x-5 when x=1
### 1 Answer by Expert Tutors
Alexander B. | Experienced and Enthusiastic Math TutorExperienced and Enthusiastic Math Tutor
0
Plug in x and you'll get (2 - 3) which is (-1)
Proceed to the next level of brackets => [x - (-1) - 4]
Subtracting a negative is the same as adding the positive => [x + 1 - 4]
Simplify to [x + (-3)], since 1 - 4 is (-3). You could also just write [x - 3]
Plug in x and you'll get [1 - 3] which is (-2)
Don't forget the 2 outside the brackets
2 * (-2) = (-4)
Solve for the second function x - 5
Plug in x and you'll get 1 - 5 which is (-4)
(-4) = (-4)
:D | 233 | 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} | 3.875 | 4 | CC-MAIN-2018-17 | longest | en | 0.928082 |
https://web2.0calc.com/questions/write-an-equation-for-the-line-through-the-points | 1,628,026,826,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046154471.78/warc/CC-MAIN-20210803191307-20210803221307-00008.warc.gz | 621,506,342 | 5,468 | +0
Write an equation for the line through the points (4,2)(4,2) and (3,4)(3,4) in point-slope form π¦=π¦0+π(π₯βπ₯0)
0
95
1
+408
Write an equation for the line through the points (4,2)(4,2) and (3,4)(3,4) in point-slope form π¦=π¦0+π(π₯βπ₯0)
Mar 16, 2021
#1
+121004
+1
Slope = ( 2 - 4) / ( 4 - 3) = -2 / 1 = -2 = m
Using either point
y = -2 ( x - 3) + 4
y = 4 + -2 ( x -3)
Mar 16, 2021 | 250 | 440 | {"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} | 3.828125 | 4 | CC-MAIN-2021-31 | latest | en | 0.330691 |
http://www.phy.olemiss.edu/~jgladden/buckle/index.html | 1,674,861,503,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764499468.22/warc/CC-MAIN-20230127231443-20230128021443-00644.warc.gz | 77,835,378 | 2,990 | # Dynamic Buckling and Fragmentation of Brittle Rods
Andrew Belmonte, The Pennsylvania State University
At the W. G. Pritchard Laboratories in the Math Department at Penn State we devised a simple experiment to investigate how a slender beam responds to an impulsive load along it's axis. High speed video uncovered some very interesting dynamics which we reported in Physical Review Letters (PRL 94, 035503 (2005)). Figure 1 shows frames of a video of a piece of pasta just after being struck by a projectile moving at 3.5 m/s. A well defined buckling wavelength emerges.
A video of breaking pasta can be viewed here: Hit with a 4 m/s projectile or a 40 m/s projectile.
We applied linear stability analysis to a model of the stress wave propagating down the rod based on the results of Saint-Venant and the fourth order dynamical buckling PDE. This produced a prefered buckling wavelength [lambda] (the one that grows fastest) and growth rate [tau_b] of that mode. The expressions for these are listed below:
where c is the speed of sound in the material, d is the diameter of the circular cross-section, lambda is the wavelength and U_0 is the impact speed of the projectile.
When we plot these scaling laws along with wavelength data obtained from the high speed video (shown below) we see excellent agreement. The only adjustment which needed to be made is to add a factor (gamma) which is either 1 if buckling sets in before the stress pulse reflects off the back end of the rod or 2 if the reflection occurs before buckling.
## Fragmentation Statistics
In a brittle rod, bending leads quickly to breaking. By shattering over 1000 pieces of pasta and measuring the length of each resulting fragment, we get the distribution shown below. It exhibits typical power law behavior except for two peaks. The center of these peaks corresponds to fragment lengths of 1/4 and 1/2 the buckling wavelength. 1/2 wavelength would be the distance between two successive curvature extrema and 1/4 wavelegth is the distance from the end of the rod and the first curvature extrema (see the still frame inset). | 459 | 2,103 | {"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-2023-06 | longest | en | 0.908492 |
http://www.jiskha.com/display.cgi?id=1208019102 | 1,498,245,111,000,000,000 | text/html | crawl-data/CC-MAIN-2017-26/segments/1498128320130.7/warc/CC-MAIN-20170623184505-20170623204505-00428.warc.gz | 558,872,446 | 4,261 | # Annuities
posted by on .
Can someone tell me if this is ordinary annuity of future or ordinary values sinking funds present value or what is it. The question is You are earning an average of 46500 and will retire in 10 years. If you put 20% of your gross average income in an ordinary annuity compounded at 7% annually, what will be the value of the annuity when you retire? I am using a texas instrument calculator for figuring and using formulas to do this HELP HELP
• Annuities - ,
You put in P = 9,300 per year for n = 10 years at interest rate r = .07
You can do this as a future value of annuity or amount of a sinking fund. N is accumulated after n years.
N = P * [(1+r)^n - 1 ] / r
N = 9300 * [ (1.07)^10 - 1 ] / .07
N = 128,493
• Annuities - ,
that is what i was wondering if i had to figure out the from the 46500 the 20 percent and use that thanks
• Annuities - ,
With my TI 83 + calculator:
type 9300
type * times key x
left paren
1.07^10 -1
right paren
enter
get 8994.5...
divide key
.07
enter
get
128492.966 | 303 | 1,033 | {"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} | 3.625 | 4 | CC-MAIN-2017-26 | latest | en | 0.918779 |
https://qualityessaywriters.com/saint-leo-mba550-midterm-exam/ | 1,718,669,231,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861741.14/warc/CC-MAIN-20240617215859-20240618005859-00855.warc.gz | 434,607,286 | 17,253 | Welcome to Quality Essay Writers
# Saint Leo – MBA550 – Midterm Exam
Saint Mba550 midterm exam
Question 1.1.The events in an experiment are ________ if only one can occur at a time. (Points : 5)
mutually exclusive
non-mutually exclusive
mutually inclusive
independent
Question 2.2.A single-channel queuing system has an average service time of 8 minutes and an average time between arrivals of 10 minutes. What is the hourly arrival rate? (Points : 5)
8
6
4
2
Question 3.3.The ________ is computed by multiplying each decision outcome under each state of nature by the probability of its occurrence. (Points : 5)
expected value
expected value of perfect information
expected opportunity loss
none of the above
Question 4.4.Employees of a local company are classified according to gender and job type. The following table summarizes the number of people in each job category.
Job Male (M) Female (F) Administrative (AD) 110 10 Salaried staff (SS) 30 50 Hourly staff (HS) 60 40
If an employee is selected at random, what is the probability that the employee is female given that the employee is a salaried staff member? (Points : 5)
.50
.60
.625 è 50/80 = 0.625
.70
Question 5.5.The manager of the Quick Stop Corner convenience store (which is open 360 days per year) sells four cases of Stein soda each day (1440 cases per year). Order costs are \$8.00 per order. The lead time for an order is three days. Annual holding costs are equal to \$57.60 per case. If the manager orders 16 cases each time she places an order, what is the average inventory level? (Points : 5)
4 cases
8 cases
12 cases
20 cases
Question 6.6.The maximin approach to decision making refers to: (Points : 5)
minimizing the maximum return.
maximizing the minimum return.
maximizing the maximum return.
minimizing the minimum return.
Question 7.7.In the basic EOQ model, if D=60 per month, Co=\$12, and Cc=\$10 per unit per month, what is the EOQ? (Points : 5)
11
12
13
14
Question 8.8.________ is a category of statistical techniques that uses historical data to predict future behavior. (Points : 5)
Qualitative methods
Regression
Time series
Quantitative methods
Question 9.9.________ is a measure of the dispersion of random variable values about the expected value or mean. (Points : 5)
Standard deviation
Sample mean
Population mean
Expected value
Question 10.10.In an EOQ model, as the carrying cost increases, the order quantity: (Points : 5)
increases.
decreases.
remains the same.
cannot be determined.
Question 11.11.A university is planning a seminar. It costs \$3000 to reserve a room, hire an instructor, and bring in the equipment. Assume it costs \$25 per student for the administrators to provide the course materials. If we know that 20 people will attend, what price should be charged per person to break even? (Points : 5)
\$120
\$150
\$175
\$200
Question 12.12.A small entrepreneurial company is trying to decide between developing two different products that they believe they can sell to two potential companies, one large and one small. If they develop Product A, they have a 50% chance of selling it to the large company with annual purchases of about 20,000 units. If the large company won’t purchase it, then they think they have an 80% chance of placing it with a smaller company, with sales of 15,000 units. On the other hand if they develop Product B, they feel they have a 40% chance of selling it to the large company, resulting in annual sales of about 17,000 units. If the large company doesn’t buy it, they have a 50% chance of selling it to the small company with sales of 20,000 units. What is the probability that Product A will being purchased by the smaller company? (Points : 5)
0.3
0.4
0.5
0.8
Question 13.13.A difficult aspect of using spreadsheets to solve management science problems is: (Points : 5)
obtaining the solution to standard management science problems.
data entry.
performing sensitivity analysis.
setting up a spreadsheet with complex models and formulas.
Question 14.14.The limits of the problem and the degree to which it pervades other units in the organization must be included during the ________ step of the management science process. (Points : 5)
observation
definition
solution
implementation
Question 15.15.Which of the following is an equation or an inequality that expresses a resource restriction in a mathematical model? (Points : 5)
Decision variable
Objective function
Constraint
Parameter
Question 16.16.Administrators at a university will charge students \$150 to attend a seminar. It costs \$3000 to reserve a room, hire an instructor, and bring in the equipment. Assume it costs \$25 per student for the administrators to provide the course materials. How many students would have to register for the seminar for the university to break even? (Points : 5)
18
20
24
30
Question 17.17.A single-server queuing system has average time between arrivals of 20 minutes and a service time of 10 minutes each. Assuming Poisson arrivals and exponential service times, the utilization factor is approximately _____. (Points : 5)
0.25
0.33
0.50
2.0
Question 18.18.________ is a linear regression model relating demand to time. (Points : 5)
Linear trend
Linear regression
Forecast demand
Linear equation
Question 19.19.A ________ probability is the probability that an event will occur given that another event has already occurred. (Points : 5)
posterior
conditional
marginal
low
Question 20.20.Bayesian analysis involves a(n) ________ probability. (Points : 5)
a priori
posterior
joint
relative frequency
## Why US?
##### 100% Confidentiality
Information about customers is confidential and never disclosed to third parties.
##### Timely Delivery
No missed deadlines – 97% of assignments are completed in time.
##### Original Writing
We complete all papers from scratch. You can get a plagiarism report.
##### Money Back
If you are convinced that our writer has not followed your requirements, feel free to ask for a refund. | 1,442 | 6,035 | {"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-2024-26 | latest | en | 0.871557 |
https://www.tutorialspoint.com/name-the-type-of-quadrilateral-formed-if-any-by-the-following-points-and-give-reasons-for-your-answer-i-1-2-1-0-1-2-3-0-ii-3-5-3-1-0-3-1-4-iii-4 | 1,701,814,981,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100568.68/warc/CC-MAIN-20231205204654-20231205234654-00668.warc.gz | 1,167,726,125 | 23,111 | # Name the type of quadrilateral formed, if any, by the following points, and give reasons for your answer.(i) $(-1, -2), (1, 0), (-1, 2), (-3, 0)$(ii) $(-3, 5), (3, 1), (0, 3), (-1, -4)$(iii) $(4, 5), (7, 6), (4, 3), (1, 2)$
To do:
We have to find the quadrilateral formed, if any, by the given points.
Solution:
Let the given points be $A(-1, -2), B(1, 0), C(-1, 2), D(-3, 0)$.
We know that,
The distance between two points $\mathrm{A}(x_{1}, y_{1})$ and $\mathrm{B}(x_{2}, y_{2})$ is $\sqrt{(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}}$.
Therefore,
$\mathrm{AB}=\sqrt{(1+1)^{2}+(0+2)^{2}}$
Squaring on both sides, we get, $\mathrm{AB}^{2}=(1+1)^{2}+(0+2)^{2}$
$=(2)^{2}+(2)^{2}$
$=4+4$
$=8$
$\mathrm{BC}^{2}=(-1-1)^{2}+(2-0)^{2}$
$=(-2)^{2}+(2)^{2}$
$=4+4$
$=8$
$\mathrm{CD}^{2}=(-3+1)^{2}+(0-2)^{2}$
$=(-2)^{2}+(-2)^{2}$
$=4+4$
$=8$
$\mathrm{DA}^{2}=(-1+3)^{2}+(-2+0)^{2}$
$=(2)^{2}+(-2)^{2}$
$=4+4$
$=8$
$\mathrm{AC}^{2}=(-1+1)^{2}+(2+2)^{2}$
$=(0)^{2}+(4)^{2}$
$=0+16$
$=16$
$\mathrm{BD}^{2}=(-3-1)^{2}+(0-0)^{2}$
$=(-4)^{2}+0$
$=16$
Here,
$AB^2=BC^2=CD^2=DA^2$ and $AC^2=BD^2$
This implies,
$AB=BC=CD=DA$ and $AC=BD$.
All the sides are equal and the diagonals are equal to each other.
Therefore, the quadrilateral formed by the points $A, B, C, D$ is a square.
(ii) Let the given points be $A (-3, 5), B (3, 1), C (0, 3), D (-1, -4)$.
We know that,
The distance between two points $\mathrm{A}(x_{1}, y_{1})$ and $\mathrm{B}(x_{2}, y_{2})$ is $\sqrt{(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}}$.
Therefore,
$\mathrm{AB}=\sqrt{(3+3)^{2}+(1-5)^{2}}$
Squaring on both sides, we get,
$\mathrm{AB}^{2}=(3+3)^{2}+(1-5)^{2}$
$=(6)^{2}+(-4)^{2}$
$=36+16$
$=52$
$\mathrm{BC}^{2}=(0-3)^{2}+(3-1)^{2}$
$=(-3)^{2}+(2)^{2}$
$=9+4$
$=13$
$\mathrm{CD}^{2}=(-1-0)^{2}+(-4-3)^{2}$
$=(-1)^{2}+(-7)^{2}$
$=1+49$
$=50$
$\mathrm{DA}^{2}=(-3+1)^{2}+(5+4)^{2}$
$=(-2)^{2}+(9)^{2}$
$=4+81$
$=85$
$\mathrm{AC}^{2}=(0+3)^{2}+(3-5)^{2}$
$=(3)^{2}+ (-2)^{2}$
$=9+4$
$=13$
In $\Delta \mathrm{ABC}$,
$\mathrm{AB}=\sqrt{52}, \mathrm{AC}=\sqrt{13}, \mathrm{BC}=\sqrt{13}$
$\mathrm{AC}+\mathrm{BC}=\sqrt{13}+\sqrt{13}=2 \sqrt{13}$
$\mathrm{AB}=\sqrt{52}=2\sqrt{13}$
$\Rightarrow \mathrm{AC}+\mathrm{BC}=\mathrm{AB}$
This implies, the points $A, B ,C$ are collinear.
Therefore, $\Delta \mathrm{ABC}$ is not possible.
Hence, $\mathrm{ABCD}$ is not a quadrilateral.
(iii) Let the given points be $A (4, 5), B (7, 6), C (4, 3), D (1, 2)$.
We know that,
The distance between two points $\mathrm{A}(x_{1}, y_{1})$ and $\mathrm{B}(x_{2}, y_{2})$ is $\sqrt{(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}}$.
Therefore,
$\mathrm{AB}=\sqrt{(7-4)^{2}+(6-5)^{2}}$
Squaring on both sides, we get,
$\mathrm{AB}^{2}=(7-4)^{2}+(6-5)^{2}$
$=(3)^{2}+(1)^{2}$
$=9+1$
$=10$
$B C^{2}=(4-7)^{2}+(3-6)^{2}$
$=(3)^{2}+(-3)^{2}$
$=9+9$
$=18$
$\mathrm{CD}^{2}=(1-4)^{2}+(2-3)^{2}$
$=(-3)^{2}+(-1)^{2}$
$=9+1$
$=10$
$\mathrm{DA}^{2}=(4-1)^{2}+(5-2)^{2}$
$=(3)^{2}+(3)^{2}$
$=9+9$
$=18$
$\mathrm{AC}^{2}=(4-4)^{2}+(3-5)^{2}$
$=(0)^{2}+(-2)^{2}$
$=0+4$
$=4$
$\mathrm{BD}^{2}=(1-7)^{2}+(2-6)^{2}$
$=(-6)^{2}+(-4)^{2}$
$=36+16$
$=52$
Here,
$AB^2=CD^2$, $BC^2=DA^2$ and $AC^2≠BD^2$
This implies,
$A B=C D, B C=D A$ and $AC≠BD$
Opposite sides are equal and diagonals are not equal.
Therefore, the quadrilateral formed by the points $A, B, C, D$ is a parallelogram.
Tutorialspoint
Simply Easy Learning
Updated on: 10-Oct-2022
32 Views
##### Kickstart Your Career
Get certified by completing the course | 1,731 | 3,526 | {"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} | 4.5625 | 5 | CC-MAIN-2023-50 | latest | en | 0.43032 |
https://www.golinuxcloud.com/add-fractions-in-java/ | 1,713,996,154,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296819971.86/warc/CC-MAIN-20240424205851-20240424235851-00341.warc.gz | 701,232,631 | 34,559 | 2 ways to add fractions in Java [Practical Examples]
In this article we will learn how to add fractions in java, fraction addition in programming has a certain technique, which is different than adding two simple datatypes, here we will learn how we can do that.
In mathematics, the concept of rationalization is used to add two fractions i.e. before we add two fractions, we make sure that the denominator of both the fractions is the same, if not, the denominator and numerator of the fraction whose denominator is different. The common denominator is called the LCM and it is multiplied with both fractions. After this, the new numerators are then added. For example, if we want to add two fractions
`⅓ +⅘`
The denominators will be made the same
`⅓*5 + ⅘ *3`
New fractions become
`5/15 + 12/15`
`17/15`
Java Environment
Before moving on, make sure you have the latest java version installed in your system. To check that type the following command on your windows command line.
`java -version`
And the output must be something like this
How to add fractions in java
Now in Java, we first find the LCM of the two fractions. Then divide the LCM all the denominators and multiply the quotient by a fraction for a common denominator and then finally add them all up.
First lets look at a code where we try to add fractions in java by simply adding them up
``````class addFractions {
public static void main(String[] args) {
double fraction1 = 1 / 2;
double fraction2 = 1 / 3;
double fraction3 = 1 / 4;
double addFractions = fraction1 + fraction2 + fraction3;
}
}``````
The output is
`0`
As you can see, it does not add up right. Now let's add fractions in java using logic.
Add Fractions in Java using Cross Multiplication
We cross multiply the numerators and denominators and find the result. Like in the code below
``````class addFractions {
public static void main(String[] args) {
// initialise numerator of first fraction
int numerator1 = 2;
// initialise numerator of second fraction
int denominator1 = 3;
// initialise numerator of second fraction
int numerator2 = 4;
// initialise denominator of second fraction
int denominator2 = 5;
// use the cross multiplication rule
int numerator3 = numerator1 * denominator2 + numerator2 * denominator1;
// find lcm
int denominator3 = denominator1 * denominator2;
// print the results.
System.out.println(numerator3 + "/" + denominator3);
}
}``````
The output of this code is :
`22/15`
Add Fractions in Java Using Fractions Class
In java a class can be created for simplicity, in this part of code, we will create Fraction class and then add fractions.
Code
``````import java.util.*;
import java.lang.*;
class Fractions {
int numerator;
int denominator;
public Fractions(int numerator, int denominator) {
this.numerator = numerator;
this.denominator = denominator;
}
int getNumerator() {
return numerator;
}
int getDenominator() {
return denominator;
}
int numerator = this.numerator * f.getDenominator() + f.getNumerator() * this.denominator;
int denominator = this.denominator * f.getDenominator();
return new Fractions(numerator, denominator);
}
void print() {
System.out.println(numerator + "/" + denominator);
}
}
// add fractions using Fraction class
public static void main(String[] args) {
// initialize fraction 1
Fractions f1 = new Fractions(1, 2);
// initialize fraction 2
Fractions f2 = new Fractions(1, 3);
// initialize fraction 3
Fractions f3 = new Fractions(1, 4);
// initialize fraction 4
Fractions f4 = new Fractions(1, 5);
// Add fractions 1 and 2
// Add fractions 3 and 4
// Add fractions 5 and 6
// Print the result
f7.print();
}
}``````
The output of this code is :
`154/120`
Conclusion
In this article different ways how to add fractions in java. First, we studied that normal addition cannot work with fractions, then we studied the cross multiplication technique, and then we created a java class to make the addition of fractions easier.
Views: 28
Azka Iftikhar
She is proficient in multiple programming languages, including C++, Golang, and Java, she brings a versatile skillset to tackle a wide array of challenges. Experienced Computer Scientist and Web Developer, she excels as a MERN Stack Expert. You can read more on her GitHub page.
Can't find what you're searching for? Let us assist you.
Enter your query below, and we'll provide instant results tailored to your needs.
If my articles on GoLinuxCloud has helped you, kindly consider buying me a coffee as a token of appreciation.
For any other feedbacks or questions you can send mail to admin@golinuxcloud.com | 1,081 | 4,604 | {"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} | 3.46875 | 3 | CC-MAIN-2024-18 | latest | en | 0.84946 |
http://www.early-retirement.org/forums/f27/when-going-out-to-eat-in-a-group-how-do-you-settle-the-bill-82403-4.html | 1,526,811,965,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794863277.18/warc/CC-MAIN-20180520092830-20180520112830-00323.warc.gz | 371,934,660 | 21,956 | Portal Forums Links Register FAQ Members List Social Groups Calendar Search Today's Posts Mark Forums Read Log in
View Poll Results: When Eating in a Group, How Do You Settle the Bill? Divide and Conquer 40 53.33% Folks put in their estimates, last person covers the rest 2 2.67% Folks put in their estimates, last person puts in estimate, then the fun begins 7 9.33% other 26 34.67% Voters: 75. You may not vote on this poll
Thread Tools Search this Thread Display Modes
06-23-2016, 11:21 AM #61
Moderator Emeritus
Join Date: Jan 2007
Location: New Orleans
Posts: 40,148
Quote:
Originally Posted by rodi
Quote:
Originally Posted by W2R I love figuring out the bill, the tax and tip, and who pays how much, in my head. F is a brilliant engineer so naturally he loves performing this little arithmetic computation in his head, too. So, even though we each have calculators with us, we do not use them. Instead we each figure out the bill independently in our heads, as a double check. This little math exercise is a great little pleasure for us as we prepare to leave, like a dinner mint but not fattening.
I like this - Fun with Math!!! My sister and I are the same way - both math nerds who have fun doing mental math.
That's something I love about this forum - - like minded INTJ math/engineering nerds.
I think I have a system figured out. First I add up what each of us ate separately, and determine the difference (usually his is more because I drink water instead of a soft drink). Then I add up the food, tax, and tip, to figure out what the rounded total should be; then subtract the rounded difference I just computed, and divide in half to get what the lesser amount should be. Then re-add the difference to get the greater amount for the other person. Then add both together to make sure the total is correct. Then check to see if F got the same result. I think my system works better than his but I could be wrong!
__________________
__________________
5/17/2018: Retired a second time, this time from my volunteer Admin duties. After 10 years of being on the team, and 40,000+ posts, the time just seemed right. It has been such fun to work with all of our Mods and Admins and I plan to stick around as a regular member.
Join the #1 Early Retirement and Financial Independence Forum Today - It's Totally Free! Are you planning to be financially independent as early as possible so you can live life on your own terms? Discuss successful investing strategies, asset allocation models, tax strategies and other related topics in our online forum community. Our members range from young folks just starting their journey to financial independence, military retirees and even multimillionaires. No matter where you fit in you'll find that Early-Retirement.org is a great community to join. Best of all it's totally FREE! You are currently viewing our boards as a guest so you have limited access to our community. Please take the time to register and you will gain a lot of great new features including; the ability to participate in discussions, network with our members, see fewer ads, upload photographs, create a retirement blog, send private messages and so much, much more! Join Early-Retirment.org For Free - Click Here
06-23-2016, 11:37 AM #62
Thinks s/he gets paid by the post
Join Date: Mar 2005
Location: yonder
Posts: 2,411
Quote:
Originally Posted by W2R That's something I love about this forum - - like minded INTJ math/engineering nerds. I think I have a system figured out. First I add up what each of us ate separately, and determine the difference (usually his is more because I drink water instead of a soft drink).. ..
OK, but do you take it into account if you eat three or four bites of F's. food?
__________________
__________________
"Expanded polystyrene (made with a blowing agent) releases the blowing agent (gas) when you put it in the toaster. So what you end up with is non-expanded polystyrene!" -- philosopher aja8888
06-23-2016, 12:17 PM #63
Give me a museum and I'll fill it. (Picasso)
Give me a forum ...
Join Date: Jun 2008
Posts: 8,457
Quote:
Originally Posted by W2R ... I love figuring out the bill, the tax and tip, and who pays how much, in my head. ...
I'm impressed. No way could I do that accurately all in my head. I usually round up in my head and add integers. For example, I'd remember the meal that I order. Say, \$12.50. So in my head the running total is \$13. Then if I order soda to drink, I'd add that as a rounded up integer. But things get tricky as sometimes I'd order a soda without taking a good look at the menu (they aren't the easiest to find). Then again, does the place give free refills? Probably, not not sure. Plus, I have to guess on the tip and tax too. My strategy now is to surrender to a tipping app and do the calculation for what I ate and drank after looking at the check. I trust that more then keeping the totals in my head. Also, like is grade school math, I can always show my work if needed .
__________________
Have you ever seen a headstone with these words
"If only I had spent more time at work" ... from "Busy Man" sung by Billy Ray Cyrus
06-23-2016, 02:21 PM #64 Give me a museum and I'll fill it. (Picasso)Give me a forum ... Join Date: May 2005 Location: Lawn chair in Texas Posts: 13,124 McDonald's always makes me pay upfront... __________________ __________________ Have Funds, Will Retire ...not doing anything of true substance...
Currently Active Users Viewing This Thread: 1 (0 members and 1 guests)
Thread Tools Search this Thread Search this Thread: Advanced Search Display Modes Linear Mode
Posting Rules You may not post new threads You may not post replies You may not post attachments You may not edit your posts BB code is On Smilies are On [IMG] code is On HTML code is OffTrackbacks are Off Pingbacks are Off Refbacks are Off Forum Rules
Similar Threads Thread Thread Starter Forum Replies Last Post HawkeyeNFO Other topics 75 09-15-2012 07:29 PM cute fuzzy bunny Other topics 13 06-08-2008 09:03 AM lazyday FIRE and Money 10 01-28-2007 07:50 AM jazz4cash FIRE and Money 29 09-23-2006 06:26 PM | 1,483 | 6,140 | {"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} | 3.25 | 3 | CC-MAIN-2018-22 | latest | en | 0.933655 |
https://gmatclub.com/forum/is-a-positive-integer-x-an-odd-number-248408.html | 1,558,840,663,000,000,000 | text/html | crawl-data/CC-MAIN-2019-22/segments/1558232258621.77/warc/CC-MAIN-20190526025014-20190526051014-00537.warc.gz | 486,703,024 | 141,106 | GMAT Question of the Day - Daily to your Mailbox; hard ones only
It is currently 25 May 2019, 20:17
### GMAT Club Daily Prep
#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.
Customized
for You
we will pick new questions that match your level based on your Timer History
Track
every week, we’ll send you an estimated GMAT score based on your performance
Practice
Pays
we will pick new questions that match your level based on your Timer History
# Is a positive integer x an odd number?
Author Message
Math Revolution GMAT Instructor
Joined: 16 Aug 2015
Posts: 7372
GMAT 1: 760 Q51 V42
GPA: 3.82
Is a positive integer x an odd number? [#permalink]
### Show Tags
01 Sep 2017, 10:10
00:00
Difficulty:
(N/A)
Question Stats:
67% (00:17) correct 33% (00:56) wrong based on 6 sessions
### HideShow timer Statistics
Is a positive integer x an odd number?
1) The smallest prime factor of x is 7.
2) The greatest prime factor of x is 7.
--== Message from the GMAT Club Team ==--
THERE IS LIKELY A BETTER DISCUSSION OF THIS EXACT QUESTION.
This discussion does not meet community quality standards. It has been retired.
If you would like to discuss this question please re-post it in the respective forum. Thank you!
To review the GMAT Club's Forums Posting Guidelines, please follow these links: Quantitative | Verbal Please note - we may remove posts that do not follow our posting guidelines. Thank you.
_________________
MathRevolution: Finish GMAT Quant Section with 10 minutes to spare
The one-and-only World’s First Variable Approach for DS and IVY Approach for PS with ease, speed and accuracy.
"Only \$149 for 3 month Online Course"
"Free Resources-30 day online access & Diagnostic Test"
"Unlimited Access to over 120 free video lessons - try it yourself"
Math Expert
Joined: 02 Sep 2009
Posts: 55274
Re: Is a positive integer x an odd number? [#permalink]
### Show Tags
01 Sep 2017, 10:25
MathRevolution wrote:
Is a positive integer x an odd number?
1) The smallest prime factor of x is 7.
2) The greatest prime factor of x is 7.
You've posted it before: is-a-positive-integer-x-an-odd-number-1-the-smallest-prime-factor-of-247395.html
--== Message from the GMAT Club Team ==--
THERE IS LIKELY A BETTER DISCUSSION OF THIS EXACT QUESTION.
This discussion does not meet community quality standards. It has been retired.
If you would like to discuss this question please re-post it in the respective forum. Thank you!
To review the GMAT Club's Forums Posting Guidelines, please follow these links: Quantitative | Verbal Please note - we may remove posts that do not follow our posting guidelines. Thank you.
_________________
Re: Is a positive integer x an odd number? [#permalink] 01 Sep 2017, 10:25
Display posts from previous: Sort by | 725 | 2,904 | {"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-2019-22 | latest | en | 0.847429 |
https://mathematica.stackexchange.com/questions/200653/calculating-the-distance-between-two-points-and-get-direction | 1,725,752,026,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700650926.21/warc/CC-MAIN-20240907225010-20240908015010-00816.warc.gz | 374,540,615 | 36,966 | # Calculating the distance between two points and get direction [closed]
I am programming a molecular dynamics program with periodic boundary conditions. Here, I have a problem:
I have a list of coordinates like
list = RandomReal[{0, 1}, {5, 2}]
And now, I want to have the distances.
I do that with
EuclideanDistance @@@ Subsets[list, {2}]
But now, I do not know the direction, so if the distance not squared was positive or negative. Does someone know, how I can obtain the sign? Because I need that for computing my periodic boundary conditions
• EuclideanDistanceis always positive. Please clarify "direction"! Commented Jun 19, 2019 at 13:57
• @UlrichNeumann So you have $\sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2}$. For me, the "direction" is the sign of $x_1 - x_2$ and $y_1 - y_2$, which can be positive or negative, so I also want to have $x_1 - x_2$ and $y_1 - y_2$ explicitly. Commented Jun 19, 2019 at 14:03
• @ Armani42 What is the "direction" if x1-x2<0 and y1-y2>0 ? Commented Jun 19, 2019 at 14:11
Here is a solution which gives you the distance and the difference
list = RandomReal[{0, 1}, {5, 2}]; (*points*)
sl= Subsets[list, {2}] (* all point pairs*)
m=Map[{Sqrt[(#[[1]] - #[[2]]).(#[[1]] - #[[2]])], #[[1]] - #[[2]]} &,\sl]
(*{{0.72911, {0.681672, -0.258698}}, {0.410144, {0.359548, 0.197341}}, ...}*)
The first element of m is the distance
m[[All, 1]] == EuclideanDistance @@@ sl
(*True*)
the last is the difference vector.
• Thank you, that really helped me :) Commented Jun 20, 2019 at 9:24 | 501 | 1,529 | {"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": 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} | 3.609375 | 4 | CC-MAIN-2024-38 | latest | en | 0.86333 |
https://www.jiskha.com/questions/952587/a-perfect-positive-or-negative-correlation-means-that-a-the-explanatory-causes-the | 1,656,331,982,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103331729.20/warc/CC-MAIN-20220627103810-20220627133810-00140.warc.gz | 918,525,336 | 5,139 | # Math
A perfect positive or negative correlation means that
A) the explanatory causes the y-variable.
B) 100% of the variation is explained.
C) we get the same regression equation if we switch the x and y variables.
D) the slope of the regression equation is 1.0
I chose B
1. 👍
2. 👎
3. 👁
4. ℹ️
5. 🚩
1. Correlation does not necessarily mean causation. I would choose c.
1. 👍
2. 👎
3. ℹ️
4. 🚩
## Similar Questions
Still need help? You can ask a new question. | 148 | 463 | {"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-2022-27 | latest | en | 0.792673 |
http://dmac.rutgers.edu/programs/challenge/vrp/tdcarp/ | 1,726,254,889,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651535.66/warc/CC-MAIN-20240913165920-20240913195920-00317.warc.gz | 7,464,158 | 8,965 | # Time Dependent Capacitated Arc Routing
##### TDCARP Problem Statement
Congestion is a major factor in city logistics. Time-dependent routing problems capture variation in travel times that may arise as a consequence of congestion, weather conditions, road closures, and other factors. Time-dependent arc routing problems have not been well-studied in the literature, which is somewhat surprising because most related applications occur in urban contexts where travel times are greatly impacted by congestion. The Implementation Challenge will study Time-dependent Capacitated Arc Routing Problems (TDCARPs) in this portion of the competition.
TDCARP is a variant of CARP that uses a time-dependent travel cost. Like CARP, TDCARP is stated on a network consisting of nodes and (undirected) edges. An edge (i,j) between nodes i and j can be used to travel over. Each edge (i,j) is associated with a distance dij, and some of the edges require demand for a service quantity qij ≥ 0. Each vehicle has a capacity Q that limits the demand it can serve along a route. TDCARP consists of finding a set of vehicle routes—represented as a sequence of deadheaded and serviced arcs (oriented edges) starting and ending at the depot—such that each edge with non-zero demand is serviced by exactly one vehicle, the total demand served on each vehicle does not exceed its capacity limit Q, and the total travel time is minimized.
TDCARP is defined over a planning horizon [0,H]. Each undirected edge (i,j) is associated with two directed arcs (i,j) and (j,i), and the amount of time it takes to traverse an arc depends on when it is traversed within the planning horizon. Given a planning horizon [0,H], we associate to each arc (i,j) a piecewise constant speed function with hij pieces. This function represents the travel speed (i.e. the distance traveled per unit of time) on this arc as a function of time of traversal. We note that the speed of a vehicle can change while it is traveling on a link based on this function.
##### TDCARP Instances
As in the CARP competition, we will focus on instances that are based on real city networks.
The TDCARP instances are as described in Vidal, Martinelli, Pham, and Hà [VMPH]. They are based on ten CARP instances (C01-C10) of Beullens, Muyldermans, Cattrysse, and Van Oudheusden [BMCV] built from Flanders road network and ten EGL instances of Brandão and Eglese [BE08] and Li and Eglese [LE96] associated with a winter gritting case study in Lancashire. (The Buellens and EGL instances are also used in the CARP Challenge.)
From each of these 20 CARP instances [VMPH] generated three instances with different degrees of time dependency: low (L), medium (M), and high (H), for a total of 60 TDCARP instances. The network information, customer demands, and vehicle capacities are the same as those of the original instance.
The TDCARP instances can be found at: https://github.com/vidalt/HGS-TDCARP/tree/master/Instances/TDCARP-withoutSP.
A description of the input file format is contained in this file.
##### Output Format
The output file format and naming convention is the same as that described for the CARP variant.
##### Algorithm Evaluation
Algorithm evaluation will take place in single round with all testing performed on the participant's own computer.
• Runs must be performed on a processor listed in PassMark.
• Solvers must run on a single thread.
• If a solver solves linear relaxations or mixed integer linear programming problems, any commercial solver may be used as long as the commercial solver is run on a single thread.
Solvers will be assessed based on their performance on all 60 TDCARP instances.
Teams are allowed to use the preprocessed time-dependent shortest path data associated with the instances that is available at https://github.com/vidalt/HGS-TDCARP/tree/master/Instances/TDCARP-withSP.
Time Limits: Solvers will be assessed based on the quality of the solutions obtained for each instance within a given amount of time. The time limits are based on the number of service requests:
• Instances with 200 or fewer service requests have a time limit of 15 minutes.
• Instances with more than 200 service requests but no more than 400 have a time limit of 30 minutes. (I.e.,the number of service requests is in the range [201,400].)
• Instances with more than 400 service requests have a time limit of 60 minutes.
These time limits are stated for a machine with a 1500 CPU mark. Time limits for a competitor's machine must be scaled to mitigate the effect of processor speed. If the CPU mark of the competitor's machine is C, then the time limit is multiplied by 1500/C for that machine.
Scoring the Solvers: Solvers will be awarded points for their performance on each instance and ranked by the total number of points over all instances.The solver that finds the best solution for an instance receives 3 points for that instance; the next-best solver receives 2 points; and the third best gets 1 point. When there are ties, the points at play are split evenly between the tied solvers.
The competitor that accumulates the most points over all instances is declared the winner and will be invited to present at the workshop, provided that their method is adequately described in the associated extended abstract. (Failure to provide this document will result in disqualification.)
Additionally, the VRP Implementation Challenge committee will review all entries with respect to additional criteria such as novelty of the approach taken, simplicity, and versatility of a solver to work well on multiple different variants. We hope to recognize multiple entries for these attributes with invitations to present.
##### Submission
When you are ready to submit either your solver output package (a zipped file) or your Challenge paper, you may do so from the Submissions page (which is part of the navigation menu on the right-hand side of this page).
Outputs are due January 30, 2022 and must be received by 23:59 Pacific Standard Time (UTC -8). | 1,309 | 6,035 | {"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-2024-38 | latest | en | 0.938349 |
http://www.vuanswer.com/2022/06/mth302-assignment-1-solution-2022.html | 1,718,667,293,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861741.14/warc/CC-MAIN-20240617215859-20240618005859-00861.warc.gz | 59,301,040 | 37,619 | # MTH302 Assignment 1 Solution 2022 - VU Answer
VU Answer Providing Correct MTH302 Assignment No 1 Solution Spring 2022. A Perfect Solved MTH302 Assignment 1 Solution 2022 File.
In the MTH302 course, you can study various important topics and lectures.
According to the VU schedule every semester they provide mth302 assignments for students to submit with solutions before the due date.
# MTH302 Assignment 1 Solution 2022
Solution File Given Below.
Q. A deposit of Rs. 6000 in a Bank account earns interest 5% compounded monthly for four years. How much amount is accumulated in the account after 4 years?
# Solution:
Deposit P = 6000 RS
.r = 5% = 5/100 = 0.05
Term t = 4 years n = 12
A = ?
r n*t
A = p 1+ n
0.05 12*4 48
A = 6000 1+
12
6000[1.0042]
6000[1.2228] = 7336.8RS
Accumulated amount A = 7336.8 RS.
Q.2
A 55-inch segment is divided into three parts whose lengths have the ratio 2 : 4 : 5. What is the length of the longest part?
# Solution:
Total length = 55-inch Ratio is = 2 : 4 : 5
Sum of ratios =11
Length of longest part = (5/11)*55 inches = 25 inches .
Q3. Suppose that you establish an IRA (Individual Retirement Account) at age 43 and you will retire after 22 years hence at age 65. You plan to make annual payments of Rs1000 into the IRA at the beginning of each year. If you assume a rate of return of 8.5 percent a year, calculate the future value of your IRA when you will retire at age 65.
# Solution:
paymentperperiod c = 1000RS
int erst rate i = 8.5% = 0.085
number of paymet n = 22
Futurevalue Fv = ?
(1+ i)n 1
Fv = c i
(1+ 0.085)22 1
5.018
Fv = 1000
0.085
= 1000 ⎢⎣ 0.085 ⎥⎦ = 1000[59.0353]
Fv = 59, 055.3 RS
Check Also More Latest Assignment Solution:
SOC101 Assignment 1 Solution 2022
ISL202 Assignment 1 Solution 2022
PSY101 Assignment 1 Solution 2022
MTH302 Assignment 1 Solution 2022
All Assignment Solution 2022 Files
Tags:
MTH302 Assignment Solution 2022
MTH302 Assignment No 1 2022 Solution
MTH302 Assignment 1 Solution 2022
MTH302 Solution 2022 Spring
Make sure you can make some changes to your solution file before submitting copy-paste solution will be marked zero.
Before submitting check your assignment question file requirement properly.
If you need some help and question about files and solutions share them with us.
We would recommend you get the idea from the file and make your own solution. | 715 | 2,394 | {"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} | 4.53125 | 5 | CC-MAIN-2024-26 | latest | en | 0.833156 |
https://www.gradesaver.com/textbooks/science/chemistry/chemistry-atoms-first-zumdahl/chapter-13-exercises-page-575b/44 | 1,725,816,176,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651013.49/warc/CC-MAIN-20240908150334-20240908180334-00775.warc.gz | 759,162,446 | 11,993 | Chemistry: Atoms First (2nd Edition)
a. $H_2O$ is the stronger base; b. $N{O_2}^-$ is the stronger base; c. $OC_6{H_5}^-$ is the stronger base;
- Identify the base with the weakest conjugate acid; it will be the stronger base. Using the table 13-2, from page 531, and knowing that: Strong acids: $K_a > 1$ $H_3O^+$: $K_a = 1.0$ a. $HCl$ has $K_a > 1$ $H_3O^+$ has $K_a = 1$ Therefore, $H_3O^+$ is the weaker acid, and $H_2O$ is the stronger base. b. $H_3O^+$ has $K_a = 1$ $HNO_2$ has $K_a = 4.0 \times 10^{-4}$ Therefore, $HNO_2$ is the weaker acid, and $N{O_2}^-$ is the stronger base. c. $HCN$ has $K_a = 6.2 \times 10^{-10}$ $HOC_6H_5$ has $K_a = 1.6 \times 10^{-10}$ Therefore, $HOC_6H_5$ is the weaker acid, and $OC_6{H_5}^-$ is the stronger base | 322 | 753 | {"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} | 2.8125 | 3 | CC-MAIN-2024-38 | latest | en | 0.746681 |
http://docs.sympy.org/0.7.2/_modules/sympy/concrete/summations.html | 1,534,562,004,000,000,000 | text/html | crawl-data/CC-MAIN-2018-34/segments/1534221213264.47/warc/CC-MAIN-20180818021014-20180818041014-00680.warc.gz | 105,759,913 | 10,151 | # Source code for sympy.concrete.summations
from sympy.core import Add, C, Derivative, Dummy, Expr, S, sympify, Wild
from sympy.concrete.gosper import gosper_sum
from sympy.functions.elementary.piecewise import piecewise_fold
from sympy.polys import apart, PolynomialError
from sympy.solvers import solve
def _free_symbols(function, limits):
"""Helper function to return the symbols that appear in a sum-like object
once it is evaluated.
"""
isyms = function.free_symbols
for xab in limits:
# take out the target symbol
if xab[0] in isyms:
isyms.remove(xab[0])
# add in the new symbols
for i in xab[1:]:
isyms.update(i.free_symbols)
return isyms
[docs]class Sum(Expr):
"""Represents unevaluated summation."""
__slots__ = ['is_commutative']
def __new__(cls, function, *symbols, **assumptions):
from sympy.integrals.integrals import _process_limits
# Any embedded piecewise functions need to be brought out to the
# top level so that integration can go into piecewise mode at the
# earliest possible moment.
function = piecewise_fold(sympify(function))
if function is S.NaN:
return S.NaN
if not symbols:
raise ValueError("Summation variables must be given")
limits, sign = _process_limits(*symbols)
# Only limits with lower and upper bounds are supported; the indefinite Sum
# is not supported
if any(len(l) != 3 or None in l for l in limits):
raise ValueError('Sum requires values for lower and upper bounds.')
obj = Expr.__new__(cls, **assumptions)
arglist = [sign*function]
arglist.extend(limits)
obj._args = tuple(arglist)
obj.is_commutative = function.is_commutative # limits already checked
return obj
@property
def function(self):
return self._args[0]
@property
def limits(self):
return self._args[1:]
@property
[docs] def variables(self):
"""Return a list of the summation variables
>>> from sympy import Sum
>>> from sympy.abc import x, i
>>> Sum(x**i, (i, 1, 3)).variables
[i]
"""
return [l[0] for l in self.limits]
@property
[docs] def free_symbols(self):
"""
This method returns the symbols that will exist when the
summation is evaluated. This is useful if one is trying to
determine whether a sum depends on a certain symbol or not.
>>> from sympy import Sum
>>> from sympy.abc import x, y
>>> Sum(x, (x, y, 1)).free_symbols
set([y])
"""
if self.function.is_zero:
return set()
return _free_symbols(self.function, self.limits)
@property
[docs] def is_zero(self):
"""A Sum is only zero if its function is zero or if all terms
cancel out. This only answers whether the summand zero."""
return self.function.is_zero
@property
[docs] def is_number(self):
"""
Return True if the Sum will result in a number, else False.
sympy considers anything that will result in a number to have
is_number == True.
>>> from sympy import log
>>> log(2).is_number
True
Sums are a special case since they contain symbols that can
be replaced with numbers. Whether the integral can be done or not is
another issue. But answering whether the final result is a number is
not difficult.
>>> from sympy import Sum
>>> from sympy.abc import x, y
>>> Sum(x, (y, 1, x)).is_number
False
>>> Sum(1, (y, 1, x)).is_number
False
>>> Sum(0, (y, 1, x)).is_number
True
>>> Sum(x, (y, 1, 2)).is_number
False
>>> Sum(x, (y, 1, 1)).is_number
False
>>> Sum(x, (x, 1, 2)).is_number
True
>>> Sum(x*y, (x, 1, 2), (y, 1, 3)).is_number
True
"""
return self.function.is_zero or not self.free_symbols
def doit(self, **hints):
#if not hints.get('sums', True):
# return self
f = self.function
for limit in self.limits:
i, a, b = limit
dif = b - a
if dif.is_Integer and dif < 0:
a, b = b, a
f = eval_sum(f, (i, a, b))
if f is None:
return self
if hints.get('deep', True):
return f.doit(**hints)
else:
return f
def _eval_summation(self, f, x):
return
def _eval_derivative(self, x):
"""
Differentiate wrt x as long as x is not in the free symbols of any of
the upper or lower limits.
Sum(a*b*x, (x, 1, a)) can be differentiated wrt x or b but not a
since the value of the sum is discontinuous in a. In a case
involving a limit variable, the unevaluated derivative is returned.
"""
# diff already confirmed that x is in the free symbols of self, but we
# don't want to differentiate wrt any free symbol in the upper or lower
# limits
# XXX remove this test for free_symbols when the default _eval_derivative is in
if x not in self.free_symbols:
return S.Zero
# get limits and the function
f, limits = self.function, list(self.limits)
limit = limits.pop(-1)
if limits: # f is the argument to a Sum
f = Sum(f, *limits)
if len(limit) == 3:
_, a, b = limit
if x in a.free_symbols or x in b.free_symbols:
return None
df = Derivative(f, x, **{'evaluate': True})
rv = Sum(df, limit)
if limit[0] not in df.free_symbols:
rv = rv.doit()
return rv
else:
return NotImplementedError('Lower and upper bound expected.')
[docs] def euler_maclaurin(self, m=0, n=0, eps=0, eval_integral=True):
"""
Return an Euler-Maclaurin approximation of self, where m is the
number of leading terms to sum directly and n is the number of
terms in the tail.
With m = n = 0, this is simply the corresponding integral
plus a first-order endpoint correction.
Returns (s, e) where s is the Euler-Maclaurin approximation
and e is the estimated error (taken to be the magnitude of
the first omitted term in the tail):
>>> from sympy.abc import k, a, b
>>> from sympy import Sum
>>> Sum(1/k, (k, 2, 5)).doit().evalf()
1.28333333333333
>>> s, e = Sum(1/k, (k, 2, 5)).euler_maclaurin()
>>> s
-log(2) + 7/20 + log(5)
>>> from sympy import sstr
>>> print sstr((s.evalf(), e.evalf()), full_prec=True)
(1.26629073187415, 0.0175000000000000)
The endpoints may be symbolic:
>>> s, e = Sum(1/k, (k, a, b)).euler_maclaurin()
>>> s
-log(a) + log(b) + 1/(2*b) + 1/(2*a)
>>> e
Abs(-1/(12*b**2) + 1/(12*a**2))
If the function is a polynomial of degree at most 2n+1, the
Euler-Maclaurin formula becomes exact (and e = 0 is returned):
>>> Sum(k, (k, 2, b)).euler_maclaurin()
(b**2/2 + b/2 - 1, 0)
>>> Sum(k, (k, 2, b)).doit()
b**2/2 + b/2 - 1
With a nonzero eps specified, the summation is ended
as soon as the remainder term is less than the epsilon.
"""
m = int(m)
n = int(n)
f = self.function
assert len(self.limits) == 1
i, a, b = self.limits[0]
s = S.Zero
if m:
for k in range(m):
term = f.subs(i, a+k)
if (eps and term and abs(term.evalf(3)) < eps):
return s, abs(term)
s += term
a += m
x = Dummy('x')
I = C.Integral(f.subs(i, x), (x, a, b))
if eval_integral:
I = I.doit()
s += I
def fpoint(expr):
if b is S.Infinity:
return expr.subs(i, a), 0
return expr.subs(i, a), expr.subs(i, b)
fa, fb = fpoint(f)
iterm = (fa + fb)/2
g = f.diff(i)
for k in xrange(1, n+2):
ga, gb = fpoint(g)
term = C.bernoulli(2*k)/C.factorial(2*k)*(gb-ga)
if (eps and term and abs(term.evalf(3)) < eps) or (k > n):
break
s += term
g = g.diff(i, 2)
return s + iterm, abs(term)
def _eval_subs(self, old, new): # XXX this should be the same as Integral's
if any(old == v for v in self.variables):
return self
[docs]def summation(f, *symbols, **kwargs):
r"""
Compute the summation of f with respect to symbols.
The notation for symbols is similar to the notation used in Integral.
summation(f, (i, a, b)) computes the sum of f with respect to i from a to b,
i.e.,
::
b
____
\
summation(f, (i, a, b)) = ) f
/___,
i = a
If it cannot compute the sum, it returns an unevaluated Sum object.
Repeated sums can be computed by introducing additional symbols tuples::
>>> from sympy import summation, oo, symbols, log
>>> i, n, m = symbols('i n m', integer=True)
>>> summation(2*i - 1, (i, 1, n))
n**2
>>> summation(1/2**i, (i, 0, oo))
2
>>> summation(1/log(n)**n, (n, 2, oo))
Sum(log(n)**(-n), (n, 2, oo))
>>> summation(i, (i, 0, n), (n, 0, m))
m**3/6 + m**2/2 + m/3
>>> from sympy.abc import x
>>> from sympy import factorial
>>> summation(x**n/factorial(n), (n, 0, oo))
exp(x)
"""
return Sum(f, *symbols, **kwargs).doit(deep=False)
def telescopic_direct(L, R, n, limits):
"""Returns the direct summation of the terms of a telescopic sum
L is the term with lower index
R is the term with higher index
n difference between the indexes of L and R
For example:
>>> from sympy.concrete.summations import telescopic_direct
>>> from sympy.abc import k, a, b
>>> telescopic_direct(1/k, -1/(k+2), 2, (k, a, b))
-1/(b + 2) - 1/(b + 1) + 1/(a + 1) + 1/a
"""
(i, a, b) = limits
s = 0
for m in xrange(n):
s += L.subs(i,a+m) + R.subs(i,b-m)
return s
def telescopic(L, R, limits):
'''Tries to perform the summation using the telescopic property
return None if not possible
'''
(i, a, b) = limits
return None
# We want to solve(L.subs(i, i + m) + R, m)
# First we try a simple match since this does things that
# solve doesn't do, e.g. solve(f(k+m)-f(k), m) fails
k = Wild("k")
sol = (-R).match(L.subs(i, i + k))
s = None
if sol and k in sol:
s = sol[k]
if not (s.is_Integer and L.subs(i,i + s) == -R):
#sometimes match fail(f(x+2).match(-f(x+k))->{k: -2 - 2x}))
s = None
# But there are things that match doesn't do that solve
# can do, e.g. determine that 1/(x + m) = 1/(1 - x) when m = 1
if s is None:
m = Dummy('m')
try:
sol = solve(L.subs(i, i + m) + R, m) or []
except NotImplementedError:
return None
sol = [si for si in sol if si.is_Integer and
(L.subs(i,i + si) + R).expand().is_zero]
if len(sol) != 1:
return None
s = sol[0]
if s < 0:
return telescopic_direct(R, L, abs(s), (i, a, b))
elif s > 0:
return telescopic_direct(L, R, s, (i, a, b))
def eval_sum(f, limits):
(i, a, b) = limits
if f is S.Zero:
return S.Zero
if i not in f.free_symbols:
return f*(b - a + 1)
if a == b:
return f.subs(i, a)
dif = b - a
definite = dif.is_Integer
# Doing it directly may be faster if there are very few terms.
if definite and (dif < 100):
return eval_sum_direct(f, (i, a, b))
# Try to do it symbolically. Even when the number of terms is known,
# this can save time when b-a is big.
# We should try to transform to partial fractions
value = eval_sum_symbolic(f.expand(), (i, a, b))
if value is not None:
return value
# Do it directly
if definite:
return eval_sum_direct(f, (i, a, b))
def eval_sum_direct(expr, limits):
(i, a, b) = limits
dif = b - a
return Add(*[expr.subs(i, a + j) for j in xrange(dif + 1)])
def eval_sum_symbolic(f, limits):
(i, a, b) = limits
if not f.has(i):
return f*(b-a+1)
# Linearity
if f.is_Mul:
L, R = f.as_two_terms()
if not L.has(i):
sR = eval_sum_symbolic(R, (i, a, b))
if sR: return L*sR
if not R.has(i):
sL = eval_sum_symbolic(L, (i, a, b))
if sL: return R*sL
try:
f = apart(f, i) # see if it becomes an Add
except PolynomialError:
pass
L, R = f.as_two_terms()
lrsum = telescopic(L, R, (i, a, b))
if lrsum:
return lrsum
lsum = eval_sum_symbolic(L, (i, a, b))
rsum = eval_sum_symbolic(R, (i, a, b))
if None not in (lsum, rsum):
return lsum + rsum
# Polynomial terms with Faulhaber's formula
n = Wild('n')
result = f.match(i**n)
if result is not None:
n = result[n]
if n.is_Integer:
if n >= 0:
return ((C.bernoulli(n+1, b+1) - C.bernoulli(n+1, a))/(n+1)).expand()
elif a.is_Integer and a >= 1:
if n == -1:
return C.harmonic(b) - C.harmonic(a - 1)
else:
return C.harmonic(b, abs(n)) - C.harmonic(a - 1, abs(n))
# Geometric terms
c1 = C.Wild('c1', exclude=[i])
c2 = C.Wild('c2', exclude=[i])
c3 = C.Wild('c3', exclude=[i])
e = f.match(c1**(c2*i+c3))
if e is not None:
c1 = c1.subs(e)
c2 = c2.subs(e)
c3 = c3.subs(e)
# TODO: more general limit handling
return c1**c3 * (c1**(a*c2) - c1**(c2+b*c2)) / (1 - c1**c2)
if not (a.has(S.Infinity, S.NegativeInfinity) or \
b.has(S.Infinity, S.NegativeInfinity)):
r = gosper_sum(f, (i, a, b))
if not r in (None, S.NaN):
return r
return eval_sum_hyper(f, (i, a, b))
def _eval_sum_hyper(f, i, a):
""" Returns (res, cond). Sums from a to oo. """
from sympy.functions import hyper
from sympy.simplify import hyperexpand, hypersimp, fraction, simplify
from sympy.polys.polytools import Poly, factor
if a != 0:
return _eval_sum_hyper(f.subs(i, i + a), i, 0)
if f.subs(i, 0) == 0:
if simplify(f.subs(i, Dummy('i', integer=True, positive=True))) == 0:
return S(0), True
return _eval_sum_hyper(f.subs(i, i + 1), i, 0)
hs = hypersimp(f, i)
if hs is None:
return None
numer, denom = fraction(factor(hs))
top, topl = numer.as_coeff_mul(i)
bot, botl = denom.as_coeff_mul(i)
ab = [top, bot]
factors = [topl, botl]
params = [[], []]
for k in range(2):
for fac in factors[k]:
mul = 1
if fac.is_Pow:
mul = fac.exp
fac = fac.base
if not mul.is_Integer:
return None
p = Poly(fac, i)
if p.degree() != 1:
return None
m, n = p.all_coeffs()
ab[k] *= m**mul
params[k] += [n/m]*mul
# Add "1" to numerator parameters, to account for implicit n! in
# hypergeometric series.
ap = params[0] + [1]
bq = params[1]
x = ab[0]/ab[1]
h = hyper(ap, bq, x)
return f.subs(i, 0)*hyperexpand(h), h.convergence_statement
def eval_sum_hyper(f, (i, a, b)):
from sympy.functions import Piecewise
from sympy import oo, And
if b != oo:
if a == -oo:
res = _eval_sum_hyper(f.subs(i, -i), i, -b)
if res is not None:
return Piecewise(res, (Sum(f, (i, a, b)), True))
else:
res1 = _eval_sum_hyper(f, i, a)
res2 = _eval_sum_hyper(f, i, b + 1)
if res1 is None or res2 is None:
return None
(res1, cond1), (res2, cond2) = res1, res2
cond = And(cond1, cond2)
if cond is False:
return None
return Piecewise((res1 - res2, cond), (Sum(f, (i, a, b)), True))
if a == -oo:
res1 = _eval_sum_hyper(f.subs(i, -i), i, 1)
res2 = _eval_sum_hyper(f, i, 0)
if res1 is None or res2 is None:
return None
res1, cond1 = res1
res2, cond2 = res2
cond = And(cond1, cond2)
if cond is False:
return None
return Piecewise((res1 + res2, cond), (Sum(f, (i, a, b)), True))
# Now b == oo, a != -oo
res = _eval_sum_hyper(f, i, a)
if res is not None:
return Piecewise(res, (Sum(f, (i, a, b)), True))
` | 4,501 | 13,658 | {"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} | 2.578125 | 3 | CC-MAIN-2018-34 | latest | en | 0.606671 |
http://stackoverflow.com/questions/7721240/how-to-extract-non-vertical-column-from-matrix-in-matlab | 1,394,988,010,000,000,000 | text/html | crawl-data/CC-MAIN-2014-10/segments/1394678703030/warc/CC-MAIN-20140313024503-00008-ip-10-183-142-35.ec2.internal.warc.gz | 137,453,810 | 15,949 | # How to extract non-vertical column from matrix in Matlab
I have matrix A and a vector b, which specifies column index of the element to be extracted for each corresponding row of the matrix.
For example, A =
``````1 2 3
4 5 6
7 8 9
``````
b = [1 3 2]'
I'd like to have c = [1 6 8]' on output. How to achieve this?
I tried A(:, b), but it doesn't work as I need.
-
possible duplicate of MATLAB indexing question – Amro Oct 14 '11 at 4:14
There may be a more elegant solution, but this works:
``````b = [1 3 2]';
[rows, cols] = size(A);
A(sub2ind([rows cols], [1 : rows]', b))
``````
-
Avoid using `[]` when you mean `()` because it'll cause unnecessary calls to `horzcat` and `vertcat` (I expect). – Nzbuu Oct 11 '11 at 9:42
Interesting. I always think of `[]` as "whatever's in here is an array". I know that's not the technical truth, but it's what I think when I'm reading/writing code. I assume that you mean that `[1 : rows]'` could be `(1 : rows)'`. In my (admittedly flawed) mental model of what's going on, I'm saying "I want a column vector from 1 to rows." Seems like the order of operations would be such that there would be no calls to horzcat there? I.e., we get something like `temp1 = 1 : rows`, `temp2 = [temp1]` (no horzcat b/c there's only one object), `temp3 = temp2'`. – dantswain Oct 11 '11 at 13:45
TBH, I'm not sure how clever the MATLAB parser is in this case. Certainly if you have an object, `o`, and you write `[o]`, then it will call the `horzcat` and `vertcat` methods. Also, it's generally worth following the M-Lint advice regardless. – Nzbuu Oct 11 '11 at 19:08
Old habits die hard :) I was really curious about this, so I did a little profiling. I wrote three functions calculating: `[1 : 10].'`, `(1 : 10).'`, and `[(1 : 10).']`. I ran the profiler on a script that called each function 100,000 times. `[(1 : 10).']` was actually consistently the fastest by about 0.1 sec.(?!) Obviously this is not at all rigorous, but for me the moral of the story is that the readability of `[1 : N]` is worth the tiny performance hit (at least for N reasonably small). Speaking only on behalf of myself, of course. – dantswain Oct 11 '11 at 19:45
``````r = size(A,1); | 664 | 2,203 | {"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-2014-10 | latest | en | 0.937844 |
https://web2.0calc.com/questions/probability_33548 | 1,713,073,893,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296816864.66/warc/CC-MAIN-20240414033458-20240414063458-00338.warc.gz | 577,665,256 | 5,689 | +0
# probability
0
225
1
Randomly choose two numbers between 0 and 1. What is the probability that their average is no greater than 2/3 and no less than 1/3?
May 8, 2022
#1
+9519
+2
We can view this geometrically. Let x and y be the chosen numbers. We can plot (x, y) inside the unit square with vertices (0, 0) and (1, 1).
The constraints are $$\dfrac13 \leq \dfrac {x + y}2 \leq \dfrac23$$ and we are basically finding the probability that the point (x, y) falls into the region described by the compound inequality. We can do so by considering the area of the region enclosed by $$\dfrac13 \leq \dfrac {x + y}2 \leq \dfrac23$$ and the unit square.
Let $$\Omega = \left\{(x, y) \in [0, 1]^2 : \dfrac13 \leq \dfrac {x + y}2 \leq \dfrac23\right\}$$. Here's a plot of $$\Omega$$:
So the probability is just $$\operatorname{Area}(\Omega) = 1 - 2\times \dfrac{\dfrac23 \times \dfrac23}2 = \dfrac59$$
May 8, 2022 | 327 | 920 | {"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} | 4.21875 | 4 | CC-MAIN-2024-18 | latest | en | 0.83201 |
http://mathhelpforum.com/calculus/95084-u-substition.html | 1,527,423,637,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794868248.78/warc/CC-MAIN-20180527111631-20180527131631-00120.warc.gz | 186,201,949 | 9,298 | 1. u substition
Suppose
Evaluate the following
im not sure what this is asking me to do, I know that if i let t/4 be u then change the end points i get the same thing..
2. Originally Posted by acosta0809
Suppose
Evaluate the following
im not sure what this is asking me to do, I know that if i let t/4 be u then change the end points i get the same thing..
If $\displaystyle u = \frac{t}{4}$ then $\displaystyle t = 4u$ so $\displaystyle dt = 4 du$ so the second integral becomes
$\displaystyle 4 \int_0^4 g(u) du$ | 152 | 519 | {"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} | 3.03125 | 3 | CC-MAIN-2018-22 | latest | en | 0.80416 |
birdaystore.com | 1,623,906,022,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623487629209.28/warc/CC-MAIN-20210617041347-20210617071347-00591.warc.gz | 6,824,734 | 6,284 | 1 Ml A 1 Gm - birdaystore.com
The millilitre ml or mL, also spelled milliliter is a metric unit of volume that is equal to one thousandth of a litre. It is a non-SI unit accepted for use with the International Systems of Units SI. It is exactly equivalent to 1 cubic centimetre cm³, or, non-standard, cc. ›› Metric conversions and more. 1 cubic meter is equal to 1000000 ml, or 1000000 g. Note that rounding errors may occur, so always check the results. Use this page to learn how to convert between milliliters and grams. Type in your own numbers in the form to convert the units! ›› Quick conversion chart of ml to g. 1 ml to g = 1 g. 5 ml to g = 5 g. 10 ml to g = 10 g. 20 ml. Convert grams to milliliters to grams. g to ml to g, density converter, calculator, tool online. Formula and explanation, conversion. However, if water with density 1 gm/cc or 1 gm/ml, then 10 grams is equivalent to 10 ml. Asked in Math and Arithmetic, Units of Measure, Weight and Mass How many kilograms equal 1liter? Assuming fresh water with density 1 gm/cc or 1 gm/ml 1 liter = 1000 milliliter 1 liter = 1000 millilter x 1 gm/ml = 1000 gm = 1 kg Answer 1 kg.
A gram is a unit of mass, like kilogram, ton or pound pound-mass. A millilitre is a unit of volume, like litre, cubic metre or or cubic feet. As such, how many millilitres are equal to 1 gram depends on the density of the material you are dealin. La densità dell'acqua è di 1 g/ml. Se la sostanza ha una densità maggiore, allora andrà a fondo se immersa in acqua. Se la sostanza ha una densità inferiore a 1 g/ml, allora galleggerà. Avvertenze. Alcuni elementi si espandono o si restringono in base alla temperatura, soprattutto se fondono, si congelano o subiscono cambiamenti analoghi. 1 cubic meter is equal to 1000000 grams, or 1000000 milliliters. Note that rounding errors may occur, so always check the results. Use this page to learn how to convert between grams and milliliters. Type in your own numbers in the form to convert the units! ›› Quick conversion chart of grams to milliliters. 1 grams to milliliters = 1. One milliliter equals 1 gram only when the density of a liquid is 1 gram per milliliter. World View Science Pets & Animals Home Science. Does 1 Milliliter Equal 1 Gram? One milliliter equals 1 gram only when the density of a liquid is 1 gram per milliliter. How Do You Convert Mg to Ml? How Many Feet Does 1. depends, in theory 1 litre = 1 kilo gram, however where medications are involved, you must read the instructions as they may be 1 gram per 500 milliliter's, or 1mg per 5 ml.
04/01/2008 · Simply because 1 ml of water is supposed to be 1 gram. Then, 1 gram divided by 100 grams ml would give you 1%. 1 ml of oil in grams How many grams of oil in 1 milliliter? How much does 1 ml of oil weigh? 1 milliliter of oil equals 0.947 gram Volume to 'Weight' Converter. Volume to Weight Weight to Volume. Inputs? Notes: the results in this calculator are rounded by default to 3 significant figures. 1 cubic meter is equal to 1000000 g, or 1000000 ml. Note that rounding errors may occur, so always check the results. Use this page to learn how to convert between grams and milliliters. Type in your own numbers in the form to convert the units! ›› Quick conversion chart of g to ml. 1 g to ml = 1 ml. 5 g to ml = 5 ml. 10 g to ml = 10 ml. 20. 1% presumably m/v glucose solution would contain 1 g of glucose per 100 ml of solution. Therefore the conversion of 1 g / 100 ml to units of mol/L requires that we divide by the molar mass and multiply by the conversion factor of ml to L. comes with the razor-sharp entrance-conclusion and bulge muscle process to provide the most excellent management within its performance. 2019 GMC Terrain Denali Changes 2019 GMC T.
Does sitting too close to a television hurt your eyes? How many people have won EGOTs? Why do some vegetables spark in the microwave? Why are police officers called "cops"? 1 ppm means 1 mg/l so 1000 ppm means 1000 mg/l or 1gm/l, FW of sodium arsenite NaAsO2 is 130 ans As is 75 75 gm of arsenite is present in 130 gm sodium arsenite 1 gm-----will be-----130/75= 1.73 gm so if 1.73 gm of sodium arsenite is present in 1 litre of water it makes 1000 ppm solution of arsenite. It is not exactly “equal to” but one ml of water has a mass of 1 g because the density of water at 25 C is 1 g/ml. 1 ml x 1g/ml= 1g. Length and distance unit conversion between meter and gigameter, gigameter to meter conversion in batch, m Gm conversion chart. Gm↔m 1 Gm = 1000000000 m Gm↔km 1 Gm = 1000000 km Gm↔cm 1 Gm = 100000000000 cm Gm↔mm 1 Gm = 1000000000000 mm Gm↔um 1 Gm = 1.
06/03/2006 · the closest you could come to say 1 ml = 1g is because of the density of water which is 1 g/ml at 4 C. in general cm^3 is the volume of a thing gram is its weight. so you may be saying that something weighs 1 grams and occupies 1 ml of space. 1 ml = 1 cubic cm. 29/05/2013 · How to Convert Milliliters mL to Grams g. Converting from milliliters mL to grams g is more complicated than plugging in a number, because it converts a volume unit, milliliters, to a mass unit, grams. This means each substance. Volume unit conversion between cubic millimeter and milliliter, milliliter to cubic millimeter conversion in batch, mm3 mL conversion chart. ENDMEMO. mm3↔mL 1 mL = 1000 mm3 mm3↔uL 1 mm3 = 1 uL mm3↔Cc 1 Cc = 1000 mm3 mm3↔Drop 1 Drop = 50 mm3. In a limited number of well-controlled studies comparing 4-12 weeks of oral therapy with 4 g of sucralfate daily to that with 1-1.2 g of cimetidine daily, 87-100% of sucralfate-treated and 83-86% of cimetidine-treated duodenal ulcers were healed as evidenced by endoscopic examination. | 1,648 | 5,645 | {"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} | 3.421875 | 3 | CC-MAIN-2021-25 | latest | en | 0.653582 |
https://fred.stlouisfed.org/series/RIWXXX001S | 1,537,872,578,000,000,000 | text/html | crawl-data/CC-MAIN-2018-39/segments/1537267161501.96/warc/CC-MAIN-20180925103454-20180925123854-00230.warc.gz | 516,661,269 | 22,487 | # Relative Importance Weight (Contribution to the total industrial production index): Manufacturing excluding motor vehicles and parts (RIWXXX001S) Excel (data) CSV (data) Image (graph) PowerPoint (graph) PDF (graph)
Observation:
Aug 2018: 68.0355
Updated: Sep 14, 2018
Units:
Percent,
Frequency:
Monthly
1Y | 5Y | 10Y | Max
EDIT LINE 1
(a) Relative Importance Weight (Contribution to the total industrial production index): Manufacturing excluding motor vehicles and parts, Percent, Seasonally Adjusted (RIWXXX001S)
The IP proportions (typically shown in the first column of the relevant tables in the G.17 release) are estimates of the industries' relative contributions to overall growth in the following year. For example, the relative importance weight of the motor vehicles and parts industry is about 6 percent. If output in this industry increased 10 percent in a month, then this gain would boost growth in total IP by 6/10 percentage point (0.06 x 10% = 0.6%).
Source Code: RIW.XXX001.S
Relative Importance Weight (Contribution to the total industrial production index): Manufacturing excluding motor vehicles and parts
Select a date that will equal 100 for your custom index:
to
#### Customize data:
Write a custom formula to transform one or more series or combine two or more series.
You can begin by adding a series to combine with your existing series.
Now create a custom formula to combine or transform the series.
Need help? []
Finally, you can change the units of your new series.
Select a date that will equal 100 for your custom index:
#### Add data series to graph:
FORMAT GRAPH
Log scale:
fullscreen
NOTES
Frequency: Monthly
#### Notes:
The IP proportions (typically shown in the first column of the relevant tables in the G.17 release) are estimates of the industries' relative contributions to overall growth in the following year. For example, the relative importance weight of the motor vehicles and parts industry is about 6 percent. If output in this industry increased 10 percent in a month, then this gain would boost growth in total IP by 6/10 percentage point (0.06 x 10% = 0.6%).
Source Code: RIW.XXX001.S
#### Suggested Citation:
Board of Governors of the Federal Reserve System (US), Relative Importance Weight (Contribution to the total industrial production index): Manufacturing excluding motor vehicles and parts [RIWXXX001S], retrieved from FRED, Federal Reserve Bank of St. Louis; https://fred.stlouisfed.org/series/RIWXXX001S, September 25, 2018.
RELATED CONTENT
RELEASE TABLES
Retrieving data.
Updating graph. | 596 | 2,587 | {"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-2018-39 | longest | en | 0.836571 |
https://www.groundai.com/project/vertex-degrees-in-planar-maps/ | 1,603,359,827,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107879362.3/warc/CC-MAIN-20201022082653-20201022112653-00378.warc.gz | 752,333,419 | 62,221 | Vertex Degrees in Planar Maps
# Vertex Degrees in Planar Maps
Gwendal Collet, Michael Drmota, Lukas Daniel Klausner
###### Abstract
We prove a general multi-dimensional central limit theorem for the expected number of vertices of a given degree in the family of planar maps whose vertex degrees are restricted to an arbitrary (finite or infinite) set of positive integers . Our results rely on a classical bijection with mobiles (objects exhibiting a tree structure), combined with refined analytic tools to deal with the systems of equations on infinite variables that arise. We also discuss some possible extension to maps of higher genus.
• Vertex Degrees in Planar Maps
Gwendal Collet, Michael Drmota, Lukas Daniel Klausner
TU Wien, Institute of Discrete Mathematics and Geometry, Wiedner Hauptstrasse 8–10, A-1040 Wien, Austria
footnotetext: Partially supported by the Austrian Science Fund FWF, Project SFB F50-02
## 1 Introduction and Results
In this paper we study statistical properties of planar maps, which are connected planar graphs, possibly with loops and multiple edges, together with an embedding into the plane. Such objects are frequently used to describe topological features of geometric arrangements in two or three spatial dimensions. Thus, the knowledge of the structure and of properties of “typical” objects may turn out to be very useful in the analysis of particular algorithms that operate on planar maps. We say that map is rooted if an edge is distinguished and oriented. It is called the root edge. The first vertex of this oriented edge is called the root-vertex. The face to the right of is called the root-face and is usually taken as the outer (or infinite) face. Similarly, we call a planar map pointed if just a vertex is distinguished. However, we have to be really careful with the model. In rooted maps the root edge destroys potential symmetries, which is not the case if we consider pointed maps.
The enumeration of rooted maps is a classical subject, initiated by Tutte in the 1960’s, see [11]. Among many other results, Tutte computed the number of rooted maps with edges, proving the formula
Mn=2(2n)!(n+2)!n!3n
which directly provides the asymptotic formula
Mn∼2√πn−5/212n.
We are mainly interested in planar maps with degree restrictions. Actually, it turns out that this kind of asymptotic expansion is quite universal. Furthermore, there is always a (very general) central limit theorem for the number of vertices of given degree.
###### Theorem 1.
Suppose that is an arbitrary set of positive integers but not a subset of , let be the class of planar rooted maps with the property that all vertex degrees are in and let denote the number of maps in with edges. Furthermore, if contains only even numbers, then set ; set otherwise.
Then there exist positive constants and with
MD,n∼cDn−5/2ρ−nD,n≡0modd. (1)
Furthermore, let denote the random variable counting vertices of degree () in maps in . Then for some constant and for , and the (possibly infinite) random vector () satisfies a central limit theorem, that is,
1√n(Xn−E(Xn)),n≡0modd, (2)
converges weakly to a centered Gaussian random variable (in ).
Note that maps where all vertex degrees are or are very easy to characterize and are not really of interest, and that actually, their asymptotic properties are different from the general case. It is therefore natural to assume that is not a subset of .
Since we can equivalently consider dual maps, this kind of problem is the same as considering planar maps with restrictions on the face valencies. This means that the same results hold if we replace vertex degree by face valency. For example, if we assume that all face valencies equal 4, then we just consider planar quadrangulations (which have also been studied by Tutte [11]). In fact, our proofs will refer just to face valencies.
Theorem 1 goes far beyond known results. There are some general results for the Eulerian case where all vertex degrees are even. First, the asymptotic expansion (1) is known for Eulerian maps by Bender and Canfield [2]. Furthermore, a central limit theorem of the form (2) is known for all Eulerian maps (without degree restrictions) [9]. However, in the non-Eulerian case there are almost no results of this kind; there is only a one-dimensional central limit theorem for for all planar maps [10].
Section 2 introduces planar mobiles which, being in bijection with pointed planar maps, will reduce our analysis to simpler objects with a tree structure. Their asymptotic behaviour is derived in Section 3, first for the simpler case of bipartite maps (i.e., when contains only even integers), then for families of maps without constraints on . Section 4 is devoted to the proof of the central limit theorem using analytic tools from [8, 9]. Finally, in Section 5 we discuss the combinatorics of maps on orientable surface of higher genus. The expressions we obtain are much more involved than in the planar case, but it is expected to lead to similar analytic results.
## 2 Mobiles
Instead of investigating planar maps themselves, we will follow the principle presented in [5], whereby pointed planar maps are bijectively related to a certain class of trees called mobiles. (Their version of mobiles differ from the definition originally given in [3]; the equivalence of the two definitions is not shown explicitly in [5], but [7] gives a straightforward proof.)
###### Definition 1.
A mobile is a planar tree – that is, a map with a single face – such that there are two kinds of vertices (black and white), edges only occur as black–black edges or black–white edges, and black vertices additionally have so-called “legs” attached to them (which are not considered edges), whose number equals the number of white neighbor vertices.
A bipartite mobile is a mobile without black–black edges.
The degree of a black vertex is the number of half-edges plus the number of legs that are attached to it.
A mobile is called rooted if an edge is distinguished and oriented.
The essential observation is that mobiles are in bijection to pointed planar maps.
###### Theorem 2.
There is a bijection between mobiles that contain at least one black vertex and pointed planar maps, where white vertices in the mobile correspond to non-pointed vertices in the equivalent planar map, black vertices correspond to faces of the map, and the degrees of the black vertices correspond to the face valencies. This bijection induces a bijection on the edge sets so that the number of edges is the same. (Only the pointed vertex of the map has no counterpart.)
Similarly, rooted mobiles that contain at least one black vertex are in bijection to rooted and vertex-pointed planar maps.
Finally, bipartite mobiles with at least two vertices correspond to bipartite maps with at least two vertices, in the unrooted as well as in the rooted case.
###### Proof.
For the proof of the bijection between mobiles and pointed maps we refer to [7], where the bipartite case is also discussed. It just remains to note that the induced bijection on the edges can be directly used to transfer the root edge together with its direction. ∎
### 2.1 Bipartite Mobile Counting
We start with bipartite mobiles since they are more easy to count, in particular if we consider rooted bipartite mobiles, see [7].
###### Proposition 1.
Let be the solution of the equation
R=tz+z∑i≥1x2i(2i−1i)Ri. (3)
Then the generating function of bipartite rooted maps satisfies
∂M∂t=2(R/z−t), (4)
where the variable corresponds to the number of vertices, to the number of edges, and , , to the number of faces of valency .
###### Proof.
Since rooted mobiles can be considered as ordered rooted trees (which means that the neighboring vertices of the root vertex are linearly ordered and the subtrees rooted at these neighboring vertices are again ordered trees) we can describe them recursively. This directly leads to a functional equation for of the form
R=tz1−z∑i≥1x2i(2i−1i)Ri−1
which is apparently the same as (3). Note that the factor is precisely the number of ways of grouping legs and edges around a black vertex (of degree ; one edge is already there).
Hence, the generating function of rooted mobiles that are rooted by a white vertex is given by . Since we have to discount the mobile that consists just of one (white) vertex, the generating function of rooted mobiles that are rooted at a white vertex and contain at least two vertices is given by
R/z−t=∑i≥1x2i(2i−1i)Ri. (5)
We now observe that the right hand side of (5) is precisely the generating function of rooted mobiles that are rooted at a black vertex (and contain at least two vertices). Summing up, the generating function of bipartite rooted mobiles (with at least two vertices) is given by
2(R/z−t).
Finally, if denotes the generating function of bipartite rooted maps (with at least two vertices) then corresponds to rooted maps, where a non-root vertex is pointed (and discounted). Thus, by Theorem 2 we obtain (4). ∎
###### Remark 1.
It can be easily checked that Formula (4) can be specialized to count , for any subset of even positive integers: It suffices to set to every such that .
### 2.2 General Mobile Counting
We now proceed to develop a mechanism for general mobile counting that is adapted from [5]. For this, we will require Motzkin paths.
###### Definition 2.
A Motzkin path is a path starting at and going rightwards for a number of steps; the steps are either diagonally upwards (), straight () or diagonally downwards (). A Motzkin bridge is a Motzkin path from to . A Motzkin excursion is a Motzkin bridge which stays non-negative.
We define generating functions in the variables and , which count the number of steps of type and , respectively. (Explicitly counting steps of type is then unnecessary, of course.) The ordinary generating functions of Motzkin bridges, Motzkin excursions, and Motzkin paths from to shall be denoted by , and , respectively.
Continuing to follow the presentation of [5] and decomposing these three types of paths by their last passage through , we arrive at the equations:
E =1+tE+uE2, B =1+(t+2uE)B, B(+1) =EB.
In what follows we will also make use of bridges where the first step is either of type or . Clearly, their generating function is given by
¯¯¯¯B=tB+uB(+1)=B(t+uE).
When Motzkin bridges are not constrained to stay non-negative, they can be seen as am arbitrary arrangement of a given number of steps . It is then possible to obtain explicit expressions for
Bℓ,m =[tℓum]B(t,u)=(l+2ml,m,m), (6) B(+1)ℓ,m =[tℓum]B(+1)(t,u)=(l+2m+1l,m,m+1), (7) ¯¯¯¯Bℓ,m =[tℓum]¯¯¯¯B(t,u)=Bℓ−1,m+B(+1)ℓ,m−1=l+ml+2m(l+2ml,m,m). (8)
Using the above, we can now finally compute relations for generating functions of proper classes of mobiles. We define the following series, where corresponds to the number of white vertices, to the number of edges, and , , to the number of black vertices of degree :
• is the series counting rooted mobiles that are rooted at a black vertex and where an additional edge is attached to the black vertex.
• is the series counting rooted mobiles that are rooted at a univalent white vertex, which is not counted in the series.
• is the series counting rooted mobiles that are rooted at a white vertes and where an additional edge is attached to the root vertex.
Similarly to the above we obtain the following equations for the generating functions of mobiles and rooted maps.
###### Proposition 2.
Let , , and be the solutions of the equation
L =z∑ℓ,mx2m+ℓ+1Bℓ,mLℓRm, Q =z∑ℓ,mxℓ+2m+2B(+1)ℓ,mLℓRm, (9) R =tz1−Q,
and let be given by
T=1+∑ℓ,mx2m+ℓ¯¯¯¯Bℓ,mLℓRm, (10)
where the numbers , , and are given by (6)–(8). Then the generating function of rooted maps satisfies
∂M∂t=R/z−t+T, (11)
where the variable corresponds to the number of vertices, to the number of edges, and , , to the number of faces of valency .
###### Proof.
The system (9) is just a rephrasement of the recursive structure of rooted mobiles. Note that the numbers and are used to count the number of ways to circumscribe a specific black vertex and considering white vertices, black vertices and “legs” as steps , and . The generating function given in (10) is then the generating function of rooted mobiles where the root vertex is black.
Finally, the equation (11) follows from Theorem 2 since corresponds to rooted mobiles with at least one black vertex where the root vertex is white and corresponds to rooted mobiles where the root vertex is black. ∎
###### Remark 2.
Note that Proposition 1 is a special case of Proposition 2. We just have to restrict to the terms corresponding to since bipartite mobiles have no black–black edges. In particular, the series for is not needed any more and the second and third equations from (9) can be used to easily eliminate in order to recover the equation (3).
## 3 Asymptotic Enumeration
In this section we prove the asymptotic expansion (1). It turns out that it is much easier to start with bipartite maps. Actually, the bipartite case has already been treated by Bender and Canfield [2]. However, we apply a slightly different approach, which will then be extended to cover the general case as well the central limit theorem.
### 3.1 Bipartite maps
Let be a non-empty subset of even positive integers different from . Then by Proposition 1 the counting problem reduces to the discussion of the solutions of the functional equation
RD=tz+z∑2i∈D(2i−1i)RiD (12)
and the generating function that satisfies the relation
∂MD∂t=2(RD/z−t). (13)
Let . Then for combinatorial reasons it follows that there only exist maps with edges for that are divisible by . This is reflected by the fact that the equation (12) can we rewritten in the form
~R=t+∑2i∈D(2i−1i)zi/d~Ri, (14)
where we have substituted . (Recall that we finally work with .)
###### Lemma 1.
There exists an analytic function with and that is defined in a neighborhood of , and there exist analytic functions , with that are defined in a neighborhood of and such that the unique solution of the equation (12) that is analytic at and can be represented as
RD=g(t,z)−h(t,z)√1−zρ(t). (15)
Furthermore, the values , , are the only singularities of the function on the disc , and there exists an analytic continuation of to the range , , .
###### Proof.
From general theory (see [8, Theorem 2.21]), we know that an equation of the form , where is a power series with non-negative coefficients, has a square-root singularity if there are positive solutions to the following system:
R0=F(1,ρ,R0),1=FR(1,ρ,R0).
It is important to observe that the solutions are inside the region of convergence of . Besides, one has to check several analytic conditions on the derivatives of evaluated at this singular point. For a more detailed proof, the reader can refer to the work of Bender and Canfield [2]. ∎
It is now relatively easy to obtain similar properties for .
###### Lemma 2.
The function that is given by (13) has the representation
MD=g2(t,z)+h2(t,z)(1−zρ(t))3/2 (16)
in a neighborhood of and , where the functions , are analytic in a neighborhood of and and we have . Furthermore, the values , , are the only singularities of the function on the disc , and there exists an analytic continuation of to the range , , .
###### Proof.
This is a direct application of [8, Lemma 2.27]. ∎
In particular it follows that has the singular representation
MD=g2(1,z)+h2(1,z)(1−zρ(1))3/2
around . The singular representations are of the same kind around , and we have the analytic continuation property. Hence it follows by usual singularity analysis (see for example [8, Corollary 2.15]) that there exists a constant such that
[zn]MD(1,z)∼cDn−5/2ρ(1)−n,n≡0modd,
which completes the proof of the asymptotic expansion in the bipartite case.
### 3.2 General Maps
We now suppose that contains at least one odd number. It is easy to observe that in this case we have for (for some ), so we do not have to deals with several singularities.
By Proposition 2 we have to consider the system of equations for , , :
LD =z∑i∈D∑mBi−2m−1,mLi−2m−1DRmD, QD =z∑i∈D∑mB(+1)i−2m−2,mLi−2m−2DRmD, (17) RD =tz1−QD,
and also the function
TD=TD(t,z)=1+∑i∈D∑m¯¯¯¯Bi−2m,mLi−2mDRmD.
###### Lemma 3.
There exists an analytic function with and that is defined in a neighborhood of , and there exist analytic functions , with that are defined in a neighborhood of and such that
RD/z−t+TD=g(t,z)−h(t,z)√1−zρ(t). (18)
Furthermore, the value is the only singularity of the function on the disc , and there exists an analytic continuation of to the range , .
###### Proof.
Instead of a single equation, we have to deal with the strongly connected system (17), which is known to have similar analytic properties (see [8, Theorem 2.33]). As in Lemma 1, the main observation is that the singular point lies within the region of convergence of the equations, which follows directly in the finite case, but gets more technical in the infinite case. ∎
Lemma 3 shows that we are precisely in the same situation as in the bipartite case (actually, it is slightly easier since there is only one singularity on the circle ). Hence we immediately get the same property for as stated in Lemma 2 and consequently the proposed asymptotic expansion (1).
## 4 Central Limit Theorem for Bipartite Maps
Based on this previous result, we now extend our analysis to obtain a central limit theorem. Actually, this is immediate if the set is finite, whereas the infinite case needs much more care.
Let be a non-empty subset of even positive integers different from . Then by Proposition 1 the generating functions and satisfy the equations
RD=tz+z∑2i∈Dx2i(2i−1i)RiD (19)
and
∂MD∂t=2(RD/z−t). (20)
If is finite, then the number of variables is finite, too, and we can apply [8, Theorem 2.33] to obtain a representation of of the form
RD=g(t,z,(x2i)2i∈D)−h(t,z,(x2i)2i∈D)√1−zρ(t,(x2i)2i∈D), (21)
a proper extension of the transfer lemma [8, Lemma 2.27] (where the variables are considered as additional parameters) leads to
MD=g2(t,z,(x2i)2i∈D)+h2(t,z,(x2i)2i∈D)(1−zρ(t,(x2i)2i∈D))3/2, (22)
and finally [8, Theorem 2.25] implies a multivariate central limit theorem for the random vector of the proposed form.
Thus, we just have to concentrate on the infinite case. Actually, we proceed there in a similar way; however, we have to take care of infinitely many variables. There is no real problem to derive the same kind of representation (21) and (22) if is infinite. Everything works in the same way as in the finite case, we just have to assume that the variables are uniformly bounded. And of course we have to use a proper notion of analyticity in infinitely many variables. We only have to apply the functional analytic extension of the above cited theorems that are given in [9]. Moreover, in order to obtain a proper central limit theorem we need a proper adaption of [9, Theorem 3]. In this theorem we have also a single equation for a generating function that encodes the distribution of a random vector in the form
y=∑nyn(E∏i∈IxX(i)ni)zn,
where for (for some constant ) which also implies that all appearing potentially infinite products are in fact finite. (In our case this is satisfied since there is no vertex of degree larger than if we have edges.) As we can see from the proof of [9, Theorem 3], the essential part is to provide tightness of the involved normalized random vector, and tightness can be checked with the help of moment conditions. It is clear that asymptotics of moments for can be calculated with the help of derivatives of , for example . This follows from the fact all information on the asymptotic behavior of the moments is encoded in the derivatives of the singularity and by implicit differentiation these derivatives relate to derivatives of . More precisely, [9, Theorem 3] says that the following conditions are sufficient to deduce tightness of the normalized random vector:
∑i∈IFxi<∞,∑i∈IF2yxi<∞,∑i∈IFxixi<∞,
Fzxi =o(1), Fzxixi =o(1), Fyyxi =o(1), Fyyxixi =o(1), Fzzxi =O(1), Fzyxi =O(1), Fzyyxi =O(1), Fyyyxi =O(1),
(i→∞),
where all derivatives are evaluated at .
The situation is slightly different in our case since we have to work with instead of . However, the only real difference between and is that the critical exponents in the singular representations (21) and (22) are different, but the behavior of the singularity is precisely the same. Note that after the integration step we can set . Now tightness for the normalized random vector that is encoded in the function follows in the same way as for . And since the singularity is the same, we get precisely the same conditions as in the case of [9, Theorem 3].
This means that we just have to check the above conditions hold for
F=F(1,z,(x2i)2i∈D,y)=z+z∑2i∈Dx2i(2i−1i)yi,
where all derivatives are evaluated at , , and . However, they are trivially satisfied since
∑i≥1(2i−1i)iKyi<∞
for all and for positive real .
###### Remark 3.
As stated in Theorem 1, the results and methods extend to the general case as well. The main idea is to reduce the (positive strongly connected) system of two equations (17) to a single functional equation, by applying [8, Theorem 3].
## 5 Maps of Higher Genus
The bijection used in Section 2 relies solely on the orientability of the surface on which the maps are embedded. Therefore it can easily be extended to maps of higher genus, i.e., embedded on an orientable surface of genus (while planar maps correspond to maps of genus ). The main difference lies in the fact that the corresponding mobiles are no longer trees but rather one-faced maps of higher genus, while the other properties still hold.
However, due to the apparition of cycles in the underlying structure of mobiles, another difficulty arises. Indeed, in the original bijection, vertices and edges in mobiles could carry labels (related to the geodesic distance in the original map), subject to local constraints. In our setting, the legs actually encode the local variations of these labels, which are thus implicit. Local constraints on labels are naturally translated into local constraints on the number of legs. But the labels have to remain consistent along each cycle of the mobiles, which gives rise to non-local constraints on the repartition of legs.
In order to deal with these additional constraints, and to be able to control the degrees of the vertices at the same time, we will now use a hybrid formulation of mobiles, carrying both labels and legs. As before, we will focus on the simpler case of mobiles coming from bipartite maps.
### 5.1 g-Mobiles
###### Definition 3.
Given , a -mobile is a one-faced map of genus – embedded on the -torus – such that there are two kinds of vertices (black and white), edges only occur as black–black edges or black–white edges, and black vertices additionally have so-called “legs” attached to them (which are not considered edges), whose number equals the number of white neighbor vertices.
Furthermore, for each cycle of the -mobile, let , and respectively be the numbers of white vertices on , of legs dangling to the left of and of white neighbours to the left of . One has the following constraint (see Figure 1):
n→=n∘+n\includegraphics[]wn (23)
The degree of a black vertex is the number of half-edges plus the number of legs that are attached to it. A bipartite -mobile is a -mobile without black–black edges. A -mobile is called rooted if an edge is distinguished and oriented.
Notice that a -mobile is simply a mobile as described in Definition 1.
###### Theorem 3.
Given , there is a bijection between -mobiles that contain at least one black vertex and pointed maps of genus , where white vertices in the mobile correspond to non-pointed vertices in the equivalent map, black vertices correspond to faces of the map, and the degrees of the black vertices correspond to the face valencies. This bijection induces a bijection on the edge sets so that the number of edges is the same. (Only the pointed vertex of the map has no counterpart.)
Similarly, rooted -mobiles that contain at least one black vertex are in bijection to rooted and vertex-pointed maps of genus .
###### Proof.
This generalization of the bijection to higher genus was first given in [6] for quadrangulations and [4] for Eulerian maps, from which we will exploit many ideas in the present section. ∎
### 5.2 Schemes of g-Mobiles
-mobiles are not as easily decomposed as planar mobiles, due to the existence of cycles. However, they still exhibit a rather simple structure, based on scheme extraction.
The -scheme (or simply the scheme) of a -mobile is what remains when we apply the following operations (see Figure 2): first remove all legs, then remove iteratively all vertices of degree 1 and finally replace any maximal path of degree-2-vertices by a single edge.
Once these operations are performed, the remaining object is still a one-faced map of genus , with black and white vertices (white–white edges can now occur), where the vertices have minimum degree 3.
To count -mobiles, one key ingredient is the fact that there is only a finite number of schemes of a given genus. Indeed, let be the number of degree vertices of a -scheme:
∑k≥3(i−2)di=∑k≥3idi−2∑k≥3di=2(#% edges−#vertices)=4g−2.
The number of vertices (respectively edges) is then bounded by (respectively ), where this bound is reached for cubic schemes (see an example in Figure 2).
To recover a proper -mobile from a given -scheme, one would have to insert a suitable planar mobile into each corner of the scheme and to substitute each edge with some kind of path of planar mobiles. Unfortunately, this cannot be done independently: Around each black vertex, the total number of legs in every corner must equal the number of white neighbors, and around each cycle, (23) must hold.
In order to make these constraints more transparent, we will equip schemes with labels on white vertices and black corners. Now, when trying to reconstruct a -mobile from a scheme, one has to ensure that the local variations are consistent with the global labelling. To be precise, the label variations are encoded as follows (see Figure 3):
• Around a black vertex of degree , let be the labels of its corners read in clockwise order:
∀i,li+1−li=⎧⎪⎨⎪⎩+1if there is a leg % between the two corresponding corners,0if there is a black neighbor,−1if there is a white neighbor.
• Along the left side of an oriented cycle, the label decreases by 1 after a white vertex or when encountering a white neighbor and increases by 1 when encountering a leg.
The above statements hold for general – as well as bipartite – mobiles. In the following, we will only consider bipartite mobiles, as they are much easier to decompose.
### 5.3 Reconstruction of Bipartite Maps of Genus g
In the following, it will be convenient to work with rooted schemes. One can then define a canonical labelling and orientation for each edge of a rooted scheme. An edge now has an origin and an endpoint . The corners around a vertex of degree are clockwisely ordered and denoted by .
Given a scheme , let be respectively the sets of white and black vertices and of white and black corners. A labelled scheme is a pair consisting of a scheme and a labelling on white vertices and black corners, with for all . Labellings are considered up to translation, as they will not affect local variations. For , an edge of , we associate a label to each extremity . If an extremity is a white vertex of label , its label is . If the extremity is a black vertex, its label is the same as the next clockwise corner of the black vertex.
Let a doubly-rooted planar mobile be a rooted (on a black or white vertex) planar mobile with a secondary root (also black or white). These two roots are the extremities of a path . The increment of the doubly-rooted mobile is then defined as , which is not necessarily 0, as the path is not a cycle.
Similarly as in [4], we present a non-deterministic algorithm to reconstruct a -mobile:
Algorithm.
(1) Choose a labelled -scheme .
(2) , choose a sequence of non-negative integers , then attach planar mobiles and legs to (the corner of ).
(3) , replace by a doubly-rooted mobile of increment
(4) On each white corner of , insert a planar mobile.
(5) Distinguish and orient an edge as the root.
###### Proposition 3.
Given , the algorithm generates each rooted bipartite -mobile whose scheme has edges in exactly ways.
###### Proof.
One can easily see that the obtained object is indeed bipartite. Attaching planar mobiles and legs added at step (2) in a corner creates new corners, such that:
• The first carries the same label as , and
• the last carries the label .
The next corner should then be labelled , due to the next white neighbor, which is precisely what we want.
In the same fashion, at step (3), a simple counting shows that each edge is replaced by a path such that the labels along it evolve according to the scheme labelling.
We thus obtain a well-formed rooted bipartite -mobile, with a secondary root on its scheme. Since the first root destroys all symmetries, there are exactly choices for the secondary root, which would give the same rooted -mobile. ∎
### 5.4 g-Mobile Counting
A doubly-rooted bipartite planar mobile can be decomposed along a sequence of elementary cells forming the path between its two roots. Its increment is simply the sum of the increments of its cells.
###### Definition 4.
An elementary cell is a half-edge connected to a black vertex itself connected to a white vertex with a dangling half-edge. The white vertex has a sequence of black-rooted mobiles attached on each side. The black vertex has legs and white-rooted mobiles on its left, white-rooted mobiles and legs on its right, and its degree is . The increment of the cell is then .
The generating series of a cell, where marks the increment, is:
P(t,z,(x2i),s)=z2R2t∑j,k,l≥0(j+kj)(k+2l−j+2l)sj−k−1x2(k+l+2)Rk+l=z2R2stˆP.
The generating series of a doubly-rooted mobile depends on the color of its roots :
S(u,v)(t,z,(x2i),s)=⎧⎪ ⎪ ⎪ ⎪ ⎪⎨⎪ ⎪ ⎪ ⎪ ⎪⎩11−Pif (u,v)=(∘,∙) or (∙,∘),zˆP1−Pif (u,v)=(∘,∘),zR2st(1−P)if (u,v)=(∙,∙).
We can now express the generating series of rooted bipartite -mobiles with scheme :
RS(t,z,(x2i))=2z∂∂z12|E|z|E|t|V∘|(Rtz)|C∘|∙∙∑(lc) labelling⎡⎢⎣∏e∈E[sincr(e)]S(e−,e+)∏v∈V∙∑i1,…,i%deg(v)≥0⎛⎝deg(v)∏k=1(2ik+lck+1−lck+1ik)⎞⎠x2(deg(v)+∑ik)⎤⎥⎦. (24)
###### Proposition 4.
The generating series for the family of rooted bipartite maps of genus , where the vertex degrees belong to , satisfies the relation:
∂M(g)D∂t=2z∑S schemeof genus gRS(t,z,(x2i\mathds1{2i∈D})). (25)
###### Proof.
This follows directly from Theorem 3 and Equation (24). ∎
## 6 Conclusion
Theorem 1 confirms the existence of a universal behaviour of planar maps. The asymptotics (with exponent ) and this central limit theorem for the expected number of vertices of a given degree are believed to hold for any “reasonable” family of maps. It has also been shown in [6, 4] that a similar phenomenom occurs for maps of higher genus: The generating series of several families (quadrangulations, general and Eulerian maps) of genus exhibit the same asymptotic exponent .
The expression obtained in Section 5 needs to be properly studied in order to obtain an asymptotic expansion. It refines previous results by controlling the degree of each vertex in the corresponding map.
## References
• [1] Cyril Banderier and Michael Drmota, Formulae and asymptotics for coefficients of algebraic functions, Combinatorics, Probability and Computing 24 (2015), no. 1, 1–53.
• [2] E.A. Bender and E.R. Canfield, Enumeration of degree restricted maps on the sphere., Planar graphs. Workshop held at DIMACS from November 18, 1991 through November 21, 1991, Providence, RI: American Mathematical Society, 1993, pp. 13–16.
• [3] J. Bouttier, P. Di Francesco, and E. Guitter, Planar maps as labeled mobiles, Electron. J. Combin. 11 (2004), no. 1, Research Paper 69, 27.
• [4] Guillaume Chapuy, Asymptotic enumeration of constellations and related families of maps on orientable surfaces, Combin. Probab. Comput. 18 (2009), no. 4, 477–516.
• [5] Guillaume Chapuy, Éric Fusy, Mihyun Kang, and Bilyana Shoilekova, A complete grammar for decomposing a family of graphs into 3-connected components, Electron. J. Combin. 15 (2008), no. 1, Research Paper 148, 39.
• [6] Guillaume Chapuy, Michel Marcus, and Gilles Schaeffer, A bijection for rooted maps on orientable surfaces, SIAM Journal on Discrete Mathematics 23 (2009), no. 3, 1587–1611.
• [7] Gwendal Collet and Éric Fusy, A simple formula for the series of bipartite and quasi-bipartite maps with boundaries, Discrete Math. Theor. Comput. Sci. (2012), 607–618.
• [8] Michael Drmota, Random trees. An interplay between combinatorics and probability., Wien: Springer, 2009.
• [9] Michael Drmota, Bernhard Gittenberger, and Johannes F. Morgenbesser, Infinite systems of functional equations and Gaussian limiting distributions., Proceeding of the 23rd international meeting on probabilistic, combinatorial, and asymptotic methods in the analysis of algorithms (AofA’12), Montreal, Canada, June 18–22, 2012, Nancy: The Association. Discrete Mathematics & Theoretical Computer Science (DMTCS), 2012, pp. 453–478.
• [10] Michael Drmota and Konstantinos Panagiotou, A central limit theorem for the number of degree- vertices in random maps., Algorithmica 66 (2013), no. 4, 741–761.
• [11] W.T. Tutte, A census of planar maps., Can. J. Math. 15 (1963), 249–271.
You are adding the first comment!
How to quickly get a good reply:
• Give credit where it’s due by listing out the positive aspects of a paper before getting into which changes should be made.
• Be specific in your critique, and provide supporting evidence with appropriate references to substantiate general statements.
• Your comment should inspire ideas to flow and help the author improves the paper.
The better we are at sharing our knowledge with each other, the faster we move forward.
The feedback must be of minimum 40 characters and the title a minimum of 5 characters | 8,617 | 34,219 | {"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-2020-45 | latest | en | 0.912216 |
https://brainmass.com/statistics/confidence-interval/statistical-confidence-intervals-and-one-sample-tests-544112 | 1,624,237,706,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623488259200.84/warc/CC-MAIN-20210620235118-20210621025118-00560.warc.gz | 141,394,709 | 76,176 | Explore BrainMass
Statistical Confidence Intervals and One-sample tests
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!
Please review the attached document and assist.
https://brainmass.com/statistics/confidence-interval/statistical-confidence-intervals-and-one-sample-tests-544112
Solution Preview
NOTE: Here, you can see clearly that the effort gone to solve the problems deserves 10+ credits. Request you to raise a new dummy posting of at least 4 credits in my favour. Thank you.
8.28
Errors made: 2, 3, 5, 3 and 5
a. How many different samples of 2 tellers are possible?
Answer: 3 different samples of 2 tellers are possible: (2,3), (2,5) and (3,5)
b. List all possible samples of size 2 and compute the mean of each.
(2,3), (2, 5), (3,5)
Samples mean
x1_bar = (2+3)/2 = 2.5
x2_bar = (2+5)/2 = 3.5
x3_bar = (3+5)/2 = 4.0
c. Compute the mean of the sample means and compare it to the population mean.
Mean of samples mean
X_bar = (x1_bar + x2_bar + x3_bar)/3 = (2.5+3.5+4.0)/3 = 10.0/3 = 3.33
Population mean,
mu = (2+3+5+3+5)/5 = 18/5 = 3.6
Comparison
X_bar - mu = 3.33 - 3.60 = -0.27
Mean of samples mean is less than population mean by 0.27.
8.31
mean, mu = 35 hours
standard deviation, sd = 5.5 hours
sample size, n = 25
a. What can you say about the shape of the distribution of the sample mean?
The sample mean is Normal distributed, --Answer
with average value, X_bar = mu = 35 hours --Answer
and variance, V = sd^2/n = 5.5^2/25 = 1.21
b. What is the standard error of the distribution of the sample mean?
Standard error, SE = sqrt(V) = sqrt(1.21) = 1.1 --Answer
c. What proportion of the samples will have a mean useful life of more than 36 hours?
x = 36
P(X >= x) == P(Z >= (x - X_bar)/SE)
= P(Z >= (36 - 35.5)/1.1)
= P(Z >= 0.45)
= 0.5000 - 0.1736 [From N-table with upper half]
= 0.3264
Hence, 0.3264 proportion or 32.64% of the ...
Solution Summary
A few problems related to probability and confidence intervals are solved here.
\$2.49 | 683 | 2,117 | {"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} | 3.9375 | 4 | CC-MAIN-2021-25 | latest | en | 0.859403 |
http://wikistack.com/bubble-sort-video-tutorial/ | 1,568,631,907,000,000,000 | text/html | crawl-data/CC-MAIN-2019-39/segments/1568514572517.50/warc/CC-MAIN-20190916100041-20190916122041-00252.warc.gz | 213,679,410 | 15,494 | Bubble sort is sorting algorithm in which 0th element is compare with 1st element and if 0th element is greater than 1st element then they are interchanged.1st element is compare with 2nd element and if 1st element is greater than 2nd element then they are interchanged. and so on. i.e comparing each pair of adjacent items and swapping them if they are in the wrong order. this process will continue until no swap is required.
Lets take a look at example given in wikipedia.
First Pass:
( 5 1 4 2 8 ) –> ( 1 5 4 2 8 ), Here, algorithm compares the first two elements, and swaps since 5 > 1.
( 1 5 4 2 8 ) –>( 1 4 5 2 8 ), Swap since 5 > 4
( 1 4 5 2 8 ) –> ( 1 4 2 5 8 ), Swap since 5 > 2
( 1 4 2 5 8 ) –> ( 1 4 2 5 8 ), Now, since these elements are already in order (8 > 5), algorithm does not swap them.
Second Pass:
( 1 4 2 5 8 ) –> ( 1 4 2 5 8 )
( 1 4 2 5 8 ) –> ( 1 2 4 5 8 ), Swap since 4 > 2
( 1 2 4 5 8 ) –> ( 1 2 4 5 8 )
( 1 2 4 5 8 ) –> ( 1 2 4 5 8 )
Now, the array is already sorted, but our algorithm does not know if it is completed. The algorithm needs one whole pass without any swap to know it is sorted.
Third Pass:
( 1 2 4 5 8 ) –> ( 1 2 4 5 8 )
( 1 2 4 5 8 ) –> ( 1 2 4 5 8 )
( 1 2 4 5 8 ) –> ( 1 2 4 5 8 )
( 1 2 4 5 8 ) –> ( 1 2 4 5 8 )
```#include<stdio.h>
int main ()
{
int arr[] = { 5, 1, 4, 2, 8 };
bool is_swap = true;
int j = 0;
int size = sizeof(arr)/sizeof(arr[0]);
int tmp;
while (is_swap)
{
is_swap = false;
j++;
for (int i = 0; i < size - j; i++)
{
if (arr[i] > arr[i + 1])
{
tmp = arr[i];
arr[i] = arr[i + 1];
arr[i + 1] = tmp;
is_swap = true;
}
}
}
int i=0;
for (i;i<size-1;i++)
printf("%d ",arr[i]);
printf("\n");
return 0;
}``` | 689 | 1,669 | {"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-2019-39 | longest | en | 0.793211 |
https://mathoverflow.net/questions/9716/whats-the-status-of-the-following-relationship-between-ramanujans-tau-funct | 1,695,523,423,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233506559.11/warc/CC-MAIN-20230924023050-20230924053050-00009.warc.gz | 449,042,212 | 27,024 | # What's the status of the following relationship between Ramanujan's $\tau$ function and the simple Lie algebras?
Qiaochu asked this in the comments to this question. Since this is really his question, not mine, I will make this one Community Wiki. In MR0522147, Dyson mentions the generating function $\tau(n)$ given by $$\sum_{n=1}^\infty \tau(n)\,x^n = x\prod_{m=1}^\infty (1 - x^m)^{24} = \eta(x)^{24},$$ which is apparently of interest to the number theorists ($\eta$ is Dedekind's function). He mentions the following formula for $\tau$: $$\tau(n) = \frac{1}{1!\,2!\,3!\,4!} \sum \prod_{1 \leq i < j \leq 5} (x_i - x_j)$$ where the sum ranges over $5$-tuples $(x_1,\dots,x_5)$ of integers satisfying $x_i \equiv i \mod 5$, $\sum x_i = 0$, and $\sum x_i^2 = 10n$. Apparently, the $5$ and $10$ are because this formula comes from some identity of $\eta(x)^{10}$. Dyson mentions that there are similar formulas coming from identities with $\eta(x)^d$ when $d$ is on the list $d = 3, 8, 10, 14,15, 21, 24, 26, 28, 35, 36, \dots$. The list is exactly the dimensions of the simple Lie algebras, except for the number $26$, which doesn't have a good explanation. The explanation of the others is in I. G. Macdonald, Affine root systems and Dedekind's $\eta$-function, Invent. Math. 15 (1972), 91--143, MR0357528, and the reviewer at MathSciNet also mentions that the explanation for $d=26$ is lacking.
So: in the last almost-40 years, has the $d=26$ case explained?
• If somebody gets the chance, please add links to the mentioned articles to make them easier to get to (at least if you're working from a university computer where you have journal access). Dec 25, 2009 at 7:25
• Interestingly, the number 26 appears also as a number of 'sporadic' finite simple groups (i.e. not fitting into the infinite series of groups of Lie type, cyclic and alternative). Dec 25, 2009 at 7:27
• One would expect 26 the critical central charge of the Virasoro algebra to be more relevant here than 26 the number of sporadic finite simple groups, I guess. Dec 25, 2009 at 12:44
• @Anton: Sorry, I thought the MR numbers sufficed. Dec 25, 2009 at 16:37
The case of $d=26$ is related to the exceptional Lie algebra $F_4$. Let me quote from the 1980 paper by Monastyrsky which was originally published as a supplement to the Russian translation of the Dyson's paper:
A more careful study of Macdonald's article reveals that the identity for the 26th power of $\eta(x)$ is not really such a mystery. It is related to the exceptional group $F_4$ of dimension 52, where the space of dual roots $F_4^V$ and the space of roots $F_4$ are not the same. Thus, there are two distinct identities associated with $F_4$, one for $\eta^{52} (x)$ and the other for $\eta^{26} (x)$. A similar situation prevails in the case of the algebra $G_2$ of dimension 14, which yields identities for $\eta^{14} (x)$ and $\eta^{7} (x)$. The identities for $\eta^{26} (x)$ and $\eta^{7} (x)$ are considerably more complicated. | 885 | 2,986 | {"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} | 2.875 | 3 | CC-MAIN-2023-40 | latest | en | 0.904116 |
http://www.brightstorm.com/tag/sine-of-a-difference-formula/ | 1,448,709,731,000,000,000 | text/html | crawl-data/CC-MAIN-2015-48/segments/1448398452385.31/warc/CC-MAIN-20151124205412-00284-ip-10-71-132-137.ec2.internal.warc.gz | 334,696,741 | 10,117 | • #### The Sine Addition Formulas - Problem 2
How to derive the sine of a difference formula and use it to find the sine of an angle.
• #### The Sine Addition Formulas - Problem 2
How to derive the sine of a difference formula and use it to find the sine of an angle.
• #### Using the Sine and Cosine Addition Formulas to Prove Identities - Problem 3
How to use the sine and cosine addition formulas to prove a difference formula for tangent.
• #### Using the Sine and Cosine Addition Formulas to Prove Identities - Problem 3
How to use the sine and cosine addition formulas to prove a difference formula for tangent.
• #### The Sine Addition Formulas - Concept
How to derive the sine of a sum formula. | 166 | 707 | {"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-2015-48 | latest | en | 0.793358 |
https://brainly.com/question/214952 | 1,485,058,918,000,000,000 | text/html | crawl-data/CC-MAIN-2017-04/segments/1484560281332.92/warc/CC-MAIN-20170116095121-00077-ip-10-171-10-70.ec2.internal.warc.gz | 799,812,254 | 8,499 | 2014-12-07T13:53:33-05:00
5x+y=10 x2=>10x+2y=20
3x-2y=6 =>3x-2y=6
13x=26
x=26/13
x=2
sub x into 5x+y=10
5(2)+y=10
10+y=10
y=10-10
y=0
solution (2,0)
check
5x+y=10+0=10
3x-2y=6+0=6
hope this helps | 134 | 208 | {"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} | 3.859375 | 4 | CC-MAIN-2017-04 | latest | en | 0.209749 |
http://slideplayer.com/slide/4633105/ | 1,709,347,937,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947475727.3/warc/CC-MAIN-20240302020802-20240302050802-00448.warc.gz | 36,745,702 | 17,442 | Cumulative Frequency and Box Plots. Learning Objectives To be able to draw a cumulative frequency curve and use it to estimate the median and interquartile.
Presentation on theme: "Cumulative Frequency and Box Plots. Learning Objectives To be able to draw a cumulative frequency curve and use it to estimate the median and interquartile."— Presentation transcript:
Cumulative Frequency and Box Plots
Learning Objectives To be able to draw a cumulative frequency curve and use it to estimate the median and interquartile range and draw a box plot
Does anybody remember how to draw a Cumulative Frequency Diagram?
Cumulative Frequency Curves Cumulative Frequency = a running total of the data Median = Half way up the Cumulative Frequency Lower Quartile (LQ) = ¼ way up Cum Freq Upper Quartile (UQ) = ¾ way up Cum Freq Inter Quartile Range = UQ - LQ
Big Worked Example The table below shows the number of minutes students were late for their fun algebra lesson. (a) Draw a Cumulative Frequency Diagram of the data (b) Use it to find the Median, Lower Quartile, Upper Quartile, and Inter Quartile Range (c) Draw a Box-Plot assuming a minimum time of o minutes and a maximum of 25 minutes Time t (mins) Number of Students Cumulative Frequency 0 < t ≤ 5 10 5 < t ≤ 10 16 10 < t ≤ 15 30 15 < t ≤ 20 22 20 < t ≤ 25 2
510 152025t mins Cum freq 20 40 60 80 x x x x x Median = Middle Value QUARTILES Lower Quartile = ¼ way Upper Quartile = ¾ way Interquartile Range 8½ 15½12½ = 15½ - 8½ = 7 mins
Does anybody have any idea what a Box Plot is?
Box Plots Box Plots (Box and Whisker Diagram) are another way of displaying data for comparison To draw a Box Plot we need FIVE pieces of information: –1. The Minimum Value –2. Lower Quartile (Q 1 ) –3. The Median (Q 2 ) –4. Upper Quartile (Q 3 ) –5. Maximum Value They are drawn in the following way: BOX = The location of the middle 50% of the data WHISKERS = How the data is spread overall 34 1 25
010 20 30405060 Box Plot from Cumulative Frequency Curve
Box Plot 510 152025t mins 0 Lowest value Upper Quartile Highest value Lower Quartile Median
Work Ex 7:1 –Page 62 –Questions 5 and 6
Download ppt "Cumulative Frequency and Box Plots. Learning Objectives To be able to draw a cumulative frequency curve and use it to estimate the median and interquartile."
Similar presentations | 624 | 2,338 | {"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-2024-10 | latest | en | 0.829361 |
https://avatest.org/2023/02/16/ross%E6%95%B0%E5%AD%A6%E5%A4%8F%E4%BB%A4%E8%90%A52023%E9%80%89%E6%8B%94-daixie/ | 1,722,895,790,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640455981.23/warc/CC-MAIN-20240805211050-20240806001050-00406.warc.gz | 93,096,459 | 17,293 | Posted on Categories:SUMaC代写, SUMaC代考, 数学代写, 数学竞赛代写, 斯坦福大学数学夏令营
# 数学代写|Ross数学夏令营2023选拔代写
Robot Rossie moves within a square room $A B C D$. Rossie moves along straight line segments, never leaving that room.
When Rossie encounters a wall she stops, makes a right-angle turn (with direction chosen to face into the room), and continues in that new direction.
If Rossie comes to one of the room’s corners, she rotates through two right angles, and moves back along her previous path.
Suppose Rossie starts at point $P$ on $A B$ and her path begins as a line segment of slope $s$.
We hope to describe Rossie’s path.
For some values of $P$ and $s$, Rossie’s path will be a tilted rectangle with one vertex on each wall of the room. (Often, this inscribed rectangle is itself a square.) In this case, Rossie repeatedly traces that stable rectangle.
(a) Suppose $s=1$ so that the path begins at a 45 degree angle.
For every starting point $P$, show: Rossie’s path is a stable rectangle.
(If $P$ is a corner point, the path degenerates to a line segment traced back and forth.)
Now draw some examples with various $P$ and $s$.
Given $P$ and $s$, does Rossie’s path always converge to a stable rectangle?
(b) First consider the case: $01$ or when $s<0$ ? Does the argument above still apply?
Let $\mathbb{Z}$ denote the set of integers. If $m$ is a positive integer, we write $\mathbb{Z}m$ for the system of “integers modulo $m$.” Some authors write $\mathbb{Z} / m \mathbb{Z}$ for that system. For completeness, we include some definitions here. The system $\mathbb{Z}_m$ can be represented as the set ${0,1, \ldots, m-1}$ with operations $\oplus$ (addition) and $\odot$ (multiplication) defined as follows. If $a, b$ are elements of ${0,1, \ldots, m-1}$, define: $a \oplus b=$ the element $c$ of ${0,1, \ldots, m-1}$ such that $a+b-c$ is an integer multiple of $m$. $a \odot b=$ the element $d$ of ${0,1, \ldots, m-1}$ such that $a b-d$ is an integer multiple of $m$. For example, $3 \oplus 4=2$ in $\mathbb{Z}_5$, $3 \odot 3=1$ in $\mathbb{Z}_4$, and $-1=12$ in $\mathbb{Z}{13}$.
To simplify notations (at the expense of possible confusion), we abandon that new notation and write $a+b$ and $a b$ for the operations in $\mathbb{Z}_m$, rather than writing $a \oplus b$ and $a \odot b$.
Let $\mathbb{Q}$ denote the system of rational numbers.
We write $4 \mathbb{Z}$ for the set of multiples of 4 in $\mathbb{Z}$. Similarly for $4 \mathbb{Z}{12}$. Consider the following number systems: $$\mathbb{Z}, \quad \mathbb{Q}, \quad 4 \mathbb{Z}, \quad \mathbb{Z}_3, \quad \mathbb{Z}_8, \quad \mathbb{Z}_9, \quad 4 \mathbb{Z}{12}, \quad \mathbb{Z}_{13} .$$
One system may be viewed as similar to another in several different ways. We will measure similarity using only algebraic properties.
(a) Consider the following sample properties:
(i) If $a^2=1$, then $a=\pm 1$.
(ii) If $2 x=0$, then $x=0$.
(iii) If $c^2=0$, then $c=0$.
Which of the systems above have properties (i), (ii), and/or (iii)?
(b) Formulate another algebraic property and determine which of those systems have that property. [Note: Cardinality is not considered to be an algebraic property.]
Write down some additional algebraic properties and investigate them.
(c) In your opinion, which of the listed systems are “most similar” to each another?
Please spend extra effort to write up this problem’s solution as an exposition that can be read and understood by a beginning algebra student. That student knows function notation and standard properties of polynomials (as taught in a high school algebra course). Your solution will be graded not only on the correctness of the math but also on the clarity of exposition.
(a) Find all polynomials $f$ that satisfy the equation:
$$f(x+2)=f(x)+2 \text { for every real number } x .$$
(b) Find all polynomials $g$ that satisfy the equation:
$$g(2 x)=2 g(x) \text { for every real number } x .$$
(c) The problems above are of the following type: Given functions $H$ and $J$, find all polynomials $Q$ that satisfy the equation:
$$J(Q(x))=Q(H(x)) \text { for every } x \text { in } S$$
where $S$ is a subset of real numbers. In parts (a) and (b), we have $J=H$ and $S$ is all real numbers, but other scenarios are also interesting. For example, the choice $J(x)=1 /(x-1)$ and $H(x)=1 /(x+1)$, generates the question:
Find all polynomials $Q$ that satisfy the equation:
$$\frac{1}{Q(x)-1}=Q\left(\frac{1}{x+1}\right)$$
for every real number $x$ such that those denominators are nonzero.
Is this one straightforward to solve?
(d) Make your own choice for $J$ and $H$, formulate the problem, and find a solution. Choose $J$ and $H$ to be non-trivial, but still simple enough to allow you to make good progress toward a solution.
(a) 假设$s=1$，使路径以45度角开始。
(如果$P$是一个角点，该路径就退化为一条来回追踪的线段)。
(b) 首先考虑以下情况：$01$或$s<0$时，Rossie的行为是什么？上面的论证是否仍然适用？ | 1,474 | 4,840 | {"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} | 3.6875 | 4 | CC-MAIN-2024-33 | latest | en | 0.837475 |
http://www.ask.com/web?qsrc=3053&o=102140&oo=102140&l=dir&gc=1&qo=popularsearches&ad=dirN&q=How+To+Convert+Decimals+Into+Fractions | 1,485,100,732,000,000,000 | text/html | crawl-data/CC-MAIN-2017-04/segments/1484560281426.63/warc/CC-MAIN-20170116095121-00501-ip-10-171-10-70.ec2.internal.warc.gz | 349,502,244 | 15,011 | Web Results
To convert a fraction into a decimal, divide the numerator by the denominator. To change terminal decimals into fractions, count the number of decimal places, put the decimal's digits over 1 followed by the proper number of zeroes.
You should simply your fraction after converting from a decimal.
## Rewriting decimals as fractions: 0.15 - Khan Academy
Sal converts 0.15 to a fraction. ... Fraction to decimal: 11/25 · Worked example: Converting a fraction (7/8) to a decimal ... Converting decimals to fractions 2 (ex 1 ).
## Convert Decimals to Fractions - Math is Fun
www.mathsisfun.com/converting-decimals-fractions.html
To convert a Decimal to a Fraction follow these steps: Step 1: Write down the decimal divided by 1, like this: decimal 1; Step 2: Multiply ... into a whole number ?)
## Convert From a Decimal To a Fraction - WebMath
www.webmath.com/dec2fract.html
Decimals and fractions represent the same thing: a number that is not exactly a ... This page will show you how to convert a decimal into its equivalent fraction.
May 23, 2012 ... Decimal to fraction, step by step, example. For all free ... Pre-Algebra 20 - Converting Repeating Decimal Numbers to Fractions - Duration: 8:28.
## Decimal to Fraction Calculator - Calculator Soup
www.calculatorsoup.com/calculators/math/decimal-to-fraction-calculator.php
Convert decimals to fractions and mixed numbers with the Decimal to Fraction Calculator. This calculator converts decimals and repeating decimals to fractions ...
## How to Convert Decimals to Fractions - Cool Math
www.coolmath.com/prealgebra/02-decimals/05-decimals-converting-decimal-to-fraction-01
This prealgebra lesson explains how to convert a decimal to a fraction.
## The Math Dude : How to Convert Decimals to Fractions :: Quick and ...
www.quickanddirtytips.com/education/math/how-to-convert-decimals-to-fractions
Feb 4, 2011 ... Before we get into the details of how to actually convert terminating and repeating decimals into fractions, we'd better make sure we understand ...
## How to Convert Decimals to Fractions - dummies
www.dummies.com/education/math/algebra/how-to-convert-decimals-to-fractions/
Converting decimals into fractions isn't hard. To convert a decimal into a fraction, you put the numbers to the right of the decimal point in the numerator (above ...
## Converting Decimals to Fractions - SOS Math
www.sosmath.com/algebra/fraction/frac6/frac6.html
For this review, we will focus on terminating fractions. Divide the fraction by 1, and then multiply the result by 1 in a form that will remove the decimal. | 610 | 2,598 | {"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-2017-04 | latest | en | 0.707726 |
http://slideplayer.com/slide/3525211/ | 1,513,063,064,000,000,000 | text/html | crawl-data/CC-MAIN-2017-51/segments/1512948515309.5/warc/CC-MAIN-20171212060515-20171212080515-00699.warc.gz | 262,031,461 | 20,879 | 4/22/2015 1 LU Decomposition Electrical Engineering Majors Authors: Autar Kaw Transforming.
Presentation on theme: "4/22/2015 1 LU Decomposition Electrical Engineering Majors Authors: Autar Kaw Transforming."— Presentation transcript:
4/22/2015 http://numericalmethods.eng.usf.edu 1 LU Decomposition Electrical Engineering Majors Authors: Autar Kaw http://numericalmethods.eng.usf.edu Transforming Numerical Methods Education for STEM Undergraduates
LU Decomposition http://numericalmethods.eng.usf.edu http://numericalmethods.eng.usf.edu
LU Decomposition LU Decomposition is another method to solve a set of simultaneous linear equations Which is better, Gauss Elimination or LU Decomposition? To answer this, a closer look at LU decomposition is needed.
Method For most non-singular matrix [A] that one could conduct Naïve Gauss Elimination forward elimination steps, one can always write it as [A] = [L][U] where [L] = lower triangular matrix [U] = upper triangular matrix http://numericalmethods.eng.usf.edu LU Decomposition
http://numericalmethods.eng.usf.edu How does LU Decomposition work? If solving a set of linear equations If [A] = [L][U] then Multiply by Which gives Remember [L] -1 [L] = [I] which leads to Now, if [I][U] = [U] then Now, let Which ends with and [A][X] = [C] [L][U][X] = [C] [L] -1 [L] -1 [L][U][X] = [L] -1 [C] [I][U][X] = [L] -1 [C] [U][X] = [L] -1 [C] [L] -1 [C]=[Z] [L][Z] = [C] (1) [U][X] = [Z] (2)
http://numericalmethods.eng.usf.edu LU Decomposition How can this be used? Given [A][X] = [C] 1.Decompose [A] into [L] and [U] 2.Solve [L][Z] = [C] for [Z] 3.Solve [U][X] = [Z] for [X]
http://numericalmethods.eng.usf.edu When is LU Decomposition better than Gaussian Elimination? To solve [A][X] = [B] Table. Time taken by methods where T = clock cycle time and n = size of the matrix So both methods are equally efficient. Gaussian EliminationLU Decomposition
http://numericalmethods.eng.usf.edu To find inverse of [A] Time taken by Gaussian Elimination Time taken by LU Decomposition n10100100010000 CT| inverse GE / CT| inverse LU 3.2825.83250.82501 Table 1 Comparing computational times of finding inverse of a matrix using LU decomposition and Gaussian elimination.
http://numericalmethods.eng.usf.edu Method: [A] Decompose to [L] and [U] [U] is the same as the coefficient matrix at the end of the forward elimination step. [L] is obtained using the multipliers that were used in the forward elimination process
http://numericalmethods.eng.usf.edu Finding the [U] matrix Using the Forward Elimination Procedure of Gauss Elimination Step 1:
http://numericalmethods.eng.usf.edu Finding the [U] Matrix Step 2: Matrix after Step 1:
http://numericalmethods.eng.usf.edu Finding the [L] matrix Using the multipliers used during the Forward Elimination Procedure From the first step of forward elimination
http://numericalmethods.eng.usf.edu Finding the [L] Matrix From the second step of forward elimination
http://numericalmethods.eng.usf.edu Does [L][U] = [A]? ?
Example: Unbalanced three phase load Three-phase loads are common in AC systems. When the system is balanced the analysis can be simplified to a single equivalent circuit model. However, when it is unbalanced the only practical solution involves the solution of simultaneous linear equations. In one model the following equations need to be solved. Find the values of I ar, I ai, I br, I bi, I cr, and I ci using LU Decomposition.
Example: Unbalanced three phase load Use Forward Elimination to obtain the [U] matrix.
Step 1 for Row 2: Example: Unbalanced three phase load for Row 3:
Example: Unbalanced three phase load for Row 5: for Row 4:
Example: Unbalanced three phase load for Row 6:
Example: Unbalanced three phase load The system of equations after the completion of the first step of forward elimination is:
Example: Unbalanced three phase load Step 2 for Row 3: for Row 4:
Example: Unbalanced three phase load for Row 6: for Row 5:
Example: Unbalanced three phase load The system of equations after the completion of the second step of forward elimination is:
Example: Unbalanced three phase load Step 3 for Row 4: for Row 5:
Example: Unbalanced three phase load for Row 6:
Example: Unbalanced three phase load The system of equations after the completion of the third step of forward elimination is:
Example: Unbalanced three phase load Step 4 for Row 5: for Row 6:
Example: Unbalanced three phase load The system of equations after the completion of the fourth step of forward elimination is:
Example: Unbalanced three phase load Step 5 for Row 6:
Example: Unbalanced three phase load The coefficient matrix at the end of the forward elimination process is the [U] matrix
Example: Unbalanced three phase load Values of the [L] matrix are the multipliers used during the Forward Elimination Procedure For a system of six equations, the [L] matrix is in the form
Example: Unbalanced three phase load From the first step of forward elimination
Example: Unbalanced three phase load From the second step of forward elimination
Example: Unbalanced three phase load From the third step of forward elimination
Example: Unbalanced three phase load From the fourth step of forward elimination
Example: Unbalanced three phase load From the fifth step of forward elimination
Example: Unbalanced three phase load The [L] matrix is
Example: Unbalanced three phase load Does [L][U] = [A]?
Example: Unbalanced three phase load Set [L][Z] = [C]
Example: Unbalanced three phase load Solve for [Z] The six equations become
Example: Unbalanced three phase load Solve for [Z]
Example: Unbalanced three phase load The [Z] matrix is
Example: Unbalanced three phase load Set [U] [I] = [Z]
Example: Unbalanced three phase load Solve for [I] The six equations become
Example: Unbalanced three phase load Solution:
http://numericalmethods.eng.usf.edu Finding the inverse of a square matrix The inverse [B] of a square matrix [A] is defined as [A][B] = [I] = [B][A]
http://numericalmethods.eng.usf.edu Finding the inverse of a square matrix How can LU Decomposition be used to find the inverse? Assume the first column of [B] to be [b 11 b 12 … b n1 ] T Using this and the definition of matrix multiplication First column of [B] Second column of [B] The remaining columns in [B] can be found in the same manner
http://numericalmethods.eng.usf.edu Example: Inverse of a Matrix Find the inverse of a square matrix [A] Using the decomposition procedure, the [L] and [U] matrices are found to be
http://numericalmethods.eng.usf.edu Example: Inverse of a Matrix Solving for the each column of [B] requires two steps 1)Solve [L] [Z] = [C] for [Z] 2)Solve [U] [X] = [Z] for [X] Step 1: This generates the equations:
http://numericalmethods.eng.usf.edu Example: Inverse of a Matrix Solving for [Z]
http://numericalmethods.eng.usf.edu Example: Inverse of a Matrix Solving [U][X] = [Z] for [X]
http://numericalmethods.eng.usf.edu Example: Inverse of a Matrix Using Backward Substitution So the first column of the inverse of [A] is:
http://numericalmethods.eng.usf.edu Example: Inverse of a Matrix Repeating for the second and third columns of the inverse Second ColumnThird Column
http://numericalmethods.eng.usf.edu Example: Inverse of a Matrix The inverse of [A] is To check your work do the following operation [ A ][ A ] -1 = [ I ] = [ A ] -1 [ A ]
Additional Resources For all resources on this topic such as digital audiovisual lectures, primers, textbook chapters, multiple-choice tests, worksheets in MATLAB, MATHEMATICA, MathCad and MAPLE, blogs, related physical problems, please visit http://numericalmethods.eng.usf.edu/topics/lu_decomp osition.html
THE END http://numericalmethods.eng.usf.edu
Download ppt "4/22/2015 1 LU Decomposition Electrical Engineering Majors Authors: Autar Kaw Transforming."
Similar presentations | 1,990 | 7,866 | {"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} | 3.9375 | 4 | CC-MAIN-2017-51 | longest | en | 0.829763 |
https://www.jiskha.com/display.cgi?id=1352877007 | 1,503,129,150,000,000,000 | text/html | crawl-data/CC-MAIN-2017-34/segments/1502886105326.6/warc/CC-MAIN-20170819070335-20170819090335-00597.warc.gz | 947,805,435 | 3,992 | # math 1
posted by .
The average of 10 numbers is 12. If one number is removed, the average is 11. What number is removed?
• math 1 -
( a1 + a2 + a3 + a4 + a5 + a6 + a7 + a8 + a9 + a10 ) / 10 = 12 Multiply both sides by 10
a1 + a2 + a3 + a4 + a5 + a6 + a7 + a8 + a9 + a10 = 120
( a1 + a2 + a3 + a4 + a5 + a6 + a7 + a8 + a9 ) / 9 = 11 Multiply both sides by 9
a1 + a2 + a3 + a4 + a5 + a6 + a7 + a8 + a9 = 99
( a1 + a2 + a3 + a4 + a5 + a6 + a7 + a8 + a9 ) + a10 = 120
99 + a10 = 120 Subtract 99 to both sides
99 + a10 - 99 = 120 -99
a10 = 21
Number 21 is removed.
• math 1 -
the total of the 10 number is 120
solve
(120-x)/9 = 11
120-x = 99
-x = -21
x = 21
## Similar Questions
1. ### math
Does anyone know how to explain mean, median, mode and range mean?
2. ### Math
The average of 3 numbers is 45. If one of the numbers is 43, what is the average of the other 2 numbers?
3. ### math
I am looking for the median, I believe it is 39 am I correct?
4. ### Math
The average of five number is 56. Four of the numbers are the same. The average of one of the four number and the fifth number is 80. What is the fifth number. 56*5=280 80*2=160 280-160=120 5/120=24 Is 24 correct?
5. ### Programming
I suppose to write a program that calculates the average of up to 50 numbers input by the user and stored in an array. The program should use a class name Statistics and have an AddNumber method that stores numbers into an arrey one …
6. ### maths
1).Last year, a shopkeeper sold an average of 7 cartons of apples from january to april. He sold an average of 9 cartons of apples from june till end of the year. What is the total number of cartons he sold in the 11 months?
7. ### Math
The mean of five numbers is 8. If one of the numbers is removed from the list, the mean becomes 7. what is the number that was removed. I could not figure this out.
8. ### math
The mean of 14 numbers is 50. Removing one of the numbers causes the mean to decrease to 43. What number was removed?
9. ### Algebra 1
The average of five consecutive integers is 13. One of the integers is removed and the sum of the remaining integers is 53. What is the value of the integer that was removed?
10. ### computer programming
Design the logic for a program that allows a user to enter any amount of numbers up to 10 or until a sentinal value of -1 is entered. Then the program displays the count of the numbers entered and numeric average of the numbers (not …
More Similar Questions | 764 | 2,478 | {"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} | 3.859375 | 4 | CC-MAIN-2017-34 | latest | en | 0.898351 |
http://mathematics.tsu.edu/math-241/ | 1,597,417,696,000,000,000 | text/html | crawl-data/CC-MAIN-2020-34/segments/1596439739328.66/warc/CC-MAIN-20200814130401-20200814160401-00443.warc.gz | 73,914,017 | 5,729 | # math 241: Calculus and Analytic Geometry I (4)
Course Prerequisites/Corequisites: MATH 241.
Course Description:Inequalities, functions, graphs, straight lines, linear equations, limits, continuity, differentiation, maximum-minimum problems, mean value theorem, related rates, and indefinite integrals. Four hours of lecture per week.
Listed as MATH 2413 in the Texas Common Course Numbering System.
Required Reading / Textbook: Calculus: Early Transcendentals. Sullivan and Miranda, W.H. Freemand and Company
Major Assignments/Exams
• Homework
• Quizzes
• Exams
• Final Exam
Course topics
CHPATER 2. Limits
• 2.1 The Idea of Limits
• 2.2 Definitions of Limits
• 2.3 Techniques for Computing Limits
• 2.4 Infinite Limits
• 2.5 Limits at Infinity
• 2.6 Continuity
• 2.7 Precise Definitions of Limits (optional)
CHPATER 3. Derivatives.
• 3.1 Introducing the Derivative
• 3.2 Rules of Differentiation
• 3.3 The Product and Quotient Rules
• 3.4 Derivatives of Trigonometric Functions
• 3.5 Derivatives as Rates of Change
• 3.6 The Chain Rule
• 3.7 Implicit Differentiation
• 3.8 Derivatives of Logarithmic and Exponential Functions
• 3.9 Derivatives of Inverse Trigonometric Functions
• 3.10 Related Rates
CHPATER 4. Applications of the Derivative
• 4.1 Maxima and Minima
• 4.2 What Derivatives Tell Us
• 4.3 Graphing Functions
• 4.4 Optimization Problems
• 4.5 Linear Approximation and Differentials
• 4.6 Mean Value Theorem
• 4.7 L’Hôpital’s Rule
• 4.8 Antiderivatives
CHPATER 5. Integration
• 5.3 Fundamental Theorem of Calculus (if time permits)
• 5.4 Working with Integrals (if time permits)
• 5.5 Substitution Rule(if time permits)
CHPATER6. Applications of Integration
• 6.2 Regions between Curves (if time permits)
The ADA of 1990 provides civil rights protection for individual with disabilities. It guarantees equal opportunities for “qualified individuals” with disabilities in all public facilities, educational programs, activities, services and benefits. The ADA upholds and maintains compliance standards to ensure institutions of higher education policies, procedures and practices are non-discriminatory. Students needing academic accommodations should contact the Office of Disability Services at 713-313-7691 | 605 | 2,241 | {"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-2020-34 | latest | en | 0.731934 |
https://la.mathworks.com/matlabcentral/answers/160612-creating-a-new-matrix-from-two-matrixes | 1,669,959,774,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710898.93/warc/CC-MAIN-20221202050510-20221202080510-00367.warc.gz | 385,240,070 | 25,320 | # Creating a new matrix from two matrixes
2 views (last 30 days)
Moe on 29 Oct 2014
Answered: James Tursa on 29 Oct 2014
Suppose I have a matrix a:
a = [9,1;1,1;3,1;2,1;5,1;6,1;8,1;1,1;2,1;5,1;8,1;2,1];
And I have another matrix b which is like some selected index number of matrix a:
b = [4;10;7]; % for example 4 means forth row in matrix a which is (2,1) here
So, I need to have a new matrix c in that way present those rows in matrix a which are indicated in matrix b and others array be equal to 0:
c = [0,0;0,0;0,0;2,1;0,0;0,0;8,1;0,0;0,0;5,1;0,0;0,0;]; % for example forth row is (2,1) because 4 row is presented in matrix b
James Tursa on 29 Oct 2014
c = zeros(size(a));
c(b,:) = a(b,:);
### Categories
Find more on Matrices and Arrays in Help Center and File Exchange
### Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!
Translated by | 324 | 923 | {"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} | 3.140625 | 3 | CC-MAIN-2022-49 | longest | en | 0.844013 |
http://www.brainia.com/essays/Res-342-Uop-Course-Tutorial/320315.html | 1,500,599,353,000,000,000 | text/html | crawl-data/CC-MAIN-2017-30/segments/1500549423629.36/warc/CC-MAIN-20170721002112-20170721022112-00545.warc.gz | 398,911,858 | 5,642 | # RES 342 UOP COURSE TUTORIAL / res342dotcom
## RES 342 UOP COURSE TUTORIAL / res342dotcom
• Submitted By: rocks537
• Date Submitted: 10/01/2014 10:14 PM
• Category: Business
• Words: 437
• Page: 2
• Views: 1
RES 342 Entire Course
For more course tutorials visit
www.res342.com
RES 342 Week 1 Individual Assignment Hypothesis Identification Article Analysis
RES 342 Week 1 Individual Assignment E-Text
RES 342 Week 2 Individual Assignment E-Text
RES 342 Week 2 Individual Quiz
RES 342 Week 2 Team Assignment One Sample Hypothesis Testing Paper
RES 342 Week 3 Individual Assignment E-Text
RES 342 Week 3 Individual Assignment Applying Analysis of Variance (ANOVA) and Nonparametric Tests Simulation
RES 342 Week 3 Team Assignment Two or More Sample Hypothesis Testing Paper
RES 342 Week 4 Individual Assignment E Text
RES 342 Week 4 Individual Assignment Parametric and Nonparametric Data Identification Assignment
RES 342 Week 4 Individual Quiz
RES 342 Week 4 Team Assignment Nonparametric Hypothesis Testing Paper
RES 342 Week 5 Individual E Text
RES 342 Week 5 Individual Assignment Applying Time Series Methodologies Simulation
RES 342 Week 5 Team Assignment Regression Paper
---------------------------------------------------------------------------------------------------------------------------------------------
RES 342 Week 4 Team Assignment Nonparametric Hypothesis Testing Paper
For more course tutorials visit
www.res342.com
Use the same research question and data from the Learning Team Meeting Two assignment.
Conduct the equivalent, nonparametric test of hypothesis using the five-step process.
Prepare a 1,050- to 1,750-word paper describing the results of the nonparametric hypothesis test. Include the following in your paper:
• Formulate a hypothesis statement regarding your research issue.
• Perform the five-step hypothesis test on the data.
• Explain which nonparametric test you used to analyze your data and why you chose that test.
• Interpret the results of your test, and explain the differences, if any, that you observe from your Week Three paper.
Include your raw data tables and the results of your... | 481 | 2,147 | {"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} | 3.515625 | 4 | CC-MAIN-2017-30 | latest | en | 0.683056 |
https://mathexamination.com/class/monotonic-function.php | 1,620,872,920,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243992721.31/warc/CC-MAIN-20210513014954-20210513044954-00287.warc.gz | 416,720,001 | 6,878 | ## Do My Monotonic Function Class
A "Monotonic Function Class" QE" is a standard mathematical term for a generalized constant expression which is used to solve differential equations and has options which are regular. In differential Class solving, a Monotonic Function function, or "quad" is used.
The Monotonic Function Class in Class kind can be expressed as: Q( x) = -kx2, where Q( x) are the Monotonic Function Class and it is a crucial term. The q part of the Class is the Monotonic Function constant, whereas the x part is the Monotonic Function function.
There are four Monotonic Function functions with correct service: K4, K7, K3, and L4. We will now look at these Monotonic Function functions and how they are fixed.
K4 - The K part of a Monotonic Function Class is the Monotonic Function function. This Monotonic Function function can also be written in partial fractions such as: (x2 - y2)/( x+ y). To resolve for K4 we increase it by the correct Monotonic Function function: k( x) = x2, y2, or x-y.
K7 - The K7 Monotonic Function Class has a service of the form: x4y2 - y4x3 = 0. The Monotonic Function function is then multiplied by x to get: x2 + y2 = 0. We then need to increase the Monotonic Function function with k to get: k( x) = x2 and y2.
K3 - The Monotonic Function function Class is K3 + K2 = 0. We then multiply by k for K3.
K3( t) - The Monotonic Function function equationis K3( t) + K2( t). We multiply by k for K3( t). Now we increase by the Monotonic Function function which offers: K2( t) = K( t) times k.
The Monotonic Function function is likewise referred to as "K4" because of the initials of the letters K and 4. K indicates Monotonic Function, and the word "quad" is pronounced as "kah-rab".
The Monotonic Function Class is among the main methods of fixing differential equations. In the Monotonic Function function Class, the Monotonic Function function is first increased by the suitable Monotonic Function function, which will provide the Monotonic Function function.
The Monotonic Function function is then divided by the Monotonic Function function which will divide the Monotonic Function function into a real part and a fictional part. This provides the Monotonic Function term.
Lastly, the Monotonic Function term will be divided by the numerator and the denominator to get the quotient. We are entrusted the right hand side and the term "q".
The Monotonic Function Class is a crucial idea to comprehend when solving a differential Class. The Monotonic Function function is simply one approach to fix a Monotonic Function Class. The approaches for resolving Monotonic Function formulas consist of: particular worth decomposition, factorization, ideal algorithm, mathematical solution or the Monotonic Function function approximation.
## Pay Me To Do Your Monotonic Function Class
If you wish to end up being acquainted with the Quartic Class, then you require to first start by looking through the online Quartic page. This page will show you how to utilize the Class by utilizing your keyboard. The description will also show you how to produce your own algebra equations to help you study for your classes.
Prior to you can comprehend how to study for a Monotonic Function Class, you should initially comprehend making use of your keyboard. You will find out how to click the function keys on your keyboard, along with how to type the letters. There are three rows of function keys on your keyboard. Each row has 4 functions: Alt, F1, F2, and F3.
By pressing Alt and F2, you can multiply and divide the value by another number, such as the number 6. By pressing Alt and F3, you can use the 3rd power.
When you press Alt and F3, you will enter the number you are attempting to multiply and divide. To multiply a number by itself, you will press Alt and X, where X is the number you wish to increase. When you push Alt and F3, you will type in the number you are attempting to divide.
This works the exact same with the number 6, other than you will only type in the two digits that are 6 apart. Lastly, when you push Alt and F3, you will utilize the fourth power. Nevertheless, when you push Alt and F4, you will utilize the real power that you have actually found to be the most proper for your issue.
By using the Alt and F function keys, you can multiply, divide, and after that use the formula for the third power. If you require to increase an odd number of x's, then you will require to go into an even number.
This is not the case if you are trying to do something complex, such as increasing two even numbers. For instance, if you want to increase an odd variety of x's, then you will require to get in odd numbers. This is specifically real if you are attempting to determine the response of a Monotonic Function Class.
If you want to convert an odd number into an even number, then you will need to press Alt and F4. If you do not know how to multiply by numbers by themselves, then you will require to use the letters x, a b, c, and d.
While you can multiply and divide by use of the numbers, they are a lot easier to use when you can look at the power tables for the numbers. You will need to do some research study when you first start to use the numbers, however after a while, it will be second nature. After you have created your own algebra formulas, you will be able to produce your own multiplication tables.
The Monotonic Function Solution is not the only method to resolve Monotonic Function equations. It is very important to find out about trigonometry, which utilizes the Pythagorean theorem, and then use Monotonic Function solutions to fix issues. With this technique, you can know about angles and how to fix issues without having to take another algebra class.
It is necessary to attempt and type as rapidly as possible, due to the fact that typing will assist you understand about the speed you are typing. This will assist you write your responses quicker.
## Pay Someone To Take My Monotonic Function Class
A Monotonic Function Class is a generalization of a direct Class. For instance, when you plug in x=a+b for a given Class, you acquire the value of x. When you plug in x=a for the Class y=c, you obtain the values of x and y, which provide you a result of c. By using this fundamental principle to all the formulas that we have actually attempted, we can now solve Monotonic Function equations for all the values of x, and we can do it rapidly and effectively.
There are lots of online resources offered that offer complimentary or affordable Monotonic Function formulas to resolve for all the values of x, including the cost of time for you to be able to take advantage of their Monotonic Function Class task help service. These resources normally do not require a subscription fee or any type of investment.
The answers provided are the result of complex-variable Monotonic Function formulas that have actually been fixed. This is likewise the case when the variable utilized is an unknown number.
The Monotonic Function Class is a term that is an extension of a linear Class. One advantage of using Monotonic Function formulas is that they are more general than the linear formulas. They are much easier to solve for all the values of x.
When the variable utilized in the Monotonic Function Class is of the kind x=a+b, it is easier to resolve the Monotonic Function Class due to the fact that there are no unknowns. As a result, there are less points on the line specified by x and a consistent variable.
For a right-angle triangle whose base indicate the right and whose hypotenuse points to the left, the right-angle tangent and curve chart will form a Monotonic Function Class. This Class has one unknown that can be discovered with the Monotonic Function formula. For a Monotonic Function Class, the point on the line defined by the x variable and a continuous term are called the axis.
The presence of such an axis is called the vertex. Considering that the axis, vertex, and tangent, in a Monotonic Function Class, are an offered, we can find all the values of x and they will sum to the provided worths. This is achieved when we use the Monotonic Function formula.
The aspect of being a continuous aspect is called the system of formulas in Monotonic Function formulas. This is often called the central Class.
Monotonic Function equations can be resolved for other values of x. One way to resolve Monotonic Function equations for other worths of x is to divide the x variable into its aspect part.
If the variable is provided as a positive number, it can be divided into its element parts to get the typical part of the variable. This variable has a magnitude that amounts to the part of the x variable that is a consistent. In such a case, the formula is a third-order Monotonic Function Class.
If the variable x is negative, it can be divided into the same part of the x variable to get the part of the x variable that is increased by the denominator. In such a case, the formula is a second-order Monotonic Function Class.
Service assistance service in solving Monotonic Function formulas. When using an online service for fixing Monotonic Function formulas, the Class will be solved quickly. | 2,054 | 9,225 | {"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-2021-21 | latest | en | 0.896047 |
https://affairscloud.com/most-important-topics-to-study-for-ibps-rrb-officer-scale-i-2018/ | 1,643,319,079,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320305288.57/warc/CC-MAIN-20220127193303-20220127223303-00341.warc.gz | 144,541,002 | 34,390 | # Most Important Topics to Study for IBPS RRB Officer Scale I 2018
Are you preparing for IBPS RRB Exams? If yes, this post is going to be of good help. Today, we bring to you the most important topics that you must study for the upcoming IBPS RRB Officer Scale I Exam 2018. This post is brought to you by Oliveboard, an online exam preparation platform for banking and government exams.
Most Important Topics from Quantitative Aptitude
Simplification / Approximation
There is high possibility that 5-10 questions on Simplification / Approximation will be asked in the Quantitative Aptitude section of IBPS RRB exams. Your calculation skills will be the deciding factor for these types of questions. The faster you calculate, the better you perform. Stick to the BODMAS rule and do practice good number of Simplification Questions of both easy and difficult level for better preparation.
Data Interpretation
Data Interpretation or DI is one of the most important topics in any bank exams and IBPS RRB is no exception. Candidates can expect at least 10 questions from this topic. Various types of DI questions can be asked in the IBPS RRB exams including Tabular DI, Line and Bar Graph. Make sure you have good concept knowledge of arithmetic topics like Ratio and Proportion, Percentage, Average and Profit and Loss, Time and Work etc. as DI questions are usually based on these concepts.
Inequality
Another important topic for IBPS RRB exam would be Inequalities. Candidates can expect 5-10 questions from Inequalities. Be well prepared with Quadratic Equations and other topics from the arithmetic sections to tackle Inequalities questions. Usually questions are based on Quadratic Equations but there are possibilities of questions being based on Average and Mensuration etc.
Number Series
Candidates can expect around 5 questions from Number Series. There are patterns of Number Series that you must consider – square pattern, prime number pattern, multiplication and difference pattern etc. To excel and score more in this topic, ensure you practice good number of number series questions.
Most Important Topics from Reasoning Ability
Seating Arrangement & Puzzles
This is one of the most important reasoning topics and candidates can expect 10 to 20 questions from this topic. Seating arrangement includes linear arrangement, circular arrangement, square arrangement and Puzzles include sequence-based, scheduling-based, comparison-based. For these types of questions, practice is the key.
Logical Reasoning
Candidates can expect 5-6 questions from logical reasoning. Logical Reasoning questions in most of the bank exams including IBPS RRB have been moderate and therefore, you should be able to solve these questions at ease. Questions are based on cause and effect, a course of action, assumptions and conclusions etc.
Syllogism
There can around 5 questions on Syllogism. Practice both the old and new pattern where conclusions are given in questions and statements are given in options.
Hope this post helps you perform better in the upcoming IBPS RRB exams. If you have any other question, feel free to mention in the comment section below.
AffairsCloud Recommends Oliveboard Mock Test
AffairsCloud Ebook - Support Us to Grow | 666 | 3,258 | {"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-2022-05 | latest | en | 0.927614 |
http://mathhelpforum.com/algebra/203706-potens-print.html | 1,529,504,251,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267863519.49/warc/CC-MAIN-20180620124346-20180620144346-00635.warc.gz | 207,366,446 | 3,695 | # potens
• Sep 19th 2012, 06:33 AM
Petrus
potens
Expression profile https://webwork.math.su.se/webwork2_...142cdc9bc1.png as a power of 5. If not so, then https://webwork.math.su.se/webwork2_...d08713f261.png
x=?
i dont really know how to especially write 3sqrt(5) on another form :S
if i got it right i can write it like 5^1/3?
• Sep 19th 2012, 06:53 AM
Wilmer
Re: potens
Numerator: 5^1 * (5^2)^3 * 5^(1/2) = 5^1 * 5^6 * 5^(1/2) = 5^(1 + 6 + 1/2) = 5^(15/2)
Let you have the fun of doing the denominator!
• Sep 19th 2012, 07:39 AM
Prove It
Re: potens
Quote:
Originally Posted by Petrus
Expression profile https://webwork.math.su.se/webwork2_...142cdc9bc1.png as a power of 5. If not so, then https://webwork.math.su.se/webwork2_...d08713f261.png
x=?
i dont really know how to especially write 3sqrt(5) on another form :S
if i got it right i can write it like 5^1/3?
Yes, \displaystyle \displaystyle \begin{align*} \sqrt[3]{5} = 5^{\frac{1}{3}} \end{align*}
• Sep 19th 2012, 07:51 AM
Petrus
Re: potens
5^(1 + 6 + 1/2) = 5^(15/2)
idk but maybe stupid question how can that become 15/2? i cant figoure it out how u did so
• Sep 19th 2012, 08:13 AM
Prove It
Re: potens
Quote:
Originally Posted by Petrus
5^(1 + 6 + 1/2) = 5^(15/2)
idk but maybe stupid question how can that become 15/2? i cant figoure it out how u did so
Sure you know there are 15 halves in 7 and a half...
• Sep 19th 2012, 08:34 AM
Petrus
Re: potens
Quote:
Originally Posted by Prove It
Sure you know there are 15 halves in 7 and a half...
ehmm sorry but i did not understand what u mean, any way u can put it on more simple words^^? (english is my weak language)
• Sep 19th 2012, 09:04 AM
Prove It
Re: potens
If you had 7 and a half pies, and if you cut each of the whole pies in half, how many halves would you then have?
• Sep 19th 2012, 09:23 AM
Petrus
Re: potens
Quote:
Originally Posted by Prove It
If you had 7 and a half pies, and if you cut each of the whole pies in half, how many halves would you then have?
over 9000 =D joke hahaha u made my day :D
• Sep 19th 2012, 09:29 AM
Wilmer
Re: potens
Quote:
Originally Posted by Petrus
ehmm sorry but i did not understand what u mean, any way u can put it on more simple words^^? (english is my weak language)
Sorry Petrus, but if you don't understand that 7.5 = 7 1/2 = 15/2,
then speaking to you in any language will not help: you need serious classroom help.
• Sep 19th 2012, 09:29 AM
Petrus
Re: potens
holy i cant finish the last one ehmm
5^1/3*5^15?
• Sep 19th 2012, 09:35 AM
Wilmer
Re: potens
Quote:
Originally Posted by Petrus
holy i cant finish the last one ehmm
5^1/3*5^15?
Do you not know this rule: a^x * a^y = a^(x + y) ?
Your 5^1/3 needs brackets: 5^(1/3) ; no brackets means 5^1 times 3
So 5^(1/3) * 5^15 = 5^(1/3 + 15) = 5^(15 1/3) = 5^(46/3)
WHAT does "potens" mean ?????
• Sep 19th 2012, 10:14 AM
Petrus
Re: potens
Quote:
Originally Posted by Wilmer
Do you not know this rule: a^x * a^y = a^(x + y) ?
Your 5^1/3 needs brackets: 5^(1/3) ; no brackets means 5^1 times 3
So 5^(1/3) * 5^15 = 5^(1/3 + 15) = 5^(15 1/3) = 5^(46/3)
WHAT does "potens" mean ?????
ty ehmm potens is on swedish i was just lazy to google translate but in english it means Exponentiation
so now i got 5^(15/2)/5^(46/3) and next i do is 5^(15/2)-(46/3)?
• Sep 19th 2012, 10:19 AM
Petrus
Re: potens
nvm finish it :D sorry for being so stupid but i am just tired and not really good on math ohh well | 1,302 | 3,428 | {"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} | 3.75 | 4 | CC-MAIN-2018-26 | latest | en | 0.87876 |
https://gmatclub.com/forum/james-took-a-3-hour-bike-ride-in-the-second-hour-he-travlle-159409.html?sort_by_oldest=true | 1,490,221,216,000,000,000 | text/html | crawl-data/CC-MAIN-2017-13/segments/1490218186353.38/warc/CC-MAIN-20170322212946-00246-ip-10-233-31-227.ec2.internal.warc.gz | 786,230,933 | 60,763 | Check GMAT Club Decision Tracker for the Latest School Decision Releases https://gmatclub.com/AppTrack
GMAT Club
It is currently 22 Mar 2017, 15:20
GMAT Club Daily Prep
Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.
Customized
for You
we will pick new questions that match your level based on your Timer History
Track
every week, we’ll send you an estimated GMAT score based on your performance
Practice
Pays
we will pick new questions that match your level based on your Timer History
Events & Promotions
Events & Promotions in June
Open Detailed Calendar
James took a 3-hour bike ride. In the second hour he travlle
Author Message
TAGS:
Hide Tags
Director
Joined: 29 Nov 2012
Posts: 895
Followers: 15
Kudos [?]: 1122 [0], given: 543
James took a 3-hour bike ride. In the second hour he travlle [#permalink]
Show Tags
09 Sep 2013, 01:22
00:00
Difficulty:
25% (medium)
Question Stats:
78% (02:21) correct 22% (01:49) wrong based on 37 sessions
HideShow timer Statistics
James took a 3-hour bike ride. In the second hour he travlled 18 miles, which was 20 percent farther than he traveled the first hour. If he traveled 25 percent farther in the third hour than he did in the second hour, how many miles did jose travel during the entire ride?
A) 54.0
B) 54.9
C) 55.5
D) 57.0
E) 63.0
[Reveal] Spoiler: OA
_________________
Click +1 Kudos if my post helped...
Amazing Free video explanation for all Quant questions from OG 13 and much more http://www.gmatquantum.com/og13th/
GMAT Prep software What if scenarios http://gmatclub.com/forum/gmat-prep-software-analysis-and-what-if-scenarios-146146.html
Verbal Forum Moderator
Joined: 10 Oct 2012
Posts: 630
Followers: 82
Kudos [?]: 1147 [0], given: 136
Re: James took a 3-hour bike ride. In the second hour he travlle [#permalink]
Show Tags
09 Sep 2013, 01:41
fozzzy wrote:
James took a 3-hour bike ride. In the second hour he travlled 18 miles, which was 20 percent farther than he traveled the first hour. If he traveled 25 percent farther in the third hour than he did in the second hour, how many miles did jose travel during the entire ride?
A) 54.0
B) 54.9
C) 55.5
D) 57.0
E) 63.0
Let the distance travelled in the first hour be x. Thus, 1.2x = 18 , x = 15. Now, the distance travelled in the 3rd hour = $$18+\frac{1}{4}*18$$ = The only option ending with a 0.5 in the decimal's place is C.
C.
_________________
Director
Joined: 29 Nov 2012
Posts: 895
Followers: 15
Kudos [?]: 1122 [1] , given: 543
Re: James took a 3-hour bike ride. In the second hour he travlle [#permalink]
Show Tags
09 Sep 2013, 01:51
1
KUDOS
Just to clarify in first hour 15 miles travelled in second hour 18 miles ( 3 additional miles) and 3rd hour 22.5 miles ( 4.5 miles travelled)
The numbers in the bracket represents the net increase.
_________________
Click +1 Kudos if my post helped...
Amazing Free video explanation for all Quant questions from OG 13 and much more http://www.gmatquantum.com/og13th/
GMAT Prep software What if scenarios http://gmatclub.com/forum/gmat-prep-software-analysis-and-what-if-scenarios-146146.html
Re: James took a 3-hour bike ride. In the second hour he travlle [#permalink] 09 Sep 2013, 01:51
Similar topics Replies Last post
Similar
Topics:
4 A truck driver drove for 2 days. On the second day, he drove 3 hours 3 27 Dec 2015, 09:30
12 If it takes 70 workers 3 hours to disassemble the exhibition rides at 9 26 Nov 2014, 07:41
22 Working alone at its constant rate, machine K took 3 hours 12 29 Nov 2012, 22:44
2 One pump drains one-half of a pond in 3 hours, and then a second pump 6 16 Oct 2011, 09:55
9 The sum of 3 hours 45 minutes and 2 hours 55 minutes is approximately 10 15 Jan 2008, 14:52
Display posts from previous: Sort by | 1,139 | 3,875 | {"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} | 3.765625 | 4 | CC-MAIN-2017-13 | longest | en | 0.885806 |
http://www.convertit.com/Go/ConvertIt/Measurement/Converter.ASP?From=NLG&To=currency | 1,566,194,117,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027314667.60/warc/CC-MAIN-20190819052133-20190819074133-00071.warc.gz | 230,985,386 | 3,912 | Partner with ConvertIt.com
New Online Book! Handbook of Mathematical Functions (AMS55)
Conversion & Calculation Home >> Measurement Conversion
Measurement Converter
Convert From: (required) Click here to Convert To: (optional) Examples: 5 kilometers, 12 feet/sec^2, 1/5 gallon, 9.5 Joules, or 0 dF. Help, Frequently Asked Questions, Use Currencies in Conversions, Measurements & Currencies Recognized Examples: miles, meters/s^2, liters, kilowatt*hours, or dC.
Conversion Result: ```Netherlands guilder on 4-28-2017 = 0.494359756819166 currency (currency)``` Related Measurements: Try converting from "NLG" to ADP (Andorran peseta on 4-28-2017), ATS (Austrian schilling on 4-28-2017), AUD (Australian dollar on 4-28-2017), EUR (European Union euro on 4-28-2017), FIM (Finnish markka on 4-28-2017), FJD (Fijian dollar on 4-28-2017), FKP (Falkland pound on 4-28-2017), FRF (French franc on 4-28-2017), GTQ (Guatemalan quetzal on 4-28-2017), ILS (Israeli new shekel on 4-28-2017), ITL (Italian lira on 4-28-2017), JPY (Japanese yen on 4-28-2017), KRW (South Korean won on 4-28-2017), MMK (Burmese kyat on 4-28-2017), NOK (Norwegian krone on 4-28-2017), PLN (Polish zloty on 4-28-2017), PTE (Portuguese escudo on 4-28-2017), SZL (Swazi lilangeni on 4-28-2017), USD (United States dollar on 4-28-2017), VND (Vietnamese dong on 4-28-2017), or any combination of units which equate to "currency" and represent currency. Sample Conversions: NLG = .66132414 AUD (Australian dollar on 4-28-2017), .98871951 BBD (Barbadian dollar on 4-28-2017), .69066824 BND (Bruneian dollar on 4-28-2017), 87.86 DJF (Djiboutian franc on 4-28-2017), 75.5 ESP (Spanish peseta on 4-28-2017), 2.7 FIM (Finnish markka on 4-28-2017), 1.03 FJD (Fijian dollar on 4-28-2017), 2.98 FRF (French franc on 4-28-2017), 3.8 GTQ (Guatemalan quetzal on 4-28-2017), 11.56 HNL (Honduran lempira on 4-28-2017), 1.79 ILS (Israeli new shekel on 4-28-2017), 63.91 JMD (Jamaican dollar on 4-28-2017), 6.6 LSL (Lesotho loti on 4-28-2017), 1.98 LTL (Lithuanian litas on 4-28-2017), 9.36 MXN (Mexican peso on 4-28-2017), 1.6 PEN (Peruvian nuevo sol on 4-28-2017), 90.97 PTE (Portuguese escudo on 4-28-2017), 4.38 SEK (Swedish krona on 4-28-2017), 17.12 THB (Thai baht on 4-28-2017), 6.6 ZAR (South African rand on 4-28-2017).
Feedback, suggestions, or additional measurement definitions?
Please read our Help Page and FAQ Page then post a message or send e-mail. Thanks! | 922 | 2,424 | {"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-2019-35 | latest | en | 0.608103 |
http://mathhelpforum.com/calculus/209595-find-minimum-function-print.html | 1,519,463,004,000,000,000 | text/html | crawl-data/CC-MAIN-2018-09/segments/1518891815500.61/warc/CC-MAIN-20180224073111-20180224093111-00725.warc.gz | 221,164,900 | 2,919 | # Find the minimum of a function.
• Dec 11th 2012, 09:59 AM
Ant
Find the minimum of a function.
Hi,
I'm trying to find the minimum of the following function:
$f(x) = \frac{x^2}{2K} - ln(2cosh(x+h))$
My working so far:
$f'(x) = \frac{x}{K} - tanh(x+h) =0$
Where we set the derivative equal to zero as we're trying to find the min.
However, I'm not sure how to solve this equation, and find an expression for x. Is it best to rewrite the hyperbolic tan in terms of e?
Thanks!
• Dec 11th 2012, 10:38 AM
ebaines
Re: Find the minimum of a function.
I don't believe there is a closed form solution to the equation $x - C tanh(x) = 0$, so instead you should use a numerical technique to find a very good approximation for the value of x.
• Dec 11th 2012, 10:41 AM
Ant
Re: Find the minimum of a function.
hmm, Okay thanks.
I think I may just be able to define $x*$ as the min and carry on. (The full task is evaluating an integral using the laplace approx) | 285 | 957 | {"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": 0, "img_math": 0, "codecogs_latex": 4, "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-2018-09 | latest | en | 0.910585 |
https://www.wyzant.com/resources/answers/722992/the-world-s-population-is-currently-2-if-the-population-doubles-every-39-ye | 1,603,294,764,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107876768.45/warc/CC-MAIN-20201021151342-20201021181342-00066.warc.gz | 985,294,453 | 14,350 | # The world's population is currently 2. If the population doubles every 39 years, what will the population be 15000 years from now?
Hello all. I am writing a fantasy novel and need to calculate population growth for realism purposes. There was another person that asked a similar question on a smaller scale, so I was plugging in the figures for my own question, but I got as far as solving for k and got stuck. Not sure how to do that. Also, I would need to know how to do this for any given year, not just 15000. Solving for k seems to be something that would be necessary each time I changed the target year, right? The equation I'm dealing with comes from the other person's question, the replies to that question bringing up the response shown below.
(Beginning of copied response with my own variables applied instead):
The population will grow according to the exponential model
P(t) = P(0)ekt
P(t) = 2·2kt
where t is the time in years and k is a constant.
You first need to determine k from the doubling time, 39 years.
Set t = 39 and P(t) = 4, then solve for k:
4 = 2·2k(39)
(End of copied response)
4 = 2·2k(39) is as far as I got. Didn't know how to proceed from there. Like I said, I know you have to solve for k. I just don't know how to do that. At risk of sounding slow, please explain as simply and with as little jargon as possible. A layman's terms explanation for how to do this is what I'm looking for. Also, this figure will be discounting any natural disasters, death rates, or any other world-issue variables. I just need a baseline figure to get started, and from there I will be able to use that figure in tandem with my own created history specific to my fantasy novel. So, again, I'm not looking for complicated answers or a discussion on world-building. I'm just looking for a baseline number to start with and the knowledge for how to get a new baseline number each time I need to change the target year. Any help anyone can give would be greatly appreciated. Thank you!
By:
Tutor
5.0 (223)
English: Professional & Public Writing + Technical Communication Major
## Still looking for help? Get the right answer, fast.
Get a free answer to a quick problem.
Most questions answered within 4 hours.
#### OR
Choose an expert and meet online. No packages or subscriptions, pay only for the time you need. | 553 | 2,346 | {"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} | 3.953125 | 4 | CC-MAIN-2020-45 | latest | en | 0.966303 |
https://www.coursehero.com/file/6401886/unit2/ | 1,513,267,613,000,000,000 | text/html | crawl-data/CC-MAIN-2017-51/segments/1512948544677.45/warc/CC-MAIN-20171214144324-20171214164324-00492.warc.gz | 720,900,544 | 24,589 | unit2 - Stat131A 1 Statistics 131A Spring 2011 Unit 2...
This preview shows pages 1–9. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
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.
Unformatted text preview: Stat131A 1 Statistics 131A, Spring 2011 Unit 2: Probability Chris Paciorek Stat131A 2 Outline of Material Introduction to Probability Discrete Random Variables and Probability Distributions Continuous Random Variables and Probability Distributions Joint Probability Distributions and Random Samples Stat131A 3 Sources I MS: Chapters 3-6, excluding 3.8, 3.9, 4.8, 4.9, 4.11, 5.7, 5.8, 5.9, 5.10, most of 6.6, 6.10, 6.11 I Basically we’re excluding some of the more technical mathematics. Stat131A 4 Introduction to Probability Randomness I Randomness means that the possible outcomes are known, but for any given observation, the outcome is uncertain. I Probability quantifies the chance of each possible outcome when the outcomes are random. I Example: I A coin flip: what are the possible outcomes? Is the outcome of a given coin flip random? I Randomly choose a person in the US and find out their age: What are the possible outcomes? Stat131A 5 Introduction to Probability Defining probability I One definition of probability is that the probability of an outcome is the long-run frequency of that outcome. I First, the pattern in the short-run can differ from what you might expect. I Next, let’s look at the long-run frequency of heads in coin-flipping. [see R demo] I Note that in a small sample, the empirical proportion does not equal 0.50 because of the randomness. Stat131A 6 Introduction to Probability Things can get complicated I What is the probability that it will rain tomorrow? Is there any long-run frequency? I What is the probability that the Sierra snowpack at the end of winter will decrease with climate change? Is there a long-run frequency? I What is the probability that an individual has a disease, not having any other information. Is this a random event? I What is the probability that a randomly selected individual in the US has a disease. Is this random? Stat131A 7 Introduction to Probability Subjective probability I The subjective view of probability is that probability summarizes a person’s uncertainty (or degree of belief) about an outcome, based on an evaluation of the information available to the person. I Example: how could we make a probability statement about the Sierra snowpack? I In this version, one way of looking at probability is that it determines how you should bet on an outcome. If your probability is 0.10 (odds of 9:1), you should need to receive at least \$9 for winning a bet in order to place a \$1 bet on the outcome. I We’ll generally work with the long-run frequency interpretation in this class, but there is an important branch of statistics called Bayesian statistics, which is grounded in a subjective probability interpretation. Stat131A 8 Introduction to Probability The sample space and events I The sample space, S , is the collection of all the possible outcomes (MS calls these ’simple events’)....
View Full Document
{[ snackBarMessage ]}
Page1 / 90
unit2 - Stat131A 1 Statistics 131A Spring 2011 Unit 2...
This preview shows document pages 1 - 9. Sign up to view the full document.
View Full Document
Ask a homework question - tutors are online | 842 | 3,767 | {"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-2017-51 | latest | en | 0.857697 |
https://es.mathworks.com/matlabcentral/answers/673253-monte-carlo-integration-of-functions-to-the-3rd-power | 1,610,814,451,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703506697.14/warc/CC-MAIN-20210116135004-20210116165004-00060.warc.gz | 323,185,905 | 24,600 | # Monte Carlo integration of functions to the 3rd power
37 views (last 30 days)
Denise Veldhuis on 2 Dec 2020
Edited: Denise Veldhuis on 6 Dec 2020
I have to create a monte carlo integration of the function above and this is what i've tried so far;
f = @(t) (-0.1).*(t.^3) + (0.58).*(t.^2).*(cosh((-1)./(t+1))) + exp((t)./(2.7)) + 9.6;
n = rand(10,1); %deze waardes moeten in de domain van f passen
upper_lim = 12;
lower_lim = 0;
y_max = 200;
N = 1000; %total number of points
N0 = 66; %number of points underneath the graph
y = (10/n).*(y_max.*(upper_lim - lower_lim))
monte_carlo = sum(y);
plot(n,y); hold on;
end
This gives me 2 straight lines but I am struggeling to translate this into the correct monte carlo formation. I have succeeded in creating a random graph of poitns which one would need but can't seem to get it coded into the function above, namely;
f = @(t)(-0.1).*(t.^3) + (0.58).*(t.^2).*(cosh((-1)./(t+1))) + exp((t)./(2.7)) + 9.6;
%yt = sort(rand(2, N)), 1, 'descend');
%t = 12*yt(1.:);
n = rand(10,1); %deze waardes moeten in de domain van f passen
N = 1000; %total number of points
t = 12*rand(1, N); %domain x = [0,12]
y = 200*rand(1, N); %y = [0,200]
plot(t,y,'.')
y_max = 200;
%N0 = ; %number of pointa underneed the graph
y1 = (n_max/n).*(y_max.*(max(t) - min(t)))
monte_carlo = sum(y1);
end
If someone can help or point me in the right direction it would be much appreciated!
Alan Stevens on 2 Dec 2020
Edited: Alan Stevens on 2 Dec 2020
Here's a simple MC calculation of the integration of your function
f = @(t) -0.1*t.^3 + 0.58*t.^2.*cosh(-1./(t+1)) + exp(t/2.7) + 9.6;
t = 0:0.1:12;
% Always a good idea to plot the function first
plot(t,f(t)),grid
%The function values lie below 20, so scatter random points
% with t values between 0 and 12, and y values between 0 and 20.
% The total rectangular area is
Arect = 12*20;
% The approximate area under the curve is given by
N = 10^4; % Number of MC trials
t = 12*rand(N,1); % Random values of t
y = 20*rand(N,1); % Random values of y
% Now find values of y that are less than or equal to the function.
ylo = y(y<=f(t));
% Approximate area
A = Arect*numel(ylo)/N;
disp(A)
(This could be made more efficient by noticing that the curve doesn't drop below 5 over the domain of interest.)
#### 1 Comment
Denise Veldhuis on 6 Dec 2020
Thank you so much! I changed some parts of it but I see what I was messing up, thank you!
R2020a
### Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!
Translated by | 864 | 2,552 | {"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-2021-04 | latest | en | 0.665782 |
https://cs.stackexchange.com/questions/444/ccs-process-for-a-drink-dispenser-with-two-different-prices/447 | 1,638,101,213,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964358520.50/warc/CC-MAIN-20211128103924-20211128133924-00477.warc.gz | 233,994,776 | 35,383 | # CCS process for a drink dispenser with two different prices
A drink dispenser requires the user to insert a coin ($\bar c$), then press one of three buttons: $\bar d_{\text{tea}}$ requests a cup of tea $e_{\text{tea}}$, ditto for coffee, and $\bar r$ requests a refund (i.e. the machine gives back the coin: $\bar b$). This dispenser can be modeled by the following CCS process:
$$M \stackrel{\mathrm{def}}= c.(d_{\text{tea}}.\bar e_{\text{tea}}.M + d_{\text{coffee}}.\bar e_{\text{coffee}}.M + r.\bar b.M)$$
A civil war raises the price of coffee to two coins, while the price of tea remains one coin. We want a modified machine that delivers coffee only after two coins, and acquiesces to a refund after either one or two coins. How can we model the modified machine with a CCS process?
• What is a CCS model/process? Are they equivalent to labeled transition systems (LTS)?
– Raphael
Mar 17 '12 at 10:55
• @Raphael CCS is a process calculus, a precursor of the pi calculus. A CCS model is just a model in CCS. I've added a Wikipedia link and a tag wiki. Mar 17 '12 at 13:24
• I think logic and programming-languages are appropriate for this question. Process algebras are studied in these areas, and for this question logic seems more appropriate one, e.g. please check the area tags here. Mar 26 '12 at 18:20
You can easily profit from warfare that way:
$$M \stackrel{\mathrm{def}} = c.( d_{\text{tea}}.\bar e_{\text{tea}}.M + r.\bar b.M + c.( d_{\text{coffee}}.\bar e_{\text{coffee}}.M + r.\bar b.\bar b.M ) )$$
note that you have to press refund to get a tea if you put too many coins. If you don't want that, you can adapt it (or maybe set up a (finite is enough) counter) :
$$M \stackrel{\mathrm{def}} = c.( d_{\text{tea}}.\bar e_{\text{tea}}.M + r.\bar b.M + c.( d_{\text{coffee}}.\bar e_{\text{coffee}}.M + d_{\text{tea}}.\bar b.\bar e_{\text{tea}}.M + r.\bar b.\bar b.M ) )$$
• I don't understand your answer. The first process you show has the price of coffee at one coin, and has the machine somehow cause the user to insert a coin. I don't see any connection with the question. The second process looks on the right track, but what's $\bar c$ supposed to do?? Mar 17 '12 at 1:11
• @Gilles: $\bar c$ gives back the money, but it would be better I you gave us another name to send back the money. Mar 17 '12 at 1:17
• @StéphaneGimenez You're right, I've added that. Mar 17 '12 at 1:20
• @Gilles and Stéphane: you are right, $\bar c$ is a very bad choice for the refund. (For example you could require the machine to be asynchronous: $r.(\bar c\mid M)$ and then the machine could take it itself so you'll need to be quick to catch your money!)
Mar 17 '12 at 1:32
• @Gilles: I chose $\bar b$ too, independently of you. I guess this is the canonical choice :-)
Mar 17 '12 at 1:39
This $M_0$ machine is more convenient than the one you propose:
$$M_0 := c.M_1$$
$$M_1 := d_{\text{tea}}.\bar e_{\text{tea}}.M_1 + r.\bar b.M_0 + c.M_2$$
$$M_{n} := d_{\text{tea}}.\bar e_{\text{tea}}.M_{n-1} + d_{\text{coffee}}.\bar e_{\text{coffee}}.M_{n-2} + r.\underbrace{\bar b.\dots\bar b.}_{n}M_0 + c.M_{n+1}$$
(But using infinite processes is like cheating).
• I like the compositional aspect here. However, I guess it is fine for the automaton to not allow more than two coins?
– Raphael
Mar 17 '12 at 10:59
• Well this also gives an idea of how to deal with coins that have different values :-) Mar 17 '12 at 13:21 | 1,067 | 3,429 | {"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} | 3.34375 | 3 | CC-MAIN-2021-49 | latest | en | 0.864518 |
https://software.sandia.gov/trac/mtgl/wiki/TrianglesAlgorithm?version=2 | 1,558,840,436,000,000,000 | text/html | crawl-data/CC-MAIN-2019-22/segments/1558232258621.77/warc/CC-MAIN-20190526025014-20190526051014-00433.warc.gz | 634,924,026 | 3,799 | Version 2 (modified by jberry, 9 years ago) (diff)
--
Algorithms in the MTGL typically accept a callback function or "visitor" from the user. This visitor will be called upon the occurrence of certain events, and the user will have an opportunity to react. It is common practice to include auxiliary data in the visitor object and update those data when events occur.
We will consider the example of finding triangles. There is MTGL code to do this: trunk/mtgl/triangles.hpp. This file contains the functor (algorithm object) "find_triangles" (along with some deprecated versions of triangles finding algorithms). Our example usage of find_triangles is found in the file trunk/tutorial/triangles1.cpp.
To compile the latter in the mtgl/tutorial directory:
``` c++ -I.. -o triangles1 triangles1.cpp
```
To run the program:
```./triangles1 < graphs/two_triangles.graph
Graph (g):
(0, 1)
(1, 2)
(2, 0)
(2, 3)
(3, 4)
(4, 2)
Graph (dg):
(0, 1)
(1, 2)
(2, 3)
(2, 0)
(3, 4)
(4, 2)
0 : { {0}(0, 1) }
1 : { {1}(1, 2) }
2 : { {2}(2, 0) {3}(2, 3) }
3 : { {4}(3, 4) }
4 : { {5}(4, 2) }
triangle eids: {5,4,3}
triangle eids: {0,1,2}
triangle: {(4,2), (3,4), (2,3)}
triangle: {(0,1), (1,2), (2,0)}
max b: 2
max d: 2
num tri: 2
0 : { {0}(0, 1) }
1 : { {1}(1, 2) }
2 : { {3}(2, 3) {2}(2, 0) }
3 : { {4}(3, 4) }
4 : { {5}(4, 2) }
triangle eids: {5,4,3}
triangle eids: {0,1,2}
triangle: {(4,2), (3,4), (2,3)}
triangle: {(0,1), (1,2), (2,0)}
max b: 2
max d: 2
num tri: 2
```
Triangle Visitor
Upon finding a triangles, the algorithm will extract the three edge id's for the edges in the triangle and pass them to a user-defined function. That function is the operator() method in a visitor object such as the one defined below. It is up to the user to determine what data (if any) to store in this object. The only requirement of the triangle algorithm API is that the operator() must take three edge id's as arguments. The example below simply extracts the id's of the endpoints of the triangle edges and prints them.
```template <class graph>
class triangle_printing_visitor {
public:
typedef typename graph_traits<graph>::size_type size_type;
typedef typename graph_traits<graph>::vertex_descriptor vertex_descriptor;
typedef typename graph_traits<graph>::edge_descriptor edge_descriptor;
typedef typename graph_traits<graph>::edge_iterator edge_iterator;
triangle_printing_visitor(graph& gg) : g(gg), edgs(edges(g)),
vid_map(get(_vertex_id_map, g)) {}
void operator()(size_type eid1, size_type eid2, size_type eid3) {
edge_descriptor e1 = edgs[eid1];
edge_descriptor e2 = edgs[eid2];
edge_descriptor e3 = edgs[eid3];
printf("triangle eids: {%lu,%lu,%lu}\n", eid1, eid2, eid3);
vertex_descriptor e1src = source(e1, g);
vertex_descriptor e1trg = target(e1, g);
vertex_descriptor e2src = source(e2, g);
vertex_descriptor e2trg = target(e2, g);
vertex_descriptor e3src = source(e3, g);
vertex_descriptor e3trg = target(e3, g);
size_type e1srcid = get(vid_map, e1src);
size_type e1trgid = get(vid_map, e1trg);
size_type e2srcid = get(vid_map, e2src);
size_type e2trgid = get(vid_map, e2trg);
size_type e3srcid = get(vid_map, e3src);
size_type e3trgid = get(vid_map, e3trg);
printf("triangle: {(%lu,%lu), (%lu,%lu), (%lu,%lu)}\n",
e1srcid, e1trgid, e2srcid, e2trgid, e3srcid, e3trgid);
}
private:
graph& g;
edge_iterator edgs;
vertex_id_map<graph> vid_map;
};
``` | 1,092 | 3,356 | {"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-2019-22 | latest | en | 0.80727 |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.