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https://artofproblemsolving.com/wiki/index.php?title=2005_AIME_I_Problems/Problem_4&diff=prev&oldid=55189 | 1,670,214,984,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446711003.56/warc/CC-MAIN-20221205032447-20221205062447-00010.warc.gz | 138,848,655 | 13,035 | # Difference between revisions of "2005 AIME I Problems/Problem 4"
## Problem
The director of a marching band wishes to place the members into a formation that includes all of them and has no unfilled positions. If they are arranged in a square formation, there are $5$ members left over. The director realizes that if he arranges the group in a formation with $7$ more rows than columns, there are no members left over. Find the maximum number of members this band can have.
## Solution
### Solution 1
If $n > 14$ then $n^2 + 6n + 14 < n^2 + 7n < n^2 + 8n + 21$ and so $(n + 3)^2 + 5 < n(n + 7) < (n + 4)^2 + 5$. If $n$ is an integer there are no numbers which are 5 more than a perfect square strictly between $(n + 3)^2 + 5$ and $(n + 4)^2 + 5$. Thus, if the number of columns is $n$, the number of students is $n(n + 7)$ which must be 5 more than a perfect square, so $n \leq 14$. In fact, when $n = 14$ we have $n(n + 7) = 14\cdot 21 = 294 = 17^2 + 5$, so this number works and no larger number can. Thus, the answer is $\boxed{294}$.
### Solution 2
Define the number of rows/columns of the square formation as $s$, and the number of rows of the rectangular formation $r$ (so there are $r - 7$ columns). Thus, $s^2 + 5 = r(r-7) \Longrightarrow r^2 - 7r - s^2 - 5 = 0$. The quadratic formula yields $r = \frac{7 \pm \sqrt{49 - 4(1)(-s^2 - 5)}}{2} = \frac{7 \pm \sqrt{4s^2 + 69}}{2}$. $\sqrt{4s^2 + 69}$ must be an integer, say $x$. Then $4s^2 + 69 = x^2$ and $(x + 2s)(x - 2s) = 69$. The factors of $69$ are $(1,69), (3,23)$; $x$ is maximized for the first case. Thus, $x = \frac{69 + 1}{2} = 35$, and $r = \frac{7 \pm 35}{2} = 21, -14$. The latter obviously can be discarded, so there are $21$ rows and $21 - 7 = 14$ columns, making the answer $294$.
### Solution 3
The number of members is $m^2+5=n(n+7)$ for some $n$ and $m$. Multiply both sides by $4$ and complete the square to get $4m^2+69=(2n+7)^2$. Thus, we have $69=((2n+7)+2m)((2n+7)-2m)$. Since we want to maximize $n$, set the first factor equal to $69$ and the second equal to $1$. Solving gives $n=14$, so the answer is $14\cdot21=294$.
## See also
2005 AIME I (Problems • Answer Key • Resources) Preceded byProblem 3 Followed byProblem 5 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 All AIME Problems and Solutions
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.
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theb-line.blogspot.com | 1,440,981,101,000,000,000 | text/html | crawl-data/CC-MAIN-2015-35/segments/1440644065464.19/warc/CC-MAIN-20150827025425-00052-ip-10-171-96-226.ec2.internal.warc.gz | 231,543,747 | 25,679 | ### let's make {countdown garlands}
I think I may have mentioned once or twice that I'm a bit of a counter. I count steps. When I take a jog, I'm always calculating what percentage I've got left to go. And if there's a big event on the horizon (trip, birthday, deadline), you'd better believe I'm counting down the days.
All of this brings us to today's project, countdown garlands. The idea struck me a few weeks back that as long as I'm counting down (a habit that can be annoying or irrelevant to some), I may as well decorate around it. I figured maybe I'm not the only one out there who likes to keep track of time, so I thought I'd share it with you. Enjoy!
let's make {countdown garlands}
This is such a simple project. For materials, you'll need
3 sheets of cardstock or photo paper
color printer
scissors
tiny (1") clothespins (one for each day you're counting down)
3-5 feet of twine or ribbon
this pdf freebie*
Of course, you can make your own numbers and photos. But if you'd like to use the ones I've shown here (taken by yours truly), just click on the image above (or here) to open a pdf file with 21 images. That's three weeks of counting. :)
Directions:
Open the pdf file and print the images on 3 pieces of cardstock or photopaper. Cut out each of the images you'll be using for your countdown. Each day has a number and a photograph. Here's number one. (Yep, the photo's upside down! You'll see why in the next step.)
Once you've cut out all of the images you'll be using, fold each one down the center so you have a 2x2" square for each one. The number will be on one side, the photo on the other. The photo and number will both be right side up now.
Find a place you'd like to hang your countdown. On or above a mantle looks great, but this would also be pretty on a door (or door frame) or a window. Or just a wall with a lot of space. Hang your twine or ribbon however you like--tiny nails or brads, or even heavy duty tape or that blue stuff we used to use to hang posters. Now, using the mini clothespins (I found mine at JoAnn's), post each of the cards to your twine with the numbered side facing you.
I told you it was simple! The project is done. Now it's just a matter of counting down. Here's a 21-day countdown at the beginning. Look at all those numbers (Well, imagine them anyway. Sorry for the lack of contrast.).
Slowly but surely. Five days into our countdown (just 16 to go!). See, we're flipping to the photo side of the card for each day that passes.
Can you feel the excitement? Just five days to go!
Et voilà! Countdown is finished. Wanna start again?
What's got you counting down? Wedding, arrival of a new baby, guests coming to visit, vacation of a lifetime, kids going back to school, a big birthday or anniversary? Share your excitement!
Heidi Jo
What a great idea! I love countdown garland, but love how instead of taking a piece away each day, you get a pretty photo to look at! Bravo! : )
mchen
Lovely way to count down to a special event... Great pictures, terrific idea! Thanks a lot — one more thing on my wedding planning to-do list, ha ha! ;)
Pink Trees and Sunshine
This is an adorable idea~~so inventive! It's an everyday Advent Calender :~)
Tintel
These are beautiful, Amy!
Thanks for sharing with us.
PS~Erin
What a fun (and beautiful) idea! Thanks!
Amy
That's a great idea!
leslie
Oh -this is such a pretty idea -I too love counting down!!! Thank you!
Hi there! I'm so glad you visited my blog today. And look, you made it all the way down here! While you're here, leave me a note. And have a lovely day.
Amy
## more photos
www.flickr.com
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# <DATA STRUCTURE><CBDS2103>
## < DATA STRUCTURE >
MATRICULATION NO. IDENTITY CARD NO. TELEPHONE NO. E-MAIL LEARNING CENTRE
: : : : :
## <861014025706001> < 861014-02-5706> <016-5539566> gayu_chopra@yahoo.com PETALING JAYA LEARNING CENTRE
NO TITLE 1 Question 1 1.0 Problem 1.1 Input 1.2 Calculation 1.3 Output 1.4 Programming (coding) 1.5 Answer 2 Question 2 2.0 Problem 2.1 Input 2.2 Output 2.3 Programming (coding) 2.4 Answer
## NO PAGE 3 3 3 3 4-6 7 8 8 8 9-11 12
<DATA STRUCTURE><CBDS2103> QUESTION 1 1.0 PROBLEM Required to calculate the average and total of sales for each a year, each of staff and also for a quarter. 1.1 INPUT Average of the staff, quarter and a year. Total of a year, staff and quarter. The overall average of sales and total of sales for a year. CALCULATION Total Quarter (total=total+sales) Average Quarter (avgtotal=total/6) Total sales a year(sum=sum+sales) Average sales a year (yearavg=sum/4) 1.3 OUTPUT The total of sales of the year. The average of sales for the year.
1.2
<DATA STRUCTURE><CBDS2103>
1.4
PROGRAMMING (CODING) #include <string.h> #include <conio.h> void main() { int i, j, sum=0,total2,total,total1; float yearavg=0, avgtotal2=0, avgtotal=0; int sales [6][5]= {{100802, 5500, 7000, 5500, 6800}, {100888, 4800, 6700, 7000, 6500}, {100188, 6890, 2000, 3500, 2500}, {100189, 4000, 5000, 5500, 6000}, {100800, 1210, 4000, 8000, 8000}, {100122, 8000, 12000, 14000, 10000}};
printf("****************************************************************** **************\n"); printf(" * * * * * * * \n"); printf("StaffID * Quarter1 * Quarter2 * Quarter3 * Quarter4 * Total * Average * \n"); printf(" * * * * * * * \n"); printf("****************************************************************** **************\n"); for(i=0;i<6;++i) { printf("\n %d",sales[i][0]);
## total=0; for(j=1;j<5;j++) { printf("%10d",sales[i][j]); total=total+sales[i][j];
4
<DATA STRUCTURE><CBDS2103> }
%d",i,total2);
## for(i=0;i<6;i++){ for(j=1;j<5;j++){ sum=sum+sales[i][j];
5
<DATA STRUCTURE><CBDS2103> yearavg= (float)sum/4; } } printf("\n\n\n *** THE TOTAL SALES OF THE YEAR IS *** : %d", sum); printf("\n\n *** THE AVERAGE SALES FOR THE YEAR IS *** : %0.2f", yearavg); getch(); }
<DATA STRUCTURE><CBDS2103> QUESTION 2 2.0 PROBLEM Consider the problem of recognizing whether a particular string is in the language, and the programs have to determining whether a give string is in L. L={w\$w : w is a possibly empty string of characters other than \$,w=reverse(w)}.
2.1
INPUT User will enter the string. OUTPUT Result will displayed once entered the string. Example : Output 1: Enter a string : ABC\$CBA Result : Inserted string is in the language Output 2: Enter a string : ABCD\$CB Result : Inserted string is not in the language
2.2
<DATA STRUCTURE><CBDS2103> 2.3 PROGRAMMING (CODING) #include <stdio.h> #include <stdlib.h> #include <conio.h> #define SIZE 10 typedef struct { int items[SIZE]; int top; }STACK;
void push(STACK *p, int element); int pop(STACK *p); void display(STACK s); int isoverflow(int top); int isempty(int top); int main() { STACK s; char str[100]; int i, frame; s.top = -1; printf("\nEnter a string: "); gets(str); for(i=0;str[i]!='\0'; i++) push(&s, str[i]); frame = 1; for(i=0;str[i]!='\$';i++) { if(str[i] != pop(&s)) { frame = 0;
9
<DATA STRUCTURE><CBDS2103> break; } } if(frame == 1) printf("\nRESULT: *** STRING IS IN THE LANGUAGE ***\n"); else printf("\nRESULT: *** STRING IS NOT IN THE LANGUAGE ***\n"); getch(); }
void push(STACK *p, int element) { if(isoverflow(p->top)) { printf("\nStack is overflow"); } else { (p->top)++; p->items[p->top] = element; } } int pop(STACK *p) { if(isempty(p->top)) { printf("\nStack is underflow"); } else { return (p->items[(p->top)--]); } } void display(STACK s) {
10
<DATA STRUCTURE><CBDS2103> int i; if(isempty(s.top)) { printf("\nStack is empty"); } else { for(i=s.top; i>=0; i--) { printf("\n%d", s.items[i]); } } }
int isoverflow(int top) { if(top == SIZE - 1) return (1); else return (0); }
int isempty(int top) { if(top == -1) return (1); else return (0); }
11
ii.
12 | 1,362 | 4,117 | {"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} | 3.59375 | 4 | CC-MAIN-2019-39 | latest | en | 0.489259 |
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# What percent is equal to one-twenty four?
Updated: 9/17/2023
Wiki User
β 8y ago
1/24 is approximately equal to 4.17%
Wiki User
β 8y ago
Wiki User
β 8y ago
124 = 12400%.
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Q: What percent is equal to one-twenty four?
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133.3%
20%
0.024 = 2.4%
### What is the 40 percent as a decimal and a fraction?
Written as a decimal fraction, 40 percent is equal to 0.4.Written as a vulgar fraction, 40 percent is equal to 4/10, which can be read either as "four over ten" or "four tenths".
4/100, or 1/25.
### Which is greater four fifths or 60 percent?
60% is equal to three fifths, so four fifths is greater
### How is four fifths equal to 0.8 and 80 percent?
The fraction 4/5 is equal to the fractions 8/10 and 80/100. 8/10 is the decimal 0.8 and 80/100 is equal to 80 percent.
### What is 4 percent of 62?
To find four percent of any number, simply divide it by twenty-five. For this number, sixty-two, you would divide 62/25 to equal 2.48 as your answer, or four percent of sixty-two.
### What is four sixths as a percentage?
Four sixths is equal to two thirds, which is equal to 66.6 recurring (that is 66.6666...) percent.4/6 reduces to 2/3 and 2/3 = 66.66666666666666%
### What is the fractional notation for 2.75 percent?
Expressed as a proper fraction in its simplest form, 2.75 percent is equal to 11/400 or eleven four-hundredths.
### What is 7.25 percent as a fraction?
Expressed as a proper fraction in its simplest form, 7.25 percent is equal to 29/400 or twenty-nine four hundredths.
### What is 11.75 percent as a fraction?
Expressed as a proper fraction in its simplest form, 11.75 percent is equal to 47/400 or forty-seven four hundredths. | 526 | 1,742 | {"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 | 0.9509 |
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# Linear Regression Beta Standard Error
## Contents
You may wonder whether it is valid to take the long-run view here: e.g., if I calculate 95% confidence intervals for "enough different things" from the same data, can I expect Privacy policy About Wikipedia Disclaimers Contact Wikipedia Developers Cookie statement Mobile view menuMinitab® 17 SupportWhat is the standard error of the coefficient?Learn more about Minitab 17 The standard deviation of the estimate of The Regression Sum of Squares is the difference between the Total Sum of Squares and the Residual Sum of Squares. Other regression methods besides the simple ordinary least squares (OLS) also exist. have a peek here
r regression standard-error lm share|improve this question edited Aug 2 '13 at 15:20 gung 74.2k19160309 asked Dec 1 '12 at 10:16 ako 383146 good question, many people know the Adjusted-R² will be described during the discussion of multiple regression. You could not use all four of these and a constant in the same model, since Q1+Q2+Q3+Q4 = 1 1 1 1 1 1 1 1 . . . . , But still a question: in my post, the standard error has (n−2), where according to your answer, it doesn't, why? https://en.wikipedia.org/wiki/Simple_linear_regression
## Standard Error Of Coefficient In Linear Regression
Got it? (Return to top of page.) Interpreting STANDARD ERRORS, t-STATISTICS, AND SIGNIFICANCE LEVELS OF COEFFICIENTS Your regression output not only gives point estimates of the coefficients of the variables in This is another issue that depends on the correctness of the model and the representativeness of the data set, particularly in the case of time series data. The column labeled Sum of Squares describes the variability in the response variable, Y. The heights were originally given in inches, and have been converted to the nearest centimetre.
Join them; it only takes a minute: Sign up Here's how it works: Anybody can ask a question Anybody can answer The best answers are voted up and rise to the Confidence intervals The formulas given in the previous section allow one to calculate the point estimates of α and β — that is, the coefficients of the regression line for the Referee did not fully understand accepted paper What to do when you've put your co-worker on spot by being impatient? Standard Error Of Beta Coefficient Formula Retrieved 2016-10-17. ^ Seltman, Howard J. (2008-09-08).
Use the standard error of the coefficient to measure the precision of the estimate of the coefficient. The F statistic, also known as the F ratio, will be described in detail during the discussion of multiple regression. The standard errors of the coefficients are in the third column. http://stats.stackexchange.com/questions/85943/how-to-derive-the-standard-error-of-linear-regression-coefficient Hence, as a rough rule of thumb, a t-statistic larger than 2 in absolute value would have a 5% or smaller probability of occurring by chance if the true coefficient were
By using this site, you agree to the Terms of Use and Privacy Policy. Standard Error Of Regression Coefficient Excel labels the two-sided P values or observed significance levels for the t statistics. Total df is n-1, one less than the number of observations. This occurs because it is more natural for one's mind to consider the orthogonal distances from the observations to the regression line, rather than the vertical ones as OLS method does.
## Standard Error Of Coefficient Multiple Regression
A 95% confidence interval for the regression coefficient for STRENGTH is constructed as (3.016 k 0.219), where k is the appropriate percentile of the t distribution with degrees of freedom equal http://support.minitab.com/en-us/minitab/17/topic-library/modeling-statistics/regression-and-correlation/regression-models/what-is-the-standard-error-of-the-coefficient/ That is, we are 99% confident that the true slope of the regression line is in the range defined by 0.55 + 0.63. Standard Error Of Coefficient In Linear Regression For example: x y ¯ = 1 n ∑ i = 1 n x i y i . {\displaystyle {\overline ∑ 2}={\frac ∑ 1 ∑ 0}\sum _ − 9^ − 8x_ Standard Error Of Beta p.462. ^ Kenney, J.
Some call R² the proportion of the variance explained by the model. navigate here Hence, if the sum of squared errors is to be minimized, the constant must be chosen such that the mean of the errors is zero.) In a simple regression model, the See the mathematics-of-ARIMA-models notes for more discussion of unit roots.) Many statistical analysis programs report variance inflation factors (VIF's), which are another measure of multicollinearity, in addition to or instead of codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1 Residual standard error: 13.55 on 159 degrees of freedom Multiple R-squared: 0.6344, Adjusted R-squared: 0.6252 F-statistic: 68.98 on What Does Standard Error Of Coefficient Mean
The total amount of variability in the response is the Total Sum of Squares, . (The row labeled Total is sometimes labeled Corrected Total, where corrected refers to subtracting the sample In case (ii), it may be possible to replace the two variables by the appropriate linear function (e.g., their sum or difference) if you can identify it, but this is not Therefore, the variances of these two components of error in each prediction are additive. Check This Out Normality assumption Under the first assumption above, that of the normality of the error terms, the estimator of the slope coefficient will itself be normally distributed with mean β and variance
Another thing to be aware of in regard to missing values is that automated model selection methods such as stepwise regression base their calculations on a covariance matrix computed in advance Standard Error Of Regression Coefficient Definition In the most extreme cases of multicollinearity--e.g., when one of the independent variables is an exact linear combination of some of the others--the regression calculation will fail, and you will need Would not allowing my vehicle to downshift uphill be fuel efficient?
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And, if (i) your data set is sufficiently large, and your model passes the diagnostic tests concerning the "4 assumptions of regression analysis," and (ii) you don't have strong prior feelings However, the standard error of the regression is typically much larger than the standard errors of the means at most points, hence the standard deviations of the predictions will often not Princeton, NJ: Van Nostrand, pp. 252–285 External links Wolfram MathWorld's explanation of Least Squares Fitting, and how to calculate it Mathematics of simple regression (Robert Nau, Duke University) v t e Standard Error Of Regression Coefficient Calculator Retrieved 2016-10-17.
It is sometimes useful to calculate rxy from the data independently using this equation: r x y = x y ¯ − x ¯ y ¯ ( x 2 ¯ − It is also possible to evaluate the properties under other assumptions, such as inhomogeneity, but this is discussed elsewhere.[clarification needed] Unbiasedness The estimators α ^ {\displaystyle {\hat {\alpha }}} and β For example: x y ¯ = 1 n ∑ i = 1 n x i y i . {\displaystyle {\overline ∑ 2}={\frac ∑ 1 ∑ 0}\sum _ − 9^ − 8x_ this contact form temperature What to look for in regression output What's a good value for R-squared?
The sample statistic is the regression slope b1 calculated from sample data. Who is the highest-grossing debut director? Simple linear regression From Wikipedia, the free encyclopedia Jump to: navigation, search This article includes a list of references, but its sources remain unclear because it has insufficient inline citations. Occasionally the fraction 1/n−2 is replaced with 1/n.
In practice, we do not usually do that. If you are regressing the first difference of Y on the first difference of X, you are directly predicting changes in Y as a linear function of changes in X, without For example, if γ = 0.05 then the confidence level is 95%. Please try the request again.
This t-statistic has a Student's t-distribution with n − 2 degrees of freedom. Coefficients Term Coef SE Coef T-Value P-Value VIF Constant 20.1 12.2 1.65 0.111 Stiffness 0.2385 0.0197 12.13 0.000 1.00 Temp -0.184 0.178 -1.03 0.311 1.00 The standard error of the Stiffness Why doesn't compiler report missing semicolon? Find the margin of error. | 1,951 | 8,357 | {"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} | 3.21875 | 3 | CC-MAIN-2017-22 | longest | en | 0.84547 |
https://se.mathworks.com/matlabcentral/cody/problems/16-return-the-largest-number-that-is-adjacent-to-a-zero/solutions/595124 | 1,606,231,757,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141176864.5/warc/CC-MAIN-20201124140942-20201124170942-00574.warc.gz | 479,818,099 | 17,113 | Cody
# Problem 16. Return the largest number that is adjacent to a zero
Solution 595124
Submitted on 11 Mar 2015 by Elizabeth
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
%% a = [1, 5, 3, 0, 2, 7, 0, 8, 9, 1 0]; b = 8; assert(isequal(nearZero(a),b))
2 Pass
%% a = [5 4 -1 0 -2 0 -5 8]; b = -1; assert(isequal(nearZero(a),b));
3 Pass
%% a = [0 3 1 0 2 9]; b = 3; assert(isequal(nearZero(a),b));
4 Pass
%% a = [1 0 2 0 3]; b = 3; assert(isequal(nearZero(a),b));
5 Pass
%% a = [0 -1]; b = -1; assert(isequal(nearZero(a),b));
6 Pass
%% a = [0 -12 0 -7 0]; b = -7; assert(isequal(nearZero(a),b));
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Start Hunting! | 320 | 860 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.296875 | 3 | CC-MAIN-2020-50 | latest | en | 0.739083 |
https://www.physicsforums.com/threads/can-this-neutrino-be-detected-via-a-cc-weak-interaction.913194/ | 1,508,250,722,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187821189.10/warc/CC-MAIN-20171017125144-20171017145144-00285.warc.gz | 966,434,374 | 16,511 | # Can this neutrino be detected via a CC weak interaction?
1. May 1, 2017
### Kara386
1. The problem statement, all variables and given/known data
A tau neutrino with energy 1GeV interacts with a stationary neutron. Can the neutrino be detected via a charged current interaction?
Take the mass of tau to be 1784MeV/c$^2$, 105MeV/c$^2$ for the muon and 939MeV/$c^2$ for the neutron.
2. Relevant equations
3. The attempt at a solution
I've got no idea what the criteria for detection via a weak interaction are, and it proved very difficult to google. Is there an energy threshold the neutrino has to reach to be detectable? As in should I be saying something like: it doesn't have 83GeV of energy, i.e. the equivalent of the mass of a W boson, so it can't create one? I know tha's wrong, the boson is virtual so doesn't have to have that mass - do I use the uncertainty principle to work out what the energy of the virtual boson would be?
I feel like I've missed some really important concept here, any help would be much appreciated. :)
2. May 1, 2017
### Orodruin
Staff Emeritus
Hint: There is an energy threshold for the CC interaction to be possible.
3. May 1, 2017
### Staff: Mentor
Don't even think about the W here. Just consider the initial versus final particles.
A virtual particle does not have a well-defined mass. In this case, the "mass" of the W can be a lot different from the mass of a real (non-virtual) W. | 373 | 1,437 | {"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} | 2.53125 | 3 | CC-MAIN-2017-43 | longest | en | 0.925679 |
https://boardgames.stackexchange.com/questions/53730/odds-of-this-happening-in-euchre/53732 | 1,702,161,728,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100972.58/warc/CC-MAIN-20231209202131-20231209232131-00333.warc.gz | 165,122,190 | 37,457 | # Odds of this happening in euchre
Last night in euchre, 4 times in a row around the table we all turned up the Jack of spades. All good shufflers and we had 2 splits and 2 no splits. I know there are many variances in determining the odds of a given situation in a euchre game, but, is there anyone whiz out there that can break it down? What are the odds of 4 different dealers flipping up the same card consecutively?
• Wikipedia say "[Euchre] is played with a deck of 24, 28, or 32 standard playing cards" You'll need to say how many cards you were playing with and if the deck included jokers. Dec 27, 2020 at 20:58
Assuming a Euchre deck with 24 cards (9, 10, J, Q, K, A of each suit, no joker) and perfect shuffling, the chance that the 21st card in the deck (the card that gets turned up) is a particular card is 1/24. Thus, if you play 4 deals, there is a (1/24)^4 = 1/331776 = 0.0003% that all four deals will turn up the Jack of Spades.
There's much higher chance of all four deals turning up the same card: (1/24)^3 = 1/13824 = 0.007%. That's because there's a 100% chance that the first deal turns up some card, and each subsequent deal has a 1/24 chance of turning up the same card as the previous deal.
If you play N deals over the course of the night, the probability that at some point all four deals turn over the same card consecutively is (N - 4) * (1/24)^3 (the sequence can start on any but the last 3 deals, and the probability of the sequence is (1/24)^3). This works out to about 0.1% if you play 20 hands and 0.7% if you play 100.
The odds of turning up a given card that occurs once in the deck are 1 in N, where N is the size of the deck. Since you say the shuffles were all good, this means the trials are all independent. So the odds of it happening four times are
(1/N)^4
If N is 24, this comes out to 1 in 331776.
If N is 32, this comes out to 1 in 1048576.
If you don't restrict it to the Jack of spades, then, obviously, you're reducing the number of trials by one: the first draw sets the card and is not a trial itself. So for drawing 4 of the same card, whatever the card, you use (1/N)^3 to calculate the odds of the subsequent three draws matching the first one.
• Worth pointing out that with a standard deck of cards (N=52), the answer is 1 in 7,311,616...not as unlikely as you might think! Across America, say every family carefully shuffles a deck of cards then selects a card at random. They write the card value down eg. `9D`. Repeat this another 3 times and compare the card values. Dozens of families will see the "miracle" of all four cards matching. Dec 28, 2020 at 15:55
• As the other answer points out, this is correct ONLY iof yiu specify the card in advance. If the question is the chanve of getting the same card four times i a row from a deck mof N cards it is (1/N)^3 not ^4, which makes a significant difference. Dec 28, 2020 at 22:18
• Yes. The explanation of the odds can be used in many different cases, sure. Dec 29, 2020 at 13:39 | 838 | 3,004 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.375 | 4 | CC-MAIN-2023-50 | latest | en | 0.965458 |
https://math.stackexchange.com/questions/2322830/double-barrier-stopping-time-density-function | 1,558,810,650,000,000,000 | text/html | crawl-data/CC-MAIN-2019-22/segments/1558232258439.57/warc/CC-MAIN-20190525184948-20190525210948-00306.warc.gz | 575,381,051 | 32,355 | # double barrier stopping time density function
We define a Brownian motion $W$, and two stopping times as follow :
$$\tau_a=\inf(t \ge 0 | W_t>a)$$ $$\tau_b=\inf(t \ge 0 | W_t<-b)$$
where $a,b >0$
We can define another stopping time as follow $$\tau=\min(\tau_a,\tau_b)$$
While the density functions of $\tau_a$ and $\tau_b$ are known (by using the Brownian motion reflection principal), how about $\tau$ ?
This is best done by first solving for $u(x,t)=\mathbb{P}_x[\tau>t]$ where $\mathbb{P}_x$ means the probability measure if start at $x$ at time $0$. One can show (or read in a book) that $u$ solves the PDE $\frac{1}{2}\frac{d^2}{dx^2}u=\frac{d}{dt}u$ (in general $Lu=\frac{d}{dt}u$ for a process with generator $L$). The boundary conditions are easier to see, $u(0,x)=1$ for all $x\in(a,b)$ and $u(a,t)=u(b,t)=0$ for $t>0$. The solution can be given as the Fourier series $u(x,t)=\sum_{n=1}^\infty\frac{2}{n\pi}(1-(-1)^n)\sin\left(\frac{n\pi(x-a)}{b-a}\right)e^{-\frac{n^2\pi^2t}{2(b-a)^2}}$. Then your density is $-\frac{d}{dt}u(x,t)=\sum_{n=1}^\infty\frac{n\pi}{(b-a)^2}(1-(-1)^n)\sin\left(\frac{n\pi(x-a)}{b-a}\right)e^{-\frac{n^2\pi^2t}{2(b-a)^2}}$. You can't do much better than this but it could be useful for simulations.
Let $f_{\tau_a}(t)$ and $f_{\tau_b}(t)$ be the density functions of $\tau_a$ and $\tau_b$. Let $F_{\tau_a}(t)$ and $F_{\tau_b}(t)$ be the corresponding CDFs, then, assuming that $\tau_a$ and $\tau_b$ are independent, we get,
$$P(\tau > t) = P(\tau_a > t, \tau_b > t) = P(\tau_a>t)P(\tau_b>t) = (1-F_{\tau_a}(t))(1-F_{\tau_b}(t))$$
$$\implies F_{\tau}(t) = 1-(1-F_{\tau_a}(t))(1-F_{\tau_b}(t))$$
$$\implies f_{\tau}(t) = f_{\tau_a}(t)(1-F_{\tau_b}(t)) +f_{\tau_b}(t)(1-F_{\tau_a}(t))$$
The $3$rd equation follows by taking derivative of $2$nd equation with respect to $t$.
• With the independence assumption , of course it makes things easier – Canardini Jun 15 '17 at 13:00 | 724 | 1,922 | {"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} | 3.84375 | 4 | CC-MAIN-2019-22 | latest | en | 0.735264 |
https://www.physicsforums.com/threads/expectation-of-a-random-variable.510314/ | 1,511,037,039,000,000,000 | text/html | crawl-data/CC-MAIN-2017-47/segments/1510934805023.14/warc/CC-MAIN-20171118190229-20171118210229-00036.warc.gz | 833,439,944 | 19,329 | # Expectation of a random variable
1. Jun 28, 2011
### Maybe_Memorie
1. The problem statement, all variables and given/known data
I'm wondering how I go about calculating the expectation of a random variable?
Is it a different process for a discrete and a continuous?
Can you show me an example? Say Poisson and expoential?
Also, in the formula
Var(X) = E[X^2] - (E[X])^2
how does one calculate E[X^2] ?
Thanks!
2. Jun 29, 2011
### I like Serena
Hi Maybe_Memorie!
To calculate an expectation you need to multiply the chance of each outcome with the value of each outcome and add everything up.
The difference between discrete and continuous is that the first uses a summation while the second uses an integral.
So:
$$E[X] = \sum P(x) \cdot x$$
and
$$E[X^2] = \sum P(x) \cdot x^2$$
So for a six sided dice you would get:
$$E[X] = \sum_{k=1}^6 P(k) \cdot k = \frac 1 6 \cdot 1 + \frac 1 6 \cdot 2 + \frac 1 6 \cdot 3 + \frac 1 6 \cdot 4 + \frac 1 6 \cdot 5 + \frac 1 6 \cdot 6 = 3 \frac 1 2$$
3. Jul 2, 2011
### I like Serena
I see you didn't respond.
Do you already know what you wanted to know?
Is perhaps the strange wiggly symbol (sigma) putting you off?
Or is there something else that you don't understand but don't want to ask about?
4. Jul 3, 2011
### Maybe_Memorie
I think I understand.
Suppose I wanted the expectation of the exponential random variable.
f(x) = the probability density function.
A = lambda
E[X] = integral of xf(x) dx
E[X] = integral of (xAe^(-Ax)) dx between + and - infinity.
Do I have to use integration by parts? In none of my classes have I come across integration with infinity as a limit
Last edited: Jul 3, 2011
5. Jul 3, 2011
### I like Serena
Ah, there you are! :)
You certainly seem to understand.
Your boundaries are slightly off though.
The exponential distribution only has non-zero values for positive x, so the bottom boundary should be zero (and not - infinity).
And yes, you have to use integration by parts.
So you are already aware of that. Good.
As for infinity as a boundary, you have not come across it ... yet!
It means you have to integrate to an arbitrary boundary, and then take the limit of that boundary to infinity.
So in your case, you'll get:
$$\begin{eqnarray} E[X] &=& \int_0^{\infty} \lambda e^{-\lambda x} \cdot x dx \\ &=& \left[ - e^{-\lambda x} \cdot x \right]_0^{\infty} - \int_0^{\infty} - e^{-\lambda x} \cdot 1 dx \\ &=& (\lim_{x \to \infty} - e^{-\lambda x} \cdot x) - (- e^{-\lambda \cdot 0} \cdot 0) - \int_0^{\infty} - e^{-\lambda x} \cdot 1 dx \\ &=& 0 - 0 - \int_0^{\infty} - e^{-\lambda x} \cdot 1 dx \\ &=& ... \\ &=& \frac 1 {\lambda} \end{eqnarray}$$
6. Jul 3, 2011
### Maybe_Memorie
Ah! Thank you very much!
I have another question. A related application.
The system S is composed of two components, C1 and C2, in parallel (S works provided at least 1 of the components works). The two components have lifetimes T1 and T2 and are exponentially distributed with the same parameter A=0.01
(i) What is the probability that T1 is greater than 100?
P(x>100) = 1 - e^(-100(0.01))
(ii) What is the probability that Ts, the lifetime of the system, is greater than 100?
Do I just add the probabilities that T1 and T2 are greater than 100?
(iii) What is the probability density function of Ts?
I'm lost here
(iv) Derive the expectation of Ts.
When I find the PDF, multiply by x and integrate between 0 and infinity with respect to x.
7. Jul 3, 2011
### I like Serena
You're welcome!
You have calculated P(x<100)....
Do you know the rule about the probability of independent events occurring simultaneously?
That is P(A and B) where events A and B are independent?
For that you first need to answer (ii), which would show what the cumulative probability function of Ts is.
Yep! :)
(But perhaps it will turn out to be easier. )
8. Jul 3, 2011
### Maybe_Memorie
Isn't it P(A and B) = P(A).P(B) ?
9. Jul 3, 2011
### I like Serena
Yes! So you should not add up the probabilities for T1 and T2, but multiply them!
10. Jul 3, 2011
### Maybe_Memorie
Thanks!
I still need to find the PDF.
Is it the same as multiplying the PDF of T1 and T2?
11. Jul 3, 2011
### I like Serena
No, to find the PDF you need to find the CDF first, and deduce from that the PDF.
Actually, the PDF is the derivative of the CDF.
For reference did you properly solve (i) and (ii)?
Because you need them, and you'd be making it easier for me if I know that you got those and understood those.
12. Jul 3, 2011
### Maybe_Memorie
For (i), the probability is 1/e, whatever that works out at...
For (ii), I'm slightly confused. The system doesn't need T1 and T2 to be greater than 100 for Ts to be > 100. It needs only one of them to be greater than 100.
So I'm not sure why I'm using P(A and B)
13. Jul 3, 2011
### I like Serena
Good!
And more in general: [itex]P(X > x) = e^{-\lambda x}[/tex]
Yes, only one needs to be greater than 100.
But this means there are 3 cases that overlap: either T1 can be greater than 100, or T2 can be greater than 100, or both can be greater than 100.
I guess I was a little sloppy with this before.
A little sharper is, that they cannot both be less than 100.
Do you know how to calculate the corresponding probability?
14. Jul 3, 2011
### Maybe_Memorie
Is it P(A) + P(B) + P(A and B)
where A = T1 > 100
B = T2 > 100
15. Jul 3, 2011
### I like Serena
Almost. ;)
The proper formula is (sum rule): P(A or B) = P(A) + P(B) - P(A and B)
16. Jul 3, 2011
### Maybe_Memorie
So the probability is (2e-1)/e^2
Is this correct?
How does this relate to the cdf?
17. Jul 3, 2011
### I like Serena
Yes, this is correct.
You just found the probability for Ts to be greater than 100.
Can you generalize for Ts greater than some x?
And then the cdf(x) for Ts is the probability that Ts is less than x.
Can you calculate that?
18. Jul 3, 2011
### Maybe_Memorie
P(Ts>x) = 2e^(-Ax) - e^(-2Ax)
so CDF = 1 - 2e^(-Ax) + e^(-2Ax)
PDF = -2Ae^(-2Ax) + 2Ae^(-Ax)
E[X] = 3/2A
Is this correct?
19. Jul 3, 2011
### I like Serena
I believe so!
Did you really have to do this for high school?
It seems to be a bit out of its scope.
20. Jul 3, 2011
### Maybe_Memorie
Thank you so much!
I'm not in high school, I'm in university in Ireland.
Due to various personal reasons I was unable to attend lectures, failed the statistics exam and now have to sit another stats exam in August to be allowed to progress to second year. so I'm going through all the exam papers and coming here with the ones I'm having difficulty with. | 1,977 | 6,559 | {"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} | 3.875 | 4 | CC-MAIN-2017-47 | longest | en | 0.907892 |
http://blog.cardinalec.com/tag/systems-of-equations/ | 1,566,807,588,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027331228.13/warc/CC-MAIN-20190826064622-20190826090622-00061.warc.gz | 30,681,774 | 11,181 | # SAT Math: Focus on Lines & Systems of Equations
This book is the second in a series that addresses the topics that appear most frequently in the Math section of the New SAT. You’ll review the properties of lines, including their graphs, and look at the different methods for solving systems of linear equations. Then you’ll apply what you know as you solve 25 problems — some done without your calculator and some with, just as it is on test day. Each problem comes with a full explanation of how to arrive at the correct answer.
We’d like to help as many students as possible achieve success on the SAT, so we’ve made the price of each of the volumes in this series just \$0.99.
The book can be purchased on Amazon by clicking on the image to the left or going here.
# SAT Math: Systems of Equations
The new SAT has shifted the focus of the math questions away from geometry and more toward algebra and functions. One of the topics you’ll want to master is systems of linear equations. If the practice tests are any indication, you can expect to see four or five of these questions on each test.
Below you’ll find a link to 10 sample system of equations questions (and a second link to the solutions).
But first, let’s do a little review.
Most systems of linear equations have a single ordered pair as their solution. It’s the point where the two lines intersect, and you can find the coordinates of that point by graphing the lines on your calculator or using the substitution or elimination methods. But what about the special cases? Let’s take a look at systems that have either an infinite number of solutions or no solution at all.
# SAT Math: More Systems of Equations Problems
If you’ve read my recent post on Solving Systems of Equations “Cleverly” and you’re looking to try a few more of these problems, you’ll find five more below. Remember, although you can solve these problems the same way you did in Algebra 1, our goal is to do them in an efficient way so that we can conserve some time that can then be used to solve other problems in the section.
1) Given the equations 3x + 8y = 24 and 2x – 13y = 26, what is the value of x – y?
(A) 8
(B) 10
(C) 12
(D) 14
(E) 16
# SAT Math: Solving Systems of Equations “Cleverly”
If you want to do well on the SAT Math test, there are certain topics (functions, right triangles involving Pythagorean Theorem, 30-60-90 and 45-45-90 triangles, problems that can be solved by plugging in numbers) that you simply must master. If you’re looking to score exceptionally well (think 700+), then you’ll also need to be very good at the types of problems that appear less frequently but often in the latter part of a section. These are often the problems that can be the difference between a very good score and a great score.
One type of problem that falls into this category is the systems of equation problem. You remember these from Algebra 1 (and probably again in Algebra 2) — multiple equations with more than one variable share a common solution that you are asked to find. A typical problem of this type that you would have seen in math class would go something like this:
Find the solution to the system 3x – 2y = 22 and 7x + 2y = 18. | 759 | 3,216 | {"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} | 4.3125 | 4 | CC-MAIN-2019-35 | latest | en | 0.949744 |
http://wiki.kerbalspaceprogram.com/wiki/Atmosphere | 1,443,962,669,000,000,000 | text/html | crawl-data/CC-MAIN-2015-40/segments/1443736673632.3/warc/CC-MAIN-20151001215753-00181-ip-10-137-6-227.ec2.internal.warc.gz | 306,868,290 | 13,290 | This wiki is currently experiencing migration problems. This is known and will be fixed at some point.
Registered users can edit this wiki.
# Atmosphere
The pressures for all atmospheres
Planets Moons
Eve Kerbin Laythe
Duna Jool
The atmosphere of a celestial body slows the movement of any object passing through it, a force known as atmospheric drag (or simply drag). An atmosphere also allows for aerodynamic lift. The celestial bodies with atmospheres are the planets Eve, Kerbin, Duna and Jool, as well as Laythe, a moon of Jool. Only Kerbin and Laythe have atmospheres that contain oxygen and thus produce intake air for jet engines to work.
Atmospheric pressure diminishes exponentially with increasing altitude. An atmosphere's scale height is the distance over which atmospheric pressure changes as a factor of e, or 2.718. For example, Kerbin's atmosphere has a scale height of 5000 m, meaning the atmospheric pressure at altitude n is 2.718 times greater than the pressure at altitude n + 5000.
Atmospheres vary in temperature, though this has no bearing on gameplay.
Atmospheres allow aerobraking and easier landing. However, an atmosphere makes taking off from a planet more difficult and increases the minimum stable orbit altitude.
## Drag
A Mk1-2 pod with a Mk16-XL parachute being slowed by drag in Kerbin's atmosphere.
In the game, the force of atmospheric drag (FD) is modeled as follows:[1]
$F_{D}=0.5\,\rho \,v^{2}\,d\,A$
where ρ is the atmospheric density (kg/m3), v is the ship's velocity (m/s), d is the coefficient of drag (dimensionless), and A is the cross-sectional area (m2).
Note that the cross-sectional area is not actually calculated in the game. It is instead assumed that it is directly proportional to the mass, which is an unrealistic simplification made by KSP. The parameter FlightGlobals.DragMultiplier indicates that the proportionality ratio is 0.008 m2/kg, so:
$A=0.008\cdot m$
where m is the ship's mass (kg).
The atmospheric density ρ is directly proportional to atmospheric pressure (p of unit atm), which is a function of altitude, the atmosphere's pressure at altitude 0 (p0), and scale height (H):
{\begin{aligned}p&=p_{0}\cdot e^{{\frac {-altitude}{H}}}\\\rho &=1.2230948554874{\frac {{\text{kg}}}{{\text{m}}^{3}\cdot {\text{atm}}}}\cdot p\end{aligned}}
where p here is in units atm, and ρ in kg/m3. The conversion factor of 1.2230948554874 kg/(m3·atm) is given by FlightGlobals.getAtmDensity(1.0), which returns the density at 1 atmosphere (sea level on Kerbin) pressure.
The coefficient of drag (d) is calculated as the mass-weighted average of the max_drag values of all parts on the ship. For most ships without deployed parachutes, d will be very near 0.2, since this is the max_drag value of the vast majority of parts. Also a group of the same part have always the same drag coefficient. Parts with a drag coefficient other than 0.2 are fuel lines, chairs, all spaceplane cockpits, all boosters, all rcs engines, bz-52 radial, regular strut, ncs adapter, all quad- and tricouplers, all nose cones, circular intake, shock cone intake, all wings, all docking ports, launch escape system, all parachutes, all wheels but the Rovemax Model XL-3, small gear bay, and mystery goo.
As an example, the coefficient of drag for a craft consisting simply of a Mk1-2 Command Pod (mass 4, drag 0.2) and a deployed Mk16-XL Parachute (mass 0.3, drag 500) is:
${\frac {4\cdot 0.2+0.3\cdot 500}{4+0.3}}=35.07$
### Improvements
The current drag and lift calculations are relatively simplistic, and as such make it relatively simple to get aircraft flying. However, a more sophisticated drag system is planned to be included in version 1.0.[2] Meanwhile, some players use the popular Ferram Aerospace Research mod which implements more realistic drag and lift models.
## Terminal velocity
The terminal velocity of an object falling through an atmosphere is the velocity at which the force of gravity is equal to the force of drag. Terminal velocity changes as a function of altitude. Given enough time, an object falling into the atmosphere will slow to terminal velocity and then remain at terminal velocity for the rest of its fall.
Terminal velocity is important because:
1. It describes the amount of velocity which a spacecraft must burn away when it is close to the ground.
2. It represents the speed at which a ship should be traveling upward during a fuel-optimal ascent.
The force of gravity (FG) is:
$F_{G}=m\cdot a=m\cdot {\frac {GM}{r^{2}}}$
where m is still the ship's mass, G is the gravitational constant, M is the mass of the planet, and r is the distance from the center of the planet to the falling object.
To find terminal velocity, we set FG equal to FD:
{\begin{aligned}m\cdot {\frac {GM}{r^{2}}}&=0.5\cdot \rho \cdot v^{2}\cdot d\cdot \left(0.008{\frac {{\text{m}}^{2}}{{\text{kg}}}}\cdot m\right)\\{\frac {GM}{r^{2}}}&=0.004{\frac {{\text{m}}^{2}}{{\text{kg}}}}\cdot \rho \cdot v^{2}\cdot d\\v&=v_{T}={\sqrt {{\frac {250{\frac {{\text{kg}}}{{\text{m}}^{2}}}\cdot GM}{r^{2}\cdot \rho \cdot d}}}}\end{aligned}}
Assuming d is 0.2 (which is a good approximation, provided parachutes are not in use), this simplifies to:
$v_{T}={\sqrt {{\frac {1250{\frac {{\text{kg}}}{{\text{m}}^{2}}}\cdot GM}{r^{2}\,\rho }}}}$
For the Mk1-2 pod and Mk16XL parachute example pictured above, the drag coefficient is 35.07, so its terminal velocity at sea level on Kerbin (which is 600 km from Kerbin's center) is:
$v_{T}={\sqrt {{\frac {250{\frac {{\text{kg}}}{{\text{m}}^{2}}}\cdot GM}{r^{2}\,\rho \cdot 35.07}}}}$
$\rho =1.2230948554874{\frac {{\text{kg}}}{{\text{m}}^{3}\cdot {\text{atm}}}}\cdot 1{\text{atm}}\cdot e^{{\frac {-0{\text{m}}}{5000{\text{m}}}}}$
$v_{T}={\sqrt {{\frac {250{\frac {{\text{kg}}}{{\text{m}}^{2}}}\cdot 6.674\cdot 10^{{-11}}{\frac {{\text{m}}^{3}}{{\text{kg}}\cdot {\text{s}}^{2}}}\cdot 5.2915793\cdot 10^{{22}}\operatorname {kg}}{(600000\operatorname {m})^{2}\cdot 1.2230948554874{\frac {{\text{kg}}}{{\text{m}}^{3}}}\cdot 35.07}}}}=7.56{\frac {{\text{m}}}{{\text{s}}}}$
### Examples
Altitude (m) vT (m/s)
Eve Kerbin Duna Jool Laythe
0 58.385 100.13 122.63 23.124 115.62
100 58.826 101.11 124.30 23.203 117.20
1000 62.952 110.33 140.36 23.932 132.48
10000 124.01 264.21 473.41 32.609 451.26
## On-rails physics
If a ship is "on rails" (meaning it's further than 2.25 km from the actively-controlled ship) and its orbit passes through a planet's atmosphere, one of two things will happen based on atmospheric pressure at the ship's altitude:
• below 0.01 atm: no atmospheric drag will occur — the ship will be completely unaffected
• 0.01 atm or above: the ship will disappear
The following table gives the altitude of this 0.01 atm threshold for each celestial body with an atmosphere:
Body Altitude (m)
Eve 40 485
Kerbin 23 333
Duna 10 842
Jool 105 870
Laythe 14 818
## Atmospheric height
The atmospheric height depends on the scale height of the celestial body and is where 0.000001th (0.0001%) of the surface pressure remains. Therefore, the atmospheric pressure at the edge of the atmosphere is relative; for example a craft in orbit around Jool can have a lower orbit (relative to the surface) because the surface pressure is higher.
$alt_{{{\text{atmospheric height}}}}=-ln\left(10^{{-6}}\right)\cdot {\text{scale height}}$
$p_{{{\text{atmospheric height}}}}=p_{0}\cdot 10^{{-6}}$
To calculate the atmospheric heights of other celestial bodies:
$alt_{{{\text{atmospheric height (real)}}}}=-ln\left({\frac {10^{{-6}}}{p_{0}}}\right)\cdot {\text{scale height}}$ | 2,197 | 7,580 | {"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": 13, "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-2015-40 | longest | en | 0.898317 |
http://gmatclub.com/forum/retail-sales-rose-8-10-of-1-percent-in-august-intensifying-10546-40.html | 1,485,062,671,000,000,000 | text/html | crawl-data/CC-MAIN-2017-04/segments/1484560281332.92/warc/CC-MAIN-20170116095121-00260-ip-10-171-10-70.ec2.internal.warc.gz | 119,457,694 | 57,006 | Retail sales rose 8/10 of 1 percent in August, intensifying : GMAT Sentence Correction (SC) - Page 3
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# Retail sales rose 8/10 of 1 percent in August, intensifying
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Re: Retail sales rose 8/10 of 1 percent in August, intensifying [#permalink]
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02 Oct 2009, 10:59
For option D, OG explanation says: "Although this option is not technically wrong; it is less clear and graceful than B".
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Re: Retail sales rose 8/10 of 1 percent in August, intensifying [#permalink]
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05 Nov 2009, 21:07
"Expectation.....would double" would go together better than "Expectation.....doubled". Since expectation is about thinking in future
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Re: Retail sales rose 8/10 of 1 percent in August, intensifying [#permalink]
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21 Aug 2010, 09:46
hello,
here's the question...
Retail sales rose 0.8 of 1 percent in August,
intensifying expectations that personal spending in
the July–September quarter more than doubled that
of the 1.4 percent growth rate in personal spending
for the previous quarter.
(A) that personal spending in the July–September
quarter more than doubled that of
(B) that personal spending in the July–September
quarter would more than double
(C) of personal spending in the July–September
quarter, that it more than doubled
(D) of personal spending in the July–September
quarter more than doubling that of
(E) of personal spending in the July–September
quarter, that it would more than double that of
i have a few questions regarding this problem. first, can someone explain the use of "double" used here? is it a verb? secondly, can someone explain why choice D is not right. thanks!
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Re: Retail sales rose 8/10 of 1 percent in August, intensifying [#permalink]
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21 Aug 2010, 17:00
Expectations of "Someone"??? Expectations of personal Spending is wrong
Whose expectation that is why D is wrong
that lefts A and B
the statement after that modifies anything before it hence A is wrong answer should be B
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Re: Retail sales rose 8/10 of 1 percent in August, intensifying [#permalink]
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26 Aug 2010, 19:10
here is the question...
Retail sales rose 0.8 of 1 percent in August,
intensifying expectations that personal spending in
the July–September quarter more than doubled that
of the 1.4 percent growth rate in personal spending
for the previous quarter.
(A) that personal spending in the July–September
quarter more than doubled that of
(B) that personal spending in the July–September
quarter would more than double
(C) of personal spending in the July–September
quarter, that it more than doubled
(D) of personal spending in the July–September
quarter more than doubling that of
(E) of personal spending in the July–September
quarter, that it would more than double that of
i am wondering if the word "double" here is a verb or an adjective. thanks!
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Re: Retail sales rose 8/10 of 1 percent in August, intensifying [#permalink]
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27 Aug 2010, 01:28
IMO B
Double is a comparative "Signal".
1) Idiom Issue: Expect that.
2) Since its Aug. i.e mid of Quater (July-Aug-Sept) we need to bring uncertainity into our speculation. So Would is appropriate.
3) We dont require "that of" becaz the second term is rate itself.
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Re: Retail sales rose 8/10 of 1 percent in August, intensifying [#permalink]
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30 Mar 2011, 12:45
+1 B
C, D, and E change the meaning.
In A, "that of" is not necessary.
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Re: Retail sales rose 8/10 of 1 percent in August, intensifying [#permalink]
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05 Jun 2011, 06:27
B for me too...
Expect should be followed by "that"...this eliminates C,D,E..
Between A and B..."more than doubled that of..." is unidiomatic...
left with B
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Re: Retail sales rose 8/10 of 1 percent in August, intensifying [#permalink]
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05 Jun 2011, 14:53
yes, with subjunctive mood (would), it should be B.
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Re: Retail sales rose 8/10 of 1 percent in August, intensifying [#permalink]
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07 Jul 2011, 21:01
Expectations that resulted from a past and continue in future would be used
for eg Retail sales rose 0.8 of 1 percent in August,intensifying expectations that personal spending in the July–September quarter would more than double
Expectations that resulted from a present and continue in future will be used
for eg According to some analysts, the gains in the stock
market reflect growing confidence that the economy
will avoid the recession that many had feared earlier in
the year and instead come in for a “soft landing,”
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Re: Retail sales rose 8/10 of 1 percent in August, intensifying [#permalink]
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01 Aug 2011, 01:39
Quote:
Retail sales rose 0.8 of 1 percent in August, intensifying expectations that personal spending in the July-Sept quarter more than doubled that of the 1.4 percent growth rate in personal spending for the previous quarter.
A. that personal spending in the July-Sept quarter more than doubled that of
B. that personal spending in the July-Sept quarter would more than double
C. of personal spending in the July-Sept quarter, that it more than doubled
D. of personal spending in the July-Sept quarter more than doubling that of
E. of personal spending in the July-Sept quarter, that it would more than double that of
idiom: expectations that vs. expectations of
expectations of is incorrect
Hence A or B
A uses incorrect tense
B uses 'would' which indicates that the situation has not occurred yet, hence the expectations. Hence, tense complies with intended meaning.
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Re: Retail sales rose 8/10 of 1 percent in August, intensifying [#permalink]
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01 Aug 2011, 01:42
Also, the use of that of is incorrect.
This year's rate has doubled that of the last year's. - correct
This year's rate has doubled the 1.8 units rate of last year - correct
This year's rate has doubled that of the 1.8 units - incorrect. 1.8 units is the rate itself.
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Re: Retail sales rose 8/10 of 1 percent in August, intensifying [#permalink]
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01 Aug 2011, 19:57
gmate2010 wrote:
flyinhair wrote:
Retail sales rose 0.8 of 1 percent in August, intensifying expectations that personal spending in the July-Sept quarter more than doubled that of the 1.4 percent growth rate in personal spending for the previous quarter.
A. that personal spending in the July-Sept quarter more than doubled that of
B. that personal spending in the July-Sept quarter would more than double
C. of personal spending in the July-Sept quarter, that it more than doubled
D. of personal spending in the July-Sept quarter more than doubling that of
E. of personal spending in the July-Sept quarter, that it would more than double that of
And, BTW, Could anyone please offer me an introduction about how to use "that of" when there is a "than"
or "as" before? thanks a lot!
C, D and E ..incorrect ..personal spending does not have expectations.. Awkward..
A. ) that personal spending in the July-Sept quarter more than doubled that of(personal spending of) the 1.4 percent growth rate in personal spending for the previous quarter --- redundant..
B. ) correct.. We need that..Also comparison is correct.
Hi All,
I also selected B ( looks best in given option)and used the same logic (as described by gmate2010) but still feel that "of" is required after double. Expert please share views."
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Re: Retail sales rose 8/10 of 1 percent in August, intensifying [#permalink]
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08 Oct 2011, 11:22
tt11234 wrote:
here is the question...
Retail sales rose 0.8 of 1 percent in August,
intensifying expectations that personal spending in
the July–September quarter more than doubled that
of the 1.4 percent growth rate in personal spending
for the previous quarter.
(A) that personal spending in the July–September
quarter more than doubled that of
(B) that personal spending in the July–September
quarter would more than double
(C) of personal spending in the July–September
quarter, that it more than doubled
(D) of personal spending in the July–September
quarter more than doubling that of
(E) of personal spending in the July–September
quarter, that it would more than double that of
i am wondering if the word "double" here is a verb or an adjective. thanks!
I did pick 'B'. But I have a question with choice B
Retail sales rose 0.8 of 1 percent in August, intensifying expectations that [highlight]personal spending[/highlight] in the July–September
quarter would more than double [highlight]the 1.4 percent growth rate[/highlight] in personal spending
for the previous quarter.
Is the comparison correct here ?
Personal spending ....... more than double the ...growth rate
Shouldnt it be
growth rate in Personal spending ...more than double the ..... growth rate
I would appreciate any help with this .
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Re: Retail sales rose 8/10 of 1 percent in August, intensifying [#permalink]
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08 Oct 2011, 17:38
1
KUDOS
Thabk wrote:
tt11234 wrote:
here is the question...
Retail sales rose 0.8 of 1 percent in August,
intensifying expectations that personal spending in
the July–September quarter more than doubled that
of the 1.4 percent growth rate in personal spending
for the previous quarter.
(A) that personal spending in the July–September
quarter more than doubled that of
(B) that personal spending in the July–September
quarter would more than double
(C) of personal spending in the July–September
quarter, that it more than doubled
(D) of personal spending in the July–September
quarter more than doubling that of
(E) of personal spending in the July–September
quarter, that it would more than double that of
i am wondering if the word "double" here is a verb or an adjective. thanks!
I did pick 'B'. But I have a question with choice B
Retail sales rose 0.8 of 1 percent in August, intensifying expectations that [highlight]personal spending[/highlight] in the July–September
quarter would more than double [highlight]the 1.4 percent growth rate[/highlight] in personal spending
for the previous quarter.
Is the comparison correct here ?
Personal spending ....... more than double the ...growth rate
Shouldnt it be
growth rate in Personal spending ...more than double the ..... growth rate
I would appreciate any help with this .
You can also look at the comparison as
Retail sales rose 0.8 of 1 percent in August, intensifying expectations that [highlight]personal spending[/highlight] in the July–September
quarter would more than double [highlight]the 1.4 percent growth rate in personal spending[/highlight]
for the previous quarter.
Now, this implies => the increased personal spending will cause the growth rate to be double of 1.4 .
Crick
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Re: Retail sales rose 8/10 of 1 percent in August, intensifying [#permalink]
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25 Nov 2011, 06:35
I also think it is B. But here the problem could be parallelism. personal spending i getting copared with 1.4 percent growth here. "That of" would have been more appropriate
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Re: Retail sales rose 8/10 of 1 percent in August, intensifying [#permalink]
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25 Nov 2011, 06:42
The answer can be straight derived from the word "expectation" which demands would
correct me if i am wrong
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Re: Retail sales rose 8/10 of 1 percent in August, intensifying [#permalink]
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25 Nov 2011, 07:37
+1 B
Its a official guide question
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Re: Retail sales rose 8/10 of 1 percent in August, intensifying [#permalink]
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25 Nov 2011, 14:57
I chose B as well. Seems the most concise..
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Re: Retail sales rose 8/10 of 1 percent in August, intensifying [#permalink]
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08 Mar 2012, 08:28
A. This answer choice states the verb "doubled" in the simple past tense, but this doesn't make any sense. The reason it doesn't make sense is because we are looking at statistics in August and we are predicting the outcome of the future. Therefore, the use of the simple past tense is incorrect here - it should be the conditional word "would."
B. At first I wasn't sure if this was the correct answer because it seemed to compare personal spending to growth rate. However, after further analysis I realized that you could include "in personal spending" as part of the comparison. This makes more sense. It also uses the conditional "would" to agree with the word "expectation."
C. I was quite confused by the use of "it" because I wasn't sure if it correctly refers to personal spending. However, I noticed that this answer choice uses "doubled" which is the simple past tense of the verb and therefore cannot be correct.
D. By using "that of" in this comparison, it is redundant because "that of" is a pronoun used to replace "personal spending of." If you notice later in the sentence, it repeats the words "personal spending," so this answer cannot be correct.
E. Again, I was unsure about the usage of the pronoun "it" and whether it correctly refers to personal spending. Regardless, I found this answer choice wordy and the use of the relative pronoun "that of" is unnecessary.
Re: Retail sales rose 8/10 of 1 percent in August, intensifying [#permalink] 08 Mar 2012, 08:28
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This assignment supports the following objectives: Calculate IRR, NPV and Payback Period Analyze the cash flows generated by mutually exclusive projects Formulate a recommendation using IRR, NPV and Payback Period as the criteria Background Suppose that your firm is considering the following two mutually exclusive projects. Both projects have the same initial cost of \$312,500 and the resulting annual cash flows for the first five years are as shown in the table below: Year Alpha Beta 0 \$ (312,500) \$ (312,500) 1 375,000 28,935 2 46,875 80,376 3 15,625 99,229 4 15,625 160,788 5 3,125 191,414 It is your job to analyze the feasibility of these two projects and to make recommendations as to which project should be undertaken. You must present a written report of your findings and your recommendations to middle managers, many of whom may be unfamiliar with some of your computations. You may choose to present the results of your calculations in table form. In your report, you must address each of the following: Analyses: a. Calculate the IRR of each project. Which project should be selected using IRR as the criterion? b. In its analyses of projects of this type, your firm uses a 14.0 percent discount rate. Compute the NPV for each project using the 14.0 percent discount rate. Which project should be chosen based on this result? c. In some cases, your firm uses the payback period to assess projects, with a cut-off point of 3 years. Calculate the payback period for each project and explain which project should be chosen using this criterion.
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• 1
##### Is Australia wider than it is now?
« on: January 03, 2019, 01:07:04 AM »
I have trouble with Australia Coordinates.
In the flat world; The distance between the meridians increases towards the South Pole.So doesn't Australia have to be wider in the flat earth?
For Australia; longitudes range from 113.65 to 153.61. (153-113 = 40 pieces of meridians)
Distance between meridians in Equator is approximately 111 kilometers.If we calculate accordingly; => 40 x 111 =4440 km
But This number must increase further for Flat Earth.Because The distance between the meridians in Australia is greater than Equator.
However If you google it, The distance between the east and west of Australia is 4030 kilometers.In this way, After the equator, The distance between the meridians have to decrease towards the South Pole.
#### JCM
• 156
##### Re: Is Australia wider than it is now?
« Reply #1 on: January 03, 2019, 04:31:41 AM »
It works in the Northern hemisphere as well. It is an easy test of the globe, equal latitudes will have equal distances as their opposite latitude in the other hemisphere. This is proposed by Rowbotham as a test of the globe if I recall as well. | 299 | 1,192 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.703125 | 3 | CC-MAIN-2019-26 | latest | en | 0.926026 |
https://kmmiles.com/189-53-km-in-miles | 1,669,639,609,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710503.24/warc/CC-MAIN-20221128102824-20221128132824-00285.warc.gz | 395,452,103 | 6,227 | kmmiles.com
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# 189.53 km in miles
## Result
189.53 km equals 117.6981 miles
You can also convert 189.53 km to mph.
## Conversion formula
Multiply the amount of km by the conversion factor to get the result in miles:
189.53 km × 0.621 = 117.6981 mi
## How to convert 189.53 km to miles?
The conversion factor from km to miles is 0.621, which means that 1 km is equal to 0.621 miles:
1 km = 0.621 mi
To convert 189.53 km into miles we have to multiply 189.53 by the conversion factor in order to get the amount from km to miles. We can also form a proportion to calculate the result:
1 km → 0.621 mi
189.53 km → L(mi)
Solve the above proportion to obtain the length L in miles:
L(mi) = 189.53 km × 0.621 mi
L(mi) = 117.6981 mi
The final result is:
189.53 km → 117.6981 mi
We conclude that 189.53 km is equivalent to 117.6981 miles:
189.53 km = 117.6981 miles
## Result approximation
For practical purposes we can round our final result to an approximate numerical value. In this case one hundred eighty-nine point five three km is approximately one hundred seventeen point six nine eight miles:
189.53 km ≅ 117.698 miles
## Conversion table
For quick reference purposes, below is the kilometers to miles conversion table:
kilometers (km) miles (mi)
190.53 km 118.31913 miles
191.53 km 118.94013 miles
192.53 km 119.56113 miles
193.53 km 120.18213 miles
194.53 km 120.80313 miles
195.53 km 121.42413 miles
196.53 km 122.04513 miles
197.53 km 122.66613 miles
198.53 km 123.28713 miles
199.53 km 123.90813 miles
## Units definitions
The units involved in this conversion are kilometers and miles. This is how they are defined:
### Kilometers
The kilometer (symbol: km) is a unit of length in the metric system, equal to 1000m (also written as 1E+3m). It is commonly used officially for expressing distances between geographical places on land in most of the world.
### Miles
A mile is a most popular measurement unit of length, equal to most commonly 5,280 feet (1,760 yards, or about 1,609 meters). The mile of 5,280 feet is called land mile or the statute mile to distinguish it from the nautical mile (1,852 meters, about 6,076.1 feet). Use of the mile as a unit of measurement is now largely confined to the United Kingdom, the United States, and Canada. | 648 | 2,292 | {"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-2022-49 | latest | en | 0.81668 |
https://www.pinterest.com/explore/algebra-2-activities/ | 1,480,787,465,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698541066.79/warc/CC-MAIN-20161202170901-00349-ip-10-31-129-80.ec2.internal.warc.gz | 1,013,323,752 | 78,620 | Pinterest • The world’s catalog of ideas
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1 | 397 | 1,938 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2016-50 | latest | en | 0.880416 |
https://www.teacherspayteachers.com/Product/Play-Dough-Tangram-Mats-Counting-with-ten-frames-and-tally-marks-2-986555 | 1,502,980,593,000,000,000 | text/html | crawl-data/CC-MAIN-2017-34/segments/1502886103316.46/warc/CC-MAIN-20170817131910-20170817151910-00396.warc.gz | 960,127,771 | 23,121 | ## Main Categories
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# Play Dough & Tangram Mats – Counting with ten frames and tally marks - 2
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Product Description
Penguin - Counting Math Mats (1-10): Can you make falling snowflakes?
+ 4 additional activities: self-correcting clip cards (1-20) & coloring cards.
Kids love to play with play dough (play doh) and tangrams puzzles. So why not combine the two and make math mats that can be used for more than just one type of activity.
This set contains play dough mats, tangram mats, ten frame task cards and tally marks task cards in color and black and white. Also included are two extra play dough task cards to encourage students to add more details to the math mats. With this set students can practice counting, fine motor skills and creativity.
Originally these math mats where designed to use with play dough or tangram pieces, but they can be used for many other fun and creative activities, such as:
• Trace and color numbers, dots, words and tally marks on the laminated task cards with a whiteboard marker instead of using play dough.
• Make ‘numbers snakes’ with the cards on the worksheets.
• Play matching games like memory with the task cards.
• Use manipulatives like buttons or matchsticks instead of play dough.
• Print the black and white math mats on white paper and draw objects, plants, people and/or animals on the background.
• Print the black and white math mats on white paper and give the penguin ´strange´ attributes like a bucket for his fish or a sombrero.
As an additional activity, two sets of self-correcting clip cards are added: ten frames & tally marks / 1-20. Clip a clothespin on the card, and then turn it over. The giraffe on the back of the card shows the correct answer. Also added are two sets of coloring cards. Use whiteboard markers to color in the correct number of dots or stripes.
Enjoy!
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\$1.50 | 489 | 2,088 | {"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-2017-34 | longest | en | 0.89046 |
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[July, 2015] On May 4, 2015, DIMACS Associate Director Eugene Fiorini was a central figure in what turned out to be a multi-state “crime wave.” The “crimes” were completely staged, but they were serious business for students participating in two different classes on mathematical forensics.
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More recently, Fiorini has developed another module on blood spatter analysis and time-of-death calculation. The modules are part of a growing portfolio of activities and materials in mathematical forensics that Fiorini has been building and sharing with teachers. In so doing, he has inspired teachers, like Jessie Yackel at Ayer-Shirley High School, to bring the topic to their students. Violeta Vasilevska, a professor of mathematics at Utah Valley University (UVU), is also helping to spread the word about mathematical forensics following an encounter with Fiorini. Vasilevska attended the 2014 Reconnect Workshop on Forensics in which Fiorini was the primary speaker. Inspired by the topic, she and her colleagues organized a conference, “Math and Forensics: Whodunit, Howdunit, Whendunit,” for high school students and teachers held at UVU in May 2015. With 150 registered participants, the conference illustrates the enthusiasm for the topic.
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Printable version of this story: [PDF]
DIMACS Homepage
Contacting the Center | 988 | 4,912 | {"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-2017-43 | latest | en | 0.95028 |
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Find if the sequence $a_n$ converges, $a_n=\frac{n!}{10^{6n}}=\frac{1}{\frac{10^6n}{n!}}$. Now if we take the limit of numerator and denominator as $n \to \infty$, we get $\frac{1}{0}$($\lim\limits_{n\to\infty} \frac{x^n}{n!}=0$ for any $x$), but division by zero is not defined, so we are stuck, but the answer is $\infty$. Where am I going wrong?
-
Fact: If $x_n\gt0$ and $x_n\to0$, then $1/x_n\to+\infty$. Can you show this fact and apply it to your setting? – Did Apr 2 '12 at 6:08
Look at the argument which establishes the limit you know (the one which tends to zero) and see if you can adapt it directly to the new case. Alternatively, what does that limit mean - you can make a certain expression as close to zero as you choose: how does that help you with the reciprocal? – Mark Bennet Apr 2 '12 at 6:08
Let $$b_n=\frac1{a_n}=\frac{10^{6n}}{n!}\;.$$ It appears from what you say in the question that you know that $$\lim_{n\to\infty}b_n=0\;.$$ It’s also clear that $b_n>0$ for all $n$. Thus, as $n\to\infty$, we have not just that $b_n\to 0$, but more specifically that $b_n\to 0^+$. But then clearly $$a_n=\frac1{b_n}\to+\infty\;.$$ At no point should you be trying to divide by $0$; that manipulation is clearly illegitimate.
$$\rm \frac{a_{\large n+10^6}}{a_{\large 10^6}}\normalsize =\frac{10^6+1}{10^6}\cdot\frac{10^6+2}{10^6}\cdots\frac{10^6+n}{10^6}>\left(1+\frac{1}{10^6}\right)^n \to \infty.$$ | 521 | 1,460 | {"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} | 4.4375 | 4 | CC-MAIN-2016-18 | latest | en | 0.762329 |
https://slideplayer.com/slide/2403863/ | 1,531,960,510,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676590362.13/warc/CC-MAIN-20180718232717-20180719012717-00631.warc.gz | 757,713,462 | 24,831 | # A Fast Variational Framework for Accurate Solid-Fluid Coupling Christopher Batty Florence Bertails Robert Bridson University of British Columbia.
## Presentation on theme: "A Fast Variational Framework for Accurate Solid-Fluid Coupling Christopher Batty Florence Bertails Robert Bridson University of British Columbia."— Presentation transcript:
A Fast Variational Framework for Accurate Solid-Fluid Coupling Christopher Batty Florence Bertails Robert Bridson University of British Columbia
Motivation Goal: Simulate fluids coupled to objects. Extend the basic Eulerian approach: – Advect fluid velocities – Add forces (eg. gravity) – Enforce incompressibility via pressure projection See eg. [Stam ‘99, Fedkiw et al. ‘01, Foster & Fedkiw ‘01, Enright et al. ‘02, etc.]
Motivation Cartesian grid fluid simulation is great! – Simple – Effective – Fast data access – No remeshing needed But…
Motivation Achilles’ heel: Real objects rarely align with grids.
Overview Three parts to our work: 1) Irregular static objects on grids 2) Dynamic & kinematic objects on grids 3) Improved liquid-solid boundary conditions
Previous Work First solution Voxelize – [Foster & Metaxas ’96] Easy! “Stairstep” artifacts Artificial viscosity Doesn’t converge under refinement!
Previous Work Better solution Subdivide nearby – [Losasso et al. ‘04] Stairs are smaller But problem remains
Previous Work Better yet Mesh to match objects – [Feldman et al. ‘05] Accurate! Needs remeshing Slower data access Trickier interpolation Sub-grid objects?
And now… back to the future? We’ll return to regular grids – But achieve results like tet meshes!
Pressure Projection Converts a velocity field to be incompressible (or divergence-free) No expansion or compression No flow into objects Images courtesy of [Tong et al. ‘03]
Pressure Projection We want the “closest” incompressible velocity field to the input. It’s a minimization problem!
Key Idea! Distance metric in the space of fluid velocity fields is kinetic energy. Minimizing KE wrt. pressure is equivalent to the classic Poisson problem! fluid nn 2 11 2 1 KEu
Minimization Interpretation Fluid velocity update is: Resulting minimization problem is: p t n uu ~ 1 2 ~ 2 1 minarg fluid p p t u
What changes? Variational principle automatically enforces boundary conditions! No explicit manipulation needed. Volume/mass terms in KE account for partial fluid cells. – Eg. Result: Easy, accurate fluid velocities near irregular objects. 2 2/12/12/1 2 1 KE iii uvol
Measuring Kinetic Energy
Discretization Details Normal equations always give an SPD linear system. – Solve with preconditioned CG, etc. Same Laplacian stencil, but with new volume terms. Classic: Variational: x uu x ppp iiiii 2/12/1 2 11 )1()1()1()2()1( x uVuV x pVpVVpV iiiiiiiiiii 2/12/12/12/1 2 12/12/12/112/1 )()()()()(
Object Coupling This works for static boundaries How to extend to… – Two-way coupling? Dynamic objects fully interacting with fluid – One-way coupling? Scripted/kinematic objects pushing the fluid
Object Coupling – Previous Work “Rigid Fluid” [Carlson et al ’04] – Fast, simple, effective – Potentially incompatible boundary velocities, leakage Explicit Coupling [Guendelman et al. ’05] – Handles thin shells, loose coupling approach – Multiple pressure solves per step, uses voxelized solve
Object Coupling – Previous Work Implicit Coupling [Klingner et al ‘06, Chentanez et al.’06] – solves object + fluid motion simultaneously – handles tight coupling (eg. water balloons) – requires conforming (tet) mesh to avoid artifacts
A Variational Coupling Framework Just add the object’s kinetic energy to the system. Automatically gives: – incompressible fluid velocities – compatible velocities at object surface fluid solid VMVu * 2 1 2 1 KE 2
A Coupling Framework Two components: 1) Velocity update: How does the pressure force update the object’s velocity? 2) Kinetic energy: How do we compute the object’s KE?
Example: Rigid Bodies 1) Velocity update: 2) Kinetic Energy: Discretize consistently with fluid, add to minimization, and solve. solid CM solid p p nXx n ˆ )(Torque ˆ Force 22 2 1 2 1 KEIωv m
Sub-Grid Rigid Bodies
Interactive Rigid Bodies
One Way Coupling Conceptually, object mass infinity In practice: drop coupling terms from matrix
Wall Separation Standard wall boundary condition is u·n = 0. – Liquid adheres to walls and ceilings! Ideally, prefer u·n ≥ 0, so liquid can separate – Analogous to rigid body contact.
Liquid Sticking Video
Wall Separation - Previous Work If ũ·n ≥ 0 before projection, hold u fixed. – [Foster & Fedkiw ’01, Houston et al ’03, Rasmussen et al ‘04] Inaccurate or incorrect in certain cases:
Wall Separation Two cases at walls: – If p > 0, pressure prevents penetration (“push”) – If p < 0, pressure prevents separation (“pull”) Disallow “pull” force: – Add p ≥ 0 constraint to minimization – Gives an inequality-constrained QP – u·n ≥ 0 enforced implicitly via KKT conditions
Liquid Peeling Video
Future Directions Robust air-water-solid interfaces. Add overlapping ghost pressures to handle thin objects, à la [Tam et al ’05]
Future Directions Explore scalable QP solvers for 3D wall- separation. Extend coupling to deformables and other object models. Employ linear algebra techniques to accelerate rigid body coupling.
Summary Easy method for accurate sub-grid fluid velocities near objects, on regular grids. Unified variational framework for coupling fluids and arbitrary dynamic solids. New boundary condition for liquid allows robust separation from walls.
Thanks!
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This page offers structured overviews of one or more selected methods. Add additional methods for comparisons by clicking on the dropdown button in the right-hand column. To practice with a specific method click the button at the bottom row of the table
Goodness of fit test
Paired sample $t$ test
Paired sample $t$ test
Independent variableIndependent variableIndependent variable
None2 paired groups2 paired groups
Dependent variableDependent variableDependent variable
One categorical with $J$ independent groups ($J \geqslant 2$)One quantitative of interval or ratio levelOne quantitative of interval or ratio level
Null hypothesisNull hypothesisNull hypothesis
• H0: the population proportions in each of the $J$ conditions are $\pi_1$, $\pi_2$, $\ldots$, $\pi_J$
or equivalently
• H0: the probability of drawing an observation from condition 1 is $\pi_1$, the probability of drawing an observation from condition 2 is $\pi_2$, $\ldots$, the probability of drawing an observation from condition $J$ is $\pi_J$
H0: $\mu = \mu_0$
Here $\mu$ is the population mean of the difference scores, and $\mu_0$ is the population mean of the difference scores according to the null hypothesis, which is usually 0. A difference score is the difference between the first score of a pair and the second score of a pair.
H0: $\mu = \mu_0$
Here $\mu$ is the population mean of the difference scores, and $\mu_0$ is the population mean of the difference scores according to the null hypothesis, which is usually 0. A difference score is the difference between the first score of a pair and the second score of a pair.
Alternative hypothesisAlternative hypothesisAlternative hypothesis
• H1: the population proportions are not all as specified under the null hypothesis
or equivalently
• H1: the probabilities of drawing an observation from each of the conditions are not all as specified under the null hypothesis
H1 two sided: $\mu \neq \mu_0$
H1 right sided: $\mu > \mu_0$
H1 left sided: $\mu < \mu_0$
H1 two sided: $\mu \neq \mu_0$
H1 right sided: $\mu > \mu_0$
H1 left sided: $\mu < \mu_0$
AssumptionsAssumptionsAssumptions
• Sample size is large enough for $X^2$ to be approximately chi-squared distributed. Rule of thumb: all $J$ expected cell counts are 5 or more
• Sample is a simple random sample from the population. That is, observations are independent of one another
• Difference scores are normally distributed in the population
• Sample of difference scores is a simple random sample from the population of difference scores. That is, difference scores are independent of one another
• Difference scores are normally distributed in the population
• Sample of difference scores is a simple random sample from the population of difference scores. That is, difference scores are independent of one another
Test statisticTest statisticTest statistic
$X^2 = \sum{\frac{(\mbox{observed cell count} - \mbox{expected cell count})^2}{\mbox{expected cell count}}}$
Here the expected cell count for one cell = $N \times \pi_j$, the observed cell count is the observed sample count in that same cell, and the sum is over all $J$ cells.
$t = \dfrac{\bar{y} - \mu_0}{s / \sqrt{N}}$
Here $\bar{y}$ is the sample mean of the difference scores, $\mu_0$ is the population mean of the difference scores according to the null hypothesis, $s$ is the sample standard deviation of the difference scores, and $N$ is the sample size (number of difference scores).
The denominator $s / \sqrt{N}$ is the standard error of the sampling distribution of $\bar{y}$. The $t$ value indicates how many standard errors $\bar{y}$ is removed from $\mu_0$.
$t = \dfrac{\bar{y} - \mu_0}{s / \sqrt{N}}$
Here $\bar{y}$ is the sample mean of the difference scores, $\mu_0$ is the population mean of the difference scores according to the null hypothesis, $s$ is the sample standard deviation of the difference scores, and $N$ is the sample size (number of difference scores).
The denominator $s / \sqrt{N}$ is the standard error of the sampling distribution of $\bar{y}$. The $t$ value indicates how many standard errors $\bar{y}$ is removed from $\mu_0$.
Sampling distribution of $X^2$ if H0 were trueSampling distribution of $t$ if H0 were trueSampling distribution of $t$ if H0 were true
Approximately the chi-squared distribution with $J - 1$ degrees of freedom$t$ distribution with $N - 1$ degrees of freedom$t$ distribution with $N - 1$ degrees of freedom
Significant?Significant?Significant?
• Check if $X^2$ observed in sample is equal to or larger than critical value $X^{2*}$ or
• Find $p$ value corresponding to observed $X^2$ and check if it is equal to or smaller than $\alpha$
Two sided:
Right sided:
Left sided:
Two sided:
Right sided:
Left sided:
n.a.$C\%$ confidence interval for $\mu$$C\% confidence interval for \mu -\bar{y} \pm t^* \times \dfrac{s}{\sqrt{N}} where the critical value t^* is the value under the t_{N-1} distribution with the area C / 100 between -t^* and t^* (e.g. t^* = 2.086 for a 95% confidence interval when df = 20). The confidence interval for \mu can also be used as significance test. \bar{y} \pm t^* \times \dfrac{s}{\sqrt{N}} where the critical value t^* is the value under the t_{N-1} distribution with the area C / 100 between -t^* and t^* (e.g. t^* = 2.086 for a 95% confidence interval when df = 20). The confidence interval for \mu can also be used as significance test. n.a.Effect sizeEffect size -Cohen's d: Standardized difference between the sample mean of the difference scores and \mu_0:$$d = \frac{\bar{y} - \mu_0}{s}$$Cohen's d indicates how many standard deviations s the sample mean of the difference scores \bar{y} is removed from \mu_0. Cohen's d: Standardized difference between the sample mean of the difference scores and \mu_0:$$d = \frac{\bar{y} - \mu_0}{s}$$Cohen's$d$indicates how many standard deviations$s$the sample mean of the difference scores$\bar{y}$is removed from$\mu_0.$n.a.Visual representationVisual representation - n.a.Equivalent toEquivalent to - • One sample$t$test on the difference scores. • Repeated measures ANOVA with one dichotomous within subjects factor. • One sample$t$test on the difference scores. • Repeated measures ANOVA with one dichotomous within subjects factor. Example contextExample contextExample context Is the proportion of people with a low, moderate, and high social economic status in the population different from$\pi_{low} = 0.2,\pi_{moderate} = 0.6,$and$\pi_{high} = 0.2$?Is the average difference between the mental health scores before and after an intervention different from$\mu_0 = 0$?Is the average difference between the mental health scores before and after an intervention different from$\mu_0 = 0$? SPSSSPSSSPSS Analyze > Nonparametric Tests > Legacy Dialogs > Chi-square... • Put your categorical variable in the box below Test Variable List • Fill in the population proportions / probabilities according to$H_0$in the box below Expected Values. If$H_0$states that they are all equal, just pick 'All categories equal' (default) Analyze > Compare Means > Paired-Samples T Test... • Put the two paired variables in the boxes below Variable 1 and Variable 2 Analyze > Compare Means > Paired-Samples T Test... • Put the two paired variables in the boxes below Variable 1 and Variable 2 JamoviJamoviJamovi Frequencies > N Outcomes -$\chi^2$Goodness of fit • Put your categorical variable in the box below Variable • Click on Expected Proportions and fill in the population proportions / probabilities according to$H_0$in the boxes below Ratio. If$H_0\$ states that they are all equal, you can leave the ratios equal to the default values (1)
T-Tests > Paired Samples T-Test
• Put the two paired variables in the box below Paired Variables, one on the left side of the vertical line and one on the right side of the vertical line
• Under Hypothesis, select your alternative hypothesis
T-Tests > Paired Samples T-Test
• Put the two paired variables in the box below Paired Variables, one on the left side of the vertical line and one on the right side of the vertical line
• Under Hypothesis, select your alternative hypothesis
Practice questionsPractice questionsPractice questions | 2,070 | 8,201 | {"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-2022-05 | latest | en | 0.849375 |
http://codeforces.com/problemset/problem/150/B | 1,652,785,432,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662517245.1/warc/CC-MAIN-20220517095022-20220517125022-00473.warc.gz | 13,581,538 | 13,289 | B. Quantity of Strings
time limit per test
2 seconds
memory limit per test
256 megabytes
input
standard input
output
standard output
Just in case somebody missed it: this winter is totally cold in Nvodsk! It is so cold that one gets funny thoughts. For example, let's say there are strings with the length exactly n, based on the alphabet of size m. Any its substring with length equal to k is a palindrome. How many such strings exist? Your task is to find their quantity modulo 1000000007 (109 + 7). Be careful and don't miss a string or two!
Let us remind you that a string is a palindrome if it can be read the same way in either direction, from the left to the right and from the right to the left.
Input
The first and only line contains three integers: n, m and k (1 ≤ n, m, k ≤ 2000).
Output
Print a single integer — the number of strings of the described type modulo 1000000007 (109 + 7).
Examples
Input
1 1 1
Output
1
Input
5 2 4
Output
2
Note
In the first sample only one string is valid: "a" (let's denote the only letter of our alphabet as "a").
In the second sample (if we denote the alphabet letters as "a" and "b") the following strings are valid: "aaaaa" and "bbbbb". | 320 | 1,192 | {"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} | 2.640625 | 3 | CC-MAIN-2022-21 | latest | en | 0.845472 |
https://www.eng-tips.com/viewthread.cfm?qid=500609 | 1,686,127,612,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224653631.71/warc/CC-MAIN-20230607074914-20230607104914-00673.warc.gz | 820,722,716 | 14,269 | ×
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(OP)
Hi,
May I know what are the other consequences if our Pump Head design larger than System Head? From my understanding, the pump will required excessive power.
For example, system head only required 30 m/98 ft but we use a pump that can delivered 50 m/164 ft at required flowrate.
#They design like this to give some margin for future expansion but after 10 years, no future expansion and no action to lower down the pump head
The system flow will usually increase. Flow will increase until system pressure drop equals pump total discharge head.
Your pump design head is significantly higher than what is needed. You can investigate if a control valve on the pump discharge will work. Control valves can be a good solution when you need 75 to 100% of pump rated head. That is still too high for your short term head prediction.
A possibly better option may be to install a VSD. VSD may work quite well, as you must lower the discharge head now by 40% or so. You could run the pump at about 75% of rated speed now and increase it using the VSD to higher speeds as more head is needed later on.
Einstein gave the same test to students every year. When asked why he would do something like that, "Because the answers had changed."
So these look like theoretical numbers. What flowrate are you actually flowing at?
My guess is that if you have no pressure or flow control on the system then you're flowing at probably about 20-25% more than you think the basis of your 30m requirment, but every system is different dependoing on how much of that 30m is a static head and how much is friction losses.
Remember - More details = better answers
Also: If you get a response it's polite to respond to it.
You need to look at your pump curve to see how much more power you are using, but if its higher flow then it's less time so total energy might be less....
Remember - More details = better answers
Also: If you get a response it's polite to respond to it.
Hi,
Probably good to review the basics. Consider the document attached to support your query.
Your Operating point is at the intersection between your pump curve and system curve.
You may want to share those curves with us.
Several options to consider:
a) Install a VFD
b) Trim the impeller.
note: Drop an email to the pump's vendor and request his/her support.
Good luck
Pierre
Great comments on the system. The system you have from your description sounds like it does not have and VFD (mentioned already) or any flow control devices like a triple-duty valve. If your system's primary loop is allowed to flow freely, do you have flow control on your terminal devices? Also, what is your pump speed (1800 rpm vs. 3600 rpm)? It might help if you have a high-speed pump with a low-speed TDH.
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Register now while it's still free! | 1,039 | 4,805 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2023-23 | longest | en | 0.947678 |
https://math.answers.com/Q/How_do_you_convert_7.79_into_a_mixed_number | 1,642,733,996,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320302715.38/warc/CC-MAIN-20220121010736-20220121040736-00378.warc.gz | 458,476,503 | 77,392 | 0
# How do you convert 7.79 into a mixed number?
Wiki User
2010-10-28 03:24:37
Best Answer
Expressed as a mixed number in its simplest form, 7.79 is equal to 7 79/100 or seven and seventy-nine hundredths.
Wiki User
2010-10-28 03:24:37
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Still have questions? | 161 | 460 | {"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} | 2.984375 | 3 | CC-MAIN-2022-05 | latest | en | 0.821543 |
https://themathblogger.blog/2012/10/20/solve-for-x-complicated/ | 1,603,759,354,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107892710.59/warc/CC-MAIN-20201026234045-20201027024045-00342.warc.gz | 554,420,868 | 22,311 | # Solve for x (complicated)
Q: Solve for x:
-2[8 – 5(2-3x) – 7x] = 4(x – |-9|)
A:
There are many ways to start with this, so I’m just going to start on the right side of the equation first:
-2[8 – 5(2-3x) – 7x] = 4(x – |-9|)
Starting on the right (piece at a time)
We know that |-9| = 9 (by definition of absolute value), so:
4(x – |-9|)
4(x – 9)
Now use the distributive property to distribute the 4 through (multiply the 4 to the x and the 9):
4x – 36
This is the best we can do on the right side. So the right side (for now) is 4x – 36.
Now let’s look at the left side:
-2[8 – 5(2 – 3x) – 7x]
We need to get rid of the inner most parentheses, so we should deal with the -5(2 – 3x) part. Distribute the -5 through:
-2[8 – 10 + 15x – 7x] <– that is what happens on the left when the -5 distributed through.
Now, clean up inside the brackets and combine like terms:
-2[-2 +8x] <— I combined the 8-10 and the 15x-7x
Now distribute the -2 through the brackets to get:
4 – 16x <– this is as far as the left side can be simplified. So, combining the left side = right side we get:
4 – 16x = 4x – 36
I’m going to add 16x to both sides (to get rid of the x on the left side):
4 = 20x – 36
Now add 36 to both sides:
40 = 20x
Divide both sides by 20 to get x by itself: | 450 | 1,293 | {"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} | 4.59375 | 5 | CC-MAIN-2020-45 | longest | en | 0.849135 |
https://www.jiskha.com/display.cgi?id=1332293568 | 1,503,190,196,000,000,000 | text/html | crawl-data/CC-MAIN-2017-34/segments/1502886105955.66/warc/CC-MAIN-20170819235943-20170820015943-00627.warc.gz | 913,892,193 | 3,983 | # Math
posted by .
They have 20 pieces of 12 foot lumber. If they need 2 screws for every 3 feet of lumber, how many screws do they need?
• Math -
20 * 12 = 240
240/3 = 80
2 * 80 = ?
• Math -
80*2=160
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I NEED THE ANSWER NOW PLZZ HELLPPP!! ASAP ASAP!! Juan bought a piece of lumber that is 18 feet long. a. Does he have enough lumber to make the five shelves? | 371 | 1,343 | {"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-2017-34 | latest | en | 0.927851 |
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# George Bernard Shaw wrote: " That any sane nation, having
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George Bernard Shaw wrote: " That any sane nation, having [#permalink]
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16. George Bernard Shaw wrote: " That any sane
nation, having observed that you could provide for
the supply of bread by giving bakers a pecuniary
interest in baking for you, should go on to give a
surgeon a pecuniary interest in cutting off your leg is
enough to make one despair of political humanity."
Shaw's statement would best serve as an illustration
in an argument criticizing which of the following?
(A) Dentists who perform unnecessary dental work
in order to earn a profit
(B) Doctors who increase their profits by specializ-
ing only in diseases that affect a large per-
centage of the population
(C) Grocers who raise the price of food in order to
increase their profit margins
(D) Oil companies that decrease the price of their
oil in order to increase their market share
(E) Bakers and surgeons who earn a profit by sup-
plying other peoples' basic needs
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### Show Tags
08 Apr 2004, 10:03
Toughie..........between A & C...would pick A!!
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### Show Tags
08 Apr 2004, 10:51
Since I don't know what "pecuniary" means, I choose A. It sounds about right.
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### Show Tags
10 Apr 2004, 18:56
OA is A.
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10 Apr 2004, 18:56
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# George Bernard Shaw wrote: " That any sane nation, having
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https://oj.wustacm.top/problem.php?id=1056&soj=1 | 1,623,799,995,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623487621627.41/warc/CC-MAIN-20210615211046-20210616001046-00393.warc.gz | 394,809,363 | 3,010 | Problem 1056. -- HangOver
## HDU_1056: HangOver
Time Limit: 1000 MS Memory Limit: 32 MB 64bit IO Format: %I64d
Submitted: 104 Accepted: 60
[Submit][Status][Web Board]
## Description
How far can you make a stack of cards overhang a table? If you have one card, you can create a maximum overhang of half a card length. (We're assuming that the cards must be perpendicular to the table.) With two cards you can make the top card overhang the bottom one by half a card length, and the bottom one overhang the table by a third of a card length, for a total maximum overhang of 1/2 + 1/3 = 5/6 card lengths. In general you can make n cards overhang by 1/2 + 1/3 + 1/4 + ... + 1/(n + 1) card lengths, where the top card overhangs the second by 1/2, the second overhangs tha third by 1/3, the third overhangs the fourth by 1/4, etc., and the bottom card overhangs the table by 1/(n + 1). This is illustrated in the figure below.
The input consists of one or more test cases, followed by a line containing the number 0.00 that signals the end of the input. Each test case is a single line containing a positive floating-point number c whose value is at least 0.01 and at most 5.20; c will contain exactly three digits.
For each test case, output the minimum number of cards necessary to achieve an overhang of at least c card lengths. Use the exact output format shown in the examples.
## Sample Input
1.00
3.71
0.04
5.19
0.00
## Sample Output
3 card(s)
61 card(s)
1 card(s)
273 card(s)
[Submit][Status][Web Board] | 436 | 1,520 | {"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} | 2.875 | 3 | CC-MAIN-2021-25 | latest | en | 0.826796 |
https://blog.jverkamp.com/2012/11/11/project-euler-6/ | 1,716,635,592,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058822.89/warc/CC-MAIN-20240525100447-20240525130447-00587.warc.gz | 120,211,499 | 5,361 | Project Euler 6
The sum of the squares of the first ten natural numbers is,
12 + 22 + … + 102 = 385
The square of the sum of the first ten natural numbers is,
(1 + 2 + … + 10)2 = 552 = 3025
Hence the difference between the sum of the squares of the first ten natural numbers and the square of the sum is 3025 - 385 = 2640.
Find the difference between the sum of the squares of the first one hundred natural numbers and the square of the sum. – PROJECT EULER #6
This almost doesn’t seem worth a post, being so straight forward, but I did say that I would work through all of the Project Euler problems. So of course I’ll write it up.
First, we need to write out a sum of squares and square of sums. We’ll go ahead and make a function of each, designed to operate on a list.
; square a number
(define (sqr x) (* x x))
; sum the squares of a list
(define (sum-of-squares ls)
(apply + (map sqr ls)))
; square the sum of a list
(define (square-of-sums ls)
(sqr (apply + ls)))
And then just create the list and subtract the two:
; diff between sum of squares and square of sums
(define (diff-of-sos n)
(define ls (range 1 (+ n 1)))
(- (square-of-sums ls)
(sum-of-squares ls)))
Test it out with the given values in the problem:
> (sum-of-squares (range 1 11))
385
> (square-of-sums (range 1 11))
3025
> (diff-of-sos 10)
2640
Good to go. Now we can scale up to the first 100:
> (time (diff-of-sos 100))
cpu time: 0 real time: 0 gc time: 0
25164150
And there you have it. The difference between the two is already over 25 million and the code is lightning fast.
Don’t worry, problem 7 is a bit more interesting, getting back into prime numbers. And before long, the problems get really interesting indeed…
As always, you can download my code for this or any Project Euler problem I’ve uploaded here. | 494 | 1,816 | {"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} | 4.34375 | 4 | CC-MAIN-2024-22 | latest | en | 0.889245 |
https://www.luogu.com.cn/blog/Juan-feng/solution-p3475 | 1,656,466,683,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103619185.32/warc/CC-MAIN-20220628233925-20220629023925-00372.warc.gz | 933,713,609 | 3,721 | # 一无所|知的小|J f
### 题解 P3475 【[POI2008]POD-Subdivision of Kingdom】
posted on 2019-03-25 11:47:00 | under 题解 |
## 小蒟蒻来水一篇退火题解
#include <iostream>
#include <cstring>
#include <cmath>
#include <cstdio>
#include <algorithm>
#define maxn 101
#define maxm 2002
#define re register
#define FOR(i, l, r) for(re int i = l; i <= r; ++i)
using namespace std;
int n, m, c, t, x, y, z, mid, ans;
int a[maxn], b[maxn][maxm], pl[maxn], ss[maxn], l[maxm], r[maxm];
double delta = 0.996;
int pd(int x) {
return pl[x] > mid;
}
void SA() {
double t = 1926.0;
while(t > 1e-14) {
int qwq = rand()%mid+1;
int qaq = rand()%mid+mid+1;
swap(pl[a[qwq]], pl[a[qaq]]);
swap(a[qwq], a[qaq]);
int nans = 0;
FOR(i, 1, m) {
if(pd(l[i]) ^ pd(r[i]))
++nans;
}
int de = nans - ans;
if(de < 0) {
ans = nans;
FOR(i, 1, mid)
ss[i] = a[i];
}
else if(exp(-de/t)*RAND_MAX <= rand()){
swap(pl[a[qwq]], pl[a[qaq]]);
swap(a[qwq], a[qaq]);
}
t *= delta;
}
}
int main() {
srand(71806291);
srand(rand());
srand(rand());
scanf("%d%d", &n, &m);
mid = n/2;
FOR(i, 1, n)
a[i] = i, pl[i] = i;
//pl代表编号为i的点的位置 a代表i位置的点的编号
FOR(i, 1, n/2)
ss[i] = i;
FOR(i, 1, m) {
scanf("%d%d", &l[i], &r[i]);
if(pd(l[i]) ^ pd(r[i]))
++ans;
}
int st = 25;
while(st--) {
SA();
}
FOR(i, 1, n/2)
printf("%d ", ss[i]);
} | 536 | 1,258 | {"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} | 3.328125 | 3 | CC-MAIN-2022-27 | latest | en | 0.192248 |
http://tuxgraphics.org/~guido/javascript/cm-inch-conversion.html | 1,558,637,786,000,000,000 | text/html | crawl-data/CC-MAIN-2019-22/segments/1558232257361.12/warc/CC-MAIN-20190523184048-20190523210048-00160.warc.gz | 209,627,311 | 2,560 | # Metric to Inch conversions with fractions
## cm, mm to inch conversion
```cm => inch and /163264
```
Formula: 1inch = 2.54cm
The cm-input field accepts a decimal point. Example: Enter 2cm and 4mm as 2.4cm
## inch to cm conversion
```inch and / 248163264 => cm
```
Formula: 1inch = 2.54cm
## Background information
In the metric world you can go to a mechanic and ask for a iron rod, 5.32 cm (or 53.2 mm) long. If you would convert that to inch then you end up with 2.094488inch. Unfortunately nobody knows what that is. If you go to a workshop with that number then they don't know what to do. A mechanic's measurement tape in the imperial system has divisions of 1/16 inch or 1/32 inch (or 1/64 inch). In other words the mechanic wants full number of inches plus fractions and not inch with a decimal point. 5.32 cm is therefore equivalent to 2 3/32 inch.
This page is operting system independent. It only requires a javascript capable webbrowser. You can install it locally on your PC, iPhone, Android-phone or netbook just by saving this page on the desktop. The page is optimized for mobile devices.
© Guido Socher, License: GPL, This software is provided without warrenty of any kind.
Version: 2019-02-18 | 335 | 1,222 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.875 | 3 | CC-MAIN-2019-22 | latest | en | 0.859591 |
https://www.teacherspayteachers.com/Product/Addition-Fact-Fluency-Make-Ten-Activities-Assessments-and-Game-Packet-2785967 | 1,513,480,630,000,000,000 | text/html | crawl-data/CC-MAIN-2017-51/segments/1512948592846.98/warc/CC-MAIN-20171217015850-20171217041850-00513.warc.gz | 836,840,032 | 25,935 | Total:
\$0.00
# Addition Fact Fluency: Make Ten Activities, Assessments, and Game Packet
Common Core Standards
Product Rating
File Type
PDF (Acrobat) Document File
20 MB|47 pages
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Product Description
FACT FLUENCY PACKET SERIES: SET 4 - MAKING TEN
This fact fluency packet contains everything you need to continue your fact fluency unit and teach all of the combinations of single digit numbers that make ten. The activities and games will engage and motivate your students and are differentiated to meet your students’ learning needs!
Contents:
• 2 Pages of Teaching Notes and Ideas
• Optional Parent Letter
• Lesson Introduction Poster
• 2 Poster Problems with Printable Teacher and Student Manipulatives
• Corresponding Poster Problem Rubric
• 3 Printable, No Prep Activities
• Printable, No Prep Scaffolding Activity
• Printable, No Prep Enrichment Activity
• 6 Quick Check Assessments
• 1 On-Level Game: Cupcake Delivery Version A
• 1 Enrichment Game: Cupcake Delivery Version B
• Answer Keys for Activities and Assessments
Check out the preview for more details!
***************************************************************************
1.OA.C.6
Add and subtract within 20, demonstrating fluency for addition and subtraction within 10. Use strategies such as counting on; making ten ; decomposing a number leading to a ten ; using the relationship between addition and subtraction; and creating equivalent but easier or known sums.
2.OA.B.2
Fluently add and subtract within 20 using mental strategies. By end of Grade 2, know from memory all sums of two one-digit numbers.
***************************************************************************
More Fact Fluency Packets from Abby’s Creative Schoolhouse
Fact Fluency Packet Series: Set 1 - Add Zero
Fact Fluency Packet Series: Set 2 - Add One and Two More
Fact Fluency Packet Series: Set 3 - Adding on to Five Packet
Fact Fluency Packet Series: Set 5 - Sums Greater Than Ten
Fact Fluency Packet Series: Set 6 - Doubles Packet
Fact Fluency Packet Series: Set 7 - Near Doubles Packet
Related Resources from Abby’s Creative Schoolhouse
Fact Fluency Interactive Student Journal
***************************************************************************
Follow me and be one of the first to know about new resources: Look for the green star next to my store logo and click it to become a follower.
***************************************************************************
Total Pages
47 pages
Included
Teaching Duration
3 days
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\$4.00 | 567 | 2,616 | {"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} | 3.546875 | 4 | CC-MAIN-2017-51 | latest | en | 0.771766 |
https://forum.roulette30.com/index.php?topic=959.0 | 1,524,762,466,000,000,000 | text/html | crawl-data/CC-MAIN-2018-17/segments/1524125948426.82/warc/CC-MAIN-20180426164149-20180426184149-00576.warc.gz | 602,605,183 | 8,036 | ### Author Topic: The progression which can stand a long sleep dozen. (Read 1999 times)
0 Members and 1 Guest are viewing this topic.
#### Jesper
• Hero Member
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##### The progression which can stand a long sleep dozen.
« on: June 09, 2016, 11:16:18 AM »
The about 50 to 100 step progression on DOZEN.
Jake a member here looks for an ideal progression on dozen, and not only him.
Jesper tactic is simple and rather rigid, but most of the time successfull!!!
We chose if we can a fair odds table (not worry about zero(s)).
We chose a table allowing for small bets (win will not allways be small).
We bet 12 units on a dozen (if you think waiting for virtual losses OK,(you lose less but probably not win more, all is due to less play, this I do not argue, it will in MHO at least not hurt)).
Bet 12 on a dozen. If hit session over.
If loss MOVE one fromthe dozen to a number or split in the dozen. We do it until we win and never the same move to number(s), unless full at the stage. It happens we are not positive in the first hit, then we go on until we win.
Should we be back in balance having twelve
straight up (after 12 spins with loss), we
progress by using chips, new 12 on the dozen and work the same way until a win or when we chicken out.
« Last Edit: June 09, 2016, 11:54:09 AM by Jesper »
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#### kav
• www.Roulette30.com
• Hero Member
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• Gender:
##### Re: The progression which can stand a long sleep dozen.
« Reply #1 on: June 09, 2016, 05:08:51 PM »
Very fresh idea.
A question: if we are losing after 12 spins, we bet both the 12 on dozen and the 12 inside?
My version.
The following users thanked this post: december, nowun, Reyth
#### Jesper
• Hero Member
• Posts: 1454
• Thanked: 753 times
• Gender:
##### Re: The progression which can stand a long sleep dozen.
« Reply #2 on: June 10, 2016, 12:51:06 AM »
Yes we add 12 every twelve spin. So when it is 12 plein we progress by 12 units on the dozen.
The following users thanked this post: december, Reyth
#### nowun
• New
• Posts: 12
• Thanked: 10 times
##### Re: The progression which can stand a long sleep dozen.
« Reply #3 on: July 19, 2016, 12:29:36 AM »
My version.
Very interesting, I play a version of this with both a dozen and a column when I need to vary my betting. Accidently discovered it trying to find a different bet. I limit sessions to approximately 100 spins and have not come across the same dozen and column sleeping simultaneously yet which is what will kill my bet. Both the dozen and column that I am betting on would have to sleep for about 15 spins. I have never seen both sleep simultaneously for that long yet, but I am sure it has happened in the past. I use idependent progressions on each position. Live dealer only.
Sometimes it can be quite boring when you BR yoyos, but if you limit target to about 20-50 units per 100 spins and don't get greedy it is doable on a regular basis. I have been in the plus for years playing this way.
The following users thanked this post: kav, december, Reyth, Sheridan44
#### Sheridan44
• Mature Member
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• Gender:
##### Re: The progression which can stand a long sleep dozen.
« Reply #4 on: July 19, 2016, 08:34:49 AM »
I sometimes have used a distant variant of this method...
It involves playing the two dozens that didn't hit last spin.... upon a theory that one of those two unhit dozens are more likely to hit (>2/3 to 1/3) on the next spin. It's supported by a single column bet as a stabilizing/recovery technique to lessen the whipsaw effect.
Examples: The 2nd dozen shows....play 1u ea on the 1st and 3rd dozens.... also a 1u bet on the 2nd column....
3rd dozen shows...... play the 1st and 2nd dozens...and the 3rd column.
In effect you play the dozens that didn't show - and the column number of the dozen that did show. The bets are quite well overlapped. It can be a bit of a grind, but less dramatic with the up and down swings.
« Last Edit: July 19, 2016, 08:44:14 AM by Sheridan44 »
The following users thanked this post: Sputnik, Reyth, Miyu | 1,142 | 4,203 | {"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} | 2.703125 | 3 | CC-MAIN-2018-17 | latest | en | 0.800003 |
https://helpseeker.net/excel-formula-to-check-if-value-is-in-a-list-vlookup-10-handy-steps-to-escape-from-a-typical-error-trap/ | 1,674,805,412,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764494974.98/warc/CC-MAIN-20230127065356-20230127095356-00611.warc.gz | 314,944,704 | 11,158 | # Excel Formula To Check If Value Is In A List VLOOKUP – 10 Handy Steps to Escape From a Typical Error Trap
You are searching about Excel Formula To Check If Value Is In A List, today we will share with you article about Excel Formula To Check If Value Is In A List was compiled and edited by our team from many sources on the internet. Hope this article on the topic Excel Formula To Check If Value Is In A List is useful to you.
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## VLOOKUP – 10 Handy Steps to Escape From a Typical Error Trap
The VLOOKUP function is generally considered to be a problematic formula. Why does this happen?
Let’s recreate a typical case…
1. The backend list is in place.
2. You have an index column to search on, for example: “product code”.
3. You write a VLOOKUP formula in less than a minute: you enter the lookup_value, you select the data set, you set the column number to retrieve and the fourth parameter (exact or approximate match).
4. All the problems start when you press enter; The lookup value is supposed to be in the backend, but you get a NA error. Your frustration increases when no amount of change fixes the error.
Then, you are out of ideas and bored.
Escape the NA Error Trap in 10 Steps…
1. Check that the index column is on the leftmost side of your lookup table.
2. Check that there are no leading and trailing spaces in the left column.
3. Check that there are no misspellings in the left column (pay attention to this when searching for texts).
4. Check that the index column’s data is the same as the lookup_value argument. For example, if you are searching for numbers, the left column should not contain numbers stored as text.
5. Check that the left column is sorted in ascending order when you perform the approximate match VLOOKUP.
6. Check that the formula is well written; All arguments must be complete. Remember to set the fourth argument to 0 for an exact match formula.
7. Check whether lookup_value is referenced correctly or lookup_value is properly hard coded.
8. Notice that lookup_value refers to the left column of the table array. For example: your lookup_value points to the “product name” column but your left index column field is “product code”.
9. Check that the table_array argument refers to the correct data set. Check this when you insert columns into the backend list or to confirm that the table_array range is entered correctly.
10. If you still receive the NA message; The backend index column does not include search value.
The most important way to escape the VLOOKUP trap is to make your index column a high priority quality control point. Don’t make the mistake of looking for an index column that is completely messy. You will have problems again and again.
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#VLOOKUP #Handy #Steps #Escape #Typical #Error #Trap
Source: https://ezinearticles.com/?VLOOKUP—10-Handy-Steps-to-Escape-From-a-Typical-Error-Trap&id=3193880 | 828 | 3,678 | {"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-2023-06 | latest | en | 0.80994 |
https://www.manchester.ac.uk/study/undergraduate/courses/2022/00558/bsc-computer-science-and-mathematics/course-details/MATH20902 | 1,657,111,798,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656104672585.89/warc/CC-MAIN-20220706121103-20220706151103-00614.warc.gz | 917,853,629 | 14,853 | This course is unavailable through clearing
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# BSc Computer Science and Mathematics / Course details
Year of entry: 2022
## Course unit details:Discrete Mathematics
Unit code MATH20902 10 Level 2 Semester 2 Department of Mathematics No
### Overview
Modern Discrete Mathematics is a broad subject bearing on everything from logic to logistics. Roughly speaking, it is a part of mathematics that touches on those subjects that Calculus and Algebra can't: problems where there is no sensible notion continuity or smoothness and little algebraic structure. The subject, which is typically concerned with finiteor at the most countablesets of objects, abounds with interesting, concrete problems and entertaining examples.
### Pre/co-requisites
Unit title Unit code Requirement type Description
Foundations of Pure Mathematics A MATH10101 Pre-Requisite Compulsory
Foundations of Pure Mathematics B MATH10111 Pre-Requisite Compulsory
### Aims
This module aims to engage students with a circle of concrete problems and applications, algorithmic techniques and basic theorems arising in graph theory.
### Learning outcomes
On completion of this unit successful students will be able to:
• Define what it means for two graphs to be isomorphic and determine, with rigorous supporting arguments, whether two (small) graphs are isomorphic.
• Explain what the chromatic number of a graph is, determine it for small graphs and apply the idea to scheduling problems.
• Say what it means for a graph to be Eulerian and determine whether small graphs or multigraphs are Eulerian.
• Say what it means for a graph to be Hamiltonian and use the Bondy-Chvátal theorem to prove that a graph is Hamiltonian.
• Construct the graph Laplacian and apply the Matrix-Tree Theorem to count the number of spanning trees or spanning arborescences contained in a graph.
• Construct the adjacency matrix of a graph and exploit the connection between powers of the adjacency matrix to count walks. Also, define the operations of tropical arithmetic, construct the weight matrix associated with a weighted graph and use its tropical matrix powers to find the lengths of shortest paths.
• Given a project defined by a set of tasks, along with their durations and prerequisites, use critical path analysis to determine how quickly the project can be completed.
• Say what it means for a graph to be planar; state and apply Kuratowski’s theorem and determine whether a graph is planar or not.
### Syllabus
Graph Theory & Applications: [22]
The basic definitions about graphs should be familiar from the Mathematical Workshop MATH10001, so after a brief review we will treat the following topics:
• Basic notions & notations, trees [5]
• Eulerian tours & Hamiltonian cycles [4]
• The Principle of Inclusion/Exclusion and the Matrix-Tree Theorem [4]
• Shortest paths & applications to scheduling [4]
Planar graphs & map colouring [5]
### Assessment methods
Method Weight
Other 20%
Written exam 80%
• Coursework; Weighting within unit 20%. This will consist of a problem set due on the last Friday before Easter and handed out two weeks earlier.
• End of semester examination; Weighting within unit 80%
### Feedback methods
Feedback tutorials will provide an opportunity for students' work to be discussed and provide feedback on their understanding. Coursework or in-class tests (where applicable) also provide an opportunity for students to receive feedback. Students can also get feedback on their understanding directly from the lecturer, for example during the lecturer's office hour.
Recommended: Dieter Jungnickel (2013), Graphs, Networks and Algorithms, 4th edition, Springer.
Harris, Hurst & Mossinghoff (2008), Combinatorics and Graph Theory, Springer.
Marcus (2008), Graph Theory: a problem oriented approach, Mathematical Association of America.
Biggs (1993), Algebraic Graph Theory, 2nd edition, CUP.
Cameron (2017), Notes on Counting: An Introduction to Enumerative Combinatorics, CUP.
### Study hours
Scheduled activity hours
Lectures 12
Tutorials 12
Independent study hours
Independent study 76
### Teaching staff
Staff member Role
Mark Muldoon Unit coordinator
The independent study hours will normally comprise the following. During each week of the taught part of the semester:
· You will normally have approximately 60-75 minutes of video content. Normally you would spend approximately 2-2.5 hrs per week studying this content independently
· You will normally have exercise or problem sheets, on which you might spend approximately 1.5hrs per week
· There may be other tasks assigned to you on Blackboard, for example short quizzes or short-answer formative exercises
· In some weeks you may be preparing coursework or revising for mid-semester tests
Together with the timetabled classes, you should be spending approximately 6 hours per week on this course unit.
The remaining independent study time comprises revision for and taking the end-of-semester assessment. | 1,092 | 5,168 | {"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-2022-27 | latest | en | 0.866248 |
https://community.powerbi.com/t5/Desktop/Calculated-Column-that-returns-Distinct-Count-with-a-date-slicer/td-p/1218634 | 1,618,614,448,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038092961.47/warc/CC-MAIN-20210416221552-20210417011552-00627.warc.gz | 292,449,614 | 144,783 | cancel
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Frequent Visitor
## Calculated Column that returns Distinct Count with a date slicer
Hi All,
This is a calculated column in Sheet (Mails) that returns 1 if someone in Sheet (Test) submitted twice (by distinct-counting the submitted on dates), 0.5, if someone submitted once, and 0 if someone didn't submitt at all.
Column =
VAR M = Mails[Mail]
RETURN
IF(
CALCULATE(
DISTINCTCOUNT(Test[SubmittedOn]),
FILTER(ALLSELECTED(Test),
Test[Submittedby] = M && Test[SubmittedOn] >= SELECTEDVALUE('Start Date'[Start Date]) && Test[SubmittedOn] <= SELECTEDVALUE('End Date'[End Date]))
)>=2,1,
IF(
CALCULATE(
DISTINCTCOUNT(Test[SubmittedOn]),
FILTER(ALLSELECTED(Test),
Test[Submittedby] = M && Test[SubmittedOn] >= SELECTEDVALUE('Start Date'[Start Date]) && Test[SubmittedOn] <= SELECTEDVALUE('End Date'[End Date]))
)=1,0.5,0
)
)
Start Date and End Time are
= GENERATESERIES(DATE(2020, 1,1), DATE(2060,12,31))
The problem is I want the returend value to be flitered by a date slicer on my dashboard, and I'm unable to do so using the DAX above.
But when I under the dates manually into the function, using Date(Year, Month, Day), it works perfectly.
Is there anyway to make this work in a calculated column?
Thanks.
1 ACCEPTED SOLUTION
Community Support
If you want to use date slicer to control the result, you need to create a measure not a calculate column.
1. We need to create a relationship between Mail table and Test table.
2. Then we can create a measure.
``````Measure =
VAR M =
MAX ( Mail[Mails] )
VAR _selected_Start =
SELECTEDVALUE ( 'Start Date'[Value] )
VAR _selected_End =
SELECTEDVALUE ( 'End Date'[Value] )
VAR _result1 =
IF (
CALCULATE (
DISTINCTCOUNT ( Test[SubmittedOn] ),
FILTER (
ALLSELECTED ( Test ),
Test[Submittedby] = M
&& Test[SubmittedOn] >= _selected_Start
&& Test[SubmittedOn] <= _selected_End
)
) >= 2,
1,
IF (
CALCULATE (
DISTINCTCOUNT ( Test[SubmittedOn] ),
FILTER (
ALLSELECTED ( Test ),
Test[Submittedby] = M
&& Test[SubmittedOn] >= _selected_Start
&& Test[SubmittedOn] <= _selected_End
)
) = 1,
0.5,
0
)
)
VAR _result2 =
IF (
CALCULATE (
DISTINCTCOUNT ( Test[SubmittedOn] ),
FILTER ( ALLSELECTED ( Test ), Test[Submittedby] = M )
) >= 2,
1,
IF (
CALCULATE (
DISTINCTCOUNT ( Test[SubmittedOn] ),
FILTER ( ALLSELECTED ( Test ), Test[Submittedby] = M )
) = 1,
0.5,
0
)
)
RETURN
IF (
ISBLANK ( _selected_Start ) || ISBLANK ( _selected_End ),
_result2,
_result1
)``````
3. At last we create two slicers based on Start Date and End Date.
If it doesn’t meet your requirement, could you please show the exact expected result based on the table that you have shared?
BTW, pbix as attached.
Best regards,
Community Support Team _ zhenbw
If this post helps, then please consider Accept it as the solution to help the other members find it more quickly.
4 REPLIES 4
Community Support
If you want to use date slicer to control the result, you need to create a measure not a calculate column.
1. We need to create a relationship between Mail table and Test table.
2. Then we can create a measure.
``````Measure =
VAR M =
MAX ( Mail[Mails] )
VAR _selected_Start =
SELECTEDVALUE ( 'Start Date'[Value] )
VAR _selected_End =
SELECTEDVALUE ( 'End Date'[Value] )
VAR _result1 =
IF (
CALCULATE (
DISTINCTCOUNT ( Test[SubmittedOn] ),
FILTER (
ALLSELECTED ( Test ),
Test[Submittedby] = M
&& Test[SubmittedOn] >= _selected_Start
&& Test[SubmittedOn] <= _selected_End
)
) >= 2,
1,
IF (
CALCULATE (
DISTINCTCOUNT ( Test[SubmittedOn] ),
FILTER (
ALLSELECTED ( Test ),
Test[Submittedby] = M
&& Test[SubmittedOn] >= _selected_Start
&& Test[SubmittedOn] <= _selected_End
)
) = 1,
0.5,
0
)
)
VAR _result2 =
IF (
CALCULATE (
DISTINCTCOUNT ( Test[SubmittedOn] ),
FILTER ( ALLSELECTED ( Test ), Test[Submittedby] = M )
) >= 2,
1,
IF (
CALCULATE (
DISTINCTCOUNT ( Test[SubmittedOn] ),
FILTER ( ALLSELECTED ( Test ), Test[Submittedby] = M )
) = 1,
0.5,
0
)
)
RETURN
IF (
ISBLANK ( _selected_Start ) || ISBLANK ( _selected_End ),
_result2,
_result1
)``````
3. At last we create two slicers based on Start Date and End Date.
If it doesn’t meet your requirement, could you please show the exact expected result based on the table that you have shared?
BTW, pbix as attached.
Best regards,
Community Support Team _ zhenbw
If this post helps, then please consider Accept it as the solution to help the other members find it more quickly.
Frequent Visitor
Genius, Thanks.
Super User IV
@NaderSaeed ,Can you share sample data and sample output in table format?
Proud to be a Super User!
Frequent Visitor
Sheet(Test)
Sheet (Mail)
The Sheet (Mail) has all the databbase for all emails, and the Sheet (Test) only has the ones who have submitted, so I'm distinct-counting the SubmittedOn Coulmn as it'd be the best indicator in our case.
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Top Kudoed Authors | 1,419 | 5,177 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.734375 | 3 | CC-MAIN-2021-17 | latest | en | 0.6994 |
https://zbmath.org/?q=an:0681.20044 | 1,620,520,259,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243988953.13/warc/CC-MAIN-20210509002206-20210509032206-00478.warc.gz | 1,155,536,066 | 8,955 | # zbMATH — the first resource for mathematics
On the theory of F-quasigroups. (Russian) Zbl 0681.20044
Webs and quasigroups, Interuniv. thematic Collect. sci. Works, Kalinin 1988, 127-130 (1988).
[For the entire collection see Zbl 0632.00006.]
A groupoid $$(Q,A)$$ is a semigroup iff the following law holds: $$A(A(x,y),z)=A(x,A(y,z))$$. $$(Q,A)$$ is a quasigroup iff the equations $$A(a,x)=b$$ and $$A(y,a)=b$$ are uniquely solvable for all $$a,b\in Q$$. A group is an associative quasigroup. Investigation of quasigroups naturally yields the following question: how near to groups are special quasigroups. D. C. Murdoch [Am. J. Math. 61, 509-522 (1939; Zbl 0020.34702)] treated this problem in the following way. In every quasigroup $$(Q,A)$$ the equation $$A(A(a,b),c)=A(a,A(b,x))$$ is uniquely solvable for all $$a,b,c\in Q$$. If the solution is denoted by $$f_{(a,b)}c$$, then $$f_{(a,b)}$$ is a permutation of the set $$Q$$ which in general depends on $$a$$ and $$b$$. Then especially the following cases are of interest: a) $$f_{(a,b)}$$ depends on $$b$$ only; b) $$f_{(a,b)}$$ depends on $$a$$ only; and c) $$f_{(a,b)}$$ does not depend neither on $$a$$ nor on $$b$$. Quasigroups for which b) holds are said to be F-quasigroups. In this article a relationship between F-quasigroups and loops (quasigroups with a neutral element) satisfying some special laws is stated.
Reviewer: J.Ušan
##### MSC:
20N05 Loops, quasigroups
##### Keywords:
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# First Grade Math Homework - 2nd Quarter
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This Math Homework for 1st graders provides a weekly spiral review of skills in Operations and Algebraic Thinking, Numbers and Operations in Base Ten, Measurement and Data, and Geometry. It is Common Core aligned.
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# PPT ON TRIANGULAR NOTCH.pptx
fluid mechanics
fluid mechanics
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### PPT ON TRIANGULAR NOTCH.pptx
1. 1. KALI CHARAN NIGAM INSTITUTE OF TECHNOLOGY, BANDA LECTURE ON TRIANGULAR NOTCH FLUID MECHANICS AND FLUID MACHINE(KME302) ( III SEMESTER ) BY : DEEPAK KUMAR YADAV ME DEPARTMENT
2. 2. DISCHARGE OVER A TRIANGULAR NOTCH OR WEIR • The expression for the discharge over a triangular notch or weir is the same. It is derived as • Width of strip • Area of strip • Discharge through the strip,
3. 3. DISCHARGE OVER A TRIANGULAR NOTCH OR WEIR • Total discharge, • For a right angled V notch , If Cd=0.6 θ = 90o Discharge
4. 4. NUMERICAL PROBLEM 1:During an experiment in a laboratory 0.05m3 of water flowing over a right-angled notch was collected in 1minute. If the head of the sill is 50mm, calculate the co-efficient of discharge. [ AKTU 2021-22] PROBLEM 2:Find the discharge over a triangular notch of angle 60o when the head over the V notch is 0.3m. Assume Cd =0.6. PROBLEM 3: Water flows over a rectangular weir 1m at a depth of 150mm and afterwards passes through a triangular right-handed weir. Take Cd for the rectangular and triangular weir as 0.62 and 0.59 respectively, find the depth over the triangular weir.
5. 5. THANK YOU | 482 | 1,479 | {"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-2023-14 | latest | en | 0.593262 |
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# Exchanging More than Complete Data
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### Exchanging More than Complete Data
1. 1. Exchanging More than Complete Data Jorge P´rez e Universidad de Chilejoint work with Marcelo Arenas (PUC-Chile) and Juan Reutter (Univ. Edinburgh)
2. 2. FatherAndy BobBob Danny Father(x, y ) → Parent(x, y )Danny Eddie Mother(x, y ) → Parent(x, y ) Parent(x, y ) ∧ Parent(y , z) → Gr-Parent(x, z) Mother Carrie Bob
3. 3. FatherAndy BobBob Danny Father(x, y ) → Parent(x, y )Danny Eddie Mother(x, y ) → Parent(x, y ) Parent(x, y ) ∧ Parent(y , z) → Gr-Parent(x, z) Mother Carrie Bob Father(x, y ) → Pateras(x, y ) Source Target Gr-Parent(x, y ) → Pappoi(x, y )
4. 4. FatherAndy BobBob Danny Father(x, y ) → Parent(x, y )Danny Eddie Mother(x, y ) → Parent(x, y ) Parent(x, y ) ∧ Parent(y , z) → Gr-Parent(x, z) Mother Carrie Bob Father(x, y ) → Pateras(x, y ) Source Target Gr-Parent(x, y ) → Pappoi(x, y ) Pateras Pappoi Andy Bob Andy Danny Bob Danny Carrie Danny Danny Eddie Bob Eddie
5. 5. FatherAndy BobBob Danny Father(x, y ) → Parent(x, y )Danny Eddie Mother(x, y ) → Parent(x, y ) Parent(x, y ) ∧ Parent(y , z) → Gr-Parent(x, z) Mother Carrie Bob Father(x, y ) → Pateras(x, y ) Source Target Gr-Parent(x, y ) → Pappoi(x, y ) Pateras(x, y ) ∧ Pateras(y , z) → Pappoi(x, z)
6. 6. FatherAndy BobBob Danny Father(x, y ) → Parent(x, y )Danny Eddie Mother(x, y ) → Parent(x, y ) Parent(x, y ) ∧ Parent(y , z) → Gr-Parent(x, z) Mother Carrie Bob Father(x, y ) → Pateras(x, y ) Source Target Gr-Parent(x, y ) → Pappoi(x, y ) PaterasAndy BobBob DannyDanny Eddie Pateras(x, y ) ∧ Pateras(y , z) → Pappoi(x, z) PappoiCarrie Danny
7. 7. Exchanging More than Complete Data Jorge P´rez e Universidad de Chilejoint work with Marcelo Arenas (PUC-Chile) and Juan Reutter (Univ. Edinburgh)
8. 8. We can exchange more than complete data ◮ In data exchange we begin with a database instance we begin with complete data ◮ What if we have a representation of a set of possible instances? ◮ We propose a new general formalism to exchange representations of possible instances and apply it to incomplete instances and knowledge bases
9. 9. Outline Formalism for exchanging representations systems Applications to incomplete instances Applications to knowledge bases What is the complexity of testing kb-solutions? How can we compute a good kb-solution? Concluding remarks
10. 10. Outline Formalism for exchanging representations systems Applications to incomplete instances Applications to knowledge bases What is the complexity of testing kb-solutions? How can we compute a good kb-solution? Concluding remarks
11. 11. Representation systems A representation system is composed of: ◮ a set W of representatives ◮ a function rep from W to sets of instances rep V −→ {I1 , I2 , I3 , . . .} = rep(V)
12. 12. Representation systems A representation system is composed of: ◮ a set W of representatives ◮ a function rep from W to sets of instances rep V −→ {I1 , I2 , I3 , . . .} = rep(V) ◮ Incomplete instances and knowledge bases are representation systems
13. 13. In classical data exchangewe consider only complete data M = (S, T, Σ) a schema mapping from S to T I ∈ Inst(S), J ∈ Inst(T) J is a solution for I under M iff (I , J) |= Σ J ∈ SolM (I )
14. 14. In classical data exchangewe consider only complete data M = (S, T, Σ) a schema mapping from S to T I ∈ Inst(S), J ∈ Inst(T) J is a solution for I under M iff (I , J) |= Σ J ∈ SolM (I ) We can extend this to set of instances: given a set X ⊆ Inst(S) SolM (X ) = SolM (I ) I ∈X
15. 15. Extending the definition to representation systems Given: ◮ a mapping M = (S, T, Σ) ◮ a representation system R = (W, rep) ◮ U, V ∈ W
16. 16. Extending the definition to representation systems Given: ◮ a mapping M = (S, T, Σ) ◮ a representation system R = (W, rep) ◮ U, V ∈ W Definition U is an R-solution of V if rep(U) ⊆ SolM (rep(V))
17. 17. Extending the definition to representation systems Given: ◮ a mapping M = (S, T, Σ) ◮ a representation system R = (W, rep) ◮ U, V ∈ W Definition U is an R-solution of V if rep(U) ⊆ SolM (rep(V)) or equivalently, ∀J ∈ rep(U), ∃I ∈ rep(V) such that J ∈ SolM (I )
18. 18. Universal solutions and strong representation systems Definition U is a universal R-solution of V if rep(U) = SolM (rep(V))
19. 19. Universal solutions and strong representation systems Definition U is a universal R-solution of V if rep(U) = SolM (rep(V)) Let C be a class of mappings, Definition R = (W, rep) is a strong representation system for C if for every M ∈ C, and V ∈ W, there exists a U ∈ W such that: rep(U) = SolM (rep(V))
20. 20. Universal solutions and strong representation systems Definition U is a universal R-solution of V if rep(U) = SolM (rep(V)) Let C be a class of mappings, Definition R = (W, rep) is a strong representation system for C if for every M ∈ C, and V ∈ W, there exists a U ∈ W such that: rep(U) = SolM (rep(V)) Strong representation systems ⇒ universal solutions for representatives in R can always be represented in the same system.
21. 21. Outline Formalism for exchanging representations systems Applications to incomplete instances Applications to knowledge bases What is the complexity of testing kb-solutions? How can we compute a good kb-solution? Concluding remarks
22. 22. Incomplete Instances We consider: ◮ Naive instances: instances with null values R(1, N1 ) R(N1 , 2)
23. 23. Incomplete Instances We consider: ◮ Naive instances: instances with null values R(1, N1 ) R(N1 , 2) ◮ Positive conditional instances: naive instances plus conditions for tuples ◮ equalities between nulls, and between nulls and constants ◮ conjunctions and disjunctions T (1, N1 ) true R(1, N1 , N2 ) N1 = N2 ∨ (N1 = 2 ∧ N2 = 3)
24. 24. Incomplete Instances We consider: ◮ Naive instances: instances with null values R(1, N1 ) R(N1 , 2) ◮ Positive conditional instances: naive instances plus conditions for tuples ◮ equalities between nulls, and between nulls and constants ◮ conjunctions and disjunctions T (1, N1 ) true R(1, N1 , N2 ) N1 = N2 ∨ (N1 = 2 ∧ N2 = 3) Semantics given by valuations of nulls (+ open world): rep(I) = {K | µ(I) ⊆ K for some valuation µ}
25. 25. Strong representation systems for st-tgds Let M = (S, T, Σ) be specified by source-to-target tgds (st-tgd: φS (¯ ) → ∃¯ψT (¯ , y ), φS and ψT are CQs) x y x ¯ (FKMP03) For every complete instance I there exists a naive instance J s.t. rep(J ) = SolM (I )
26. 26. Strong representation systems for st-tgds Let M = (S, T, Σ) be specified by source-to-target tgds (st-tgd: φS (¯ ) → ∃¯ψT (¯ , y ), φS and ψT are CQs) x y x ¯ (FKMP03) For every complete instance I there exists a naive instance J s.t. rep(J ) = SolM (I ) Proposition Naive instances are not a strong representation system for st-tgds Idea Mng(N, Alice) Mng(x, y ) → Reports(x, y ) Mng(x, x) → Self-Mng(x)
27. 27. Strong representation systems for st-tgds Let M = (S, T, Σ) be specified by source-to-target tgds (st-tgd: φS (¯ ) → ∃¯ψT (¯ , y ), φS and ψT are CQs) x y x ¯ (FKMP03) For every complete instance I there exists a naive instance J s.t. rep(J ) = SolM (I ) Proposition Naive instances are not a strong representation system for st-tgds Idea Mng(N, Alice) Mng(x, y ) → Reports(x, y ) ? Mng(x, x) → Self-Mng(x)
28. 28. Positive conditional instances are enough Theorem Positive conditional instances are a strong representation system for st-tgds.
29. 29. Positive conditional instances are enough Theorem Positive conditional instances are a strong representation system for st-tgds. Mng(N, Alice) Mng(x, y ) → Reports(x, y ) Mng(x, x) → Self-Mng(x)
30. 30. Positive conditional instances are enough Theorem Positive conditional instances are a strong representation system for st-tgds. Mng(N, Alice) Mng(x, y ) → Reports(x, y ) Reports(N, Alice) true Mng(x, x) → Self-Mng(x)
31. 31. Positive conditional instances are enough Theorem Positive conditional instances are a strong representation system for st-tgds. Mng(N, Alice) Mng(x, y ) → Reports(x, y ) Reports(N, Alice) true Mng(x, x) → Self-Mng(x) Self-Mng(Alice)
32. 32. Positive conditional instances are enough Theorem Positive conditional instances are a strong representation system for st-tgds. Mng(N, Alice) Mng(x, y ) → Reports(x, y ) Reports(N, Alice) true Mng(x, x) → Self-Mng(x) Self-Mng(Alice) N = Alice
33. 33. Positive conditional instances are enough Theorem Positive conditional instances are a strong representation system for st-tgds. Mng(N, Alice) Mng(x, y ) → Reports(x, y ) Reports(N, Alice) true Mng(x, x) → Self-Mng(x) Self-Mng(Alice) N = Alice Corollary Positive conditional instances are a strong representation system for SO-tgds.
34. 34. Positive conditional instances areexactly the needed representation system Theorem Positive conditional instances are minimal: ◮ equalities between nulls ◮ equalities between constant and nulls ◮ conjunctions and disjunctions are all needed for a strong representation system of st-tgds
35. 35. Positive conditional instancesmatch the complexity bounds of classical data exchange Theorem For positive conditional instances and (fixed) st-tgds ◮ Universal solutions can be constructed in polynomial time. ◮ Certain answers of unions of conjunctive queries can be computed in polynomial time.
36. 36. Outline Formalism for exchanging representations systems Applications to incomplete instances Applications to knowledge bases What is the complexity of testing kb-solutions? How can we compute a good kb-solution? Concluding remarks
37. 37. The semantics of knowledge basesis given by sets of instances Knowledge base over S: (I , Γ) s.t. I ∈ Inst(S) Γ a set of rules over S Semantics: finite models Mod(I , Γ) = {K ∈ Inst(S) | I ⊆ K and K |= Γ}
38. 38. We can apply our formalismto knowledge bases (I2 , Γ2 ) is a kb-solution for (I1 , Γ1 ) under M iff Mod(I2 , Γ2 ) ⊆ SolM (Mod(I1 , Γ1 ))
39. 39. Outline Formalism for exchanging representations systems Applications to incomplete instances Applications to knowledge bases What is the complexity of testing kb-solutions? How can we compute a good kb-solution? Concluding remarks
40. 40. We study the following decision problem Check-KB-Sol: Input: M = (S, T, Σ) with Σ st-tgds (I1 , Γ1 ) knowledge base over S with Γ1 tgds (I2 , Γ2 ) knowledge base over T with Γ2 tgds Output: Is (I2 , Γ2 ) a kb-solution of (I1 , Γ1 ) under M?
41. 41. We study the following decision problem Check-KB-Sol: Input: M = (S, T, Σ) with Σ st-tgds (I1 , Γ1 ) knowledge base over S with Γ1 tgds (I2 , Γ2 ) knowledge base over T with Γ2 tgds Output: Is (I2 , Γ2 ) a kb-solution of (I1 , Γ1 ) under M? Theorem Check-KB-Sol is undecidable (even for fixed M).
42. 42. We study the following decision problem Check-KB-Sol: Input: M = (S, T, Σ) with Σ st-tgds (I1 , Γ1 ) knowledge base over S with Γ1 tgds (I2 , Γ2 ) knowledge base over T with Γ2 tgds Output: Is (I2 , Γ2 ) a kb-solution of (I1 , Γ1 ) under M? Theorem Check-KB-Sol is undecidable (even for fixed M). Undecidability is a consequence of using ∃ in knowledge bases Check-Full-KB-Sol: Γ1 , Γ2 full-tgds (no ∃)
43. 43. The complexity of Check-Full-KB-Sol Theorem Check-Full-KB-Sol is EXPTIME-complete.
44. 44. The complexity of Check-Full-KB-Sol Theorem Check-Full-KB-Sol is EXPTIME-complete. Theorem If M = (S, T, Σ) is fixed Check-Full-KB-Sol is ∆P [O(log n)]-complete. 2 ∆P [O(log n)] 2 ≡ P NP with only a logarithmic number of calls to the NP oracle.
45. 45. Hardness is obtained by reducingthe Max-Odd-Clique problem Given a graph G with n nodes: S: E (·, ·), Succ(·, ·), Clique(·)
46. 46. Hardness is obtained by reducingthe Max-Odd-Clique problem Given a graph G with n nodes: S: E (·, ·), Succ(·, ·), Clique(·) T: E ′ (·, ·), Succ ′ (·, ·), Clique ′ (·)
47. 47. Hardness is obtained by reducingthe Max-Odd-Clique problem Given a graph G with n nodes: S: E (·, ·), Succ(·, ·), Clique(·) T: E ′ (·, ·), Succ ′ (·, ·), Clique ′ (·) I1 and I2 are copies of G , plus a successor relation over [1, n]
48. 48. Hardness is obtained by reducingthe Max-Odd-Clique problem Given a graph G with n nodes: S: E (·, ·), Succ(·, ·), Clique(·) T: E ′ (·, ·), Succ ′ (·, ·), Clique ′ (·) I1 and I2 are copies of G , plus a successor relation over [1, n] Γ1 : “E has a clique of even size x ≤ n” → Clique(x)
49. 49. Hardness is obtained by reducingthe Max-Odd-Clique problem Given a graph G with n nodes: S: E (·, ·), Succ(·, ·), Clique(·) T: E ′ (·, ·), Succ ′ (·, ·), Clique ′ (·) I1 and I2 are copies of G , plus a successor relation over [1, n] Γ1 : “E has a clique of even size x ≤ n” → Clique(x) Γ2 : “E ′ has a clique of odd size x ≤ n” → Clique ′ (x)
50. 50. Hardness is obtained by reducingthe Max-Odd-Clique problem Given a graph G with n nodes: S: E (·, ·), Succ(·, ·), Clique(·) T: E ′ (·, ·), Succ ′ (·, ·), Clique ′ (·) I1 and I2 are copies of G , plus a successor relation over [1, n] Γ1 : “E has a clique of even size x ≤ n” → Clique(x) Γ2 : “E ′ has a clique of odd size x ≤ n” → Clique ′ (x) Σ: Clique(x) ∧ Succ(x, y ) → Clique ′ (y )
51. 51. Hardness is obtained by reducingthe Max-Odd-Clique problem Given a graph G with n nodes: S: E (·, ·), Succ(·, ·), Clique(·) T: E ′ (·, ·), Succ ′ (·, ·), Clique ′ (·) I1 and I2 are copies of G , plus a successor relation over [1, n] Γ1 : “E has a clique of even size x ≤ n” → Clique(x) Γ2 : “E ′ has a clique of odd size x ≤ n” → Clique ′ (x) Σ: Clique(x) ∧ Succ(x, y ) → Clique ′ (y ) (I2 , Γ2 ) is a kb-solution of (I1 , Σ1 ) iff the maximum size of a clique in G is odd.
52. 52. Outline Formalism for exchanging representations systems Applications to incomplete instances Applications to knowledge bases What is the complexity of testing kb-solutions? How can we compute a good kb-solution? Concluding remarks
53. 53. Minimal knowledge bases as good kb-solutions What are good knowledge-base solutions? ◮ first choice: universal kb-solutions, but ◮ there are some other kb-solutions desirable to materialize
54. 54. Minimal knowledge bases as good kb-solutions What are good knowledge-base solutions? ◮ first choice: universal kb-solutions, but ◮ there are some other kb-solutions desirable to materialize X ≡min Y: X and Y coincide in the minimal instances
55. 55. Minimal knowledge bases as good kb-solutions What are good knowledge-base solutions? ◮ first choice: universal kb-solutions, but ◮ there are some other kb-solutions desirable to materialize X ≡min Y: X and Y coincide in the minimal instances (I2 , Γ2 ) is a minimal kb-solution of (I1 , Γ1 ) under M if Mod(I2 , Γ2 ) ≡min SolM (Mod(I1 , Γ1 )) We consider minimal kb-solutions as the good solutions
56. 56. Two requirements to constructminimal knowledge-base solutions Given (I1 , Γ1 ) and Σ, when constructing a minimal-knowledge base solution (I2 , Γ2 ) we would want: 1. Γ2 to only depend on Γ1 and Σ Γ2 is safe for Γ1 and Σ
57. 57. Two requirements to constructminimal knowledge-base solutions Given (I1 , Γ1 ) and Σ, when constructing a minimal-knowledge base solution (I2 , Γ2 ) we would want: 1. Γ2 to only depend on Γ1 and Σ Γ2 is safe for Γ1 and Σ Definition Γ2 is safe for Γ1 and Σ, if for every I1 there exists I2 s.t. (I2 , Γ2 ) is a minimal kb-solution of (I1 , Γ1 )
58. 58. Two requirements to constructminimal knowledge-base solutions Given (I1 , Γ1 ) and Σ, when constructing a minimal-knowledge base solution (I2 , Γ2 ) we would want: 2. Γ2 to be as informative as possible thus minimizing the size of I2
59. 59. Two requirements to constructminimal knowledge-base solutions Given (I1 , Γ1 ) and Σ, when constructing a minimal-knowledge base solution (I2 , Γ2 ) we would want: 2. Γ2 to be as informative as possible thus minimizing the size of I2 Γ2 is optimum-safe if for every other safe set Γ′ Γ2 |= Γ′
60. 60. Safe sets: data independent target knowledge bases Unfortunately, optimum-safe sets are not (always) FO-expressible Theorem There exist sets Γ1 (full & acyclic) and Σ (full) s.t. no FO sentence is optimum-safe for Γ1 and Σ
61. 61. Safe sets: data independent target knowledge bases Unfortunately, optimum-safe sets are not (always) FO-expressible Theorem There exist sets Γ1 (full & acyclic) and Σ (full) s.t. no FO sentence is optimum-safe for Γ1 and Σ In our paper: An algorithm to compute optimum-safe set in SO
62. 62. Safe sets: data independent target knowledge bases Unfortunately, optimum-safe sets are not (always) FO-expressible Theorem There exist sets Γ1 (full & acyclic) and Σ (full) s.t. no FO sentence is optimum-safe for Γ1 and Σ In our paper: An algorithm to compute optimum-safe set in SO Can we do better? not optimum but useful set? (non-trivial safe set in a good language)
63. 63. We can compute a good safe setfor acyclic tgds knowledge bases Input: Γ1 acyclic full-tgds, Σ full-tgds (from S to T)
64. 64. We can compute a good safe setfor acyclic tgds knowledge bases Input: Γ1 acyclic full-tgds, Σ full-tgds (from S to T) 1. Unfold Γ1 to get Γ+ 1
65. 65. We can compute a good safe setfor acyclic tgds knowledge bases Input: Γ1 acyclic full-tgds, Σ full-tgds (from S to T) 1. Unfold Γ1 to get Γ+ 1 2. Construct Σ− (from T to S) as follows: for every α(¯ ) → R(¯) in Γ+ x x 1
66. 66. We can compute a good safe setfor acyclic tgds knowledge bases Input: Γ1 acyclic full-tgds, Σ full-tgds (from S to T) 1. Unfold Γ1 to get Γ+ 1 2. Construct Σ− (from T to S) as follows: for every α(¯ ) → R(¯) in Γ+ x x 1 Rewrite α(¯) in the target of Σ to obtain β(¯ ) x x
67. 67. We can compute a good safe setfor acyclic tgds knowledge bases Input: Γ1 acyclic full-tgds, Σ full-tgds (from S to T) 1. Unfold Γ1 to get Γ+ 1 2. Construct Σ− (from T to S) as follows: for every α(¯ ) → R(¯) in Γ+ x x 1 Rewrite α(¯) in the target of Σ to obtain β(¯ ) x x Add β(¯) → R(¯) to Σ− x x
68. 68. We can compute a good safe setfor acyclic tgds knowledge bases Input: Γ1 acyclic full-tgds, Σ full-tgds (from S to T) 1. Unfold Γ1 to get Γ+ 1 2. Construct Σ− (from T to S) as follows: for every α(¯ ) → R(¯) in Γ+ x x 1 Rewrite α(¯) in the target of Σ to obtain β(¯ ) x x Add β(¯) → R(¯) to Σ− x x 3. Let Γ2 = Σ− ◦ Σ
69. 69. We can compute a good safe setfor acyclic tgds knowledge bases Input: Γ1 acyclic full-tgds, Σ full-tgds (from S to T) 1. Unfold Γ1 to get Γ+ 1 2. Construct Σ− (from T to S) as follows: for every α(¯ ) → R(¯) in Γ+ x x 1 Rewrite α(¯) in the target of Σ to obtain β(¯ ) x x Add β(¯) → R(¯) to Σ− x x 3. Let Γ2 = Σ− ◦ Σ Theorem Γ2 is safe for Γ1 and Σ (Γ2 is a set of full-tgds with =)
70. 70. In the paper: a complete strategy for computingminimal knowledge-base solutions Given Γ1 and Σ: ◮ We compute Γ2 safe for Γ1 and Σ ◮ We then compute minimal set Γ∗ (implied by Γ1 ) such that for every I1
71. 71. In the paper: a complete strategy for computingminimal knowledge-base solutions Given Γ1 and Σ: ◮ We compute Γ2 safe for Γ1 and Σ ◮ We then compute minimal set Γ∗ (implied by Γ1 ) such that for every I1 chaseΓ∗ (I1 )
72. 72. In the paper: a complete strategy for computingminimal knowledge-base solutions Given Γ1 and Σ: ◮ We compute Γ2 safe for Γ1 and Σ ◮ We then compute minimal set Γ∗ (implied by Γ1 ) such that for every I1 chaseΣ ( chaseΓ∗ (I1 ))
73. 73. In the paper: a complete strategy for computingminimal knowledge-base solutions Given Γ1 and Σ: ◮ We compute Γ2 safe for Γ1 and Σ ◮ We then compute minimal set Γ∗ (implied by Γ1 ) such that for every I1 chaseΣ ( chaseΓ∗ (I1 )), Γ2
74. 74. In the paper: a complete strategy for computingminimal knowledge-base solutions Given Γ1 and Σ: ◮ We compute Γ2 safe for Γ1 and Σ ◮ We then compute minimal set Γ∗ (implied by Γ1 ) such that for every I1 chaseΣ ( chaseΓ∗ (I1 )), Γ2 is a minimal kb-solution for (I1 , Γ1 ).
75. 75. Outline Formalism for exchanging representations systems Applications to incomplete instances Applications to knowledge bases What is the complexity of testing kb-solutions? How can we compute a good kb-solution? Concluding remarks
76. 76. We can exchange more than complete data We propose a general formalism to exchange representation systems ◮ Applications to incomplete instances ◮ Applications to knowledge bases
77. 77. We can exchange more than complete data We propose a general formalism to exchange representation systems ◮ Applications to incomplete instances ◮ Applications to knowledge bases Next step: apply our general setting to the Semantic Web ◮ Semantic Web data have nulls (blank nodes) ◮ Semantic Web specifications have rules (RDFS, OWL)
78. 78. Exchanging More than Complete Data Jorge P´rez e Universidad de Chilejoint work with Marcelo Arenas (PUC-Chile) and Juan Reutter (Univ. Edinburgh)
79. 79. Outline Formalism for exchanging representations systems Applications to incomplete instances Applications to knowledge bases What is the complexity of testing kb-solutions? How can we compute a good kb-solution? Concluding remarks | 6,379 | 21,030 | {"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-2017-39 | latest | en | 0.417639 |
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### Questions
2. 2. WWW.SMUSOLVEDASSIGNMENTS.COM (Formulas-1 Marks, Calculation/solution – 3 Marks, Interpretation – 1 Mark) Q4. Index number acts as a barometer for measuring the value of money. What are the characteristics of an index number? State its utility. (Meaning – 2 marks, Utility – 2 marks, Characteristics – 6 marks) Q5. Business forecasting acquires an important place in every field of the economy. Explain the objectives and theories of Business forecasting. (Meaning – 2 marks, Objectives – 3 marks, Theories – 5 marks) Q6. The weekly wages of 1000 workers are normally distributed around a mean of Rs. 70 and a standard deviation of Rs. 5. Estimate the number of workers whose weekly wages will be: a. Between 70 and 72 b. Between 69 and 72 c. More than 75 d. Less than 63 (Formula – 2 marks, Calculation/Solution/Interpretation-8 marks) MB0041 – Financial and Management Accounting Q1. Accounting is one of the oldest, structured management information system. Give the meaning of accounting and book keeping? Explain the objectives of accounting? (Meaning of accounting 2 marks; Meaning of book keeping 3 marks; objectives of accounting 5 marks) Q2. Explain GAAP and write down the relationship between accounting principles, accounting concepts, and accounting conventions. Explain all the five accounting concepts with an example. (Meaning of GAAP 2 marks; Relationship between accounting principles, accounting concepts, and accounting conventions 3 marks; Explanation of five accounting concepts with examples 5 marks) Q3. List down the classification of accounts according to accounting equation approach. Give the meaning and examples for all the types of accounts. (listing 2 marks; meaning of 5 types of accounts 4 marks; examples for all 5 types of accounts 4 marks) Q4. What is cash book? Differentiate between other subsidiary books and cash book. (Meaning of cash book 2 marks; Differences of subsidiary book and cash book 4 differences (each difference carries 2 marks) 8 marks) Q5. The following items are found in the trial balance of M/s Sharada Enterprise on 31st December, 2000. Sundry Debtors Rs.160000 Bad Debts written off Rs 9000 Discount allowed to Debtors Rs. 1800 Reserve for Bad and doubtful Debts 31-12-1999 Rs. 16500 Reserve for discount on Debtors 31-12-1999 Rs. 3200 You are required to provide the bad and doubtful debts at 5% and for discount on debtors at 2%. Show the adjustments for bad debts, bad debts reserve, discount account, and provision for discount on debtors. Hint: RBD to be provided = 500 Reserve for discount to be provided now =1640 (Calculation of amount debited to P/L a/c towards RBD 3 marks ; calculation of amount debited to P/L a/c towards reserve for discount on debtors 5 marks; conclusion with summary 2 marks) Q6. What is management accounting? Explain the roles of management accounting and write down about any 2 functions of management accounting. (meaning of management accounting 2marks ; explanation on roles of management accounting 4 marks ; functions of any 2 management accounting (any 2 each carries 2 marks) 4 marks) 10 Marks MB0042 – Managerial Economics Q1. Discuss the practical application of Price elasticity and Income elasticity of demand. (Practical application of price elasticity -5 marks; practical application of income elasticity of demand - 5 marks) 10 marks Q2. Explain the profit maximisation model in detail. (Main propositions- 4; Explanation- 3 marks; Assumption- 3 marks) 10 marks Q3. Describe the objectives of pricing Policies. [ List of objectives-(any 10 objectives- 1 mark for each objective)-10 marks] 10 marks
3. 3. WWW.SMUSOLVEDASSIGNMENTS.COM Q4. Define Fiscal Policy and the instruments of Fiscal policy. Definition- 2 marks , Instruments of Fiscal Policy-8 marks) 10 marks Q5. Explain the kinds and the basis of Price discrimination under monopoly. (Kinds of Price discrimination-3, Basis Of Price Discrimination-7 Marks) 10 marks Q6. Define the term Business Cycle and also explain the phases of business or trade cycle in brief. (Definition of Business cycle – 2 marks, Phases of business or trade cycle- 8 marks) 10 marks MB0043 – Human Resource Management Q1. Define Human resource planning (HRP). Explain the objectives and process of HRP. (Definition of HRP -2, List the objectives -2, Explanation of process -3 Description of steps in HRP -3) 10 marks Q2. What are the factors affecting recruitment? What are the sources of recruitment? (Meaning - 3, Description of factors - 3, Explanation of Internal sources- 2, Explanation of external sources - 2) 10 marks Q3. What are the main objectives of training? Explain on-the job and off the job training. (Meaning of training - 1, Objectives of training - 3, Explanation of „on the job‟ training - 3, explanation of „off-the job‟ training - 3) 10 marks Q4. Define performance management. Write a brief note on 360 degree appraisal. (Definition of PM- 2, Meaning of 360 degree appraisal- 3, Feedback from multiple sources 5) 10 marks Q5. What is meant by job analysis? Explain its purpose and methods. (Meaning of job analysis - 2, Explanation of purpose - 3, Explanation of methods of job analysis - 5) 10 marks Q6. What are the benefits and objectives of employee welfare measures?(Meaning of employee welfare - 2, Explanation of benefits - 4, Objectives - 4) SEM 3 COMMONSUMMER 2013 MB0050 – Research Methodology Q1. Explain the process of problem identification with an example. (Process – 7 marks, Example – 3 marks) 10 marks Q2. Interview method involves a dialogue between the Interviewee and the Interviewer. Explain the interview method of data collection. What are the uses of this technique? What are the different types of interviews? (Explanation – 4 marks, Uses – 3 marks, Types – 3 marks) 10 marks Q3. A study of different sampling methods is necessary because precision, accuracy, and efficiency of the sample results depend on the method employed for selecting the sample. Explain the different types of Probability and Non-Probability sampling designs. (Probability sampling designs – 5 marks, Non- probability designs – 5 marks) 10 marks Q4. a. Differentiate between descriptive and inferential analysis of data. (Differences – 5 marks)b. Explain with examples various measures of Central Tendency. (Explanation – 5 marks) 10 marks Q5. The chi-square test is widely used in research. Discuss the various applications of chisquare test. Under what conditions is this test applicable? (Meaning – 3 marks, Applications – 4 marks, Conditions – 3 marks) 10 marks Q6. What is analysis of variance? What are the assumptions of the technique? Give a few examples where this technique could be used. (Meaning – 3 marks, Assumptions – 4 marks, Examples – 3 marks) 10 marks MB0051 –Legal aspects of Business Q1. It is important for any person to know law as ignorance of law is no excuse. Modern Indian law has been derived from some sources. Discuss the primary and secondary sources of Indian law. (primary sources- 5 marks, secondary sources- 5 marks) 10 marks
4. 4. WWW.SMUSOLVEDASSIGNMENTS.COM Q2. We all enter into many contracts in a day knowingly or unknowingly. Explain the definition of a valid contract. How are contracts classified? (definition - 5 marks, classification – 5 marks) 10 marks Q3. The parties to bailment have certain rights and duties. Discuss the duties of both parties i.e. the bailor and bailee. ( duties of bailor- 5 marks, duties of bailee- 5 marks) 10 marks Q4. A contract comprises of reciprocal promises. In a contract of sale who is an unpaid seller? Discuss the remedies for breach of contract under Sale of Goods Act, 1930. (unpaid seller- 2 marks, breach of contract- 8 marks) 10 marks Q5. The Companies Act, 1956 deals with the formation and transaction of business of a company. Discuss the features of a company. Also explain the process of formation of a company. ( features- 3 marks, process of formation- 7 marks) 10 marks Q6. With Information Technology Act, 2000, India has a set of cyber laws to provide legal infrastructure for e commerce. Discuss the objectives and limitations of this Act. (objectives3 marks, limitations- 7 marks) 10 marks SEM 3 BANKINGSUMMER 2013 MA0036 – Financial system and Commercial Banking Q1. The key functions of financial system are to provide a link between savers and investors. What are the key functions of financial market? (explanation of functions of financial markets - 10 marks) 10 marks Q2. Explain the meaning and the purpose of financial instruments. (meaning - 5 marks; functions of financial instruments - 5 marks) 10 marks Q3. Discuss the role played by brokers and primary dealers in the process of intermediation? (brokers role-5 marks; primary dealers role - 5marks) 10 marks Q4. The Bank for International Settlement (BIS) was established in 1930 with its headquarters in Basel, Switzerland. Explain Basel Concordant. (Basel origin- 2marks; Basel Concordant- 8 marks) 10 marks Q5. The RBI, apart from the role of regulator and supervisor of payment systems, also plays the role of a settlement bank apart from being a catalyst, an operator and a user. Discuss the role played by RBI in technological upgradation? (explain RBI- 4 marks; explain the role played by RBI in technological upgradation-6 marks) 10 marks Q6. Explain the challenges and issues regarding the perspective in banking industry. (challenges-5 marks; issues-5 marks) 10 marks MA0037 – Banking Related Laws and Practices Q1. Define the term banking. What are the permitted businesses for a banking company as per BR Act 1949? (explanation of term Banking - 4 marks; features of business of banking – 6 marks) 10 marks Q2. When is a negotiable instrument considered as dishonoured? What steps should be taken by the holder? (explanation of Negotiable Instrument-2 marks; features of dishonouring a negotiable instrument-5 marks; action to be taken by the holder- 3 marks) 10 marks Q3. Certain goods of A were bailed with B. B omitted to lock up the goods bailed while he has taken care to lock up similar goods of his own. Who is liable to whom? (bailee meaning3 marks; duties of bailee -7marks) 10 marks Q4. Clayton‟s case is considered to be one of the most essential legal decisions in banking laws that established the principle of the order of application of credits against debits, in running accounts like overdraft. Explain Clayton‟s case. (explanation of Clayton‟s case- 7 marks; usage -3 marks) 10 marks Q5. Write about constitutional validity of the DRT Act.(explain the DRT act-3 marks; explanation of constitutional validity DRT Act- 7 marks) 10 marks Q6. “Cooperative principles” means the cooperative principles specified in the First Schedule of the Multi-State Co-operatives Act, 2002. Explain cooperative principles. (explanation of cooperatives- 4 marks; cooperative principles- 6 marks) 10 marks
5. 5. WWW.SMUSOLVEDASSIGNMENTS.COM MA0038 -Bank Management Q1. Intermediation is the process of linking savers of money with those who are in need of money. Explain the intermediation process of banks. (explanation of intermediation process5 marks; role played by banks-5 marks) 10 marks Q2. Business loans form the core of credit portfolio of banks .What are the basic objectives which the banks pursue while pricing their business loans? (explanation of objectives of pricing the business loans- 5 marks; objectives of banks- 5marks marks) 10 marks Q3. Explain non-interest income and non-interest expenses. (explanation of non-interest income - 5marks; explanation of non-interest expenses - 5marks) 10 marks Q4. Banks investment consists of different types of instruments. Explain the composition of investments. (explanation of investments- 3 marks; composition of investments- 7 marks) 10 marks Q5. What are the guidelines prescribed by the RBI to open a foreign bank branch in India? (meaning of foreign bank-3 marks; guidelines prescribed by RBI -7 marks) 10 marks Q6. Explain profitability analysis models. (explanation of different models- 2 marks; explanation of profitability analysis- 6 marks; benefits-2 marks) 10 marks MA0039 –Retail Banking Q1.The Banking Laws (Amendment) Act 1983 introduced section 45 ZA in the Banking Regulation Act, 1949, which facilitates applicability of nomination to all deposit accounts. What are the benefits of nomination to a depositor?(explanation of nomination-4 marks; benefits-6 marks) 10 marks Q2.Electronic clearing services include both credit and debit. It is regulated by RBI. Explain Electronic Clearing Service (ECS).(explanation of ECS debit - 5 marks+ ECS - Credit - 5 marks 10 marks Q3.Banks need to implement KYC guidelines for all prospective customers before entertaining new business. Explain KYC guidelines.(introduction of KYC guidelines-3 marks; explanation of KYC guidelines - 7 marks) 10 marks Q4.Cross selling is an act of selling a range of additional products to a customer who has already availed of a particular product or service from the seller or the service provider. Explain Cross selling.(explanation of cross selling- 5 marks; benefits-5 marks) 10 marks Q5.The services extended by banks through technology enabled channels are cost effective and increase the profitability of the bank. Explain Internet banking.(explanation– 5 marks; benefits- 5marks) 10 marks Q6.Explain inter bank settlements.(introduction- 3 marks; explanation- 7 marks) 10 marks SEM 3 FINANCESUMMER 2013 MF0010 – Security Analysis and Portfolio Management Q1. Explain the characteristics of investment. Differentiate between investment and speculation. (Characteristics of Investment 5 marks ; Difference between investment and speculation 5 marks) 10 marks Q2. What do you understand risk and measurement of risk? Explain the factors that affect risk. (Explanation to risk 2 marks; measurement to risk 2 marks ; factors that affect risk 6 marks) 10 marks Q3. Compare and contrast the fundamental and technical analysis (Differences between fundamental and technical analysis 4 differences each carries 2 marks - 8 marks; Conclusion 2 marks) 10 marks Q4. Write the assumptions of CAPM. Explain the limitations of CAPM. (Assumptions of CAPM 5 marks; Limitations of CAPM 5 marks) 10 marks Q5. Write about emerging markets. Explain the risks involved in international investing. (Introduction of emerging markets 2 marks; Features of emerging markets 2 marks; Risks involved in international investing 6 marks) 10 marks
6. 6. WWW.SMUSOLVEDASSIGNMENTS.COM Q6. What is economy analysis? Explain the factors to be considered in economy analysis. (Introduction of economy analysis 2 marks ; factors in economy analysis 8 marks) 10 Marks MF0011 – Mergers and Acquisitions Q1. Write the types of mergers and acquisitions. Explain the steps to a successful merger. (Explanation on types of mergers and acquisitions 5marks; Steps to a successful merger 5marks) 10 marks Q2. Explain the process of merger. Write down the goals of a merger. (Process of merger 5marks; Goals of a merger 5marks) 10 marks Q3. What is creating synergy? Explain the prerequisites for the creation of synergy. (Introduction of creating synergy 2marks; Pre requisites for the creation of synergy( all the 4 points to be explained each carries 2 marks) 8marks) 10 marks Q4. Give the meaning of Divesture. List and explain the reasons for divesture. (Meaning of divesture 2marks; Listing of reasons for divesture 3marks; Explanation of reasons for divesture 5marks) 10 marks Q5. Explain the key rules of Employee Stock Ownership Plans. Discuss the two types of ESOPs. (Key rules of ESOP 5marks ; Explanation on two types of ESOP 5marks) 10 marks Q6. Explain the following with examples : Exchange rates (3marks) External advantages in different products (3marks) Role of government policies (4 marks) 10 Marks MF0012 – Taxation Management Q1.Explain the objectives of tax planning. Discuss the factors to be considered in tax planning.(Objectives of tax planning 5 marks; Factors in tax planning 5 marks) 10 marks Q2.Explain the categories in Capital assets.Mr. C acquired a plot of land on 15th June, 1993 for 10,00,000 and sold it on 5th January, 2010 for 41,00,000. The expenses of transfer were 1,00,000.Mr. C made the following investments on 4th February, 2010 from the proceeds of the plot.a) Bonds of Rural Electrification Corporation redeemable after a period of three years, 12,00,000.b) Deposits under Capital Gain Scheme for purchase of a residential house 8,00,000 (he does not own any house).Compute the capital gain chargeable to tax for the AY2010-11.(Explanation of categories of capital assets 4 marks ; Calculation of indexed cost of acquisition 2 marks; Calculation of long term capital gain 2 marks; calculation of taxable long term capital gain 2 marks) 10marks Q3.X Ltd. has Unit C which is not functioning satisfactorily. The following are the details of its fixed assets: Asset Date of acquisition Book value (Rs. lakh) Land 10th February, 2003 30 Goodwill (raised in books on 31st 10 March, 2005) Machinery 5th April, 1999 40 Plant 12th April, 2004 20 The written down value (WDV) is Rs. 25 lakh for the machinery, and Rs.15 lakh for the plant. The liabilities on this Unit on 31st March, 2011 are Rs.35 lakh. The following are two options as on 31st March, 2011: Option 1: Slump sale to Y Ltd for a consideration of 85 lakh. Option 2: Individual sale of assets as follows: Land Rs.48 lakh, goodwill Rs.20 lakh, machinery Rs.32 lakh, Plant Rs.17 lakh. The other units derive taxable income and there is no carry forward of loss or depreciation for the company as a whole. Unit C was started on 1st January, 2005. Which option would you choose, and why? (Computation of capital gain for both the options 4 marks; Computation of tax liability for both the options 4 marks ; Conclusion 2 marks) 10marks Q4.What do you understand by customs duty? Explain the taxable events for imported,warehoused and exported goods. List down the types of duties in customs.An
7. 7. WWW.SMUSOLVEDASSIGNMENTS.COM importer imports goods for subsequent sale in India at \$10,000 on assessable value basis. Relevant exchange rate and rate of duty are as follows: Particulars Date Exchange Rate Rate of Basic Declared by CBE&C Customs Duty Date of submission 25th February, 2010 Rs.45/\$ 8% of bill of entry Date of entry inwards 5th March, 2010 Rs.49/\$ 10% granted to the vessel Calculate assessable value and customs duty.(Meaning and explanation of customs duty 2 marks; Explanation of taxable events for imported,warehoused and exported goods 3 marks; Listing of duties in customs 2 marks; Calculation of assessable value and customs duty 3marks) 10marks Q5.Explain the Service Tax Law in India and concept of negative list. Write about the exemptions and rebates in Service Tax Law.(Explanation of Service Tax Law in India 5 marks; explanation of concept of negative list 2marks; Explanation of exemptions and rebates in Service Tax Law 3 marks) 10marks Q6.Explain major considerations in capital structure planning. Write about the dividend policy and factors affecting dividend decisions.(Explanation of factors of capital structure planning 6 marks; Explanation of dividend policy 2 marks; factors affecting dividend decisions 2 marks) 10marks MF0013 – Internal Audit and Control Q1. Discuss, in brief, the advantages and limitations of auditing. (Advantages of auditing 5 marks; Limitations of auditing 5 marks) 10 marks Q2. Explain the key objectives of a good internal audit system. Write down the essentials for effective internal auditing. (Objectives of good internal audit system 5 marks ; Essentials of effective internal auditing 5 marks) 10 marks Q3. List the required qualifications of an internal auditor. Describe the role of internal auditor in the company‟s management. (Listing of qualifications for internal auditor 5 marks; role of internal auditor 5 marks) 10 marks Q4. Explain the basic principles of governing internal control. (Basic principles of governing internal control 10 marks) 10 marks Q5. Discuss the specific problems of Electronic Data Processing (EDP) relating to internal control. (Explanation of all problems of EDP 10 marks) 10 marks Q6. Explain the factors for having the effective internal control system for a bank. (Explanation of various aspects of having the effective internal control system 10 marks) SEM 3 HUMAN RESOURCE(HR) SUMMER 2013 MU0010 – Manpower Planning and Resourcing Q1. Explain the need for manpower planning. What are the obstacles in manpower planning? (Meaning of manpower planning- 1, explanation of need - 3, description of obstacles - 6) 10 marks Q2. What are the objectives of human resource accounting (HRA)? What are the methods of human resource accounting? Explain the cost based approach of HRA. (Explanation of objectives - 3, listing the methods - 1, description of cost based approach - 2, explanation of four cost concepts - 4) 10 marks Q3. What are the objectives of recruitment? Explain recruitment process. (Meaning of recruitment- 2, explanation of objectives - 3, explanation of recruitment process- 5) 10 marks Q4. Describe the benefits of induction program. Explain the types of induction program. (Meaning of induction- 2, explanation benefits of induction program - 2, Listing the types - 1, explanation of types - 4, example - 1) 10 marks
8. 8. WWW.SMUSOLVEDASSIGNMENTS.COM Q5. What do you mean by career management? Explain career planning process. (Meaning of career management - 2, focus of career management process – 1, Meaning of career planning- 2, Process of career planning - 5) 10 marks Q6. Mention the causes of employee turnover. Explain the employee exit process. (Listing 2, Meaning of employee exit process - 2, explanation - 6) MU0011 – Management and Organisational Development Q1. Explain the importance of Organisational development to managers. Describe the characteristics of Organisational Development. (meaning -2, importance -3, characteristics of OD – 5) Q2. What is meant by Organisational change? What are the various strategies for change? Explain the positive model of planned change. (meaning – 2, listing the strategies – 1, explanation of four strategies – 3, explanation of stages - 4) Q3. Define the term „ethical dilemma‟. What are the various ethical dilemmas? Explain. (meaning of ethics – 1, definition of ethical dilemma – 2, examples of ethical mistakes – 1, explanation of ethical dilemmas – 6) Q4. What are the two major types of human process interventions? Describe the role negotiation technique? (listing – 2, explanation of the two human process interventions- 3, meaning negotiation technique- 2, steps of role negotiation technique-3) Q5. What approaches have been set by OD practitioners while setting up the goal program? What are the reasons for setting goals? Explain the advantages and barriers of goal setting.(meaning of goal- 1, explanation of approaches while setting goal – 3, reasons - 2, advantages -2, barriers of goal setting- 2) Q6. What is meant by „learning organisation‟? What are the characteristics of learning organisation? Describe the seven steps of initiating Organisational learning.(definition – 2, characteristics – 2, listing the seven steps -1, explanation of seven steps– 5) MU0012 – Employee Relations Management Q1. Define conflict management. What are the causes of workplace conflicts? What are the various strategies to be adopted for resolving conflicts? (definition of conflict management – 3, explanation of causes of workplace conflicts – 3, explanation of strategies – 4) 10 marks Q2. What is meant by organisational culture? What are the elements and dimensions of organisational culture? (meaning of organisational culture – 3, explanation of elements – 3, dimensions – 4) 10 marks Q3. Explain the importance and features of Human Resource Information systems (HRIS).(meaning – 1, example - 2, importance – 3, features of HRIS – 4) 10 marks Q4. What is meant by grievance redressal? Explain the three stages of Grievance redressal. What precautions are to be taken while handling grievances? (meaning of grievance & grievance redressal – 2, explanation of three stages – 4, precautions – 4) Q5. What are trade unions? What are the objectives and activities of trade unions? (meaning – 2, explanation objectives – 3, activities – 5) Q6. Write a brief note on stress and employee wellness in organisations. (meaning of stress – 2, explanation of symptoms– 1, factors that increases stress – 2, explanation employee wellness – 4, example – 1) MU0013 – HR Audit Q1. Define Human Resource (HR) Audit. What is the need for HR Audit? What are the various approaches to HR Audit? (definition– 2, need for HR audit- 2, explanation of HR audit approach by Walker- 3, common approach to HR audit- 3) 10 marks Q2. Write a brief note on staffing. How does employee orientation programs help employees? What are the characteristics of good employee orientation programs? (meaning of staffing- 2, explanation of staffing – 3, use of orientation programs- 2, characteristics– 3)10 marks
9. 9. WWW.SMUSOLVEDASSIGNMENTS.COM Q3. What is HR Scorecard? Explain the reason for implementing HR Scorecard. (meaning of HR scorecard- 3, explanation of reasons – 7) 10 marks Q4. Define competency management. Explain the two frameworks of competency management. (definition – 3, listing two frameworks – 1, explanation of two frameworks -6) 10 marks Q5. Write a brief note on workplace policies and practices. (explanation on workplace policies and practices – 1, safeguarding employee information – 2, performance management – 2, safe work environment – 2 and auditing workplace behaviours – 3) 10 marks Q6. What are the areas to be concentrated on for HR Audit? Prepare a questionnaire for conducting an audit for manpower planning. (explanation of areas – 4, preparation of questionnaire – 6) SEM 3 MARKETINGSUMMER 2013 MK0010 – Sales, Distribution and Supply Chain Management Q1. How to manage a company‟s sales force? [Managing the sales force(5 processes-each carry two marks)-10 marks] 10 marks Q2. Explain Gap analysis with SERVQUAL model. [Five aspects of service quality- 5 marks; gap analysis- 5 marks] 10 marks Q3. Write a short notes on: A. Elements of Physical Distribution(any four) (Any four elements-four marks) B. Patterns of Distribution [Patterns of Distribution(each pattern carry 2 marks)- 6 marks] 4+6 = 10 marks Q4. Explain three components of Supply chain management. (Definition of supply chain management- 2 marks; components – 8 marks) 10 marks Q5. Define Aggregate Planning and its strategies to meet demand and supply. [Definition of Aggregate planning- 1 mark; Strategies(each carry 3 marks)-9 marks] 10 marks Q6. Explain the challenges faced by International Sales Managers (Explanation of challenges- 10 marks) MK0011 – Consumer Behaviour Q1. Explain the components of learning and also classical conditioning theory in brief.(Components- 3 marks; theory- 7 marks) 10 marks Q2. Briefly discuss the decision-making models (any three). (Decision making models- 10 marks) 10 marks Q3. Write a short notes on the following: A. Forms of Motivational conflict (forms- 3 marks) B. Forms of Defense mechanisms (forms -7 marks) 7+3 marks =10 marks Q4. Describe the levels of consumer decision making while buying. (Levels – two for 3 marks and two for 2 marks) 10 marks Q5. Discuss the influences of the reference group and Applications of reference group in a company. (Influences – 6 marks; Applications – 4 marks) 10 marks Q6. Explain the five categories of Adopters in innovation process. (Categories – each 2 marks) MK0012 – Retail Marketing Q1. Define e-tailing. Explain the future of electronic retailing (Definition- 2 marks; Future of electronic retailing- 8 marks) 10 marks Q2. Explain the factors which are leading to the growth of retail sector. (Listing and explanation – 10 marks) 10 marks Q3. Describe the tools of Integrated marketing communication. Definition of Integrated marketing communication- 1 mark; Indirect marketing tool- 4 marks; Direct marketing tool- 5 marks) 10 marks Q4. Discuss the Retail pricing strategies. (Explanation- 1 mark; Retail Pricing Strategies- 9 marks) 10 marks
10. 10. WWW.SMUSOLVEDASSIGNMENTS.COM Q5. Write a short notes on: A. Types of retail store location with examples(any five) (Types3 marks; examples- 2 marks) B. Factors affecting retail store location(any five) (Factors – 5 marks) 5+5 = 10 marks Q6. Write a short notes on: A. Classification of retail consumers based on shopping. (Classification- 6 marks) B. Types of Buying behaviour (Types- 4 marks) MK0013 –Marketing Research Q1. Write a short notes on: A. Various types of Research(any five) (Types – 5 marks) B. Characteristics of Research (characteristics- 5 marks) 5+5 = 10 marks Q2. Explain the process of sampling and classification of non-probability sampling techniques. (Definition of sampling- 1 mark; process of sampling - 5 marks; classification of nonprobability sampling- 4 marks) 10 marks Q3. Discuss the types of Research Design in brief. (Definition of Research Design- 2 marks; Types – 8 marks) 10 marks Q4. Explain the various types of Consumer and Business to Business (B2B) market research. (Types of Consumer market research- 6 marks; Types of B2B market research- 4 marks) 10 marks Q5. Explain the various methods used to collect primary data in brief. (Definition of Primary data- 2; methods – 8 marks) 10 marks Q6. Explain various methods of central tendency with formula and example for each. Mean (formula – 1 mark, example- 1 mark), Median (formula – 1 mark, example- 1 mark), Mode (formula – 1 mark, example- 1 mark) SEM 3 INFORMATION TECHNOLOGY( IT) SUMMER 2013 MI0033 –Software Engineering Q1. Waterfall Model, V-Model and Spiral Model are of the software development processes. Companies are using these models to have a systematic and defined approach in software development. Which of the four phases are involved in the software development process? (listing the 4 types- 4 marks, explaining the types- 4 marks, examples-2 marks) 10 marks Q2. Suppose you have assigned the task of measuring the software product, how would you use the size oriented and function oriented metrics in your task? (explanation of size oriented metrics- 4 marks, explanation of function oriented metrics- 4marks, examples for each with problem-2 marks) 10 marks Q3. a. You are a SCM manager in a software company. How will you establish a software configuration management process in a company? (definition- 1 marks, 4 procedures in software configuration management process-4 marks) b. You are appointed as a software developer in a software company and you have been asked by your project manager to check details of the bugs from the previous version. How will you gather details associated with the various bugs in the previous version? (definition – 1 mark, typical work cycle-4 marks) 5+5 marks Q4. a. If you are a software engineer you must be expert in the field of software, hardware and also database. It has been listed by the industry professionals that time and effort are the most important factors in the system analysis stage. Briefly explain the steps used in in system analysis. ( listing the steps- 2 marks, explanation-4 marks) b. If you are a software developer in small company, how will you conduct a specification review? (explanation-4 marks) Q5. What are the different methods of software prototypes and tools? (definition of software prototype-1 mark, 3 types of techniques- 3 marks, explaining tools- 4 marks, benefits of software prototype- 2 marks) 10 marks Q6. List and explain any 5 principles of design fundamentals (Listing-5 marks, explanation- 5 marks) MI0034 –Database Management Systems
11. 11. WWW.SMUSOLVEDASSIGNMENTS.COM Q1. Suppose the employee name, employee id, designation, salary, attendance and address of any employee has to be stored in a database. You can store these data in a sequential address book or it can be stored on a hard disk, using a computer and software like Microsoft Excel. Using this example define a database. List and explain the various procedures carried on in a DBMS with a detailed example of the database. (defining a database- 1 mark, listing the 3 procedures – 3 marks, explanation- 3 marks, one examples for each of them - 3 marks)10 marks Q2. What are the different types of interfaces provided by DBMS? (listing the 6 types: 3 marks, listing the properties of these types-7 marks) 10 marks Q3. Level 2 cache has got higher latency than Level 1 by 2 times to 10 times in 512 KiB or more. Its value is nearer to kilobyte. This is one of the levels of memory hierarchy. Define memory hierarchy. What are the other levels in memory hierarchy? Explain in one life each for each of them. (defining memory hierarchy- 1 mark, listing the 6 levels -3 marks, explanation- 6 marks) 10 marks Q4. Indexes are usually defines on a single field of a file called an indexing field. List and describe the different types of indexes (listing the 5 types of indexes-3marks, explanation in detail- 5 marks, examples- 2 marks ) 10 marks Q5. Consider a book is written by a particular author. And you have to explain to some one the relationship that exists between the author and the book. Normally you can draw a diagram and show the relation. These diagrams are called entity-relationship diagram in which book is one entity, author is one entity, and the relationship that exists between the two entities is written. Likewise explain the various notations used to represent the ER diagram. (listing the notations with diagrams – 5 marks, explaining each one of them in one line with example each-5 marks ) Q6. Consider a banking database (select the tables and fields of your choice). Now apply all the operations of relational algebra and find the result. (creating database-2 marks, listing the operations of relational algebra- 3 marks, showing the result for all operations- 5 marks) MI0035 –Computer Networks Q1. Videoconferencing is used to conduct meetings with the people who are located in far distance. Videoconferencing is an example of which type of computer networks? Explain in detail the different types of networks with example. (identifying the correct type – 2 marks, list the different types of networks – 1 mark, explaining the types- 6 marks, examples-1 mark) 10 marks Q2. Explain the three principles of data communication system. (explanation-5 marks, diagram – 2 marks, examples- 3 marks) 10 marks Q3. Data compression technique makes the file size smaller so that it can easily get transmitted over any types of network such as internet, intranet or a local area network. Explain the different types of data compression (listing the different types with example – 5 marks, explanation- 5 marks) 5+5 marks Q4. A protocol helps in establishing the communication between two systems. Protocols are mainly divided into asynchronous and synchronous protocols. List and describe in detail the different types of protocols that are classified under the two main protocols. (list and correct differentiating of protocols under the two main types – 5 marks, explanation- 5 marks) 10 marks Q5. How are the different topologies of computer networks arranged? Explain the working of each of them (listing the different topologies with examples – 2 marks, explanation- 4 marks, diagrams- 4 marks) 10 marks Q6. a. Elaborate the arrangement of IP addressing with an example (explanation- 3 marks, examples- 2 marks) b. List the various best internet service providers in India with respect to their efficiency. (listing – 2 marks, explaining – 3 marks) MI0036 –Business Intelligence Tools
13. 13. WWW.SMUSOLVEDASSIGNMENTS.COM marks; description of one ERP software for each type of ERP – 2marks; conclusion – 1 mark) 10 marks Q2. For a successful MRP system three types of information are very essential. Describe the types of information. (Master Production Schedule – 3 marks ;Bill of Material – 4 marks ;Inventory Records – 3 marks) 10 marks Q3. Explain the features, benefits and Limitations of ERP Inventory Management. (Explanation ERP inventory management – 1 mark; features – 3 marks ;benefits – 3 marks; limitations – 3 marks) 10 marks Q4. What are the activities of Human Resources management systems? (explanation of activities – 10 marks) 10 marks Q5. Explain any two benefits of an ERP implementation? (5 marks for each benefit) 10 marks Q6. Who is an ERP vendor and what are his roles? (Explanation of ERP vendor – 4 marks; role – 6 marks) 10 marks OM 0012 – SUPPLY CHAIN MANAGEMENT Q1. It is necessary for Supply Chain managers to identify the obstacles to co-ordination in the Supply Chain so that they can take suitable actions that help achieve co-ordination. Explain the major categories of obstacles. (Incentive obstacles; Information processing obstacles; Operational obstacles; Pricing obstacles; Behavioural obstacles – 10 marks, i.e. 2 marks each) 10 Marks Q2. Write a note on assessment tool. (Planning assessment- 1 mark; Sample Schedule for Conducting an Assessment – 3 marks; Description/Using an of assessment tool – 4 marks; Completing assessment tool-2 marks) 10 Marks Q3. How can differential advantage be achieved through Supply Chain Management? (Competitive-forces approach – 2 marks; capabilities approach – 6 marks ; conclusion – 2 marks) 10 Marks Q4. Discuss the five basic components of Supply Chain Management. (5 components X 2 marks =unit 10) 10 Marks Q5. Explain Relationship Marketing‟s impact on firms. (7 impacts X 1.25 = 8.75 marks; conclusion – 1.25 marks) 10 Marks Q6. The Global Supply Chain Forum (GSCF) framework consists of eight supply chain management processes. Explain them. (8 process X 1.25 marks) OM 0013 – ADVANCED PRODUCTION Q1. What is Flexible Manufacturing System (FMS)? How does it help in improving the manufacturing process? (Definition of FMS - 1 mark; Characteristics- 2 marks; Explanation of system designed to produce components – 3 marks; How there is flexibility in the manufacturing process – 2 marks; Components of FMS – 2 marks) 10 marks Q2. What is Logical Process Modelling and give the comparison between Logical and Physical Modelling? (definition of logical process modeling- 1 mark; steps- 2 marks; explanation of Logical Process Model - 3 marks; logical modelling formats- 1 mark; Comparison Between Logical and Physical Modelling- 3 marks) 10 marks Q3. Write short notes of Risk management. (Meaning of Risk management risk – 2 marks; management life cycle including diagram-3 marks; risk reduction by noting changes to existing risks and updating in the risk management mode- 3 marks; role or project risk manager and individuals in managing risk- 2 marks) 10 marks Q4. List out the inventory decision rules for MRP and the benefits and drawbacks of MRP. (Inventory decision rules – 6 marks; benefits – 2 marks; drawbacks – 2 marks) 10 marks Q5. The v4L learning principles are perfectly blended across all Toyota supply chain management processes to systematically concentrate on the v4L balance. Explain them. (Variety; Velocity; Variability; Visibility – 10 i.e. 2.5 marks each) Q6. Explain the key elements of Just In Time. (Key elements – 9 marks, i.e. 1 mark each; conclusion – 1 mark)
14. 14. WWW.SMUSOLVEDASSIGNMENTS.COM SEM 3 PROJECT MANAGEMENT(PM) SUMMER 2013 PM 0010 – INTRODUCTION TO PROJECT MANAGEMENT Q1. Describe the strategy planning tools of Ansoff matrix and BCG matrix.(Ansoff matrix : use and factors it considers – 1 mark, explanation – 2marks, limitation -1 mark; BCG matrix : use - 1 mark, explanationincluding 4 types of SBU‟s – 3 marks, limitations - 2 marks)10 marks (4 forAnsoff matrix +6for BCG matrix) Q2. Describe the approaches used to screen projects.(3 approaches - each 3 marks i.e. 3 X 3marks = 9 marks ; conclusion – 1mark)10 marks Q3. Explain any 3 parameters analysed during technical analysis of a project.(any 3 parameters – 3 marks each i.e. 3 X 3marks = 9 marks;conclusion- 1 mark)10 marks Q4. Write short notes on Cost Breakdown Structure(CBS).(Explanation of Cost Break down Structure(CBS) including details itprovides, categories of CBS , cost baseline – 3 marks ; Characteristics ofCost Breakdown Structure -2; Five major forms of cost breakdownstructure- 3 marks, Principle of cost breakdown structure- 2 marks) 10 marks Q5. Briefly explain the different steps or methodologies of project risk 10 marksmanagement?(1.67 marks for each step, i.e 6 X 1.67 marks= approx. 10 marks) Q6. Briefly describe the key project contracts under SPV (Special PurposeVehicle) for infrastructure projects.(Key project contracts - 3 marks each, i.e 3 X 3 marks =9 marks;conclusion-1 mark) PM0011 – PROJECT PLANNING AND SCHEDULING Q1. The PMBOK Guide addresses four elements related to scope. List andexplain them.(Listing of 4 elements – 1 mark, Explanation of elements - 9 marks)10 marks Q2. Write short notes of PERT (Explanation/definition of PERT- 1 mark; Four requirements of PERT – 2marks ; Basic features and functions of PERT techniques – 2 marks;PERT System of Time Estimates – 3 marks; PERT Computations – 2marks )10 marks Q3. Explain the various planning processes which are part of the riskmanagement knowledge area.(each process 2.5 marks, i.e. 4 processes X 2.5 marks =10 marks)10 marks Q4. What is an expert system? Explain different parts involved in an expertsystem.(Definition of expert system, use, essential ingredient – 2 marks; parts ofan expert system – 8 marks)10 marks Q5. Explain four different types of predecessor and the reasons why one taskmay be dependent on another.(Four different types of predecessor- 2 marks each i.e 2 X 4 = 8 marks;reasons why one task may be dependent on another – 2 marks) Q6. What is delay analysis and explain its methodology?(Delay analysis definition, critical and non-critical delay – 2 marks;Methodology – 8 marks) PM 0012 – PROJECT FINANCE AND BUDGETING Q1. There are several elements which you can take into consideration, while budgeting a project. Explain these elements. (4 elements X 2.5 marks =10 marks) 10 marks Q2. Explain the different methods/sources to finance a project? (5 methods/sources X 2 marks =10 marks) 10 marks Q3. Describe any 5 considerations that are crucial in the design of the financing plan for a project. (any 5 considerations X 2 marks = 10 marks) 10 marks Q4. Discuss some of the tools and techniques of Cost Management. (Cost aggregation- 1 mark; Reserve analysis -1 mark; Expert judgment – 1 mark; Historical relationships- 1 mark; Funding limit reconciliation – 2 marks ;Cost performance baseline – 2 marks ; Project funding baseline- 2 marks) 10 marks Q5. Explain the various key determinants of initial project cost. (8 key determinants X1.25 marks =10 marks) 10 marks Q6. Explain any 5 risks associated with project evaluation. (5 risks X2 marks= 10 marks) PM 0013 – MANAGING HUMAN RESOURCES IN PROJECTS | 10,328 | 43,293 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.59375 | 3 | CC-MAIN-2017-30 | latest | en | 0.867288 |
http://www.engineeringexpert.net/Engineering-Expert-Witness-Blog/2016/12 | 1,726,668,887,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651899.75/warc/CC-MAIN-20240918133146-20240918163146-00382.warc.gz | 43,660,727 | 9,370 | ## Archive for December, 2016
### Pulleys Make Santa’s Job Easier
Monday, December 19th, 2016
Happy Holidays from EngineeringExpert.net, LLC and the Engineering Expert Witness Blog. Pulleys Make Santa’s Job Easier Copyright 2016 – Philip J. O’Keefe, PE Engineering Expert Witness Blog ____________________________________
### Friction Reduces Pulleys’ Mechanical Advantage
Tuesday, December 13th, 2016
The presence of friction in mechanical designs is as guaranteed as conflict in a good movie, and engineers inevitably must deal with the conflicts friction produces within their mechanical designs. But unlike a good movie, where conflict presents a positive, engaging force, friction’s presence in pulleys results only in impediment, wasting energy and reducing mechanical advantage. We’ll investigate the math behind this phenomenon in today’s blog. Friction Reduces Pulleys’ Mechanical Advantage A few blogs back we performed a work input-output analysis of an idealized situation in which no friction is present in a compound pulley. The analysis yielded this equation for mechanical advantage, MA = d2 ÷ d1 (1) where d2 is the is the length of rope Mr. Toga extracts from the pulley in order to lift his urn a distance d1 above the ground. Engineers refer to this idealized frictionless scenario as an ideal mechanical advantage, IMA, so equation (1) becomes, IMA = d2 ÷ d1 (2) We also learned that in the idealized situation mechanical advantage is the ratio of the urn’s weight force, W, to the force exerted by Mr. Toga, F, as shown in the following equation. See our past blog for a refresher on how this ratio is developed. IMA = W ÷ F (3) In reality, friction exists between a pulley’s moving parts, namely, its wheels and the rope threaded through them. In fact, the more pulleys we add, the more friction increases. The actual amount of lifting force required to lift an object is a combination of FF , the friction-filled force, and F, the idealized friction-free force. The result is FActual as shown here, FActual = F + FF (4) The real world scenario in which friction is present is known within the engineering profession as actual mechanical advantage, AMA, which is equal to, AMA = W ÷ FActual (5) To see how AMA is affected by friction force FF, let’s substitute equation (4) into equation (5), AMA = W ÷ (F + FF) (6) With the presence of FF in equation (6), W gets divided by the sum of F and FF . This results in a smaller number than IMA, which was computed in equation (3). In other words, friction reduces the actual mechanical advantage of the compound pulley. Next time we’ll see how the presence of FF translates into lost work effort in the compound pulley, thus creating an inequality between the work input, WI and work output WO. Copyright 2016 – Philip J. O’Keefe, PE Engineering Expert Witness Blog ____________________________________ | 721 | 3,338 | {"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} | 3.546875 | 4 | CC-MAIN-2024-38 | latest | en | 0.880926 |
https://preprod.bigthink.com/the-future/robot-tuna-tail/ | 1,680,093,268,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296948976.45/warc/CC-MAIN-20230329120545-20230329150545-00531.warc.gz | 535,810,206 | 28,662 | The Future
# What robots can learn from fish and fancy math
A new tuna robot leads the way to more agile underwater robots and drones.
Key Takeaways
• Today's submersible robots and drones tend to be optimized for traveling through water at a single speed.
• Fish speed up and slow down by adjusting the rigidity of their tails, and researchers have cracked the formula for doing this.
• An experimental robot tuna is the first robot that can dart through water or float along slowly.
Underwater robots are useful for examining objects in the deep sea, from shipwrecks to new lifeforms. But they suffer from some limitations. For instance, they tend to go one speed, and their operators have to plonk them in the water near the desired location. Diver propulsion vehicles — basically, underwater scooters — are fast, but they would zoom right by something that warranted a better look.
What is needed is an underwater robot that can zip through the water to a remote location and then slow down for a closer examination. This is harder than it might seem.
However, researchers at the University of Virginia have figured out how to make a robotic craft that does just that by mimicking the rigidity of a fish’s tail and discovering and applying a beautifully simple mathematical formula. The result is a robot that is nearly as good as a fish at speeding up and slowing down. Their work is published inScience Robotics.
## A fishy riddle is solved
“Having one tail stiffness is like having one gear ratio on a bike,” says co-author Dan Quinn. “You’d only be efficient at one speed. It would be like biking through San Francisco with a fixed-gear bike; you’d be exhausted after just a few blocks.”
Quinn and his colleague Qiang Zhong say that it is assumed that fish alter the rigidity of their tails to meet their varying need for speed. Unfortunately, this is difficult to confirm or measure as a fish swims, so the requisite rigidity required for efficient travel at different speeds was unknown. So, Quinn and Zhong combined fluid dynamics and biomechanics to develop a model that could provide an answer and, as Zhong says, “uncover the long-existing mystery about how tail stiffness affects swimming performance.”
The solution to the puzzle was unexpectedly elegant. “Surprisingly,” Quinn revealed, “a simple result came out of all the math: Stiffness should increase with swimming speed squared.”
## A tuna takes a test
Zhong, Quinn, and their tuna robotCredit: University of Virginia
To test their findings, the pair built a tuna-like submersible robot with a programmable artificial tendon with which its tail stiffness could be tuned for different speeds as it moved through a test water channel. “What happened,” recalls Quinn, “is that suddenly our robot could swim over a wider range of speeds while using almost half as much energy as the same robot with a fixed-stiffness tail. The improvement was really quite remarkable.”
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Zhong credits their multidisciplinary approach, saying, “We are not just focused on theory analysis, but also on proposing a practical guide for tunable stiffness. Our proposed tunable stiffness strategy has proved effective in realistic swimming missions, where a robot fish achieved high speed and high efficiency swimming simultaneously.”
Having demonstrated their theory with their tunable tuna, they are interested in transplanting their tail technology to aquatic robot creatures ranging in size from tadpoles to dolphins. “I don’t think we’ll run out of projects any time soon,” says Quinn. “Every aquatic animal we’ve looked at has given us new ideas about how to build better swimming robots. And there are plenty more fish in the sea.”
The duo are currently developing an undulating robot based on stingrays.
Related
The automated McDonald’s has a staff comparable to other stores. But the crew members are all focused on making and packaging orders instead of delivering them.
If dogs are out in coats and boots, how are the squirrels feeling?
He was also a eugenicist — but at least he could draw pretty pictures.
But they’re still lovable.
Up Next
Why I was prepared to hate The Structure of Scientific Revolutions but ended up loving it. | 872 | 4,311 | {"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-2023-14 | latest | en | 0.971502 |
https://www.physicsforums.com/threads/intensity-of-sound.144924/ | 1,519,549,023,000,000,000 | text/html | crawl-data/CC-MAIN-2018-09/segments/1518891816178.71/warc/CC-MAIN-20180225070925-20180225090925-00186.warc.gz | 867,395,574 | 15,006 | # Intensity of Sound
1. Nov 22, 2006
### needhelp83
What is the intensity of sound at the pain level of 120 dB? Compare this to that of a whisper at 20 dB?
b=10 log (I/ I0)
120 dB=10 log (I/1)
10^12=(I/ 1)
I= 10^12 W/m^2
b=10 log (I/I0)
20 dB=10 log (I/1)
10^2=(I/1)
I=10^2 W/m^2
The intensity of 120 dB would be 10^12 W/m^2 compared to the whisper with the intensity at 10^2 W/m^2 which equals 10^10 more intense.
Sound good?
2. Nov 22, 2006
### OlderDan
How did I0 become 1? Where did you get that from?
3. Nov 22, 2006
### needhelp83
This is the pain threshold
4. Nov 22, 2006
### needhelp83
This should look better:
SL=10 log(Sl/10)
I=Io10^(sl/10) = (1.0*10-12 W/m^2) = 1 W/m^2
SL=10 log(Sl/10)
I=Io10^(sl/10) = (1.0*10-12 W/m^2) = 1.0*10^-10 W/m^2
5. Nov 22, 2006
### OlderDan
It sounds to me like you are trying to use the answer to find the answer. The things I highlighted in red in your quote are not equalities. I think your first equation was OK, but I_o is not the pain threshold; it is the threshoold of hearing, the 1.0*10-12 W/m^2. If you had used that instead of 1 in your original equation, you would have found the correct value for I. The value you got initially would knock your head off.
http://hyperphysics.phy-astr.gsu.edu/hbase/sound/intens.html | 464 | 1,291 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.578125 | 4 | CC-MAIN-2018-09 | longest | en | 0.893879 |
https://john.toebes.com/index.php/projects/electronics/29-gnomesort | 1,620,968,164,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243991737.39/warc/CC-MAIN-20210514025740-20210514055740-00175.warc.gz | 346,748,032 | 6,040 | As part of doing the code for our new picmas tree, I needed to come up with a way to sort a small array of numbers on the Microchip 16F685 processor that we are using. I've looked through a few different sort algorithms and came across the Gnome Sort algorithm on Wikipedia. It was originally created by Dick Grune although his web page is only available on archive.org. The algorithm is ideal for this small processor as it is simple, predictable and uses virtually no memory. The pseudo code for this function is basically:
`function gnomeSort(a[0..size-1]) { i := 1 j := 2 while i < size { if a[i-1] >= a[i] { i := j j := j + 1 } else { swap a[i-1] and a[i] i := i - 1 if i = 0 { i := 1 } } }} `
Converting this to the 16F685 is a little more complicated as you only have a single pointer register, but since you are only comparing values next to each other you can load FSR once and just increment/decrement to get to the values. Two other creative tricks that I applied to reduce the code and memory footprint is to reverse the destructive comparison to recover the value to swap and then to use the typical three XOR back and forth to swap the values without an intermediary.
`;; PARAMETERS; SORT_ARRAY - Array to be sorted; SORT_LENGTH - Number of bytes in the array to sort; SORT_J - Temporary variable to hold the end sort pointer;; NOTES on implementation; This uses the typical XOR trick to swap the two array elements; FSR points to the second of the two array elements that we are comparing.; SORT MOVLW SORT_ARRAY+1 MOVWF FSR ; i = 1 INCF FSR,W MOVWF SORT_J ; j := 2SORT_1 ; while i < size MOVFW INDF ; a[i] DECF FSR SUBWF INDF,W SKPNC GOTO SORT_2 ; if a[i-1] >= a[i] MOVFW SORT_J ; i := j MOVWF FSR SUBLW SORT_ARRAY+SORT_LENGTH-1 SKPC RETURN INCF SORT_J,F ; j := j + 1 GOTO SORT_1SORT_2 ; else SUBWF INDF,W ; Recover a[i] XORWF INDF,W ; Swap a[i] and value for a[i-1] XORWF INDF,F XORWF INDF,W INCF FSR ; point to a[i] MOVWF INDF ; store a[i] DECF FSR ; i := i - 1 MOVLW SORT_ARRAY SUBWF FSR,W ; if i = 0 SKPNZ INCF FSR,F ; i := 1 GOTO SORT_1 ; } END` | 686 | 2,642 | {"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-2021-21 | latest | en | 0.876974 |
http://www.mathisfunforum.com/viewtopic.php?pid=27844 | 1,506,403,315,000,000,000 | text/html | crawl-data/CC-MAIN-2017-39/segments/1505818695066.99/warc/CC-MAIN-20170926051558-20170926071558-00694.warc.gz | 486,653,833 | 3,603 | Discussion about math, puzzles, games and fun. Useful symbols: ÷ × ½ √ ∞ ≠ ≤ ≥ ≈ ⇒ ± ∈ Δ θ ∴ ∑ ∫ π -¹ ² ³ °
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## #1 2006-02-18 03:06:25
Math Student
Guest
### Tricky Pythagoras' Theorem Question
Please can you help me with this question?
I've got 3 lengths
x
x+7
and x+8
x+8 is the hypotenuse.
I need to find the length of each side, I've tried but still can't figure it out, please help! Thanks in advance! ^_^
## #2 2006-02-18 03:17:39
ganesh
Registered: 2005-06-28
Posts: 23,216
### Re: Tricky Pythagoras' Theorem Question
Since Pythogoras theorem is applicable, the triangle is right-angled. As (x+8) is the hypotenuse,
x²+ (x+7)²= (x+8)²
x²+x²+14x + 49 = x²+16x + 64
2x²+14x+49-x²-16x-64=0
x²-2x-15=0
x = [2 ± √(4+60)]/2 = (2 ± 8)/2 = 5 or -3.
The lenghts of the sides are 5, 12 and 13, and these form a Pythogorean triple.
It is no good to try to stop knowledge from going forward. Ignorance is never better than knowledge - Enrico Fermi.
Nothing is better than reading and gaining more and more knowledge - Stephen William Hawking.
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## #3 2006-02-18 03:27:44
irspow
Member
Registered: 2005-11-24
Posts: 457
### Re: Tricky Pythagoras' Theorem Question
Using the theorem;
c = √(a² + b²) gives;
x+8 = √(x² + (x+7)²);
x+8 = √(x² + x² + 14x + 49);
x+8 = √(2x² + 14x + 49);
Squaring both sides gives;
x² + 16x + 64 = 2x² + 14x + 49;
0 = x² - 2x - 15;
0 = (x-5)(x+3);
This is only true when x = 5 or -3;
Since x represents one of the sides it cannot be negative;
Therefore x = 5;
So the triangle has sides of 5, 12, and 13.
edit*
Sorry ganesh, you are a lot faster than me! lol.
Last edited by irspow (2006-02-18 03:29:06)
Offline | 655 | 1,700 | {"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} | 3.90625 | 4 | CC-MAIN-2017-39 | latest | en | 0.84047 |
http://www.efunda.com/glossary/formulas/beams/simply_supported--uniformly_distributed_load--single_span--wide_flange_steel_i_beam--w10_x_54.cfm | 1,519,261,000,000,000,000 | text/html | crawl-data/CC-MAIN-2018-09/segments/1518891813832.23/warc/CC-MAIN-20180222002257-20180222022257-00281.warc.gz | 447,266,138 | 12,039 | Wide Flange Steel I Beam: W10 × 54
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Glossary » Beams » Simply Supported » Uniformly Distributed Load » Single Span » Wide Flange Steel I Beam » W10 × 54
For a simply supported beam in a single span, the maximum displacement and the maximum normal stress occur at the center of the beam.
The tabulated data listed in this page are calculated based on the area moment of inertia (Ixx = 303 in4) for the W10 × 54 Wide Flange Steel I Beam and the typical Young's modulus (E = 3.046 × 107 psi) of steels. Note that the typical yielding stress of steels can range from 1.015 × 104 to 2.970 × 105 psi. The purpose of this page is to give a rough estimation of the load-bearing capacity of this particular beam, rather than a guideline for designing actual building structures. Please check your local building codes for regulatory requirements.
Note: The weight of the beam itself is not included in the calculation.
Steel I Beam: W10 × 54 (Wide Flange 10 inch tall × 54 lbf/ft) L = 3.00 ft (lbf/ft) (in) (psi) 54.0 1.07 × 10-5 12.1 60.0 1.18 × 10-5 13.5 70.0 1.38 × 10-5 15.7 80.0 1.58 × 10-5 18.0 90.0 1.78 × 10-5 20.2 100 1.97 × 10-5 22.5 200 3.95 × 10-5 45.0 300 5.92 × 10-5 67.4 400 7.90 × 10-5 89.9 500 9.87 × 10-5 112 600 1.18 × 10-4 135 700 1.38 × 10-4 157 800 1.58 × 10-4 180 900 1.78 × 10-4 202 1000 1.97 × 10-4 225 2000 3.95 × 10-4 450 3000 5.92 × 10-4 674 4000 7.90 × 10-4 899 5000 9.87 × 10-4 1120 6000 0.00118 1350 7000 0.00138 1570 8000 0.00158 1800 9000 0.00178 2020 1.00 × 104 0.00197 2250 2.00 × 104 0.00395 4500 3.00 × 104 0.00592 6740 4.00 × 104 0.00790 8990 5.00 × 104 0.00987 1.12 × 104 6.00 × 104 0.0118 1.35 × 104 7.00 × 104 0.0138 1.57 × 104 8.00 × 104 0.0158 1.80 × 104 9.00 × 104 0.0178 2.02 × 104 1.00 × 105 0.0197 2.25 × 104 2.00 × 105 0.0395 4.50 × 104 3.00 × 105 0.0592 6.74 × 104 4.00 × 105 0.0790 8.99 × 104 5.00 × 105 0.0987 1.12 × 105 6.00 × 105 0.118 1.35 × 105 7.00 × 105 0.138 1.57 × 105 8.00 × 105 0.158 1.80 × 105 9.00 × 105 0.178 2.02 × 105 1.00 × 106 0.197 2.25 × 105 1.32 × 106 0.261 2.97 × 105
Steel I Beam: W10 × 54 (Wide Flange 10 inch tall × 54 lbf/ft) L = 5.00 ft (lbf/ft) (in) (psi) 54.0 8.23 × 10-5 33.7 60.0 9.14 × 10-5 37.5 70.0 1.07 × 10-4 43.7 80.0 1.22 × 10-4 50.0 90.0 1.37 × 10-4 56.2 100 1.52 × 10-4 62.4 200 3.05 × 10-4 125 300 4.57 × 10-4 187 400 6.10 × 10-4 250 500 7.62 × 10-4 312 600 9.14 × 10-4 375 700 0.00107 437 800 0.00122 500 900 0.00137 562 1000 0.00152 624 2000 0.00305 1250 3000 0.00457 1870 4000 0.00610 2500 5000 0.00762 3120 6000 0.00914 3750 7000 0.0107 4370 8000 0.0122 5000 9000 0.0137 5620 1.00 × 104 0.0152 6240 2.00 × 104 0.0305 1.25 × 104 3.00 × 104 0.0457 1.87 × 104 4.00 × 104 0.0610 2.50 × 104 5.00 × 104 0.0762 3.12 × 104 6.00 × 104 0.0914 3.75 × 104 7.00 × 104 0.107 4.37 × 104 8.00 × 104 0.122 5.00 × 104 9.00 × 104 0.137 5.62 × 104 1.00 × 105 0.152 6.24 × 104 2.00 × 105 0.305 1.25 × 105 3.00 × 105 0.457 1.87 × 105 4.00 × 105 0.610 2.50 × 105 4.76 × 105 0.725 2.97 × 105
Steel I Beam: W10 × 54 (Wide Flange 10 inch tall × 54 lbf/ft) L = 10.0 ft (lbf/ft) (in) (psi) 54.0 0.00132 135 60.0 0.00146 150 70.0 0.00171 175 80.0 0.00195 200 90.0 0.00219 225 100 0.00244 250 200 0.00488 500 300 0.00731 749 400 0.00975 999 500 0.0122 1250 600 0.0146 1500 700 0.0171 1750 800 0.0195 2000 900 0.0219 2250 1000 0.0244 2500 2000 0.0488 5000 3000 0.0731 7490 4000 0.0975 9990 5000 0.122 1.25 × 104 6000 0.146 1.50 × 104 7000 0.171 1.75 × 104 8000 0.195 2.00 × 104 9000 0.219 2.25 × 104 1.00 × 104 0.244 2.50 × 104 2.00 × 104 0.488 5.00 × 104 3.00 × 104 0.731 7.49 × 104 4.00 × 104 0.975 9.99 × 104 5.00 × 104 1.22 1.25 × 105 6.00 × 104 1.46 1.50 × 105 7.00 × 104 1.71 1.75 × 105 8.00 × 104 1.95 2.00 × 105 9.00 × 104 2.19 2.25 × 105 1.00 × 105 2.44 2.50 × 105 1.19 × 105 2.90 2.97 × 105
Steel I Beam: W10 × 54 (Wide Flange 10 inch tall × 54 lbf/ft) L = 15.0 ft (lbf/ft) (in) (psi) 54.0 0.00666 303 60.0 0.00741 337 70.0 0.00864 393 80.0 0.00987 450 90.0 0.0111 506 100 0.0123 562 200 0.0247 1120 300 0.0370 1690 400 0.0494 2250 500 0.0617 2810 600 0.0741 3370 700 0.0864 3930 800 0.0987 4500 900 0.111 5060 1000 0.123 5620 2000 0.247 1.12 × 104 3000 0.370 1.69 × 104 4000 0.494 2.25 × 104 5000 0.617 2.81 × 104 6000 0.741 3.37 × 104 7000 0.864 3.93 × 104 8000 0.987 4.50 × 104 9000 1.11 5.06 × 104 1.00 × 104 1.23 5.62 × 104 2.00 × 104 2.47 1.12 × 105 3.00 × 104 3.70 1.69 × 105 4.00 × 104 4.94 2.25 × 105 5.00 × 104 6.17 2.81 × 105 5.29 × 104 6.52 2.97 × 105
Steel I Beam: W10 × 54 (Wide Flange 10 inch tall × 54 lbf/ft) L = 20.0 ft (lbf/ft) (in) (psi) 54.0 0.0211 539 60.0 0.0234 599 70.0 0.0273 699 80.0 0.0312 799 90.0 0.0351 899 100 0.0390 999 200 0.0780 2000 300 0.117 3000 400 0.156 4000 500 0.195 5000 600 0.234 5990 700 0.273 6990 800 0.312 7990 900 0.351 8990 1000 0.390 9990 2000 0.780 2.00 × 104 3000 1.17 3.00 × 104 4000 1.56 4.00 × 104 5000 1.95 5.00 × 104 6000 2.34 5.99 × 104 7000 2.73 6.99 × 104 8000 3.12 7.99 × 104 9000 3.51 8.99 × 104 1.00 × 104 3.90 9.99 × 104 2.00 × 104 7.80 2.00 × 105 2.97 × 104 11.6 2.97 × 105
Steel I Beam: W10 × 54 (Wide Flange 10 inch tall × 54 lbf/ft) L = 25.0 ft (lbf/ft) (in) (psi) 54.0 0.0514 843 60.0 0.0571 937 70.0 0.0667 1090 80.0 0.0762 1250 90.0 0.0857 1400 100 0.0952 1560 200 0.190 3120 300 0.286 4680 400 0.381 6240 500 0.476 7800 600 0.571 9370 700 0.667 1.09 × 104 800 0.762 1.25 × 104 900 0.857 1.40 × 104 1000 0.952 1.56 × 104 2000 1.90 3.12 × 104 3000 2.86 4.68 × 104 4000 3.81 6.24 × 104 5000 4.76 7.80 × 104 6000 5.71 9.37 × 104 7000 6.67 1.09 × 105 8000 7.62 1.25 × 105 9000 8.57 1.40 × 105 1.00 × 104 9.52 1.56 × 105 1.90 × 104 18.1 2.97 × 105
Steel I Beam: W10 × 54 (Wide Flange 10 inch tall × 54 lbf/ft) L = 30.0 ft (lbf/ft) (in) (psi) 54.0 0.107 1210 60.0 0.118 1350 70.0 0.138 1570 80.0 0.158 1800 90.0 0.178 2020 100 0.197 2250 200 0.395 4500 300 0.592 6740 400 0.790 8990 500 0.987 1.12 × 104 600 1.18 1.35 × 104 700 1.38 1.57 × 104 800 1.58 1.80 × 104 900 1.78 2.02 × 104 1000 1.97 2.25 × 104 2000 3.95 4.50 × 104 3000 5.92 6.74 × 104 4000 7.90 8.99 × 104 5000 9.87 1.12 × 105 5110 10.1 1.15 × 105
Steel I Beam: W10 × 54 (Wide Flange 10 inch tall × 54 lbf/ft) L = 40.0 ft (lbf/ft) (in) (psi) 54.0 0.337 2160 60.0 0.374 2400 70.0 0.437 2800 80.0 0.499 3200 90.0 0.562 3600 100 0.624 4000 200 1.25 7990 300 1.87 1.20 × 104 400 2.50 1.60 × 104 500 3.12 2.00 × 104 600 3.74 2.40 × 104 700 4.37 2.80 × 104 800 4.99 3.20 × 104 900 5.62 3.60 × 104 1000 6.24 4.00 × 104 1620 10.1 6.46 × 104
Steel I Beam: W10 × 54 (Wide Flange 10 inch tall × 54 lbf/ft) L = 50.0 ft (lbf/ft) (in) (psi) 54.0 0.823 3370 60.0 0.914 3750 70.0 1.07 4370 80.0 1.22 5000 90.0 1.37 5620 100 1.52 6240 200 3.05 1.25 × 104 300 4.57 1.87 × 104 400 6.10 2.50 × 104 500 7.62 3.12 × 104 600 9.14 3.75 × 104 662 10.1 4.13 × 104
Steel I Beam: W10 × 54 (Wide Flange 10 inch tall × 54 lbf/ft) L = 60.0 ft (lbf/ft) (in) (psi) 54.0 1.71 4860 60.0 1.90 5390 70.0 2.21 6290 80.0 2.53 7190 90.0 2.84 8090 100 3.16 8990 200 6.32 1.80 × 104 300 9.48 2.70 × 104 319 10.1 2.87 × 104
Steel I Beam: W10 × 54 (Wide Flange 10 inch tall × 54 lbf/ft) L = 70.0 ft (lbf/ft) (in) (psi) 54.0 3.16 6610 60.0 3.51 7340 70.0 4.10 8570 80.0 4.68 9790 90.0 5.27 1.10 × 104 100 5.85 1.22 × 104 172 10.1 2.11 × 104
Steel I Beam: W10 × 54 (Wide Flange 10 inch tall × 54 lbf/ft) L = 80.0 ft (lbf/ft) (in) (psi) 54.0 5.39 8630 60.0 5.99 9590 70.0 6.99 1.12 × 104 80.0 7.99 1.28 × 104 90.0 8.99 1.44 × 104 100 9.99 1.60 × 104 101 10.1 1.62 × 104
Steel I Beam: W10 × 54 (Wide Flange 10 inch tall × 54 lbf/ft) L = 90.0 ft (lbf/ft) (in) (psi) 54.0 8.64 1.09 × 104 60.0 9.60 1.21 × 104 63.1 10.1 1.28 × 104 | 4,649 | 7,803 | {"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} | 2.984375 | 3 | CC-MAIN-2018-09 | longest | en | 0.349622 |
https://quant.stackexchange.com/questions/74901/estimating-the-knockout-probability-of-a-discretely-observed-autocall-note/75094 | 1,718,902,386,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861957.99/warc/CC-MAIN-20240620141245-20240620171245-00403.warc.gz | 430,799,374 | 40,652 | # Estimating the knockout probability of a discretely observed autocall note
For simplicity, let's suppose the underlier follows a Geometric Brownian Motion $$S_t\sim\text{GBM}(\mu, \sigma), t\ge 0$$ with $$S_0=1$$. A discretely-observed binary autocall note is a derivative structure with observation dates $$t_1 and pays the investor a high coupon $$c$$ on the first observation date on which the underlier $$S_t$$ exceeds some preset barrier $$K$$. In mathematical notations, define the knockout date $$\tau=\inf\{t_i,i=1,\cdots,n\mid S_{t_i}>K\}$$ We are given the below tasks:
1. Estimate the distribution of $$\tau$$, i.e. calculate $$P(\tau=t_i)$$ and $$P(\tau=\infty)$$
2. Or at least calculate $$P(\tau < \infty)$$, if the first task proves too challenging.
Practically, MC simulation is clearly the way to go. But let's say we're more interested on the mathematical side of this problem and focus not so much on pragmatism. Are there any exact or approximate solutions that permit a simple numerical implementation? Thanks.
• The problem is definitely tractable when $\mu=0$ and $t_1=0$. For then $p(\tau > t_i) = 1 - 2p(S_t_i>K)$ by the reflection theorem. [sorry for format- can't seem to fix the mathjax]
– dm63
Commented Apr 1, 2023 at 14:43
• @dm63 we can assume $\mu=0$ but not $t_1=0$. In fact for a one-year long contract observations often start in the third or even sixth month i.e. $t_1=1/4$ or $1/2$.
– Vim
Commented Apr 2, 2023 at 8:02
• @dm63 you seem to have forgot to put the sub of $S$ between braces: S_{t_i} is the recognisable format. So your formula renders as $p(\tau > t_i) = 1 - 2p(S_{t_i}>K)$
– Vim
Commented Apr 2, 2023 at 8:07
There is a closed-form formula for the probability $$\mathbb{P}(\tau = t_i)$$.
First, we remind that $$S_t=S_0\cdot \exp\left(\left(\mu-\frac{1}{2}\sigma^2 \right)t+\sigma W_t \right)$$ For $$i=1$$, it's easy that \begin{align} \mathbb{P}(\tau = t_1) &= \mathbb{P}(S_{t_1}> K ) \\ &=\mathbb{P}\left(W_{t_1}> \frac{\ln \left(\frac{K}{S_0}\right) -\left(\mu-\frac{1}{2}\sigma^2 \right)t_1}{\sigma} \right)\\ &=\color{red}{\Phi\left( \frac{-\ln \left(\frac{K}{S_0}\right) +\left(\mu-\frac{1}{2}\sigma^2 \right)t_1}{\sigma \sqrt{t_1}} \right)} \end{align} where $$\Phi(\cdot)$$ the probability distribution function of the univariate standard normal distribution $$\mathcal{N} (0,1)$$.
For $$i \ge 2$$, we have \begin{align} \mathbb{P}(\tau = t_i) &= \mathbb{P}(\bigcap_{0 \leq k \leq i-1} \{S_{t_k}\le K \} \cap \{S_{t_i}> K \} ) \\ &= \mathbb{P}\left(\bigcap_{0 \leq k \leq i-1} \left\{W_{t_k} \le \frac{\ln \left(\frac{K}{S_0}\right) -\left(\mu-\frac{1}{2}\sigma^2 \right)t_k}{\sigma} \right\} \cap \left\{W_{t_i}> \frac{\ln \left(\frac{K}{S_0}\right) -\left(\mu-\frac{1}{2}\sigma^2 \right)t_i}{\sigma} \right\} \right) \tag{1}\\ \end{align} We notice that the vector $$(W_{t_1}, W_{t_2},...,W_{t_i})$$ is a $$i$$-variate normal distribution with zero mean and the covariance matrix $$\mathbf{\Sigma} \in \mathbb{R}^{i\times i}$$ defined by $$\Sigma_{hk} = Cov (W_{t_h},W_{t_k}) = \min \{t_h,t_k\} \qquad \text{for }1\le h,k\le i \tag{2}$$
By denoting $$\Phi_i(\mathbf{L},\mathbf{U};\mathbf{0}_i,\mathbf{\Sigma} )$$ the probability distribution function of the $$i$$-variate normal distribution $$\mathcal{N}_i (\mathbf{0}_i,\mathbf{\Sigma})$$ with
• zero mean $$\mathbf{0}_i$$,
• covariance matrix $$\mathbf{\Sigma}$$ defined by $$(2)$$
• from the lower bound $$\mathbf{L}$$ to the upper bound $$\mathbf{U}$$ with $$\mathbf{L}, \mathbf{U} \in \mathbb{R}^{i}$$ $$L_k=\begin{cases} -\infty & \text{if 0\le k\le i-1 }\\ \frac{\ln \left(\frac{K}{S_0}\right) -\left(\mu-\frac{1}{2}\sigma^2 \right)t_k}{\sigma} & \text{if k = i }\\ \end{cases}$$ $$U_k=\begin{cases} \frac{\ln \left(\frac{K}{S_0}\right) -\left(\mu-\frac{1}{2}\sigma^2 \right)t_k}{\sigma} & \text{if 0\le k\le i-1 }\\ +\infty & \text{if k = i }\\ \end{cases}$$
Then, from $$(1)$$, we have
$$\mathbb{P}(\tau = t_i) = \color{red}{\Phi_i(\mathbf{L},\mathbf{U};\mathbf{0}_i,\mathbf{\Sigma} ) }$$
• Thanks this is a neat representation. For what it's worth, a simple scipy implementation can be found here.
– Vim
Commented Apr 2, 2023 at 8:20
• I think $$\Sigma_{hk} = Cov (W_{t_h},W_{t_k}) = \color{red}{\min} \{t_h,t_k\} \qquad \text{for }1\le h,k\le i$$ though, rather than $\max$. Is this a typo?
– Vim
Commented Apr 2, 2023 at 8:51
• Applying your formula to the scenario where $\mu=5\%, \sigma=25\%,t_k=\frac{6+(k-1)}{12},k=1,\cdots,6$ gives the following probabilities in percentage terms: $$[45.45, 6.29, 4.1, 3.07, 2.43, 1.99]$$ which I believe is sound.
– Vim
Commented Apr 2, 2023 at 8:58
• @Vim yes, it’s a typo. I corrected it
– NN2
Commented Apr 2, 2023 at 10:19
• Excellent solution !
– dm63
Commented Apr 2, 2023 at 13:36 | 1,807 | 4,767 | {"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": 36, "wp-katex-eq": 0, "align": 2, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.46875 | 3 | CC-MAIN-2024-26 | latest | en | 0.794582 |
https://bookdown.org/robertjcarroll/ProbStatsFSU/prob1.html | 1,660,907,864,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882573667.83/warc/CC-MAIN-20220819100644-20220819130644-00262.warc.gz | 162,454,830 | 16,828 | # Chapter 2 Probability Spaces
This chapter, the first of two, introduces our formalism for simple statistical spaces, which are the natural conclusion of the simplest chance regularity—that of the random experiment. At the end of these two lectures, we will have a complete theory of chance regularity stemming from random experiments.9
Having just described the forest in such happy terms, let me apologize now: you are about to be bludgeoned by trees. Do not lose sight of the forest!
## 2.1 Introductory Remarks
### 2.1.1 The Random Experiment
We aim to build a formalism for the ideal type of chance mechanism: the random experiment. In words, a random experiment is a chance mechanism with three properties:
1. all possible outcomes are known a priori;
2. the outcome of any given trial is not known a priori but there exists a perceptible regularity of occurrence associated with the outcomes; and
3. the trials can be repeated under identical conditions.
A simple statistical space—which, again, is just a formalization of the abstract ideal of a random experiment—is comprised of two elements:
1. A probability space that enumerates:
1. The possible outcomes of the experiment;
2. The events of interest; and
3. The probabilities assigned to each of those events.
2. A simple sampling space that enumerates the relationships among the observations.
Now, most of us work primarily with observational data, which often fail to live up to the three standards of a random experiment. It is rare for observations to be identical under repeated trials; what does a “repeated trial” mean for those of us interested in why civil wars start? Or voting—you’re not supposed to vote twice, right? So, perhaps our simple statistical space is inadequate to discuss anything realistic?
To this, I offer two responses.
1. First, experimental data is increasingly popular in the discipline; and
2. Second, even if most data fail to live up to the lofty standards of a random experiment, it remains that most analyses implicitly operate with the question “what would the ideal experiment for this phenomenon be?” in the background. At the very least, good analyses operate with that question in the background.
### 2.1.2 Goals of the Lecture
Today, we want to have a formalism for what a probability space is. We therefore will be formalizing the first two aspects of a random experiment: the outcomes set, the events of interest, and the probability function. As you can imagine, this will first require a mathematical digression on sets and functions. So, here comes the…
## 2.2 Mathematical Digression on Sets and Functions
### 2.2.1 Sets
The field of statistics depends on the theory of probability, and the theory of probability itself depends on the theory of sets. Set theory, in turn, is taken as primitive for all mathematical analysis but requires careful answers to important philosophical questions that go well beyond our scope here.10 Our main goal for now is to gather all the set theory needed to define probability functions in an acceptable way.
So what is a set? Speaking very loosely, a set is a collection of distinct objects,11 which we call elements. For example, consider
$S = \left\{\clubsuit, \diamondsuit, \heartsuit, \spadesuit\right\}.$
Here $$S$$ is a set, and $$\clubsuit$$, $$\diamondsuit$$, $$\heartsuit$$, and $$\spadesuit$$ are its elements. We denote this inclusion by, say, $$\clubsuit \in S$$, which we read “$$\clubsuit$$ is an element of $$S$$”, or usually just “$$\clubsuit$$ is in $$S$$”. If we want to say that some entity is not an element of $$S$$, we do so by with, for example, $$\maltese \not\in S$$. Note that the order of the elements does not matter for the definition of a set.
The elements of a set may themselves be sets. In such cases, I will try to call them collections and will try to use calligraphic letters (e.g. $$\mathcal{A}$$) instead of standard letters (e.g. $$A$$). For example, let’s break down the suits of cards by color: \begin{align*} S_B &= \{\clubsuit, \spadesuit\}, \\ S_R &= \{\diamondsuit, \heartsuit\}, \end{align*} where $$S_B$$ captures the black suits and $$S_R$$ captures the red suits.12 Then we might form the collection \begin{align*} \mathcal{S} &= \{S_B, S_R\}, \\ &= \{\{\clubsuit, \spadesuit\}, \{\diamondsuit, \heartsuit\}\}, \end{align*}
so that $$S_B \in \mathcal{S}$$ and $$S_R \in \mathcal{S}$$ but $$\clubsuit \not\in \mathcal{S}$$.
As a general rule, we will denote sets with capital lettrs like $$S$$ or $$A$$ and elements with lower-case letters like $$s$$ or $$a$$. Thus, $$s \in S$$ and $$a \in A$$ are both very common things to see, but $$A \in a$$ is not.
You can imagine that it is often inconvient and sometimes impossible to introduce a set by enumerating all of its element like we just did above. We often define sets based on some property, such as with
$[0,1] = \left\{x : 0 \leq x \leq 1 \right\}.$ This is read “$$[0,1]$$ is the set of all $$x$$ such that $$0$$ is less than or equal to $$x$$, which is less than or equal to $$1$$.” The square brackets indcitate that these are weak inequalities; for strict inequalities, we use parentheses: $(0,1) = \left\{x : 0 < x < 1 \right\}.$
#### 2.2.1.1 Some Special Sets
In particular, there are some sets of numbers that are relevant for our purposes. We have:
1. The natural numbers, $$\mathbb{N} = 1,2,3,\ldots$$;
2. The integers, $$\mathbb{Z} = ..., -2, -1, 0, 1, 2, \ldots$$;
3. The rational numbers, $\mathbb{Q} = \left\{x : \exists~p,q \in \mathbb{Z}~\text{such that}~ x = \frac{p}{q}\right\},$ where $$\exists$$ is read “there exists.” Thus, the definition above reads “$$\mathbb{Q}$$ is the set of all numbers $$x$$ such that there exist integers $$p$$ and $$q$$ where $$x = \frac{p}{q}$$.” Not so bad, right?
4. The real numbers, denoted $$\mathbb{R}$$. These are more difficult to define.13 These include the rational numbers and the limits of all of the sequences of rational numbers. To give the usual classic example, $$\sqrt{2}$$ is the limit of a sequence of rational numbers, namely $1, 1.4, 1.41, 1.414, 1.4142, 1.41421, 1.414213, \ldots ,$ but it is straightforward to show that $$\sqrt{2}$$ not a rational number.14 Basically, the reals are any finite number you can think of and many that you cannot. It is tempting to write $\mathbb{R} = \left\{x : -\infty < x < \infty\right\},$ and you see that sometimes, but it rubs me the wrong way. We also might focus on the non-negative reals, $\mathbb{R}_+ = \{x \in \mathbb{R} : x \geq 0\},$ and likewise for the strictly positive reals $$\mathbb{R}_{++}$$ when this inequality holds strictly.
One of the most important sets is the one with no elements: the empty set. We denote it $$\emptyset$$.
#### 2.2.1.2 Basic Operations
There are several relevant operations that work on pairs of sets, say $$A$$ and $$B$$. For starters, we say $$A$$ is a subset of $$B$$—written $$A \subset B$$—if $$a \in A$$ implies that $$a \in B$$, too.15 For example, let \begin{align*} A &= \{1,2,3\}, \\ B &= \{1,2,3,4\}. \end{align*} Here $$A \subset B$$ but $$B \not\subset A$$. Two sets are equal if each is a subset of the other; for example, letting \begin{align*} C &= \{1,2,3,4,5\}, \\ D &= \{5,4,3,2,1\}, \end{align*}
we have $$C \subset D$$ (because every element of $$C$$ is also an element of $$D$$) and $$D \subset C$$ (because every element of $$D$$ is also an element of $$C$$) and thus that $$C = D$$.16 If we want to emphasize that such is not the case, we might write $$A \subsetneq B$$, read “$$A$$ is a subset of $$B$$, but is not equal to $$B$$,” or more compactly “$$A$$ is a proper subset of $$B$$.”
For a given set $$A$$, the set of all subsets is called the power set and is denoted $$2^A$$. For example, if $$A = \{1,2,3\}$$, then $2^A = \{\emptyset, \{1\}, \{2\}, \{3\}, \{1,2\},\{1,3\},\{2,3\}, \{1,2,3\}\}.$ Note the inclusion of the empty set; it is a subset of all nonempty sets. Note also that, since $$A \subset A$$, it is included in the power set, too. If you count the number of elements in $$A$$ and in $$2^A$$, you’ll understand the notation $$2^A$$.
We now define the union of two sets: \begin{align*} A \cup B &= \{x : x \in A~\text{or}~x \in B\}, \end{align*} and the intersection of two sets: \begin{align*} A \cap B &= \{x : x \in A~\text{and}~x \in B\}. \end{align*} For example, let \begin{align*} A &= \{\text{Mingus}, \text{Gillespie}, \text{Ellington}\}, \\ B &= \{\text{Monk}, \text{Evans}, \text{Ellington}\}. \end{align*} Then \begin{align*} A \cup B &= \{\text{Mingus}, \text{Gillespie}, \text{Ellington}, \text{Monk}, \text{Evans}\}, \\ A \cap B &= \{\text{Ellington}\}. \end{align*} Two sets $$A$$ and $$B$$ are disjoint or mutually exclusive if $$A \cap B = \emptyset$$. For example, let \begin{align*} E &= \{x \in \mathbb{Z}: x~\text{is an even number}\}, \\ O &= \{x \in \mathbb{Z}: x~\text{is an odd number}\}. \end{align*}
Clearly, $$E \cap O = \emptyset$$. Here it is also true that $$E \cup O = \mathbb{Z}$$. If disjoint subsets of a set equal that set when joined together via a union, we refer to them as a partition. Here $$E$$ and $$O$$ form a partition over $$\mathbb{Z}$$.
We can extend unions and intersections to allow for more than two sets. Let $$\mathcal{A}$$ be a collection with elements $$A_i, i = 1,2,\ldots$$, that is, $\mathcal{A} = \{A_1,A_2,\ldots\}.$ Here the dots indicate that the sets continue on without end. Then we have \begin{align*} \bigcup \mathcal{A} &= \bigcup_{i=1}^\infty A_i = \{a : a \in A_i~\text{for some}~A_i \in \mathcal{A}\}, \\ \bigcap \mathcal{A} &= \bigcap_{i=1}^\infty A_i = \{a : a \in A_i~\text{for all}~A_i \in \mathcal{A}\}. \end{align*}
Other operations work with respect to some reference set or universe of discourse. To build up this idea, define $X \setminus A = \{x \in X : x \not\in A\}.$ Letting $$X$$ serve as the reference set (which is often understood from the context), we might write $$\overline{A} = X \setminus A$$, which we read as the complement of $$A$$.
From here, we move on to de Morgan’s laws. Let $$I = \{1,2,3,\ldots\}$$ serve as our index set. Now we have \begin{align*} \overline{\bigcup_{i \in I} A_i} &= \bigcap_{i \in I} \overline{A}_i, \\ \overline{\bigcap_{i \in I} A_i} &= \bigcup_{i \in I} \overline{A}_i. \end{align*}
So, the complement of the union of a collection is just the intersection of the individual complements, and the complement of the intersection of a collection is just the union of the individual complements.
#### 2.2.1.3 Cartesian Productions
Given two sets $$A$$ and $$B$$, the Cartesian product $$A \times B$$ is just $A \times B = \{(a,b) : a \in A~\text{and}~b \in B\}.$ So, it is the set of all ordered pairs. More generally, given a collection of sets $$\mathcal{A}$$, we have $\prod \mathcal{A} = A_1 \times A_2 \times A_3 \times \cdots.$ We will also use the symbol $$\prod$$ to denote multiplication of numbers, but you will be able to tell which usage is intended from the context.
#### 2.2.1.4 Cardinality and the Infinities
We often are concerned with the size of a set. The most intuitive way to work here is to think about the number of elements. We refer to this as a set’s cardinality. For a set $$A$$, the resulting cardinality is $$|A|$$. For example, the cardinality of $$S = \{A,B,C\}$$ is $$|S| = 3$$. We call sets with a single element singletons. For example, $$X = \{a\}$$ is a singleton, as $$|X| = 1$$. Note that $$a$$ and $$\{a\}$$ are not the same thing, strictly speaking, as $$a \in X$$ but $$\{a\} \subset X$$ (indeed, $$\{a\} = X$$, since $$X \subset \{a\}$$, too).
The most straightforward kind of set is one like the example immediately above: there is a finite number of elements, and so we say the set is finite. Otherwise, it is infinite.
Now consider the natural numbers, $$\mathbb{N} = \{1,2,3,\ldots\}$$. This is not a finite set, as there is no largest natural number. Its cardinality is therefore infinite, and so $$\mathbb{N}$$ is an infinite set. In particular, we will say that $$\mathbb{N}$$ is countably infinite. Indeed, we will say that any set $$S$$ is countably infinite if we can assign each of its elements to a natural number. For example, the odd numbers, the even numbers, and the integers are all countably infinite—you can assign each of their elements a unique natural number.
But not all infinite sets are countable; some are larger. Letting $$\aleph_1$$ denote the number of elements in $$\mathbb{R}$$ and $$\aleph_2$$ denote the number of elements in $$\mathbb{N}$$, we have $$\aleph_1 > \aleph_2$$. To get an intuition as to why this is, consider two real numbers, $$x,y \in \mathbb{R}$$. Without any loss of generality, suppose that we were trying to count the reals and that we assigned $$x$$ an index of 1 and $$y$$ an index of 2. Could we safely proceed on to other numbers knowing that $$x$$ and $$y$$ are addressed? No; we’ve skipped over infinite real numbers. There’s no way to order them so that we can assign each a natural number.17 The real numbers $$\mathbb{R}$$ are what we call uncountably infinite and thus are an uncountable set.
Note that $$A= [0,1]$$ and $$B=[0,2]$$ are both uncountably infinite. Wouldn’t most of us say that $$A$$ is smaller than $$B$$? This just goes to show that cardinality is a somewhat limited way to talk about size. It is, however, the most useful. We could use other approaches—like those you see in measure theory—to come up with intuitive ways to say something like $$B$$ is longer than $$A$$, or in many dimensions that $$B$$ is bigger than $$A$$.
### 2.2.2 Functions
You are probably accustomed to statements of the form $$f(x) = x^2$$. We will need a somewhat more general way to think about functions. I will start off somewhat loosely. We are often concerned with attaching elements of one set with elements of another set. Sometimes, you see this referred to as a “marriage” between the sets. So, loosely speaking, we can say that a function, call it $$f$$, is a relation among two sets, call them $$A$$ and $$B$$, that satisfies the restriction that for each $$a \in A$$, there is a single element $$b \in B$$ such that $$(x,y) \in f$$. We call the sets $$A$$ and $$B$$ the domain and co-domain, respectively.
What is a relation? A relation between sets $$A$$ and $$B$$ is any subset of their Cartesian product, $$A \times B$$. A function is a special kind of relation that ensures that each element of $$A$$ is paired and that it is assigned a single element of $$B$$.18
When we introduce a function, we usually do so like this: “define the function $$f: A \rightarrow B$$.” This means that $$A$$ is the domain and $$B$$ is the co-domain. You can then define $$f(x) = x^2$$ from there if you would so like.
We sometimes concern ourselves with some other relevant sets defined by $$f$$. For example, the image of $$C \subset A$$ given $$f$$ is $$f(C) = \{f(c) \in B : c \in C\}$$. Thus, if we define $$f: \mathbb{R} \rightarrow \mathbb{R}$$ where $$f(x) = x^2$$, it turns out that $$f(\mathbb{R}) = \mathbb{R}_+$$, since all squares are non-negative. The pre-image of $$D \subset B$$ is the same in reverse: $$f^{-1}(D) = \{a \in A : f(a) \in D\}$$.
A function that assigns each distinct element of the domain to a distinct element of the co-domain is called injective or one-to-one. That is, $$f$$ is injective if $$x,y \in A$$ with $$x \neq y$$ implies that $$f(x) \neq f(y)$$. A function that uses every element of the co-domain is surjective or onto. That is, $$f$$ is surjective if $$f(A) = B$$. A function that is both injective and surjective is called bijective.
## 2.3 Probability Spaces
Remember that a random experiment is a chance mechanism that satisfies:
1. all possible outcomes are known a priori;
2. in any particular trial the outcome is not known a priori but there exists a discernible regularity of occurrence associated with the outcomes; and
3. it can be repeated under identical conditions.
We will formalize the first two of these requirements today. On the subject of formalization, let’s call a random experiment $$\mathscr{E}$$.
### 2.3.1 Condition One: The Outcomes Set
We will collect all of the possible distinct outcomes of an experiment into an outcomes set, which we will usually denote $$S$$. For example, if the experiment is to toss a coin twice and note the outcome, then $S = \{(HH), (HT), (TH), TT\}.$
Outcomes sets can be countably infinite. If the experiment is to toss a coin until the first heads comes up, then $S = \{(H), (TH), (TTH), (TTTH), (TTTTH), \ldots\}.$ This is countably infinite, as each of these outcomes can be assigned a natural number, but (in theory) you could toss an arbitrary number of tails prior to the first heads.
Outcomes sets can be uncountably infinite, too. If the experiment is to turn on a lightbulb and keep it on until it burns out, then $S = \mathbb{R}_{+},$ since it cannot be turned on for “negative” time.
So, the requirement of a random experiment is that each of the distinct possible outcomes be known in advance. We then collect these into the outcomes set $$S$$.
### 2.3.2 Condition Two: The Events of Interest
It could well be that we don’t care about the elementary outcomes per se. For example, consider the experiment of tossing two coins and counting the number of heads. As noted before, we have $S = \{(HH), (HT), (TH), (TT)\}.$
Now consider the event of $$n$$ heads being tossed, where $$n \in \{0,1,2\}$$. I will label these $$E_n$$. Each will be given a subset of the outcomes set: \begin{align*} E_0 &= \{(TT)\}, \\ E_1 &= \{(HT), (TH)\}, \\ E_2 &= \{(HH)\}. \end{align*} Or maybe we care about whether the coins match: \begin{align*} E_{\text{match}} &= \{(HH), (TT)\}. \end{align*} Or maybe we care about whether at least one head is tossed: \begin{align*} E_{\geq 1} &= \{(HH), (HT), (TH)\}. \end{align*}
So, an event is just a subset of the outcomes set. When we discussed the power set, we noted that $$\emptyset$$ and $$S$$ are both subsets of $$S$$. We refer to $$\emptyset$$ as the impossible event and S as the sure event, though these are just empty nomenclature right now.
To be explicit about things, if I say $$s \in S$$, I mean that $$s$$ is being thought of as an elementary outcome. Conversely, if I say $$\{s\} \subset S$$, I’m talking about it as an event. Events are not outcomes! Outcomes are not events! Events are combinations of outcomes!
#### 2.3.2.1 Rôle of the Events of Interest
We now must make more precise just what we mean by events of interest. At the very least, we now know that events are just subsets of the outcomes set $$S$$.
To keep the drama to a minimum, let me reveal what you might have guessed by now: the events of interest will serve as the domain of a function that assigns probabilities. So, we may wish to know the probability of, say, the event getting at least one heads in two tosses of a coin. Implicit in this, however, is the probability of not getting at least one heads in two tosses. That is, if we wish for an event $$E$$ to be included in the domain of our probability fucntion, we must also have $$\overline{E}$$ included, as well.
Similarly, let’s think about two separate events—say, getting zero heads and getting two heads in the toss of a coin. Call these $$E_0$$ and $$E_2$$ as before. Along with the respective probabilities of each, we will also want the probability of their union—the probability that $$E_0$$ happens or that $$E_2$$ happens.
In other words, once we have defined the events of interest, we have implicitly implied secondary events of interest that address both the complement and the unions of events. This will play a key role in our ability to specify the events of interest proper.
#### 2.3.2.2 The Power Set and Associated Difficulties
It is tempting to build the most general theory of probability that we can.19 And hey, if each event is a subset of the outcomes space, then why not just define probability functions with a domain of $$2^S$$—the power set of the outcomes space, which contains all possible subsets of $$S$$ and thus all possible events of interest?
For example, suppose that we are interested in the outcomes of tossing a coin twice.20 So, that means the outcomes set is $S = \{(HH), (HT), (TH), (TT)\}.$ Very good. So, let’s think about what this means for the power set, shall we?
1. So, we know that $$\emptyset$$ and $$S$$ are elements, so there’s two elements;
2. We also need all the singleton options: $$(HH)$$, $$(HT)$$, $$(TH)$$, $$(TT)$$, which brings our running total up to six elements;
3. And, we need all possible pairs of these: $$((HH), (HT))$$, $$((HH), (TH))$$, $$((HH), (TT))$$, $$((HT), (TH))$$, $$((HT), (TT))$$, and $$((TH),(TT))$$, so that we’re up to 12 elements; and
4. All the possible triples: $$((HH),(HT),(TH))$$, $$((HH),(HT),(TT))$$, $$((HH),(TH),(TT))$$, $$((HT),(TH),(TT)$$, so that we’re at 16 elements, which happens to be $$2^4$$.
That isn’t so bad—with a little work, we could probably assign reasonable probabilities to sixteen things, right? Two responses:
1. For starters, even with simple problems like the one discussed above, things get unwieldy quite quickly. If we toss the coin three times instead of twice, we end up with a power set with 256 elements. Indeed, the number of elements goes up really quickly in the number of tosses. To see why, note first that there are $$2^n$$ possible outcomes for $$n$$ tosses of a coin—so that $$|S| = 2^n$$. And then, the number of elements in the power set is $$2^{|S|}$$, so that the number of elements in the power set for $$n$$ tosses is $$2^{2^n}$$.21 Four tosses? 65,536 elements of the power set. Five tosses? Over four and a quarter million elements. Working at a speed of one event per second around the clock, it would take over five years to enumerate every possible event in the case of six tosses. Now think to yourself: there are a little under 200 countries in the world, each tossing a civil war coin. There are tens of millions of voters, each tossing a Democrat-Republican-Other die—oh my word that means the base number is three instead of two. Even with finite numbers—reasonable ones that we can understand!—the power set gets too big too fast.
2. …and now the bad news! It turns out that, if $$S$$ is countable, then the power set is at least possible to define probabilities on, even if it’s not practical. But, if $$S$$ is uncountably infinite, then it may be that $$2^S$$—which still exists—cannot serve as the domain of a probability function, whether we’re willing to put in the years or not.
So, we have one practical reason and one purely mathematical reason to not use $$2^S$$ in the general case.
#### 2.3.2.3 The Workaround
Because of these problems, we need to find a way to wrangle the events of interest into something more workable. It turns out that we can focus our attention on collections of subsets of $$S$$ that satisfy certain properties. We will refer to collections that satisfy these properties as algebras or, if they satisfy stronger conditions, as $$\sigma$$-algebras.
A nonempty collection $$\mathfrak{F}$$ of subsets of a set $$S$$ is an algebra of sets if it is closed22 under finite unions and complementation—that is, if $$E_1,E_2 \in \mathfrak{F}$$, then it must be that $$E_1 \cup E_2 \in \mathfrak{F}$$ and $$\overline{E}_1 \in \mathfrak{F}$$. It is a $$\sigma$$-algebra if it is closed under countable unions, so that if $$E_1,E_2,\ldots \in \mathfrak{F}$$, then $$E_1 \cup E_2 \cup \cdots \in \mathfrak{F}$$. Clearly all $$\sigma$$-algebras are algebras, but not all algebras are $$\sigma$$-algebras.
So, as an example: suppose we’re tossing a coin three times, so that $S = \{(HHH),(HHT),(HTT),(HTH),(TTT),(TTH),(THT),(THH)\}.$ But, suppose that we’re interested only in $$A_1 = \{(HHH)\}$$ and $$A_2 = \{(TTT)\}$$, the two events where three of a kind are tossed. We might define the relevant algebra as: $\mathfrak{F} = \{\emptyset,S,A_1,A_2,(A_1 \cup A_2),\overline{A}_1,\overline{A}_2,(\overline{A}_1 \cap \overline{A}_2)\}.$ I’ll ask you to verify that this is an algebra in your problem set.
Note that, for countable sets $$S$$, $$2^S$$ is necessarily a $$\sigma$$-algebra. However, if $$S$$ is uncountable, then the smallest $$\sigma$$-algebra on $$S$$ is a proper subset of the power set—that is, there are subsets of $$S$$ that are not in the relevant $$\sigma$$-algebra. It turns out that these exclusions provide the necessary mathematical structure for us to be able to define a probability function.23
### 2.3.3 The Probability Function
So, let’s define that now. Given an outcomes set $$S$$ and a $$\sigma$$-algebra $$\mathfrak{F}$$, a function $$P: \mathfrak{F} \rightarrow [0,1]$$ is a probability function if it satisfies the following three axioms:
1. $$P(S) = 1$$ for any outcomes set $$S$$;
2. $$P(A) \geq 0$$ for any $$A \in \mathfrak{F}$$; and
3. For a countable sequence of mutually exclusive events $$A_1,A_2,\ldots$$, where $$A_i \cap A_j = \emptyset$$ for all $$i \neq j$$, we have $P\left(\bigcup_{i=1}^\infty A_i\right) = \sum_{i=1}^\infty P(A_i).$ These are called Kolmogorov’s Axioms in honor of Andrey Kolmogorov, who first axiomatized probability this way. These are not the only axioms used—in particular, some probability theories use only finite unions—but they are easily the most common.
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DeGroot, Morris H., and Mark J. Schervish. 2012. Probability and Statistics. Fourth. Boston, MA: Addison-Wesley.
Duggan, John. 2013. “Basic Concepts in Mathematical Analysis: A Tourist Brochure.” http://www.johnduggan.net.
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Friedman, Milton. 1953. “The Methodology of Positive Economics.” In Essays in Positive Economics, edited by Milton Friedman, 3–43. University of Chicago Press.
Simon, Carl P., and Lawrence Blume. 1994. Mathematics for Economists. New York: W.W. Norton & Company.
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1. After having built such a beautiful apparatus, we will quickly realize its faults; in particular, it is of no use for discussing data. Great. We will therefore concern ourselves with morphing the simple statistical space into something that can address data: the simple statistical model. From there, we will be able to create extensions to talk about all kinds of data. Be patient!
2. Consider, for example, the mind-bending Russell’s Paradox, in which we try to form the set of all sets that do not include themselves.
3. We have just begged the question—what the hell is a collection?! Seriously, you don’t want to go there.
4. Why does LaTeX show these as white? Weird.
5. Indeed, one of the main early tasks in an analysis class is to construct the real numbers using axioms. We will avoid that; you’re welcome.
6. I will prove by contradiction. Suppose $$\sqrt{2}$$ were rational. This would mean that there exist some integers $$p$$ and $$q$$ such that $$\sqrt{2} = \frac{p}{q}$$. We do not affect any results by assuming that $$p$$ and $$q$$ have no common factors, as we can just cancel them out from the numerator and denominator. Squaring both sides, we have $$2 = \frac{p^2}{q^2}$$, implying $$2q^2 = p^2$$. This means $$p^2$$ is even, which holds only when $$p$$ is even. This means $$p^2$$ must be divisible by 4. This means $$q$$, and thus $$q^2$$, must also be even. But this means $$p$$ and $$q$$ have a common factor—namely, 2. We therefore have contradicted a premise, so it must be that $$\sqrt{2}$$ is rational.
7. Implies? What is implies? By this, we mean that if we know it is true that the first statement is true, then we also know that the second statement is true. The sentence has no bite for situations where the first statements is false.
8. The order of the elements in a set does not matter.
9. The ordering part here is what matters, rather than the “skipping infinity of them” part. Note that the rational numbers $$\mathbb{Q}$$ are countable, even though you “skip infinity of them” to get from one to another. Without loss of generality, let me just work with the positive rationals. Take any positive rational and call it $$\frac{a}{b}$$, where $$a$$ and $$b$$ have no common factors. Then we can assign each of them the unique integer $$2^a3^b$$.
10. Functions are by far the most common kind of relation you will work with. However, in game theory we will discuss preference relations that satisfy certain properties, and we will also generalize functions into correspondences that can assign multiple elements of the co-domain to any element of the domain, so that $$y \in f(x)$$ instead of $$y = f(x)$$. But let’s not get ahead of ourselves.
11. Remember: simplicity is good, as it means that we can explain lots of things with few assumptions.
12. By now, I am as tired as you are of flipping two hypothetical coins, and I wonder to myself: just what sin did I commit for Dante to send me to the coin-flipping circle of the Inferno?
13. You know that numbers are getting too big too fast when people have to develop a special notation to handle them.
14. We say a collection is closed under some operator if the output of the operator when each of the collection’s elements is input are also in the family. So, for example, since the sum of two integers is itself an integer, the integers are closed under addition.
15. Actually, these are what you need to ensure that you can define a measure. Probability theory is just an application of measure theory. | 8,443 | 30,073 | {"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} | 3.8125 | 4 | CC-MAIN-2022-33 | longest | en | 0.914832 |
https://edis.ifas.ufl.edu/publication/fr426 | 1,695,634,414,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233508959.20/warc/CC-MAIN-20230925083430-20230925113430-00703.warc.gz | 256,654,091 | 49,438 | # Timber Inventory: A Primer for Landowners
John Dooner and Michael Andreu
## Introduction
An integral part of any forest management program and long-term economic plan is an estimation of the current volume and value of a timber stand. Foresters approximate the amount of lumber or fiber within a stand to inform their clients on how to manage their assets and maximize revenue. The process of determining timber inventory typically involves sampling a portion of the forest and expanding these samples to calculate stand-level estimates. Commonly referred to as timber cruising, this exercise begins with planning and determining the sample allocation and type of plots, followed by basic measurement and analysis of a single tree for an estimate of stem volume. This paper outlines the process of timber cruising from initial tree-level measurements to the final total stand-level estimates and various methods for conducting the cruise.
## Tree Measurements
### Diameter Measurements
Amongst forest measurements, the most common and easily obtained tree measurement is diameter at breast height (DBH). DBH is defined as average outside bark diameter when measured on the uphill side of a tree at 4.5 feet (Avery and Burkhart 2002). DBH is commonly measured with a diameter tape (D-tape), which is designed to allow users to measure circumference at 4.5 feet (1.3 meters) and convert to a measurement of diameter (diameter = circumference/pi [where pi = 3.14]). The converted diameter readings are printed on the D-tape for quick and efficient field work. The following video shows techniques for using a D-tape: https://www.youtube.com/watch?v=B67QPJa2pbM&list=PLr5M0QvUoAel1dKnwLvt9EF5lEtzolyTw&index=4. DBH can also be measured using calipers or Biltmore sticks.
If the bole or stem of a tree were shaped like a true cylinder, the calculation of volume in the stem would be simple. However, most tree stems taper as you move higher in the tree. Therefore, upper diameters along the stem are also important components of volume equations. The stem diameter at the small end of the first log (16 feet in the eastern United States) is often critical for determining the rate of taper in the stem and is used in many of the traditional log rules to generate volume estimates.
Because tree bark can be quite thick on some species, diameters inside the bark are used in many estimations of volume and often require the determination of the tree bark's contribution to diameter. To subtract the area occupied by the bark, the bark thickness is measured and subtracted from the diameter outside the bark. Ratios between outside and inside bark diameters at DBH are often determined and applied to outside bark diameter readings taken from points along the upper portions of the stem.
### Height
Height is the second essential element for estimating tree volume. Total height or merchantable height can be measured when cruising. Total height is measured to the top of the tree, but merchantable height is often a more informative measurement for determining the available usable wood in a tree. For most pine species in the eastern United States, merchantable heights can be measured at log lengths (16 ft) and half-log lengths (8 ft); valuable hardwood species, when used as sawtimber, can be cut to log lengths of 10, 12, 14, 16, or 18 feet. Both total and merchantable heights can be measured with various tools, but clinometers are generally most efficient. Clinometers use geometric functions to determine tree height by measuring angles from eye level to the tree base and crown at a fixed distance from the tree. Tree height can then be calculated based on the tangent of these measurements. These videos will help you learn to use a clinometer to measure tree height: www.youtube.com/watch?v=q4LhwniXAJI&list=PLr5M0QvUoAel1dKnwLvt9EF5lEtzolyTw&index=14
Merchantable heights are a function of stem quality, as well as upper-stem diameter limits, which determine the last point at which the stem is usable. Accuracy in these determinations requires considerable experience in estimating upper-stem diameters from ground level with the naked eye. Foresters and land managers must become proficient in this practice to obtain accurate volume estimates.
### Form
The final variable needed to calculate total tree volume is a measurement of the amount of stem taper in a tree. In general, tree diameter decreases with increasing tree height (like a cone). The environment in which a tree exists can influence the rate at which this taper occurs. For example, trees grown in close proximity to competitors (e.g., in a plantation) will tend to taper less than those grown in open areas. The characterization of this rate of taper is referred to as form class and is primarily expressed using the Girard form class. To determine Girard form class, the inside bark diameter at the top of the first log is expressed as a percentage of DBH. For example, if a tree has a 16.0-inch DBH and is 12.8 inches (inside bark diameter) at the top of the first log, the form class would be 80 (i.e., 100*12.8/16 = 100*0.8). This upper diameter measurement is taken at a height of 17.3 feet to accommodate a one-foot stump and 0.3 feet of log trimming. Form class measurements for southern pine plantations generally range from 78 to 82. Although this four-percent difference in form class seems miniscule, a three-percent change to merchantable volume can be attributed to a one-percent difference in form class (Avery and Burkhart 2002). In other words, even a small mistake in form class can result in a large error when merchantable volume is calculated. To ensure accuracy when determining form class for a stand, it is recommended that a small subsample of trees be carefully measured to determine the form class for the entire stand.
## Determining Tree Volume and Weight
The measurements discussed in the paragraphs above must be accompanied by a method for generating actual tree volumes. To standardize an estimate for how much wood is being transferred in a transaction for buyers and sellers, log rules were developed (e.g., Doyle, Scribner, International) to estimate the amount of usable wood in a log that is available for lumber production. Tables derived from these log rule equations were established for quick and efficient calculations of wood volumes generated from a cut log, measured in board feet. One board foot is defined as 12 inches x 12 inches x 1 inch thick, or 144 cubic inches. More recently, a large portion of the forest industry utilizes tonnage, or weight of timber, as the primary trading parameter, especially for pine products. Much of the coastal plain of the Southeast, where southern pine species (e.g., loblolly, slash, shortleaf, and longleaf pines) are predominately grown, uses tonnage for buying and selling timber. Estimates of tonnage can be obtained through equations designed to convert volume estimates into weight.
### Log Rules
Log rules use measurements like inside bark diameter and length of a log to determine wood volume in a log and then attempt to account for wood deductions lost in the manufacturing of lumber. The volume estimates generated by the log rule are commonly given in board feet, or the amount of usable one-foot by one-foot by one-inch sections in a log. Although numerous log rules have been developed, the Doyle, Scribner Decimal C, and International ¼ rules are most commonly used in the United States. The two older log rules, Doyle (developed in 1825) and Scribner (developed in 1846), tend to be less accurate when compared to the International ¼ log rule (developed in 1906).
To illustrate the implementation of log rule equations, a sixteen-foot log with a diameter (D) measuring 14 inches and an inside bark diameter of 10 inches at the small end of the log will be used. By inputting these variables into the equations in the table above, the subsequent board feet estimated by each log rule can be calculated. According to the International ¼ rule, this log produces roughly 71.5 board feet. When calculating the volume of the same log using Scribner Decimal C and Doyle rules, it will only produce 55 board feet and 36 board feet, respectively. For this reason, it is critical to know when you are buying or selling timber which log rule is being used to calculate volume. It can have a great impact on the final timber value.
To facilitate the efficient application of various log rules, tables have been created to allow foresters use measurements of DBH and height and apply the log rule of choice. To increase the accuracy and effectiveness of these tables, Girard form class is often coupled with a particular log rule in the formation of these tables. As a result, foresters can determine the stand's form class and then locate the table for that particular form class that also uses the log rule of choice. The log table below uses the International ¼ log rule and a Girard form class 80.
### Weight Equations
Today, pine tends to be sold in tons vs. volume (board feet for sawtimber, cords for pulpwood), so weight equations have been developed based on the specific gravity and moisture content of wood. According to Avery and Burkhart (2002), the following equation can be used to determine the weight of a cubic foot of wood across all species: Density = specific gravity * 62.4 (1 + [% moisture content/100]).
This process can be demonstrated by analyzing an acre of slash pine (Pinus elliottii) timber with specific gravity and moisture content in a real-world example provided by Avery and Burkhart (2002). The measured specific gravity of slash pine was 0.52 and the moisture content 120%. By inputting these numbers into the equation above, the weight of a cubic foot of slash pine timber equals 71.4 pounds, (0.52*62.4 [1 + (120/100)]). Now the number of cubic feet that will be harvested in the acre of slash pine can be multiplied by the weight of a single cubic foot. If this acre produced 864 cubic feet of solid wood, the corresponding weight equals 61,689.6 pounds (71.4 pounds/cubic foot *864 cubic feet). This number would commonly be expressed in tons as roughly 31 tons per acre.
Other quick conversion factors have been developed to streamline the process of converting volume to weight. The South Carolina Forestry Commission estimates that one cord (128 ft3) of pine pulpwood is roughly 5,350 pounds, or 2.675 tons. One thousand board feet of pine sawtimber is estimated to be between 7.50 and 7.75 tons. While these simple conversions are useful for rough estimates of harvestable tonnage, precise measurements can only be obtained from accurate measurements of specific gravity and moisture content. Small changes in moisture contents can create large miscalculations of estimated weights. That said, moisture content can vary in a tree daily, and more significantly on a seasonal basis, so the heuristics (rules of thumb) above are commonly used.
## Stand Level Inventory
The previous sections focused on individual trees and methods for determining wood volume in those trees. To apply this knowledge to stand-level estimations, foresters must evaluate a sample of trees within a stand. The volume of each of the sample trees can be determined, and average volume for the stand can be estimated using an expansion factor that projects the sampled averages to stand-level metrics. This process is used to analyze stands for a variety of objectives, including financial planning, stand management, habitat suitability, and development of silvicultural prescriptions.
To determine the method by which the timber estimate or "cruise" will be implemented, several factors must be considered. Cruises can be conducted with varying levels of accuracy depending on how the results will be used and the amount of money that will be invested in the cruise. The more individual trees measured in the cruise, the more closely the results will portray the actual stand. However, sampling individual trees requires time, resulting in increased costs. While accuracy levels can vary, the cruising technique must measure enough of the population to reflect a statistically significant estimate of the true values of the stand. Without reasonable precision, the results and the time spent gathering the input data are of little value. Finally, the technique by which the cruise will be conducted must be determined. This includes decisions such as plot type, size, and sampling design. Should variable- or fixed-radius plots be used? Would stratified random samples or grid systems work better for this situation? These types of questions must be answered to design the proper cruising plan for meeting predetermined objectives.
### Cruise Type
When determining the type of cruise to design for a particular forest, the stand characteristics must be taken into consideration. Differing soils, habitats, terrain, and density can all influence the composition of a timber stand and affect the cruising technique that is most suitable. Simple random samples may be necessary when the variability is relatively consistent throughout the stand. To achieve a truly random selection, it is essential that, to quote Avery and Burkhart, "the selection of a particular unit be in no way influenced by the other units that have been selected or will be selected" (2002). Sampling locations may fall adjacent to one another with this method, as long as the minimum number of samples is statistically efficient and samples are random and representative of the stand.
For forests with stands that contain more clustered variability, for example, a 100-acre block that has three distinct stand types, a stratified random sample may be more suitable. Simply put, stratified samples are like many small simple random samples within a stand. An example of the utilization of this method is outlined by analyzing a forest with stands encompassing a sandhill, a cypress pond, and flatwoods habitats. Rather than establish plots through a simple random sample across the entire forest, a stratified sample creates three distinct sampling areas: sandhill, cypress, and flatwoods. By sampling these strata independently of each other, a more accurate estimate of each stand can be gathered, and therefore a more accurate assessment of the entire forest.
The tendency to grow many timber species in plantation settings provides the opportunity to use another systematic sampling method that is set up through the development of a grid system. These uniform timber stands vary little in spacing and density across the stand, but random sampling is still a necessary practice. Cruises that implement grid techniques are defined by set distances between plot locations. For example, a 2x5 chain grid cruise (chain = 66 ft) is commonly used on small stands (< 50 acres) to determine plot centers and corresponds to one plot per acre. The size of the tract of land you are cruising and the desired intensity (i.e., the % of the land measured) will help determine what the grid spacing will be for any particular cruise. Common grid increments include 2x5 – 1 plot/acre, 5x10 – 1 plot/5 acres, 5x20 – 1 plot/10 acres, 10x20 – 1 plot/20 acres. Clearly, as the size of the grid spacing increases, each plot begins to represent a greater area and, subsequently, more trees. This is one component of the expansion factor that will allow one to go from individual plot-level calculations of volume to stand-level estimates.
### Plot Size and Type
Determining the size of sample plots is largely related to the stand characteristics of the area being cruised. Plots should be adequately sized to manage labor and time inputs while effectively accounting for the variability in the stand. For example, if the area being sampled is a seed tree cut twelve years after harvesting, the stand will be comprised of many small trees and a few scattered large ones. A few small plots may not adequately represent the area because the large, scattered trees may be excluded or poorly represented in the sample. As a result, either plot size or the number of plots must be increased to accurately sample the area. Stand density also plays a major role in determining plot size. Denser stands usually require smaller plots because several individuals must be measured in each sample plot, and they are closer to the plot center. Furthermore, this allows the cruiser to easily access the sample specimens. For open grown stands, larger plots may be more easily managed and can reduce sampling time and effort while still accounting for the variability within the stand. Generally, the denser and less variable a stand is, the smaller the plot size needed. Open forests, such as natural longleaf pine (Pinus palustris) habitats, may require plots as large as 1/5 acre because of the distances between individual trees.
For fixed-area plots, which are plots with a predetermined area, each tree measured represents multiple trees on that given acre. For example, if using a 1/5-acre plot, then each tree measured on that plot represents 5 trees on that acre. In the case of a 1/10-acre plot each tree represents 10 trees. Therefore, it is easy to see how important it can be to be diligent in measuring only trees that are actually "in" or within the boundaries of the plot, because adding or missing a tree can contribute greatly to error in estimates. As a general rule of thumb, with fixed-area plots you will measure 10–15 trees.
The expansion factor for fixed-area plots is thus made up of two variables, the number of acres each plot represents and the number of trees each measured tree represents. To illustrate, each tree measured on a 1/10-acre sized plot would represent 10 trees. A 1/10-acre plot with three trees of 6-, 12-, and 28-inch DBH would be expanded to represent 30 trees per acre. Specifically, it would represent 10 6-inch DBH trees, 10 12-inch DBH trees, and 10 28-inch DBH trees). If the spacing of these plots was on a 5×10 chain grid (1 plot/5 acres), this plot (3 measured trees) would represent a total of 150 trees in this 5-acre stand. Volume estimations can be determined through the same process. If the volume estimate for the 1/10-acre plot was 650 board feet, corresponding per-acre volume would be 6,500 board feet/acre (10 * 650 board feet). Because the plot represents 5 acres (2×5 chain grid), the plot represents 6,500 board feet/acre times 5 acres or a total of 32,500 board feet. The same process is used across all fixed-area plots in the stand to calculate a total volume estimate.
Not all cruising techniques use a predetermined plot size. Rather, alternative techniques use variable-area plots in which plot size is a factor of the size of the sample trees. An instrument known as prism is often used to identify individual sample trees in variable-area plots leading some to refer to this technique as prism cruising. By creating a visual illusion in which a section of the tree is offset from the rest of the stem, the cruiser can determine if the tree lies within the plot. The larger the tree diameter, the greater the distance the tree can lie from the plot center while still being included in the plot and vice versa for smaller diameters. The result of this method is that a greater number of large trees are measured in the sample in relation to small trees. Larger trees are responsible for a greater percentage of the volume (and often are the most valuable ones), so this method targets those trees that contribute most to total volume estimations (Wenger 1984). Expansion factors for prism plots are based on the basal area factor (BAF) of the prism (or the instrument used) and the size of the tree measured. The following video is useful for learning more about prism cruising: www.youtube.com/watch?v=nhzTpvET_7M&list=PLr5M0QvUoAel1dKnwLvt9EF5lEtzolyTw
### Cruising for Statistical Precision
While designing a cruising plan, it is necessary to set some level of accuracy as the goal for the cruise. This accuracy level is determined by factors ranging from the forester's opinion to the objective for the cruise and, most importantly, the budget. A cruise that is more accurate than necessary will expend more time, effort, and funding than necessary to accomplish the goal of the cruise. The key point is that both inaccurate and overly accurate cruising designs result in a waste of resources. A number of equations can be used by foresters to determine the number of plots needed for a statistically accurate cruise, so professional foresters must be adept at understanding the statistical basis for estimating inventory and the financial constraints associated with the time and labor required to meet the objective.
### Product Classes
When conducting a cruise, it is not only necessary to determine the volume within a stand but to analyze the quality of the product as well. Two different stands may have similar amounts of total standing volumes but vary greatly in the value of the crop that has been grown. For example, a stand that has been thinned and is now composed of fewer large trees may have only slightly more cubic feet than an unthinned stand with many small trees. However, the former stand is valued much higher because of the superiority of the product. Accurately identifying products in a timber stand is an important component of any timber cruise.
In the Southeast, most of the softwood traded in the markets can be divided into three major classes: pulpwood, chip-n-saw, and sawtimber. Two other common product categories are poles and plylogs. Sawtimber, poles, and plylogs require specific diameter, form and straightness and tend to command a higher value. Pulpwood is generally composed of the smallest trees or larger trees that exhibit substantial deformities such as knots, crooks, and cankers. Southern pines grown for pulpwood can be grown on the shortest rotation lengths and can be harvestable as early as age 12 on the most productive sites, but are usually harvested between ages 15 and 20. Generally, trees sold as pulpwood range from 4 to 8 inches DBH and are the primary product in many areas of the Southeast. The value of pulpwood tends to be the lowest of all classes.
Chip-n-saw ranges from about 8 to 10 inches DBH and represents the smallest trees used to manufacture lumber. These trees are used in the production of 2×4 boards, primarily used in the framing of houses. Chip-n-saw and pulpwood are the usual product classes harvested in the first and second thinnings of a timber stand. Subsequent thinnings and final harvests provide opportunities for harvesting the larger sawtimber products.
Trees 10 inches and greater can be harvested as sawtimber provided the tree bole meets quality specifications. Sawtimber trees can be used in the manufacture of both grade lumber and veneer used in plywood. The longer rotations required for growing sawtimber or other higher-valued products are rationalized by the nearly tripled stumpage price when compared with pulpwood.
To determine the value of a timber stand through cruising, the categorization of these different products is essential for accurate estimations of potential profit. The importance of product distinction necessitates the use of an experienced forester when contracting the cruising of a timber stand. If the quality of the standing timber is valued incorrectly based on product classes, the expected revenue from the harvesting operation could be severely miscalculated.
## Summary
Ultimately, timber cruising combines several disciplines to provide insight into the amount of volume and relative value of a timber stand. Each component, whether biological, statistical, or economic is dependent on the others for synthesizing data and providing beneficial results from a cruise. Furthermore, to generate accurate estimates of timber volumes and potential revenue for financial planning or management guidelines, an experienced forester with extensive knowledge of evaluating individual trees and a thorough understanding of timber market dynamics should be consulted.
## Literature Cited
Avery, T. E., and H. E. Burkhart. 2002. Forest Measurements (5th ed.). New York, NY: McGraw- Hill.
Bond, B. 2011. "Understanding Log Scales and Log Rules." Retrieved from https://extension.tennessee.edu/publications/Documents/PB1650.pdf.
Freese, F. 1962. "Elementary Forest Sampling." U.S. Department of Agriculture Handbook 232. Washington, D.C.: Government Printing Office.
Husch, B., C. I. Miller, and T. W. Beers. 1982. Forest Mensuration (3rd ed.). New York, NY: John Wiley & Sons.
South Carolina Forestry Commission. (n.d.). "Understanding Timber as a Commodity." Retrieved from https://www.scfc.gov/resources/public-information/landowner-resources/understanding-timber/.
Wenger, K. F. (ed.). 1984. Forestry Handbook (2nd ed.). New York, NY: John Wiley & Sons.
Table 1.
Log rule equations for Doyle, Scribner Decimal C, and International ¼; D = inside bark diameter at small end of log, BF = board feet and L = log length.
Table 2.
Volume estimates (measured in board feet) by DBH; International ¼ log rule, Girard form class 80.
### Publication #FOR357
Date: 3/2/2020
#### Andreu, Michael G.
University of Florida
#### Field Methods
This document is FOR357, one of a series of the School of Forest, Fisheries, and Geomatics Sciences, UF/IFAS Extension. Original publication date February 2020. Visit the EDIS website at https://edis.ifas.ufl.edu for the currently supported version of this publication.
John Dooner, forester, Southern Forestry Consultants, Inc.; and Michael Andreu, associate professor, School of Forest, Fisheries, and Geomatics Sciences, UF/IFAS Extension, Gainesville, FL 32611
### Contacts
• Michael Andreu | 5,431 | 25,808 | {"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-2023-40 | longest | en | 0.910276 |
https://electronics.stackexchange.com/questions/611034/can-a-pulsating-signal-in-which-the-direction-of-the-current-may-not-reverse-rad | 1,701,429,437,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100286.10/warc/CC-MAIN-20231201084429-20231201114429-00880.warc.gz | 274,500,161 | 44,523 | # Can a pulsating signal in which the direction of the current may not reverse radiate from an antenna?
Normally an antenna uses a sinusoidal wave whose positive half-period corresponds with positive direction of the current and vice versa. A pulsating signal/current is a signal in which the direction of the current may not reverse. Is such a signal still able to radiate from a dipole or other antennas?
• Well, I am having trouble seeing where the charges would go. They can't just pile up in the antenna. So I think they will have to stop and reverse. Mar 7, 2022 at 5:15
• But I also think that any time an antenna is driven with time varying voltage it can radiate. Mar 7, 2022 at 5:50
• Then again there are some antennas, such as folded dipoles, which would pass DC current readily (very low impedance at DC). So maybe the question does make more sense than I initially gave it credit for. Mar 7, 2022 at 17:42
A mundane example is DC with AC ripple superimposed on it like that in virtually every closed loop DC circuit that exists. Current never reverses but EMI is radiated nonetheless. The current changes (increases and decreases) but never reverses direction.
What you need are changes in charge flow, not bidirectionality but if you have an antenna that is open-ended and thus a "dead-end" for charge flow then you need bidirectionality unless you want to forever increase the voltage so charge continues to flow into and accumulate in the antenna (until a giant arc is produced*).
EDIT: *@Lorenzo Donati -- Codidact.com says
Not necessarily. If the charge buildup is sufficiently slow some charge could escape the antenna by other means than arcing (corona discharge, for example). So if the antenna is pulsed at a slow rate an equilibrium could still be reached without arcing. –
4 hours ago
• "until a giant arc is produced" Not necessarily. If the charge buildup is sufficiently slow some charge could escape the antenna by other means than arcing (corona discharge, for example). So if the antenna is pulsed at a slow rate an equilibrium could still be reached without arcing. Mar 7, 2022 at 9:53
• Some antennas such as folded dipoles and also slot antennas have low impedance at DC. Mar 7, 2022 at 18:46
• Isn't there some superposition principle that could be applied as well? Mar 7, 2022 at 22:59
Start with a charged particle, in vacuum at rest in an inertial frame. The electric field lines converge at the particle's position radially. Now, move the particle over, and bring it back to rest. The electric field lines converge at its new position. However, far away, the electric field lines still point toward the old position. The boundary between the old field configuration and the new configuration expands at the speed of light. On the boundary, the electric field lines are kinked to match the old configuration with the new. That's an electromagnetic impulse.
Now, you may move the particle again, in whatever direction, creating a new kink. The new direction may be the same as the old, opposite, perpendicular, whatever. The details of the kinks will depend on the direction, but the charge will radiate regardless. The direction of the current need not reverse. However, this process requires acceleration of the radiating charge, so even if the direction of the current doesn't change, its magnitude must change.
• So an electron flying straight through space at constant speed generates waves as it enters new space and one in a closed loop generates them too because to change directions to close the loop requires acceleration? Mar 7, 2022 at 16:15
• @DKNguyen An electron flying at constant velocity generates no waves. Acceleration, either change of speed or direction, is necessary. Mar 7, 2022 at 16:27
• Isn't it still reaching areas of space where the electric field has not yet reoriented itself to the electron? Mar 7, 2022 at 16:27
• @DKNguyen with no acceleration, there is a frame in which the electron doesn't move. If there are no waves in some frame, there are no waves in any frame. If you follow this out, as Einstein did, it leads to special relativity. Mar 7, 2022 at 16:34
• So not quite as straightforward as the field lines in space reorienting themselves to the electron has it passes by. Mar 7, 2022 at 16:40
"Pulsating" implies "alternating current" (alternating between "on" and "off" is alternating current. Think "square wave". or "rectangular wave" which is a more accurate term. Although each is made of a fundamental sine wave and its harmonics.)
Then think about the law of Conservation of Energy. Energy can neither be created or destroyed. Only converted to other forms of energy.
Then think about Faraday's laws of magnetics. How does an antenna work? What happens when current flows in a conductor? (Hint: the energy is converted into a magnetic field expanding outward from the wire at the velocity of light which is abbreviated as "c".)
What happens when the energizing current ceases to flow in the conductor? (Hint: The magnetic field is converted into electric current when it collapses back into the conductor at c, and "induces" a reverse current in the conductor.)
And don't forget c! It takes time for a pulse to travel from the feedpoint to the end of the conductor, and back when the pulse ends. The length of the conductor determines the length of time, which in turn determines the resonant frequency of the conductor. That's why antennas for different frequencies are made in different lengths.
You original question is a very good one, and you will have a great experience duplicating the thought processes of the pioneers such as Maxwell, Lorenz, Faraday, Einstein and many more. Work to master the math and you will become enlightened with a whole new understanding of how the universe actually works! | 1,302 | 5,812 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.234375 | 3 | CC-MAIN-2023-50 | latest | en | 0.94707 |
http://mathhelpforum.com/new-users/212633-help-please-print.html | 1,508,761,034,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187825900.44/warc/CC-MAIN-20171023111450-20171023131450-00100.warc.gz | 216,861,471 | 2,821 | • Feb 5th 2013, 03:12 PM
dcmath
I have no idea because of the ambiguous nature of the question whether or not the next week includes 7 days or just 3 ''In preparation for the 10km mini marathon , maray starts training. she plans to run 1.5km four days this week. Each week she plans to increase the distance she runs each day by 0.5km
(i) write down the distance she plans to run in the second , 3rd and 4th week of her training programme
• Feb 5th 2013, 03:58 PM
Soroban
Hello, dcmath!
I assume she runs four days every week.
Quote:
In preparation for the 10km mini marathon, maray starts training.
She plans to run 1.5km four days this week.
Each week she plans to increase the distance she runs each day by 0.5km
(i) Write down the distance she plans to run in the 2nd, 3rd and 4th week of her training program.
$\begin{array}{ccccccc}\text{Week 1} & 4\times 1.5 &=& 6\text{ km} \\ \text{Week 2} & 4 \times 2.0 &=& 8\text{ km} \\ \text{Week 3} & 4 \times 2.5 &=& 10\text{ km} \\ \text{Week 4} & 4 \times 3.0 &=& 12\text{ km} \\ \vdots & \vdots && \vdots \end{array}$ | 361 | 1,074 | {"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": 1, "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-2017-43 | longest | en | 0.891517 |
https://www.askiitians.com/forums/8-grade-maths/a-vessel-is-filled-with-liquid-3-parts-of-which-a_141022.htm | 1,631,919,208,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780055808.78/warc/CC-MAIN-20210917212307-20210918002307-00660.warc.gz | 692,373,808 | 36,163 | #### Thank you for registering.
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# A vessel is filled with liquid, 3 parts of which are water and 5 parts syrup. How much of the mixture must be drawn off and replaced with water so that the mixture may be half water and half syrup.Logically how it is to be done. Thanks
Harsh Patodia IIT Roorkee
5 years ago
Water / Syrup = 3x/5x
Let as say y litres is taken out and replaced. Water taken out 3y/8 and syrup 5y/8
Acc to question
(3x- 3y/8 +y )/(5x-5y/8) = 1/1 (Since both are equal)
3x+5y/8 = 5x-5y/8
2x=5y/4
y = 8x/5
Total mixture 8x
Part that should be taken out is y/x = 1/5 th of original mixture. | 310 | 922 | {"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-2021-39 | latest | en | 0.911316 |
https://www.saranextgen.com/homeworkhelp/doubts.php?id=55649 | 1,709,279,373,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947475203.41/warc/CC-MAIN-20240301062009-20240301092009-00064.warc.gz | 970,818,143 | 5,603 | # The line passing the points and the line joining the points intersect at a) b) c) d) None of these
## Question ID - 55649 | Toppr Answer The line passing the points and the line joining the points intersect at a) b) c) d) None of these
3537
(d)
The equations of the lines joining
and are respectively
…(i)
and, …(ii)
For the point of intersection, the equations (i) and (ii) should give the same value of
Hence, equating the coeff.of vectors and in the two expressions for , we get
and
Solving first two equations, we get
These values of m and n also satisfy the third equation
Hence, the lines intersect
Putting the value of m in (i), we obtain that the position vector of the point of intersection as | 195 | 728 | {"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-2024-10 | latest | en | 0.857931 |
https://igraph.org/r/html/1.2.5/layout_with_dh.html | 1,675,804,815,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764500641.25/warc/CC-MAIN-20230207201702-20230207231702-00363.warc.gz | 320,361,651 | 5,257 | # R igraph manual pages
Use this if you are using igraph from R
## The Davidson-Harel layout algorithm
### Description
Place vertices of a graph on the plane, according to the simulated annealing algorithm by Davidson and Harel.
### Usage
layout_with_dh(
graph,
coords = NULL,
maxiter = 10,
fineiter = max(10, log2(vcount(graph))),
cool.fact = 0.75,
weight.node.dist = 1,
weight.border = 0,
weight.edge.lengths = edge_density(graph)/10,
weight.edge.crossings = 1 - sqrt(edge_density(graph)),
weight.node.edge.dist = 0.2 * (1 - edge_density(graph))
)
with_dh(...)
### Arguments
graph The graph to lay out. Edge directions are ignored. coords Optional starting positions for the vertices. If this argument is not NULL then it should be an appropriate matrix of starting coordinates. maxiter Number of iterations to perform in the first phase. fineiter Number of iterations in the fine tuning phase. cool.fact Cooling factor. weight.node.dist Weight for the node-node distances component of the energy function. weight.border Weight for the distance from the border component of the energy function. It can be set to zero, if vertices are allowed to sit on the border. weight.edge.lengths Weight for the edge length component of the energy function. weight.edge.crossings Weight for the edge crossing component of the energy function. weight.node.edge.dist Weight for the node-edge distance component of the energy function. ... Passed to layout_with_dh.
### Details
This function implements the algorithm by Davidson and Harel, see Ron Davidson, David Harel: Drawing Graphs Nicely Using Simulated Annealing. ACM Transactions on Graphics 15(4), pp. 301-331, 1996.
The algorithm uses simulated annealing and a sophisticated energy function, which is unfortunately hard to parameterize for different graphs. The original publication did not disclose any parameter values, and the ones below were determined by experimentation.
The algorithm consists of two phases, an annealing phase, and a fine-tuning phase. There is no simulated annealing in the second phase.
Our implementation tries to follow the original publication, as much as possible. The only major difference is that coordinates are explicitly kept within the bounds of the rectangle of the layout.
### Value
A two- or three-column matrix, each row giving the coordinates of a vertex, according to the ids of the vertex ids.
### Author(s)
Gabor Csardi csardi.gabor@gmail.com
### References
Ron Davidson, David Harel: Drawing Graphs Nicely Using Simulated Annealing. ACM Transactions on Graphics 15(4), pp. 301-331, 1996.
layout_with_fr, layout_with_kk for other layout algorithms.
Other graph layouts: add_layout_(), component_wise(), layout_as_bipartite(), layout_as_star(), layout_as_tree(), layout_in_circle(), layout_nicely(), layout_on_grid(), layout_on_sphere(), layout_randomly(), layout_with_fr(), layout_with_gem(), layout_with_graphopt(), layout_with_kk(), layout_with_lgl(), layout_with_mds(), layout_with_sugiyama(), layout_(), merge_coords(), norm_coords(), normalize()
### Examples
set.seed(42)
## Figures from the paper
g_1b <- make_star(19, mode="undirected") + path(c(2:19, 2)) +
path(c(seq(2, 18, by=2), 2))
plot(g_1b, layout=layout_with_dh)
g_2 <- make_lattice(c(8, 3)) + edges(1,8, 9,16, 17,24)
plot(g_2, layout=layout_with_dh)
g_3 <- make_empty_graph(n=70)
plot(g_3, layout=layout_with_dh)
g_4 <- make_empty_graph(n=70, directed=FALSE) + edges(1:70)
plot(g_4, layout=layout_with_dh, vertex.size=5, vertex.label=NA)
g_5a <- make_ring(24)
plot(g_5a, layout=layout_with_dh, vertex.size=5, vertex.label=NA)
g_5b <- make_ring(40)
plot(g_5b, layout=layout_with_dh, vertex.size=5, vertex.label=NA)
g_6 <- make_lattice(c(2,2,2))
plot(g_6, layout=layout_with_dh)
g_7 <- graph_from_literal(1:3:5 -- 2:4:6)
plot(g_7, layout=layout_with_dh, vertex.label=V(g_7)\$name)
g_8 <- make_ring(5) + make_ring(10) + make_ring(5) +
edges(1,6, 2,8, 3, 10, 4,12, 5,14,
7,16, 9,17, 11,18, 13,19, 15,20)
plot(g_8, layout=layout_with_dh, vertex.size=5, vertex.label=NA)
g_9 <- make_lattice(c(3,2,2))
plot(g_9, layout=layout_with_dh, vertex.size=5, vertex.label=NA)
g_10 <- make_lattice(c(6,6))
plot(g_10, layout=layout_with_dh, vertex.size=5, vertex.label=NA)
g_11a <- make_tree(31, 2, mode="undirected")
plot(g_11a, layout=layout_with_dh, vertex.size=5, vertex.label=NA)
g_11b <- make_tree(21, 4, mode="undirected")
plot(g_11b, layout=layout_with_dh, vertex.size=5, vertex.label=NA)
g_12 <- make_empty_graph(n=37, directed=FALSE) +
path(1:5,10,22,31,37:33,27,16,6,1) + path(6,7,11,9,10) + path(16:22) +
path(27:31) + path(2,7,18,28,34) + path(3,8,11,19,29,32,35) +
path(4,9,20,30,36) + path(1,7,12,14,19,24,26,30,37) +
path(5,9,13,15,19,23,25,28,33) + path(3,12,16,25,35,26,22,13,3)
plot(g_12, layout=layout_with_dh, vertex.size=5, vertex.label=NA)
[Package igraph version 1.2.5 Index] | 1,450 | 4,879 | {"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} | 2.609375 | 3 | CC-MAIN-2023-06 | latest | en | 0.817777 |
https://people.maths.bris.ac.uk/~matyd/GroupNames/288/C6%5E2.116C2%5E3.html | 1,544,676,321,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376824448.53/warc/CC-MAIN-20181213032335-20181213053835-00452.warc.gz | 713,169,324 | 5,868 | Copied to
clipboard
## G = C62.116C23order 288 = 25·32
### 111st non-split extension by C62 of C23 acting via C23/C2=C22
Series: Derived Chief Lower central Upper central
Derived series C1 — C3×C6 — C62.116C23
Chief series C1 — C3 — C32 — C3×C6 — C62 — C6×Dic3 — C2×C6.D6 — C62.116C23
Lower central C32 — C3×C6 — C62.116C23
Upper central C1 — C22 — C23
Generators and relations for C62.116C23
G = < a,b,c,d,e | a6=b6=e2=1, c2=d2=a3, ab=ba, ac=ca, dad-1=a-1, ae=ea, cbc-1=b-1, bd=db, be=eb, cd=dc, ece=b3c, ede=b3d >
Subgroups: 1538 in 331 conjugacy classes, 68 normal (12 characteristic)
C1, C2, C2 [×2], C2 [×8], C3 [×2], C3, C4 [×4], C22, C22 [×2], C22 [×20], S3 [×20], C6 [×6], C6 [×11], C2×C4 [×8], C23, C23 [×10], C32, Dic3 [×4], C12 [×4], D6 [×64], C2×C6 [×6], C2×C6 [×11], C22⋊C4 [×4], C22×C4 [×2], C24, C3⋊S3 [×4], C3⋊S3 [×2], C3×C6, C3×C6 [×2], C3×C6 [×2], C4×S3 [×8], C2×Dic3 [×4], C2×C12 [×4], C22×S3 [×34], C22×C6 [×2], C22×C6, C2×C22⋊C4, C3×Dic3 [×4], C2×C3⋊S3 [×8], C2×C3⋊S3 [×10], C62, C62 [×2], C62 [×2], D6⋊C4 [×4], C6.D4 [×2], C3×C22⋊C4 [×2], S3×C2×C4 [×4], S3×C23 [×3], C6.D6 [×4], C6×Dic3 [×4], C22×C3⋊S3 [×2], C22×C3⋊S3 [×4], C22×C3⋊S3 [×4], C2×C62, S3×C22⋊C4 [×2], C6.D12 [×2], C3×C6.D4 [×2], C2×C6.D6 [×2], C23×C3⋊S3, C62.116C23
Quotients: C1, C2 [×7], C4 [×4], C22 [×7], S3 [×2], C2×C4 [×6], D4 [×4], C23, D6 [×6], C22⋊C4 [×4], C22×C4, C2×D4 [×2], C4×S3 [×4], C22×S3 [×2], C2×C22⋊C4, S32, S3×C2×C4 [×2], S3×D4 [×4], C6.D6 [×2], C2×S32, S3×C22⋊C4 [×2], C2×C6.D6, Dic3⋊D6 [×2], C62.116C23
Permutation representations of C62.116C23
On 24 points - transitive group 24T673
Generators in S24
```(1 2 3 4 5 6)(7 8 9 10 11 12)(13 14 15 16 17 18)(19 20 21 22 23 24)
(1 17 5 15 3 13)(2 18 6 16 4 14)(7 19 9 21 11 23)(8 20 10 22 12 24)
(1 10 4 7)(2 11 5 8)(3 12 6 9)(13 22 16 19)(14 23 17 20)(15 24 18 21)
(1 21 4 24)(2 20 5 23)(3 19 6 22)(7 18 10 15)(8 17 11 14)(9 16 12 13)
(7 21)(8 22)(9 23)(10 24)(11 19)(12 20)```
`G:=sub<Sym(24)| (1,2,3,4,5,6)(7,8,9,10,11,12)(13,14,15,16,17,18)(19,20,21,22,23,24), (1,17,5,15,3,13)(2,18,6,16,4,14)(7,19,9,21,11,23)(8,20,10,22,12,24), (1,10,4,7)(2,11,5,8)(3,12,6,9)(13,22,16,19)(14,23,17,20)(15,24,18,21), (1,21,4,24)(2,20,5,23)(3,19,6,22)(7,18,10,15)(8,17,11,14)(9,16,12,13), (7,21)(8,22)(9,23)(10,24)(11,19)(12,20)>;`
`G:=Group( (1,2,3,4,5,6)(7,8,9,10,11,12)(13,14,15,16,17,18)(19,20,21,22,23,24), (1,17,5,15,3,13)(2,18,6,16,4,14)(7,19,9,21,11,23)(8,20,10,22,12,24), (1,10,4,7)(2,11,5,8)(3,12,6,9)(13,22,16,19)(14,23,17,20)(15,24,18,21), (1,21,4,24)(2,20,5,23)(3,19,6,22)(7,18,10,15)(8,17,11,14)(9,16,12,13), (7,21)(8,22)(9,23)(10,24)(11,19)(12,20) );`
`G=PermutationGroup([(1,2,3,4,5,6),(7,8,9,10,11,12),(13,14,15,16,17,18),(19,20,21,22,23,24)], [(1,17,5,15,3,13),(2,18,6,16,4,14),(7,19,9,21,11,23),(8,20,10,22,12,24)], [(1,10,4,7),(2,11,5,8),(3,12,6,9),(13,22,16,19),(14,23,17,20),(15,24,18,21)], [(1,21,4,24),(2,20,5,23),(3,19,6,22),(7,18,10,15),(8,17,11,14),(9,16,12,13)], [(7,21),(8,22),(9,23),(10,24),(11,19),(12,20)])`
`G:=TransitiveGroup(24,673);`
48 conjugacy classes
class 1 2A 2B 2C 2D 2E 2F 2G 2H 2I 2J 2K 3A 3B 3C 4A ··· 4H 6A ··· 6F 6G ··· 6Q 12A ··· 12H order 1 2 2 2 2 2 2 2 2 2 2 2 3 3 3 4 ··· 4 6 ··· 6 6 ··· 6 12 ··· 12 size 1 1 1 1 2 2 9 9 9 9 18 18 2 2 4 6 ··· 6 2 ··· 2 4 ··· 4 12 ··· 12
48 irreducible representations
dim 1 1 1 1 1 1 2 2 2 2 2 4 4 4 4 4 type + + + + + + + + + + + + + + image C1 C2 C2 C2 C2 C4 S3 D4 D6 D6 C4×S3 S32 S3×D4 C6.D6 C2×S32 Dic3⋊D6 kernel C62.116C23 C6.D12 C3×C6.D4 C2×C6.D6 C23×C3⋊S3 C22×C3⋊S3 C6.D4 C2×C3⋊S3 C2×Dic3 C22×C6 C2×C6 C23 C6 C22 C22 C2 # reps 1 2 2 2 1 8 2 4 4 2 8 1 4 2 1 4
Matrix representation of C62.116C23 in GL6(𝔽13)
1 12 0 0 0 0 1 0 0 0 0 0 0 0 12 0 0 0 0 0 0 12 0 0 0 0 0 0 1 0 0 0 0 0 0 1
,
1 0 0 0 0 0 0 1 0 0 0 0 0 0 12 0 0 0 0 0 0 12 0 0 0 0 0 0 12 1 0 0 0 0 12 0
,
8 0 0 0 0 0 0 8 0 0 0 0 0 0 1 10 0 0 0 0 5 12 0 0 0 0 0 0 0 1 0 0 0 0 1 0
,
0 5 0 0 0 0 5 0 0 0 0 0 0 0 12 3 0 0 0 0 8 1 0 0 0 0 0 0 1 0 0 0 0 0 0 1
,
12 0 0 0 0 0 0 12 0 0 0 0 0 0 1 0 0 0 0 0 5 12 0 0 0 0 0 0 1 0 0 0 0 0 0 1
`G:=sub<GL(6,GF(13))| [1,1,0,0,0,0,12,0,0,0,0,0,0,0,12,0,0,0,0,0,0,12,0,0,0,0,0,0,1,0,0,0,0,0,0,1],[1,0,0,0,0,0,0,1,0,0,0,0,0,0,12,0,0,0,0,0,0,12,0,0,0,0,0,0,12,12,0,0,0,0,1,0],[8,0,0,0,0,0,0,8,0,0,0,0,0,0,1,5,0,0,0,0,10,12,0,0,0,0,0,0,0,1,0,0,0,0,1,0],[0,5,0,0,0,0,5,0,0,0,0,0,0,0,12,8,0,0,0,0,3,1,0,0,0,0,0,0,1,0,0,0,0,0,0,1],[12,0,0,0,0,0,0,12,0,0,0,0,0,0,1,5,0,0,0,0,0,12,0,0,0,0,0,0,1,0,0,0,0,0,0,1] >;`
C62.116C23 in GAP, Magma, Sage, TeX
`C_6^2._{116}C_2^3`
`% in TeX`
`G:=Group("C6^2.116C2^3");`
`// GroupNames label`
`G:=SmallGroup(288,622);`
`// by ID`
`G=gap.SmallGroup(288,622);`
`# by ID`
`G:=PCGroup([7,-2,-2,-2,-2,-2,-3,-3,56,64,422,219,1356,9414]);`
`// Polycyclic`
`G:=Group<a,b,c,d,e|a^6=b^6=e^2=1,c^2=d^2=a^3,a*b=b*a,a*c=c*a,d*a*d^-1=a^-1,a*e=e*a,c*b*c^-1=b^-1,b*d=d*b,b*e=e*b,c*d=d*c,e*c*e=b^3*c,e*d*e=b^3*d>;`
`// generators/relations`
×
𝔽 | 3,363 | 4,921 | {"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-2018-51 | longest | en | 0.441245 |
https://www.answers.com/Q/What_is_the_force_of_gravity | 1,621,037,217,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243991829.45/warc/CC-MAIN-20210514214157-20210515004157-00078.warc.gz | 649,555,566 | 67,048 | 0
Gravity
# What is the force of gravity?
###### Answered 2011-11-21 00:32:41
It is the force of attraction existing between any two objects in the Universe that
possess mass. It is the most important and weakest force. It is the weakest force
because its effect is seen only when there are relatively large masses involved.
e.g
Sun(huge mass)
Earth(large mass)
Force of gravity keeps Earth in orbit around the sun.
Earth (large mass)
we (tiny mass compared to the Earth).
Force of gravity keeps us bound to the Earth.
It is an attractive force that is proportional to the product of the masses of the two objects
and inversely proportional to the square of the distance between their centers.
🙏
0
🤨
0
😮
0
😂
0 | 181 | 725 | {"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-2021-21 | latest | en | 0.956534 |
https://worksheetsbag.com/mcq-chapter-11-binomial-theorem-class-11-mathematics/ | 1,718,842,170,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861853.72/warc/CC-MAIN-20240619220908-20240620010908-00052.warc.gz | 544,118,088 | 22,277 | # MCQ Chapter 11 Binomial Theorem Class 11 Mathematics
Please refer to Binomial Theorem MCQ Questions Class 11 Mathematics below. These MCQ questions for Class 11 Mathematics with answers have been designed as per the latest NCERT, CBSE books, and syllabus issued for the current academic year. These objective questions for Binomial Theorem will help you to prepare for the exams and get more marks.
## Binomial Theorem MCQ Questions Class 11 Mathematics
Please see solved MCQ Questions for Binomial Theorem in Class 11 Mathematics. All questions and answers have been prepared by expert faculty of standard 11 based on the latest examination guidelines.
### MCQ Questions Class 11 Mathematics Binomial Theorem
Question: Let[ x] denote the greatest integer less than or equal to x. If x = (√3+1)5 then [x ] x is equal to
(a) 75
(b) 50
(c) 76
(d) 152
D
Question: 1+1/3x+1.4/3.6 x+1·4·7/3·6·9 x3+… is equal to
(a) x
(b) (1+x)1/3
(c) (1-x)1/3
(d) (1-x)-1/3
D
Question: 1+2.1/3.2 +2.5/3.6(1/2)2 + 2·5 ·8/3·6·9·(1/2)3+…is equal to
(a) 21/3
(b) 31/4
(c) 41/3
(d) 31/3
C
Question: The coefficient of xn in the polynomial (x+nCo)(x+3nC1)(x+5nC2)…[x+(2n+1)nCn] ) is
(a) n· 2n
(b) n· 2n+1
(c) (n+1)2
(d) n·2n+1
C
Question:
B
Question: The number of terms in the expansion of (a+ b+ c)n will be
(a) n + 1
(b) n + 3
(c) (n+1 ) (n+2)/2
(d) None of the above
C
Question: The coefficient of x7 in (1+3x- 2x3)10 ) is equal to
(a) 62640
(b) 26240
(c) 64620
(d) None of these
A
Question: Let Tn denotes the number of triangles which can be formed using the vertices of a regular polygon of n sides. If Tn +1-Tn=21, then n is equal to
(a) 5
(b) 7
(c) 6
(d) 4
B
Question:
B
Question: The value of x, for which the 6th term in the expansion of
C
Question: The coefficient of x2m+1 in the expansion of
(a) 3
(b) 2
(c) 1
(d) 0
C
Question:
C
Question: Ifn-1Cr=(k2-3).n Cr+1,then k is belongs to
a) (-∞, -2]
(b) [2,∞)
(c) [- √3,√3]
(d) (√3 ,2]
D
Question:
D
Question:
A
Question: If sum of the coefficients of the first, second and third terms of the expansion of [x2+1/x] is 46, then the coefficient of the term that does not contain x is
(a) 84
(b) 92
(c) 98
(d) 106
A
Question: The number 101100 – is divisible by
(a) 100
(b) 1000
(c) 10000
(d) 100000
(a,b,c)
Question: The last digit of
(a) 4 C3
(b) 8C7
(c) 8
(d) 4
(a,d)
Question: Which of the following is/are correct?
(a) 10150-9950> 10050
(b) 10150– 10050> 99 50
(c) (1000) 1000> (1001)999
(d) (1001)999> (1000) 1000
(a,b,c)
The 2nd, 3rd and 4th terms in the expansion of( ) x a n + are 240, 720 and 1080, respectively.
Question: The value of (x-a) n can be
(a) 64
(b) -1
(c) -32
(d) None of these
B
Question: The sum of odd numbered terms is
(a) 1664
(b) 2376
(c) 1562
(d) 1486
C
Question: The value of least term in the expansion is
(a) 16
(b) 160
(c) 32
(d) 81
C
Assertion and Reason
Each of these questions contains two statements: Statement (Assertion) and Statement II(Reason).
Each of these questions also has four alternative choices, only one of which is the correct answer. You have to select one of the codes (a), (b), (c) and (d) given below.
(a) Statement I is true, Statement II is true; Statement II is a correct explanation for Statement I.
(b) Statement I is true, Statement II is true; Statement II is not a correct explanation for Statement I.
(c) Statement I is true; Statement II is false.
(d) Statement I is false; Statement II is true.
Question:
D
Question: Three consecutive binomial coefficients are given.
Statement I They cannot be in GP and HP.
Statement II They always are in AP.
B
Question: Statement I (√2+ 1)n can be expressed as of √N- √N -1) for all N > 1 and n is positive integer.
Statement II (√2- 1)n can be expressed as A+ B √2, where A and Bare integers and is positive integer.
B
QuestionStatement I The number of terms in the expansion of
[x+1/x+1]n is 2n +1
Statement II
B
Question:
D
Question:
A
Question: Statement I The term independent of x in the expansion of (x+1/x+2)m is (4m)!/2m!)2·
Statement II The coefficient of x6 in the expansion of (1+x)n is nC6
D
Question: Statement I Greatest term in the expansion of (1+x)12, when x=11/10 is 7th.
B
Question: Statement I If n is an odd prime, then the integral part of (√5+2)n-2n+1 is divisible by 2n
Statement II If n is prime, then nC1,nC2,nCn-1 must be divisible by n. | 1,565 | 4,376 | {"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} | 3.796875 | 4 | CC-MAIN-2024-26 | latest | en | 0.757599 |
http://reference.wolfram.com/legacy/v5_2/book/section-3.4.9 | 1,386,809,448,000,000,000 | text/html | crawl-data/CC-MAIN-2013-48/segments/1386164128316/warc/CC-MAIN-20131204133528-00025-ip-10-33-133-15.ec2.internal.warc.gz | 159,238,552 | 12,023 | ## 3.4.9 Equations and Inequalities over Domains
Mathematica normally assumes that variables which appear in equations can stand for arbitrary complex numbers. But when you use Reduce, you can explicitly tell Mathematica that the variables stand for objects in more restricted domains.
Reduce[expr, vars, dom] reduce eqns over the domain dom Complexes complex numbers Reals real numbers Integers integers
Solving over domains.
Reduce by default assumes that x can be complex, and gives all five complex solutions.
In[1]:= Reduce[x^6 - x^4 - 4x^2 + 4 0, x]
Out[1]=
But here it assumes that x is real, and gives only the real solutions.
In[2]:= Reduce[x^6 - x^4 - 4x^2 + 4 0, x, Reals]
Out[2]=
And here it assumes that x is an integer, and gives only the integer solutions.
In[3]:= Reduce[x^6 - x^4 - 4x^2 + 4 0, x, Integers]
Out[3]=
A single polynomial equation in one variable will always have a finite set of discrete solutions. And in such a case one can think of Reduce[eqns, vars, dom] as just filtering the solutions by selecting the ones that happen to lie in the domain dom.
But as soon as there are more variables, things can become more complicated, with solutions to equations corresponding to parametric curves or surfaces in which the values of some variables can depend on the values of others. Often this dependence can be described by some collection of equations or inequalities, but the form of these can change significantly when one goes from one domain to another.
This gives solutions over the complex numbers as simple formulas.
In[4]:= Reduce[x^2 + y^2 1, {x, y}]
Out[4]=
To represent solutions over the reals requires introducing an inequality.
In[5]:= Reduce[x^2 + y^2 1, {x, y}, Reals]
Out[5]=
Over the integers, the solution can be represented as equations for discrete points.
In[6]:= Reduce[x^2 + y^2 1, {x, y}, Integers]
Out[6]=
If your input involves only equations, then Reduce will by default assume that all variables are complex. But if your input involves inequalities, then Reduce will assume that any algebraic variables appearing in them are real, since inequalities can only compare real quantities.
Since the variables appear in an inequality, they are assumed to be real.
In[7]:= Reduce[{x + y + z 1, x^2 + y^2 + z^2 < 1}, {x, y, z}]
Out[7]=
Complexes polynomial 0, Root[ ... ] Reals Root[ ... ] < < Root[ ... ], Root[ ... ] Integers arbitrarily complicated
Schematic building blocks for solutions to polynomial equations and inequalities.
For systems of polynomials over real and complex domains, the solutions always consist of a finite number of components, within which the values of variables are given by algebraic numbers or functions.
Here the components are distinguished by equations and inequations on x.
In[8]:= Reduce[x y^3 + y 1, {x, y}, Complexes]
Out[8]=
And here the components are distinguished by inequalities on x.
In[9]:= Reduce[x y^3 + y 1, {x, y}, Reals]
Out[9]=
While in principle Reduce can always find the complete solution to any collection of polynomial equations and inequalities with real or complex variables, the results are often very complicated, with the number of components typically growing exponentially as the number of variables increases.
With 3 variables, the solution here already involves 8 components.
In[10]:= Reduce[x^2 y^2 z^2 1, {x, y, z}]
Out[10]=
As soon as one introduces functions like Sin or Exp, even equations in single real or complex variables can have solutions with an infinite number of components. Reduce labels these components by introducing additional parameters. By default, the n parameter in a given solution will be named C[n]. In general you can specify that it should be named f[n] by giving the option setting GeneratedParameters -> f.
The components here are labeled by the integer parameter .
In[11]:= Reduce[Exp[x] 2, x, GeneratedParameters -> (Subscript[c, #]&)]
Out[11]=
Reduce can handle equations not only over real and complex variables, but also over integers. Solving such Diophantine equations can often be a very difficult problem.
Describing the solution to this equation over the reals is straightforward.
In[12]:= Reduce[x y 8, {x, y}, Reals]
Out[12]=
The solution over the integers involves the divisors of 8.
In[13]:= Reduce[x y 8, {x, y}, Integers]
Out[13]=
Solving an equation like this effectively requires factoring a large number.
In[14]:= Reduce[{x y 7777777, x > y > 0}, {x, y}, Integers]
Out[14]=
Reduce can solve any system of linear equations or inequalities over the integers. With linear equations in variables, parameters typically need to be introduced. But with inequalities, a much larger number of parameters may be needed.
Three parameters are needed here, even though there are only two variables.
In[15]:= Reduce[{3x - 2y > 1, x > 0, y > 0}, {x, y}, Integers]
Out[15]=
With two variables, Reduce can solve any quadratic equation over the integers. The result can be a Fibonacci-like sequence, represented in terms of powers of quadratic irrationals.
Here is the solution to a Pell equation.
In[16]:= Reduce[{x^2 13 y^2 + 1, x > 0, y > 0}, {x, y}, Integers]
Out[16]=
The actual values for specific C[1] as integers, as they should be.
In[17]:= FullSimplify[% /. Table[{C[1] -> i}, {i, 4}]]
Out[17]=
Reduce can handle many specific classes of equations over the integers.
Here Reduce finds the solution to a Thue equation.
In[18]:= Reduce[x^3 - 4 x y^2 + y^3 1, {x, y}, Integers]
Out[18]=
Changing the right-hand side to 3, the equation now has no solution.
In[19]:= Reduce[x^3 - 4 x y^2 + y^3 3, {x, y}, Integers]
Out[19]=
Equations over the integers sometimes have seemingly quite random collections of solutions. And even small changes in equations can often lead them to have no solutions at all.
For polynomial equations over real and complex numbers, there is a definite decision procedure for determining whether or not any solution exists. But for polynomial equations over the integers, the unsolvability of Hilbert's Tenth Problem demonstrates that there can never be any such general procedure.
For specific classes of equations, however, procedures can be found, and indeed many are implemented in Reduce. But handling different classes of equations can often seem to require whole different branches of number theory, and quite different kinds of computations. And in fact it is known that there are universal integer polynomial equations, for which filling in some variables can make solutions for other variables correspond to the output of absolutely any possible program. This then means that for such equations there can never in general be any closed-form solution built from fixed elements like algebraic functions.
If one includes functions like Sin, then even for equations involving real and complex numbers the same issues can arise.
Reduce here effectively has to solve an equation over the integers.
In[20]:= Reduce[Sin[Pi x]^2 + Sin[Pi y]^2 + (x^2 + y^2 - 25)^2 0, {x, y}, Reals]
Out[20]=
Reduce[eqns, vars, Modulus->n] find solutions modulo n
Handling equations involving integers modulo n.
Since there are only ever a finite number of possible solutions for integer equations modulo n, Reduce can systematically find them.
This finds all solutions modulo 4.
In[21]:= Reduce[x^5 y^4 + x y + 1, {x, y}, Modulus -> 4]
Out[21]=
Reduce can also handle equations that involve several different moduli.
Here is an equation involving two different moduli.
In[22]:= Reduce[Mod[2x + 1, 5] Mod[x, 7] && 0 < x < 50, x]
Out[22]=
Reduce[expr, vars, dom] specify a default domain for all variables Reduce[{, ... , , ... }, vars] explicitly specify individual domains for variables
Different ways to specify domains for variables.
This assumes that x is an integer, but y is a real.
In[23]:= Reduce[{x^2 + 2y^2 1, x Integers, y Reals}, {x, y}]
Out[23]=
Reduce normally treats complex variables as single objects. But in dealing with functions that are not analytic or have branch cuts, it sometimes has to break them into pairs of real variables Re[z] and Im[z].
The result involves separate real and imaginary parts.
In[24]:= Reduce[Abs[z] 1, z]
Out[24]=
Here again there is a separate condition on the imaginary part.
In[25]:= Reduce[Log[z] a, {a, z}]
Out[25]=
Reduce by default assumes that variables that appear algebraically in inequalities are real. But you can override this by explicitly specifying Complexes as the default domain. It is often useful in such cases to be able to specify that certain variables are still real.
Reduce by default assumes that x is a real.
In[26]:= Reduce[x^2 < 1, x]
Out[26]=
This forces Reduce to consider the case where x can be complex.
In[27]:= Reduce[x^2 < 1, x, Complexes]
Out[27]=
Since x does not appear algebraically, Reduce immediately assumes that it can be complex.
In[28]:= Reduce[Abs[x] < 1, x]
Out[28]=
Here x is a real, but y can be complex.
In[29]:= Reduce[{Abs[y] < Abs[x], x Reals}, {x, y}]
Out[29]=
FindInstance[expr, {, , ... }, dom] try to find an instance of the in dom satisfying expr FindInstance[expr, vars, dom, n] try to find n instances Complexes the domain of complex numbers Reals the domain of real numbers Integers the domain of integers Booleans the domain of booleans (True and False)
Finding particular solutions in domains.
Reduce always returns a complete representation of the solution to a system of equations or inequalities. Sometimes, however, you may just want to find particular sample solutions. You can do this using FindInstance.
If FindInstance[expr, vars, dom] returns {} then this means that Mathematica has effectively proved that expr cannot be satisfied for any values of variables in the specified domain. When expr can be satisfied, FindInstance will normally pick quite arbitrarily among values that do this, as discussed for inequalities in Section 3.4.8.
Particularly for integer equations, FindInstance can often find particular solutions to equations even when Reduce cannot find a complete solution. In such cases it usually returns one of the smallest solutions to the equations.
This finds the smallest integer point on an elliptic curve.
In[30]:= FindInstance[{x^2 y^3 + 12, x > 0, y > 0}, {x, y}, Integers]
Out[30]=
One feature of FindInstance is that it also works with Boolean expressions whose variables can have values True or False. You can use FindInstance to determine whether a particular expression is satisfiable, so that there is some choice of truth values for its variables that makes the expression True.
This expression cannot be satisfied for any choice of p and q.
In[31]:= FindInstance[p && ! (p || ! q), {p, q}, Booleans]
Out[31]=
But this can.
In[32]:= FindInstance[p && ! (! p || ! q), {p, q}, Booleans]
Out[32]=
THIS IS DOCUMENTATION FOR AN OBSOLETE PRODUCT.
SEE THE DOCUMENTATION CENTER FOR THE LATEST INFORMATION. | 2,736 | 10,956 | {"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} | 4.28125 | 4 | CC-MAIN-2013-48 | latest | en | 0.867395 |
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A217481 Decimal expansion of sqrt(2*Pi)/4. 2
6, 2, 6, 6, 5, 7, 0, 6, 8, 6, 5, 7, 7, 5, 0, 1, 2, 5, 6, 0, 3, 9, 4, 1, 3, 2, 1, 2, 0, 2, 7, 6, 1, 3, 1, 3, 2, 5, 1, 7, 4, 6, 6, 8, 5, 1, 5, 2, 4, 8, 4, 5, 7, 9, 1, 5, 7, 4, 8, 0, 8, 9, 4, 0, 8, 5, 5, 7, 3, 4, 1, 3, 6, 5, 1, 9, 6, 0, 4, 9, 3, 7, 3, 6, 6, 4, 8, 9, 5, 9, 5, 9, 4, 5, 1, 4, 3, 1, 6, 5, 2, 9, 0, 0, 2 (list; constant; graph; refs; listen; history; text; internal format)
OFFSET 0,1 COMMENTS Equals Integral_{x>=0} sin(x^2) dx. The generalizations are Integral_{x>=0} exp(i*x^n) dx = 0.6266570686577501... + i*0.6266570686577501... for n=2, 0.7733429420779898... + i*0.4464897557846246... for n=3, 0.8374066967690864... + i*0.3468652110238094... for n=4, 0.8732303655178185... + i*0.2837297451053993... for n=5, and Gamma(1/n)*i^(1/n)/n in general, where i is the imaginary unit. - R. J. Mathar, Nov 14 2012 Mean of cycle length (and of tail length) in Pollard rho method for factoring n is sqrt(2*Pi)/4*sqrt(n). - Jean-François Alcover, May 27 2013 LINKS Muniru A Asiru, Table of n, a(n) for n = 0..2000 I. S. Gradsteyn, I. M. Ryzhik, Table of integrals, series and products, (1980), page 420 (formulas 3.757.1, 3.757.2). Wikipedia, Fresnel Integral FORMULA From A.H.M. Smeets, Sep 22 2018: (Start) Equals Integral_{x >= 0} sin(4x)/sqrt(x) dx [see Gradsteyn and Ryzhik]. Equals Integral_{x >= 0} cos(4x)/sqrt(x) dx [see Gradsteyn and Ryzhik]. (End) EXAMPLE equals 0.62665706865775012560394132120276131... = A019727 / 4 = sqrt(A019675). MAPLE evalf(sqrt(2*Pi))/4 ; MATHEMATICA First@ RealDigits[N[Sqrt[2 Pi]/4, 105]] (* Michael De Vlieger, Sep 24 2018 *) PROG (Maxima) fpprec : 100; ev(bfloat(sqrt(2*%pi)))/4; /* Martin Ettl, Oct 04 2012 */ (Sage) ((sqrt(2*pi))/4).n(digits=100) # Jani Melik, Oct 05 2012 (PARI) sqrt(2*Pi)/4 \\ Altug Alkan, Sep 08 2018 (MAGMA) SetDefaultRealField(RealField(100)); R:= RealField(); Sqrt(2*Pi(R))/4; // G. C. Greubel, Sep 30 2018 CROSSREFS Sequence in context: A011005 A292178 A281961 * A318300 A278146 A221716 Adjacent sequences: A217478 A217479 A217480 * A217482 A217483 A217484 KEYWORD cons,nonn AUTHOR R. J. Mathar, Oct 04 2012 STATUS approved
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Last modified January 16 23:44 EST 2019. Contains 319206 sequences. (Running on oeis4.) | 1,173 | 2,763 | {"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-2019-04 | latest | en | 0.593266 |
http://www.mountpleasantprimary.co.uk/january-20th | 1,656,334,037,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103331729.20/warc/CC-MAIN-20220627103810-20220627133810-00181.warc.gz | 101,359,439 | 9,875 | Mount Pleasant Primary School
# January 20th
We have been working hard on our number bomds to 100 this week- this means being able to calcuate mentally sums that have 100 in them, for example:-
23 + ? = 100
? + 56 = 100
100 - ? = 26
Please watch this video for how we have taught this in school this week:
You can use sweets or counters for this homework. There is also a link to a game below too.
The Magicians Trick
Count out 100 counters/ sweets and lay them out on the table. Drag some towards you and ask your child to estimhow many you have taken without counting. This will revise their estimation skills. Then cover the ones you have taken with a tea towel or cloth and ask pupils to calculate how many yu have taken without looking. Pupils should count the sweets/counters they can see and then calculate how many are hidden mentally . Some children have been able to do this without having to resort to using a number line but some pupils have used number lines and bead strings to help them with this in school. If your child says they have used these resources in school, ask them to use a number line and jottings on paper to help them. Their jottings should look like this:-
Say they have counted 45 sweets they can see.
50 100 Jump up to 100. Pupils should know that if 5 + 5 = 10 then 50 ( 5 tens ) + 50 ( 4 tens ) = 100 beacuse 5 tens and 5 tens make 10 tens which is 100.
Add these numbers together 5 + 50 = 55 so there should be 55 sweets that are hidden.
To revise measuring skills, pupils could also use a 100cm length of ribbon, cover a portion of it and let pupils use another tape measure to measure the portion of the ribbon they can see and then calculate the portion that is unseen. they can then measure the unseen portion to check. Similarly, pupils can use coins for this activity, they are then revising counting money too. A multi-skilled activity that revises many skills!
For an interactive game to revise bonds to 100, try this:-
http://www.topmarks.co.uk/maths-games/hit-the-button
I would love to see some evidence, whether it be pictures or calculations and pupils can record using their Reading Journals, if they wish. A few pupils have excellent evidence of homework they have completed- well done! | 541 | 2,301 | {"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} | 4.34375 | 4 | CC-MAIN-2022-27 | latest | en | 0.948913 |
http://devmaster.net/posts/10295/transform-matrix-from-3ds-max-to-right-handed | 1,394,537,346,000,000,000 | text/html | crawl-data/CC-MAIN-2014-10/segments/1394011183468/warc/CC-MAIN-20140305091943-00067-ip-10-183-142-35.ec2.internal.warc.gz | 48,523,936 | 7,032 | 0
101 Jul 22, 2006 at 03:46
Is there a way I can convert a model exported from 3ds max (positive X to the right, positive Y into the distance, and positive Z up) to a standard right-handed model (positive X to the right, positive Y up, and positive Z towards the camera) using a matrix?
I know how to do this by manipulating each vertex manually, but how can I do it with a matrix?
Thanks a lot!
#### 8 Replies
0
165 Jul 22, 2006 at 04:47
It’s easy. You’ll want to multiply each vertex by a 3x3 matrix. Now the rule for matrices is that when you multiply one by the vector <1, 0, 0> you get a vector equal to the first column of the matrix; multiplying by <0, 1, 0> gives you the second column; and multiplying by <0, 0, 1> gives you the third column. So all you have to do is figure out what you want those three vectors (in the 3DS Max coordinate system) to come out as in your coordinate system. Then constructing the appropriate matrix is easy. In this case:
<1, 0, 0> (in 3DS Max) –> <1, 0, 0> in your coordinates
<0, 1, 0> –> <0, 0, -1>
<0, 0, 1> –> <0, 1, 0>
So the appropriate matrix would be
[ 1 0 0 ]
[ 0 0 1 ]
[ 0 -1 0 ]
0
101 Jul 23, 2006 at 04:25
[ 1 0 0 ]
[ 0 0 1 ]
[ 0 1 0 ]
and reverse the cull order, but it works great! Thanks!
0
101 Jul 24, 2006 at 17:41
3dsmax coordinate system is already right handed, so you shouldn’t have to reverse the face vertex order. You just have to do a 90 degree rotation around the x axis to convert your objects. The matrix Reedbeta gave you does this rotation.
The matrix you are using is not good. It will just interchange y and z, which is not good (it’s a reflection, and that’s why you had to reverse the face vertex order)
And by the way, the usual standard is: x axis points left, y axis points up, z points forward.
0
165 Jul 24, 2006 at 22:37
@Faelenor
the usual standard is: x axis points left, y axis points up, z points forward.
That doesn’t sound like the standard coordinate system to me…I think SnprBob86 has it right. x -> right, y -> up, z -> backward for OpenGL (the camera points along the negative z axis in eye space). Direct3D uses a left-handed system that has x -> right, y -> up, and z -> forward.
So, SnprBob86’s matrix is correct for D3D, as long as he hasn’t set up D3D’s projection matrix to flip the Z axis (as some people do, to get a right-handed system in D3D).
0
101 Jul 25, 2006 at 14:27
@Reedbeta
That doesn’t sound like the standard coordinate system to me…I think SnprBob86 has it right. x -> right, y -> up, z -> backward for OpenGL (the camera points along the negative z axis in eye space). Direct3D uses a left-handed system that has x -> right, y -> up, and z -> forward.
Well, it depends. From the camera’s point of view, you are right. But usualy, objects, especially characters, are modeled with their z axis pointing towards. That was the case everywhere I worked. You just have to take the z axis of the object to know in which direction it’s facing. But well, maybe that’s not the standard, but it’s probably more convenient for AI.
@Reedbeta
So, SnprBob86’s matrix is correct for D3D, as long as he hasn’t set up D3D’s projection matrix to flip the Z axis (as some people do, to get a right-handed system in D3D).
He explicitely wanted a right handed system, so he’s obviously wrong!
0
101 Nov 21, 2008 at 01:16
@Faelenor
3dsmax coordinate system is already right handed, so you shouldn’t have to reverse the face vertex order.
That is true for the UI and for MaxScript. But you need to be careful about how you are accessing the values. This is taken from MaxScript Help Doc:
<node>.transform : Matrix3 – node’s main transformation matrix
Rotation in the internal transformation matrices is left-handed in contradiction to the 3ds Max user interface and MAXScript. Take care when mixing rotation derived from these matrices and rotation used in rotation-related functions or from rotation properties.
0
165 Nov 21, 2008 at 01:40
Thanks pha3z, but this thread is over two years old. ;)
0
101 Nov 21, 2008 at 02:32
Hey Reed,
These threads come up on google searches. I won’t be the last person to visit them. ;) I’ll leave a trail for those who come after me.
BTW, Here’s an awesome page that demistifies all the confusion between the way matrices are stored and translated in OpenGL vs. DirectX. | 1,199 | 4,333 | {"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-2014-10 | longest | en | 0.904949 |
https://de.mathworks.com/matlabcentral/cody/players/7655218/solved | 1,607,194,023,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141748276.94/warc/CC-MAIN-20201205165649-20201205195649-00543.warc.gz | 244,742,646 | 21,066 | Cody
# Katherine Borger
Rank
Score
1 – 45 of 45
#### Problem 2034. Finding fourier transform of a given vector
Created by: Gnaneswar Nadh satapathi
#### Problem 21. Return the 3n+1 sequence for n
Created by: Cody Team
Tags 3n+1, sample
Created by: Tim
#### Problem 434. Return the Fibonacci Sequence
Created by: Matt Fig
Tags fibonacci
#### Problem 240. Project Euler: Problem 6, Natural numbers, squares and sums.
Created by: Doug Hull
#### Problem 232. Project Euler: Problem 2, Sum of even Fibonacci
Created by: Doug Hull
#### Problem 12. Fibonacci sequence
Created by: Cody Team
#### Problem 230. Project Euler: Problem 1, Multiples of 3 and 5
Created by: Doug Hull
#### Problem 605. Whether the input is vector?
Created by: Sangeeta
#### Problem 2631. Flip the vector from right to left
Created by: Pritesh Shah
#### Problem 624. Get the length of a given vector
Created by: Yudong Zhang
#### Problem 135. Inner product of two vectors
Created by: AMITAVA BISWAS
#### Problem 247. Arrange Vector in descending order
Created by: Vishwanathan Iyer
#### Problem 233. Reverse the vector
Created by: Vishwanathan Iyer
#### Problem 33. Create times-tables
Created by: Cody Team
Tags matrices
#### Problem 645. Getting the indices from a vector
Created by: Doug Hull
#### Problem 10. Determine whether a vector is monotonically increasing
Created by: Cody Team
#### Problem 7. Column Removal
Created by: Cody Team
#### Problem 1035. Generate a vector like 1,2,2,3,3,3,4,4,4,4
Created by: Binbin Qi
Tags zeros, for, ones
#### Problem 1024. Doubling elements in a vector
Created by: Abdelhak ARESMOUK
#### Problem 42651. Vector creation
Created by: ruta bhat
Tags easy, vector
#### Problem 3076. Create a vector
Created by: Carlton
Tags vector, uab
#### Problem 1107. Find max
Created by: Marco
Tags find, vector, matrix
#### Problem 649. Return the first and last character of a string
Created by: @bmtran (Bryant Tran)
#### Problem 568. Number of 1s in a binary string
Created by: Srivardhini
Tags binary, sum
#### Problem 1087. Magic is simple (for beginners)
Created by: Jean-Marie Sainthillier
#### Problem 5. Triangle Numbers
Created by: Cody Team
Tags math, triangle, nice
#### Problem 641. Make a random, non-repeating vector.
Created by: Doug Hull
#### Problem 262. Swap the input arguments
Created by: Steve Eddins
#### Problem 174. Roll the Dice!
Created by: @bmtran (Bryant Tran)
#### Problem 23. Finding Perfect Squares
Created by: Cody Team
#### Problem 838. Check if number exists in vector
Created by: Nichlas
#### Problem 2015. Length of the hypotenuse
Created by: Tanya Morton
#### Problem 189. Sum all integers from 1 to 2^n
Created by: Dimitris Kaliakmanis
#### Problem 19. Swap the first and last columns
Created by: Cody Team
#### Problem 149. Is my wife right?
Created by: the cyclist
Tags easy, silly, fun
#### Problem 167. Pizza!
Created by: the cyclist
Tags fun, pizza, good
#### Problem 6. Select every other element of a vector
Created by: Cody Team
#### Problem 2. Make the vector [1 2 3 4 5 6 7 8 9 10]
Created by: Cody Team
Tags basic, basics, colon
#### Problem 1. Times 2 - START HERE
Created by: Cody Team
Tags intro, math, easy
#### Problem 1545. Return area of square
Created by: Marek Kuklis
#### Problem 26. Determine if input is odd
Created by: Cody Team
#### Problem 1702. Maximum value in a matrix
Created by: Gnaneswar Nadh satapathi
#### Problem 3. Find the sum of all the numbers of the input vector
Created by: Cody Team
#### Problem 8. Add two numbers
Created by: Cody Team
1 – 45 of 45 | 1,012 | 3,632 | {"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} | 3.03125 | 3 | CC-MAIN-2020-50 | latest | en | 0.721397 |
https://convertoctopus.com/1923-kilometers-per-hour-to-knots | 1,675,650,676,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764500303.56/warc/CC-MAIN-20230206015710-20230206045710-00020.warc.gz | 208,576,923 | 7,644 | ## Conversion formula
The conversion factor from kilometers per hour to knots is 0.53995680345662, which means that 1 kilometer per hour is equal to 0.53995680345662 knots:
1 km/h = 0.53995680345662 kt
To convert 1923 kilometers per hour into knots we have to multiply 1923 by the conversion factor in order to get the velocity amount from kilometers per hour to knots. We can also form a simple proportion to calculate the result:
1 km/h → 0.53995680345662 kt
1923 km/h → V(kt)
Solve the above proportion to obtain the velocity V in knots:
V(kt) = 1923 km/h × 0.53995680345662 kt
V(kt) = 1038.3369330471 kt
The final result is:
1923 km/h → 1038.3369330471 kt
We conclude that 1923 kilometers per hour is equivalent to 1038.3369330471 knots:
1923 kilometers per hour = 1038.3369330471 knots
## Alternative conversion
We can also convert by utilizing the inverse value of the conversion factor. In this case 1 knot is equal to 0.00096307852313932 × 1923 kilometers per hour.
Another way is saying that 1923 kilometers per hour is equal to 1 ÷ 0.00096307852313932 knots.
## Approximate result
For practical purposes we can round our final result to an approximate numerical value. We can say that one thousand nine hundred twenty-three kilometers per hour is approximately one thousand thirty-eight point three three seven knots:
1923 km/h ≅ 1038.337 kt
An alternative is also that one knot is approximately zero point zero zero one times one thousand nine hundred twenty-three kilometers per hour.
## Conversion table
### kilometers per hour to knots chart
For quick reference purposes, below is the conversion table you can use to convert from kilometers per hour to knots
kilometers per hour (km/h) knots (kt)
1924 kilometers per hour 1038.877 knots
1925 kilometers per hour 1039.417 knots
1926 kilometers per hour 1039.957 knots
1927 kilometers per hour 1040.497 knots
1928 kilometers per hour 1041.037 knots
1929 kilometers per hour 1041.577 knots
1930 kilometers per hour 1042.117 knots
1931 kilometers per hour 1042.657 knots
1932 kilometers per hour 1043.197 knots
1933 kilometers per hour 1043.737 knots | 558 | 2,133 | {"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-2023-06 | latest | en | 0.735869 |
qavehunijuwuc.novarekabet.com | 1,623,549,560,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623487598213.5/warc/CC-MAIN-20210613012009-20210613042009-00436.warc.gz | 444,605,016 | 8,455 | Last edited by Brashakar
Saturday, October 10, 2020 | History
2 edition of Five-figure mathematical tables ... also trigonometrical functions and their logs of angles found in the catalog.
Five-figure mathematical tables ... also trigonometrical functions and their logs of angles
Edwin Chappell
# Five-figure mathematical tables ... also trigonometrical functions and their logs of angles
## by Edwin Chappell
Written in English
Subjects:
• Logarithms,
• Trigonometry -- Tables, etc.
• Classifications
LC ClassificationsQA55 C53
The Physical Object
Pagination320p.
Number of Pages320
ID Numbers
Open LibraryOL16495947M
Trigomometry originated as the study of certain mathematical relations originally defined in terms of the angles and sides of a right triangle, i.e., one containing a right angle (90°). Six basic relations, or trigonometric functions, are defined. If. Trig Table of Common Angles; angle (degrees) 0 30 45 60 90 = 0; angle (radians) 0 PI/6 PI/4 PI/3 PI/2.
Plane trigonometry with tables. [Paul L Evans] triangles and their application to vectors --Graphic representation of the trigonometric functions --Functions of large angles --Functions of two angles --Oblique \u00A0\u00A0\u00A0 schema:description\/a> \" Introduction -- Trigonometric functions -- Logarithms -- Relations between. Solving problems like these uses precalculated values of the trigonometric ratios to match the lengths with the appropriate angles and vice versa. Up until the 's, these values were printed in tables that were included in the back of every textbook (along with tables of logarithms), but have recently been programmed into calculators using.
Problem: Which trigonometric functions are independent of the distance between a point and the origin (when the terminal side of an angle in standard position contains that point)? Tangent and cotangent Problem: If the sine of an angle is negative, which other trigonometric functions will. College Trigonometry. This note explains the following topics: Foundations of Trigonometry, Angles and their Measure, The Unit Circle: Cosine and Sine, Trigonometric Identities, Graphs of the Trigonometric Functions, The Inverse Trigonometric Functions, Applications of Trigonometry, Applications of Sinusoids, The Law of Sines and cosines, Polar Form of Complex Numbers.
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### Five-figure mathematical tables ... also trigonometrical functions and their logs of angles by Edwin Chappell Download PDF EPUB FB2
In view of this there is an urgent need for tables of the natural values of the trigonometric functions with a constant number of significant figures which substantially guarantees roughly the- same relative accuracy for all angles.
The present tables, together with the following, already published by Fizmatgiz: Fil'e-figure Tables (L. Trigonometry in the modern sense began with the Greeks. Hipparchus (c. – bce) was the first to construct a table of values for a trigonometric considered every triangle—planar or spherical—as being inscribed in a circle, so that each side becomes a chord (that is, a straight line that connects two points on a curve or surface, as shown by the inscribed triangle ABC in.
In view of this there is an urgent need for tables of the natural values of the trigonometric functions with a constant number of significant figures which substantially guarantees roughly the- same relative accuracy for all angles. The present tables, together with the following, already published by Fizmatgiz: Fil'e-figure Tables (L.
S /5(5). In mathematics, the trigonometric functions (also called circular functions, angle functions or goniometric functions) are real functions which relate an angle of a right-angled triangle to ratios of two side lengths.
They are widely used in all sciences that are related to geometry, such as navigation, solid mechanics, celestial mechanics, geodesy, and many others. Trigonometry (from Greek trigōnon, "triangle" and metron, "measure") is a branch of mathematics that studies relationships between side lengths and angles of field emerged in the Hellenistic world during the 3rd century BC from applications of geometry to astronomical studies.
The Greeks focused on the calculation of chords, while mathematicians in India created the earliest. In mathematics, the trigonometric functions (also called circular functions, angle functions or goniometric functions) are functions of an angle.
They relate the angles of a triangle to the lengths of its sides. Trigonometric functions are important in the study of triangles and modeling periodic phenomena, among many other applications.
Trigonometry Table 0 to Trigonometry is a branch in Mathematics, which involves the study of the relationship involving the length and angles of a triangle. It is generally associated with a right-angled triangle, where one of the angles is always 90 degrees.
It has a vast number of applications in other fields of Mathematics. Trigonometry comes from the two roots, trigonon (or “triangle”) and metria (or “measure”). The study of trigonometry is thus the study of measurements of triangles.
What can we measure in a triangle. The first objects that come to mind may be the lengths of the sides, the angles of the triangle, or the area contained in the triangle. We first explore trigonometric functions that.
2 Chapter 1 • Right Triangle Trigonometry § (a) Two acute angles are complementary if their sum equals In other words, if 0 ≤ ∠ A,∠B≤90 then ∠A and ∠ Bare complementary if ∠ +∠ = (b) Two angles between 0 and are supplementary if their sum equals In other words, if 0 ≤∠ A,∠B≤ then ∠ and ∠B are supplementary if ∠A+∠B= You can use this table of values for trig functions when solving problems, sketching graphs, or doing any number of computations involving trig.
The values here are all rounded to three decimal places. θ sinθ cosθ tanθ cotθ secθ cscθ 0° Undefined Undefined 1° 2° [ ]. It is, for example, a nice exercise to create a table of values of sine for multiples of \$3^\circ\$. For example, Ptolemy (2nd century AD) essentially created a table of values for each half degree angles.
Methods evolved and people found approximations to trigonometric functions. Trigonometric table with all 6 trigonometric functions. Free Mathematics Tutorials. Home; Trigonometric Tables.
Below are trigonometric tables of all 6 trigonometric functions, with angles in degrees and radians. Copies of these tables can be downloaded. Download Trigonometric Table. 13 Problems Involving Trig Function Values in Quadrants II, III, and IV 14 Problems Involving Angles of Depression and Inclination Chapter 2: Graphs of Trig Functions 15 Basic Trig Functions 17 Characteristics of Trigonometric Function Graphs 19 Table of Trigonometric Function Characteristics 20 Sine Function.
Trigonometric functions are the dependences between angles and numbers in a rectangle expressed in own units of measure. Own unit of measure is one of characteristics of object, accepted as a unit of measure.
In a rectangle it is possible to allocate three main types of trigonometric functions. The angles used most often in trig have trig functions with convenient exact values. Other angles don’t cooperate anywhere near as nicely as these popular ones do.
A quick, easy way to memorize the exact trig-function values of the most common angles is to construct a table, starting with the sine function and working with [ ].
Short answer: The main reason is the simplification of reducing multiplication and division to addition and subtraction. Historical aspects: One application which is heavily based upon trigonometric formulas is Spherical realm e.g. important for astronomy and geodesy used logarithmic tables of trigonometric functions right from the beginning since logarithms have been published.
Here we're calculating the values at different angles by using built-in C trigonometry functions like sin(), cos() and tan() which are available in math.h header file.
Then using a for loop we're calculating values of sine, cos and tan at different angles and each angle is 30 degree apart, starting from 0. The trigonometric table was the reason for most digital development to take place at this rate today as the first mechanical computing devices found application through careful use of trigonometry.
The Trigonometric ratios table gives us the values of standard trigonometric angles such as. He also gave trigonometric tables of values of the sine function to four sexagesimal digits (equivalent to 8 decimal places) for each 1° of argument with differences to be added for each 1/60 of 1°.
[citation needed] Ulugh Beg also gives accurate tables of sines and tangents correct to 8 decimal places around the same time. [citation needed].
In this section we will give a quick review of trig functions. We will cover the basic notation, relationship between the trig functions, the right triangle definition of the trig functions. We will also cover evaluation of trig functions as well as the unit circle (one of the most important ideas from a trig class!) and how it can be used to evaluate trig functions.
In trigonometry, trigonometric ratios are derived from the sides of a right-angled are six 6 ratios such as sine, cosine, tangent, cotangent, cosecant, and secant. You will learn here to build a trigonometry table for these ratios for some particular angles, such as 0 °, 30 °, 45 °, 60 °, 90°.Play Project TRIG at Math Playground!
Learn about angles and projectile motion. Graphs of trig functions are used to model situations in real life involving populations, waves. bVX0-zncj9qJ3G1_r18rkIpQL02X-Oi6tWViR4g4-vwDVmU50WZA-4bRZMjM2TXmc88PAkJ1g0jIembnEbM skip to main content. Use right triangle trigonometry to describe periodic behavior.Trigonometric Functions: An angle having measure greater than but less than is called an acute angle.
Consider a right angled triangle ABC with right angle at B. The side which is opposite to right angle is known as hypotenuse, the side opposite to angle A is called perpendicular for angle A and the side opposite to third angle is called base for angle A. | 2,715 | 12,380 | {"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} | 3.765625 | 4 | CC-MAIN-2021-25 | latest | en | 0.861813 |
https://gizmodo.com/the-equation-that-can-help-predict-zombie-migration-pat-1670469257 | 1,627,392,640,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046153391.5/warc/CC-MAIN-20210727103626-20210727133626-00368.warc.gz | 284,526,728 | 43,445 | We come from the future
We come from the future
# The Equation That Can Help Predict Zombie Migration Patterns
The scenario: News reports say that there's been a zombie outbreak at the city hospital. The undead are pouring into the streets, biting hapless citizens. How much time do you have to gather supplies, get out of town and set up a fortified site, before the horde catches up with you? As always, there's a way to calculate this, and it comes from physics.
There have been quite a few mathematical models of zombie infection in recent years — partially for fun, but also to test out methodologies that could be modified and applied to actual, non-zombie outbreaks.
But, Thomas Woolley, at Oxford University's Mathematical Institute, believes these earlier studies share a common flaw: they looked at time- and population-dependent interactions between humans and the undead, without considering geography. "Zombies do not move homogeneously," he explains in a recently published paper, in which he is the lead author. "This allows zombies to shuffle around, giving a much more realistic picture of an invasion."
Woolley and his colleagues also take issue with models that assume zombies and humans are well mixed, meaning that zombies can be found everywhere there are humans:
Realistically, the initial horde of zombies will be localized to areas containing dead humans, such as cemeteries and hospitals. In addition, because humans and zombies are not initially separated, humans are not able to run and hide in order to try and preserve themselves. It is a well-documented fact that zombies are deadly but slow-moving. Due to their slow movements, it is quite possible that, given sufficient warning, we would be able to outrun the zombies and produce a defensible blockade where humans could live safely. In order to do so, it would be useful to know just how long an infestation of zombies would take to reach our defenses; this would give us an estimate of how long we would have to scavenge for supplies and weaponry in order that we may protect ourselves from these oncoming, undead predators.
The movements of the undead are commonly described as small irregular steps. That immediately reminded the authors of a principle in physics known as diffusion — or more specifically, the "random walk."
Diffusion describes the spread of particles through random motion from regions of higher concentration to regions of lower concentration. Although the concept originated in physics, it has applications in multiple fields of study, including biology and chemistry.
In molecular diffusion, the moving entities are small molecules that are self-propelled by thermal energy. They move at random because they frequently collide, and as they do so, they become less concentrated in a single area.
So, for example, we have this simple sketch (above). An initial group of zombies is placed in the top left corner, close together, their initial directions denoted by the small black arrows. After a time, their random motion will cause them to spread themselves out over the domain.
Applying diffusion equations to zombies allowed the researchers to estimate the density of zombies at various points in place and time. This figure (below) shows the time in minutes until the first zombies arrive at your location, for various rates of diffusion and distances. If, for example, the initial zombie outbreak is 90 meters away and they have a diffusion rate of 100m2/min, then they'll catch up with you in around 26 minutes.
Mathematically, this also demonstrates why we should flee from zombies, instead of doing something foolhardy like standing our ground and fighting back the undead:
If we were to double the distance between ourselves and the zombies, then the time for the zombies to reach us would approximately quadruple. However, if we were to slow the zombies down by half, then the time taken would only double. Since we want to delay interaction with the zombies for as long as possible… it is much better to expend energy traveling away from the zombies than it is to try and slow them down. Without some form of projectile weaponry or chainsaw, killing zombies is particularly difficult, as they do not stop until their brains are destroyed.
Of course, this is a short-term survival strategy. Eventually, the zombies will reach your fortified location and increase in numbers as they expand their population by chomping on stragglers. At that point, your only option is to hope your supplies last long enough, before the military or Brad Pitt launch a successful counterattack.
In the event that the zombies penetrate your fortress, the mathematicians offer an important piece of advice. You need to replicate the conditions of the initial outbreak by finding ways to slow them down while you put as much distance between you and them as possible. "Thus an effective fortification should have plenty of obstructions that a human could navigate but a decaying zombie, who may not be so athletic, would find challenging."
### DISCUSSION
By
Ken
a few things that always bothered me, even after getting past the dead bodies don't survive very long bit: 1)If zombies eat brains, and the only way to kill a zombie is to destroy it's brain... then doesn't that mean victims of zombies just can't become zombies due to their brain being destroyed when they are killed.
2) why do zombies need to eat brains, or people at all if they can't actually die from starvation? Seriously if you are an immortal undead, then you don't actually need to eat to live/unlive.
3) couldn't zombies feed off of each other. as the only way to kill a zombie is to destroy the brain and they eat brains, doesn't that mean that every zombie is a walking meal to every other zombie.
4) if zombie blood or bodily fluids infect other creatures and make them zombies, shouldn't we worry about undead mosquitoes and bed bugs, and fleas. The try to feed on a movie zombie target, end up infected, and then swarm living beings burrowing through eyes ears mouth and nose to try and get brains. You can't bug spray them to death, and simply landing on you and biting infects you, while trying to squish them simply exposes you to the risk of them getting a bit in.
there was a nother one I had, but it escapes me right now. | 1,275 | 6,322 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.78125 | 3 | CC-MAIN-2021-31 | latest | en | 0.966288 |
https://www.jobilize.com/course/section/practice-makes-perfect-add-whole-numbers-by-openstax?qcr=www.quizover.com | 1,607,104,901,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141740670.93/warc/CC-MAIN-20201204162500-20201204192500-00713.warc.gz | 739,233,317 | 28,805 | 1.2 Add whole numbers (Page 5/6)
Page 5 / 6
Key concepts
• Addition Notation To describe addition, we can use symbols and words.
Operation Notation Expression Read as Result
Addition $+$ $3+4$ three plus four the sum of $3$ and $4$
• The sum of any number $a$ and $0$ is the number. $a+0=a$ $0+a=a$
• Changing the order of the addends $a$ and $b$ does not change their sum. $a+b=b+a$ .
1. Write the numbers so each place value lines up vertically.
2. Add the digits in each place value. Work from right to left starting with the ones place. If a sum in a place value is more than 9, carry to the next place value.
3. Continue adding each place value from right to left, adding each place value and carrying if needed.
Practice makes perfect
In the following exercises, translate the following from math expressions to words.
$5+2$
five plus two; the sum of 5 and 2.
$6+3$
$13+18$
thirteen plus eighteen; the sum of 13 and 18.
$15+16$
$214+642$
two hundred fourteen plus six hundred forty-two; the sum of 214 and 642
$438+113$
In the following exercises, model the addition.
$2+4$
$2+4=6$
$5+3$
$8+4$
$8+4=12$
$5+9$
$14+75$
$14+75=89$
$15+63$
$16+25$
$16+25=41$
$14+27$
In the following exercises, fill in the missing values in each chart.
1. $0+13$
2. $13+0$
1. 13
2. 13
1. $0+5,280$
2. $5,280+0$
1. $8+3$
2. $3+8$
1. $11$
2. $11$
1. $7+5$
2. $5+7$
$45+33$
$78$
$37+22$
$71+28$
$99$
$43+53$
$26+59$
$85$
$38+17$
$64+78$
$142$
$92+39$
$168+325$
$493$
$247+149$
$584+277$
$861$
$175+648$
$832+199$
$1,031$
$775+369$
$6,358+492$
$6,850$
$9,184+578$
$3,740+18,593$
$22,333$
$6,118+15,990$
$485,012+619,848$
$1,104,860$
$368,911+857,289$
$24,731+592+3,868$
$29,191$
$28,925+817+4,593$
$8,015+76,946+16,570$
$101,531$
$6,291+54,107+28,635$
Translate Word Phrases to Math Notation
In the following exercises, translate each phrase into math notation and then simplify.
the sum of $13$ and $18$
$13+18=31$
the sum of $12$ and $19$
the sum of $90$ and $65$
$90+65=155$
the sum of $70$ and $38$
$33$ increased by $49$
$33+49=82$
$68$ increased by $25$
$250$ more than $599$
$250+599=849$
$115$ more than $286$
the total of $628$ and $77$
$628+77=705$
the total of $593$ and $79$
$1,482$ added to $915$
$915+1,482=2,397$
$2,719$ added to $682$
In the following exercises, solve the problem.
Home remodeling Sophia remodeled her kitchen and bought a new range, microwave, and dishwasher. The range cost $\text{1,100},$ the microwave cost $\text{250},$ and the dishwasher cost $\text{525}.$ What was the total cost of these three appliances?
The total cost was $1,875. Sports equipment Aiden bought a baseball bat, helmet, and glove. The bat cost $\text{299},$ the helmet cost $\text{35},$ and the glove cost $\text{68}.$ What was the total cost of Aiden’s sports equipment? Bike riding Ethan rode his bike $14$ miles on Monday, $19$ miles on Tuesday, $12$ miles on Wednesday, $25$ miles on Friday, and $68$ miles on Saturday. What was the total number of miles Ethan rode? Ethan rode 138 miles. Business Chloe has a flower shop. Last week she made $19$ floral arrangements on Monday, $12$ on Tuesday, $23$ on Wednesday, $29$ on Thursday, and $44$ on Friday. What was the total number of floral arrangements Chloe made? Apartment size Jackson lives in a $7$ room apartment. The number of square feet in each room is $238,120,156,196,100,132,$ and $225.$ What is the total number of square feet in all $7$ rooms? The total square footage in the rooms is 1,167 square feet. Weight Seven men rented a fishing boat. The weights of the men were $175,192,148,169,205,181,$ and $\text{225}$ pounds. What was the total weight of the seven men? Salary Last year Natalie’s salary was $\text{82,572}.$ Two years ago, her salary was $\text{79,316},$ and three years ago it was $\text{75,298}.$ What is the total amount of Natalie’s salary for the past three years? Natalie’s total salary is$237,186.
Home sales Emma is a realtor. Last month, she sold three houses. The selling prices of the houses were $\text{292,540},\text{505,875},$ and $423,699.$ What was the total of the three selling prices?
In the following exercises, find the perimeter of each figure.
The perimeter of the figure is 44 inches.
The perimeter of the figure is 56 meters.
The perimeter of the figure is 71 yards.
The perimeter of the figure is 62 feet.
Everyday math
Calories Paulette had a grilled chicken salad, ranch dressing, and a $\text{16-ounce}$ drink for lunch. On the restaurant’s nutrition chart, she saw that each item had the following number of calories:
Grilled chicken salad – $320$ calories
Ranch dressing – $170$ calories
$\text{16-ounce}$ drink – $150$ calories
What was the total number of calories of Paulette’s lunch?
The total number of calories was 640.
Calories Fred had a grilled chicken sandwich, a small order of fries, and a $\text{12-oz}$ chocolate shake for dinner. The restaurant’s nutrition chart lists the following calories for each item:
Grilled chicken sandwich – $420$ calories
Small fries – $230$ calories
$\text{12-oz}$ chocolate shake – $580$ calories
What was the total number of calories of Fred’s dinner?
Test scores A students needs a total of $400$ points on five tests to pass a course. The student scored $82,91,75,88,\text{and}\phantom{\rule{0.2em}{0ex}}70.$ Did the student pass the course?
Yes, he scored 406 points.
Elevators The maximum weight capacity of an elevator is $1150$ pounds. Six men are in the elevator. Their weights are $210,145,183,230,159,\text{and}\phantom{\rule{0.2em}{0ex}}164$ pounds. Is the total weight below the elevators’ maximum capacity?
Writing exercises
How confident do you feel about your knowledge of the addition facts? If you are not fully confident, what will you do to improve your skills?
Self check
After completing the exercises, use this checklist to evaluate your mastery of the objectives of this section.
After reviewing this checklist, what will you do to become confident for all objectives?
where we get a research paper on Nano chemistry....?
nanopartical of organic/inorganic / physical chemistry , pdf / thesis / review
Ali
what are the products of Nano chemistry?
There are lots of products of nano chemistry... Like nano coatings.....carbon fiber.. And lots of others..
learn
Even nanotechnology is pretty much all about chemistry... Its the chemistry on quantum or atomic level
learn
da
no nanotechnology is also a part of physics and maths it requires angle formulas and some pressure regarding concepts
Bhagvanji
hey
Giriraj
Preparation and Applications of Nanomaterial for Drug Delivery
revolt
da
Application of nanotechnology in medicine
what is variations in raman spectra for nanomaterials
ya I also want to know the raman spectra
Bhagvanji
I only see partial conversation and what's the question here!
what about nanotechnology for water purification
please someone correct me if I'm wrong but I think one can use nanoparticles, specially silver nanoparticles for water treatment.
Damian
yes that's correct
Professor
I think
Professor
Nasa has use it in the 60's, copper as water purification in the moon travel.
Alexandre
nanocopper obvius
Alexandre
what is the stm
is there industrial application of fullrenes. What is the method to prepare fullrene on large scale.?
Rafiq
industrial application...? mmm I think on the medical side as drug carrier, but you should go deeper on your research, I may be wrong
Damian
How we are making nano material?
what is a peer
What is meant by 'nano scale'?
What is STMs full form?
LITNING
scanning tunneling microscope
Sahil
how nano science is used for hydrophobicity
Santosh
Do u think that Graphene and Fullrene fiber can be used to make Air Plane body structure the lightest and strongest. Rafiq
Rafiq
what is differents between GO and RGO?
Mahi
what is simplest way to understand the applications of nano robots used to detect the cancer affected cell of human body.? How this robot is carried to required site of body cell.? what will be the carrier material and how can be detected that correct delivery of drug is done Rafiq
Rafiq
if virus is killing to make ARTIFICIAL DNA OF GRAPHENE FOR KILLED THE VIRUS .THIS IS OUR ASSUMPTION
Anam
analytical skills graphene is prepared to kill any type viruses .
Anam
Any one who tell me about Preparation and application of Nanomaterial for drug Delivery
Hafiz
what is Nano technology ?
write examples of Nano molecule?
Bob
The nanotechnology is as new science, to scale nanometric
brayan
nanotechnology is the study, desing, synthesis, manipulation and application of materials and functional systems through control of matter at nanoscale
Damian
Is there any normative that regulates the use of silver nanoparticles?
what king of growth are you checking .?
Renato
What fields keep nano created devices from performing or assimulating ? Magnetic fields ? Are do they assimilate ?
why we need to study biomolecules, molecular biology in nanotechnology?
?
Kyle
yes I'm doing my masters in nanotechnology, we are being studying all these domains as well..
why?
what school?
Kyle
biomolecules are e building blocks of every organics and inorganic materials.
Joe
A soccer field is a rectangle 130 meters wide and 110 meters long. The coach asks players to run from one corner to the other corner diagonally across. What is that distance, to the nearest tenths place.
Jeannette has $5 and$10 bills in her wallet. The number of fives is three more than six times the number of tens. Let t represent the number of tens. Write an expression for the number of fives.
What is the expressiin for seven less than four times the number of nickels
How do i figure this problem out.
how do you translate this in Algebraic Expressions
why surface tension is zero at critical temperature
Shanjida
I think if critical temperature denote high temperature then a liquid stats boils that time the water stats to evaporate so some moles of h2o to up and due to high temp the bonding break they have low density so it can be a reason
s.
Need to simplify the expresin. 3/7 (x+y)-1/7 (x-1)=
. After 3 months on a diet, Lisa had lost 12% of her original weight. She lost 21 pounds. What was Lisa's original weight? | 2,772 | 10,246 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 146, "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} | 4.65625 | 5 | CC-MAIN-2020-50 | latest | en | 0.716391 |
https://www.physicsforums.com/threads/friction-vs-work.438433/ | 1,556,230,251,000,000,000 | text/html | crawl-data/CC-MAIN-2019-18/segments/1555578742415.81/warc/CC-MAIN-20190425213812-20190425235812-00100.warc.gz | 744,864,408 | 14,762 | # Friction vs Work
#### pur3geet7
1. The problem statement, all variables and given/known data
An 18.5 kg block is dragged over a rough,
horizontal surface by a constant force of 129 N
acting at an angle of angle 29.3◦ above the
horizontal. The block is displaced 64.8 m,
and the coefficient of kinetic friction is 0.115. The
acceleration of gravity is 9.8 m/s2 .
How can I figure out the magnitude of the work done by the force of friction!
2. Relevant equations
W = F*D
3. The attempt at a solution
I attempted to do -129 times the displacment but, its not correct :(
Related Introductory Physics Homework News on Phys.org
#### The legend
"Friction vs Work"
### Physics Forums Values
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• Topics based on mainstream science
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• Solo and co-op problem solving | 246 | 992 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.875 | 3 | CC-MAIN-2019-18 | latest | en | 0.824264 |
http://www.codeproject.com/Articles/62482/A-Simple-Geo-Fencing-Using-Polygon-Method?fid=1562874&df=10000&mpp=50&sort=Position&spc=Relaxed&tid=3946754 | 1,386,196,814,000,000,000 | text/html | crawl-data/CC-MAIN-2013-48/segments/1386163037829/warc/CC-MAIN-20131204131717-00085-ip-10-33-133-15.ec2.internal.warc.gz | 278,206,890 | 19,446 | # A Simple Geo Fencing Using Polygon Method
By , 7 Mar 2010
Votes of 3 or less require a comment
## Introduction
One of the important feature of GPS Tracking software using GPS Tracking devices is geo-fencing and its ability to help keep track of assets. Geo-fencing allows users of a GPS Tracking Solution to draw zones (i.e., a Geo Fence) around places of work, customer’s sites and secure areas. These geo fences when crossed by a GPS equipped vehicle or person can trigger a warning to the user or operator via SMS or Email.
## Geo Fence
A Geo fence is a virtual perimeter on a geographic area using a location-based service, so that when the geo fencing device enters or exits the area, a notification is generated. The notification can contain information about the location of the device and might be sent to a mobile telephone or an email account. Reference: http://en.wikipedia.org/wiki/Geofence.
## Background
For geo-fencing, I used Polygonal geo-fencing method where a polygon is drawn around the route or area. Using this method, GPS Tracking devices can be tracked either inside or outside of the polygon.
## Determining a Point
The function will return `true `if the point X,Y is inside the polygon, or `false `if it is not. If the point is exactly on the edge of the polygon, then the function may return `true `or `false`. Thanks for the article “Determining Whether A Point Is Inside A Complex Polygon”.
```public bool FindPoint(double X, double Y)
{
int sides = this.Count() - 1;
int j = sides - 1;
bool pointStatus = false;
for (int i = 0; i < sides; i++)
{
if (myPts[i].Y < Y && myPts[j].Y >= Y ||
myPts[j].Y < Y && myPts[i].Y >= Y)
{
if (myPts[i].X + (Y - myPts[i].Y) /
(myPts[j].Y - myPts[i].Y) * (myPts[j].X - myPts[i].X) < X)
{
pointStatus = !pointStatus ;
}
}
j = i;
}
return pointStatus;
}```
## Creating a Polygon
On the map, draw a polygon to the area which is to be geo-fenced and capture the corner points of the polygon and store into XML file (see: PolygonPoints.XML). `loadData() `function will create a polygon using defined corner points in the XML file.
```private void loadData()
{
DataSet ds = new DataSet();
foreach (DataRow dr in ds.Tables[0].Rows)
{
Point p = new Point();
//Convert Latitude into degrees
String Lat = dr[0].ToString();
double LatSec = Double.Parse(Lat.Substring(4, 4)) / 6000;
double LatMin = (Double.Parse(Lat.Substring(2, 2)) + LatSec) / 60;
p.X = Double.Parse(Lat.Substring(0, 2)) + LatMin;
//Convert Longitude into degrees
String Long = dr[1].ToString();
double LongSec = Double.Parse(Long.Substring(5, 4)) / 6000;
double LongMin = (Double.Parse(Long.Substring(3, 2)) + LongSec) / 60;
p.Y = Double.Parse(Long.Substring(0, 3)) + LongMin;
}
} ```
## Sample Code
When you run and enter latitude and longitude outside the polygon, then a message shows point not found in the route and otherwise it shows point found in the route.
```PolyGon myRoute = new PolyGon(points);
bool stat = myRoute.FindPoint(Double.Parse(txtLat.Text.ToString()),
Double.Parse(txtLang.Text.ToString()));
if(stat)
{
lblResult.Text = "Point found in the route";
}
else
Chief Technology Officer Vajra Infratech Pvt. Ltd.,
India
No Biography provided
View All Threads First Prev Next
Geofencing? Polygon? bulleh 4-Jul-11 0:40
Re: Geofencing? Polygon? RajuBhupathi 4-Jul-11 23:41
Re: Geofencing? Polygon? bulleh 7-Jul-11 1:22
Re: Geofencing? Polygon? bulleh 20-Jul-11 0:22
Re: Geofencing? Polygon? borno 5-Aug-11 8:40
Re: Geofencing? Polygon? bulleh 5-Aug-11 23:38
Re: Geofencing? Polygon? zinobu 24-Aug-11 7:30
Last Visit: 31-Dec-99 18:00 Last Update: 4-Dec-13 7:40 Refresh 1 | 1,001 | 3,640 | {"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-2013-48 | latest | en | 0.760267 |
http://quizlet.com/6680534/geometry-flash-cards/ | 1,386,855,586,000,000,000 | text/html | crawl-data/CC-MAIN-2013-48/segments/1386164587361/warc/CC-MAIN-20131204134307-00068-ip-10-33-133-15.ec2.internal.warc.gz | 144,886,796 | 19,433 | # Geometry
## 38 terms · Chapter one
### line
A endpoint that never ends or begins, 2 arrowheads
### line segment
a set of points with an end and a beggining
### ray
conists of points with a beggining and no ending
### intersection
the set of points that figures have in common
### coordinate
the real number that corresponds to a point
### distance
the space between two points
### congruent segments
line segments that have the same length
### midpoint
the point that splits it in half
### segment bisector
a line segement that splits a segment equally in half
### midpoint formula
(X1+X2/ 2, Y1+Y2/2)
### distance formula
(square root) (X2-X1)2 +(Y1-Y2)2
### angle
two different rays with the same endpoint
### sides
the rays of the angle
### vertex
the endpoints of an angle
180` angle
### acute
smaller then 90 degrees
### right
an angle equal to 90 degrees
### obtuse
an angle greater than 90 degrees
### congruent angles
two angles with the same measure
### angle bisector
a ray that splits an angle equally in half
### complementary angles
two angles that equal to 90 degrees
### supplementary angles
two angles that equal 180 degrees
two angles that share a common vertex and side, but have no common interior points
### linear pair
two adjacent angles with common sides that are opposite rays
### vertical angles
two angles with sides that form two pairs of opposite rays
### polygon
a closed plane figure with the following properties
### sides
a shape formed by three or more line segments
### vertex
the endpoint of a side
### convex
if no line that contains a side of the polygon contains a point in the interior of the polygon.
not convex
### equilateral
all sides are congruent
### equilangular
all angles in the interior of the polygon are congruent
### regular
is a convex polygon that is both equilateral and equalangular
2l+2w= P
P=a+b+b
A=1/2bh
3.14d=C
A=3.14r2 | 482 | 1,951 | {"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} | 3.765625 | 4 | CC-MAIN-2013-48 | latest | en | 0.836096 |
https://cpep.org/chemistry/2866866-a-car-on-a-25-m2-hydraulic-lift-platform-weighs-15000-n-if-the-force-o.html | 1,653,780,251,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652663021405.92/warc/CC-MAIN-20220528220030-20220529010030-00208.warc.gz | 255,637,174 | 7,696 | 18 April, 22:34
# A car on a 25-m2 hydraulic lift platform weighs 15,000 N. If the force on the smaller piston required to lift it is 1/100 its weight, what is the area of the smaller piston?
+2
1. 18 April, 22:51
0
The area is 2.5 m²
Explanation:
Step 1: Data given
A car has a 25 m² hydraulic lift platform that weighs 15000 N
The smaller piston required a force to lift = 1/100 its weight
Step 2: Calculate the area of the smaller piston
Pressure = Force / Area
F1 / A1 = F2 / A2
⇒ with F1 = weight of the hydraulic lift = 15000 N
⇒ with A1 = area of the hydraulic lift = 25 m²
⇒ with F2 = 1/100 of it's weight = 1500 N
A2 = F2 / (F1/A1)
A2 = 1500 (1500/25)
A2 = 2.5
The area is 2.5 m² | 250 | 704 | {"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} | 4.40625 | 4 | CC-MAIN-2022-21 | latest | en | 0.891945 |
https://www.projecttopics.com/questions/mathematics/a-bag-contains-16-red-balls-and-20-blue-balls-only-how-many-white-balls/ | 1,721,733,399,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763518029.87/warc/CC-MAIN-20240723102757-20240723132757-00475.warc.gz | 824,960,675 | 20,682 | Home » A bag contains 16 red balls and 20 blue balls only. How many white balls…
# A bag contains 16 red balls and 20 blue balls only. How many white balls…
A bag contains 16 red balls and 20 blue balls only. How many white balls must be added to the bag so that the probability of randomly picking a red ball is equal to (frac{2}{5})
• A.
4
• B.
20
• C.
24
• D.
40 | 105 | 367 | {"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-2024-30 | latest | en | 0.894551 |
https://www.coursehero.com/tutors-problems/Statistics-and-Probability/19607655-the-problem-is-attached-please-show-work/ | 1,575,933,411,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575540525598.55/warc/CC-MAIN-20191209225803-20191210013803-00180.warc.gz | 669,214,452 | 27,310 | U Transfer Credits | N X | UMGC Campus X USyllabus - STAT 20( x 5 DownloadAttachm X 5 STAT 200 Midterm X Homework Help - X G The four measurer X + X...
View the step-by-step solution to:
Question
the problem is attached. please show work
U Transfer Credits | N X | UMGC Campus
X
5 STAT 200 Midterm X
Homework Help - X
G The four measurer X
+
X
< -> C @ File | C:/Users/keish/OneDrive/Documents/Desktop/STAT%20200%20Midterm%20Exam%20Fall%202019%206968.pdf
STAT 200 Wildtenin Cxaint Tall 2017 0700.put
3.
The density of people per square kilometer for African countries is in the table
below.
15
16 81
3
62 367
42 123
8
12
29
70
39
83
26
51
79
6 157 105
42
45
72
72
37
4
36 134
12
3
630 563
72
29
3
13
176 341
415 187
65
194
75
16
41
18
69
49 103
5 143
2
18
31
Complete the following frequency distribution using 8 classes. Express the
relative frequency to two decimal places. (Show all work. Just the answer,
without supporting work, will receive no credit.)
Class limit
Frequency
Relative
Cumulative
Frequency
Frequency
2-80
81-159
160-238
239-317
318-396
397-475
476-554
555-633
Page 2 of 4
Show all
X
O Type here to search
W
9:36 PM
10/13/2019
First, assign each value in the set of data into the appropriate classes they belong to. For instance, 15 falls within the... View the full answer
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Browse Documents | 509 | 1,648 | {"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-2019-51 | latest | en | 0.760249 |
https://www.weegy.com/?ConversationId=D0LZNRI4&Link=i&ModeType=2 | 1,596,711,355,000,000,000 | text/html | crawl-data/CC-MAIN-2020-34/segments/1596439736902.24/warc/CC-MAIN-20200806091418-20200806121418-00119.warc.gz | 888,745,396 | 10,075 | Forces that are equal in size but opposite in direction are called unbalanced forces. Please select the best answer from the choices provided T F
Forces that are equal in size but opposite in direction are called unbalanced forces. FALSE.
s
Updated 5/25/2016 3:56:24 AM
Rating
3
Forces that are equal in size but opposite in direction are called unbalanced forces. FALSE.
Confirmed by Andrew. [5/25/2016 7:19:13 AM]
The velocity a rocket must reach to establish an orbit around Earth is called
Updated 12/15/2017 2:38:21 AM
The velocity a rocket must reach to establish an orbit around Earth is called Escape velocity.
Only the mass of the planet and radius of the orbit are needed to calculate the orbital speed of a satellite. Please select the best answer from the choices provided T F
Updated 7/29/2015 2:39:36 PM
Only the mass of the planet and radius of the orbit are needed to calculate the orbital speed of a satellite. TRUE.
Confirmed by Andrew. [7/29/2015 2:43:34 PM]
Inertia is the tendency of an object to resist any change in its motion. Please select the best answer from the choices provided T F
Updated 5/1/2015 7:56:27 AM
Inertia is the tendency of an object to resist any change in its motion. True
Confirmed by Kaysha [5/1/2015 8:04:03 AM]
Friction always acts in a direction ______ to the direction of motion. a. equal c. perpendicular b. opposite
Weegy: Friction always act in a direction opposite to the direction of motion. (More)
Updated 5/1/2015 9:38:54 AM
What kind of wavelength do television signals have? a. Short c. Long b. Intermediate d. Very rapid
Weegy: Television transmits long wavelength signals. (More)
Updated 4/27/2015 8:48:56 AM
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Home | Contact | Blog | About | Terms | Privacy | © Purple Inc. | 789 | 2,652 | {"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-2020-34 | latest | en | 0.909476 |
https://parade.com/455668/marilynvossavant/learning-with-spreadsheets/ | 1,618,429,038,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038078021.18/warc/CC-MAIN-20210414185709-20210414215709-00548.warc.gz | 554,762,875 | 70,755 | Robert Innes in Sebastopol, California, writes:
Your important column on the teaching of algebra to children(December 6, 2015) does not mention the most powerful modern tool invented to reveal clearly the concept of algebra: The spreadsheet. It is pure algebra because cells contain algebraic equations that reference other cells as placeholders for numbers. The beauty of spreadsheets for children is that once numbers are entered in the input cells, the computer–not the child–does the routine arithmetic. Understanding algebra in action becomes a look-see operation. Basic addition, multiplication, subtraction and addition can be taught in kindergarten with beads.
For example children can convince themselves empirically that (a+b).(a-b) = (a^2 – b^2) no matter what numbers are entered in the a and b entry cells by coding (a+b).(a-b) in one cell, (a^2 – b^2) in another cell, and the difference between them in a third cell. The algebraic cells can have text labels to make the spreadsheet easily readable. For the high school student, indexing with one or two indices is a walkover: Enter a formula in a cell of a table and copy it to all the other cells of the table, which fills with numbers like magic. This is a lot clearer than the stuff you and I went through in school, which does not easily take root in the minds of children without great mathematical ability, so learning algebra becomes a drag for the rest of the children.
Unfortunately there are no good spreadsheets for children that I know of. Spreadsheets designed for business have too many bells and whistles that would confuse children. For decades of writing scientific and engineering data reduction code, all my code begins in spreadsheets. They are excellent tools to develop ideas, which then provide a means of checking the computer code for bugs. I have always wished that someone of influence would get spreadsheets into the schools. Now I know who she is–you, Marilyn!
Marilyn responds:
Imagine implementing your suggestion in school systems where parents already are lobbying against the inclusion of algebra! Good luck!
A member has started a discussion.
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https://trac.sagemath.org/ticket/26478?cversion=0&cnum_hist=8 | 1,618,452,021,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038082988.39/warc/CC-MAIN-20210415005811-20210415035811-00197.warc.gz | 677,707,565 | 9,032 | Opened 3 years ago
Closed 2 years ago
# clean graph_plot_js.py, graph_list.py and graph_input.py
Reported by: Owned by: dcoudert major sage-8.5 graph theory py3, graph tscrim David Coudert Travis Scrimshaw N/A 2cf89df 2cf89df2339e0a1cb2e98fb0af9bc3ddeb0a993f
clean the files (PEP8) and simplify some tests
### comment:1 Changed 3 years ago by dcoudert
• Branch set to public/26478_cleaning
New commits:
bfe7c08 `clean plot_graph_js.py` 874eeb8 `clean graph_list.py` 8c579b0 `clean graph_input.py`
### comment:2 Changed 3 years ago by dcoudert
• Description modified (diff)
• Status changed from new to needs_review
The most significant changes are in `graph_input.py` as I changed some tests.
### comment:3 Changed 3 years ago by tscrim
For this change:
```- if not loops and any(u in neighb for u,neighb in six.iteritems(M)):
- if loops is False:
- u = next(u for u,neighb in six.iteritems(M) if u in neighb)
- raise ValueError("The graph was built with loops=False but input M has a loop at {}.".format(u))
- loops = True
- if loops is None:
- loops = False
+ if any(not isinstance(M[u], dict) for u in M):
+ raise ValueError("input dict must be a consistent format")
+
+ if not loops:
+ for u, neighb in six.iteritems(M):
+ if u in neighb:
+ if loops is False:
+ raise ValueError("the graph was built with loops=False but input M has a loop at {}".format(u))
+ loops = True
+ break
+ if loops is None:
+ loops = False
```
the `any` call is actually faster. Granted, your change will result in a faster error, but I think an error message being here is unexpected. Hence it is allowed to be slow. If someone really wants the extra speed and is using that test to separate cases, they should run their own test instead of catching the error (which is faster). So I am inclined to revert this.
Also, strictly speaking, the items for `INPUT:` are not sentences and do not end in a period (although some of them are sufficiently long with multiple parts that require it). While I was letting this slide when it was previously there, the ones you are adding should not have it.
All of the other changes LGTM.
### comment:4 Changed 3 years ago by git
• Commit changed from 8c579b08a5ac05ce2e4d840840bda79a5dad5654 to 90f89b08b54400fd66e9b62c625aa1b2c823f313
Branch pushed to git repo; I updated commit sha1. New commits:
39aa69d `trac #26478: reviewers comments in graph_input` 3826f1e `trac #26478: reviewers comments in graph_list` 90f89b0 `trac #26478: reviewers comments in graph_plot_js`
### comment:5 Changed 3 years ago by dcoudert
The following lines are apparently incompatible with Python 3 (see patchbot).
```+ if any(u in neighb for u,neighb in six.iteritems(M)):
+ u = next(u for u,neighb in six.iteritems(M) if u in neighb)
```
```+ if any(u in neighb for u, neighb in six.iteritems(D)):
+ u = next(u for u, neighb in six.iteritems(M) if u in neighb)
```
Is there a special trick or should I turn these lines to for loops ?
### comment:6 Changed 3 years ago by chapoton
No, no. It's just the patchbot prefering "iteritems" to "six.iteritems". Just change the imports accordingly.
And once again, it is never mandatory to have all green lights from the patchbot, as the patchbot is not smart..
### comment:7 Changed 3 years ago by git
• Commit changed from 90f89b08b54400fd66e9b62c625aa1b2c823f313 to 93ea7bdd587ab946b3c68ebd9ef52d3407704ae0
Branch pushed to git repo; I updated commit sha1. New commits:
93ea7bd `trac #26478: fix six import in graph_input.py`
### comment:8 Changed 3 years ago by dcoudert
Good to know. As I'm not Python 3 expert, I though it could be something else.
Version 0, edited 3 years ago by dcoudert (next)
### comment:9 Changed 2 years ago by dcoudert
• Milestone changed from sage-8.4 to sage-8.5
### comment:10 Changed 2 years ago by tscrim
Doctest failures:
```File "src/sage/graphs/digraph.py", line 399, in sage.graphs.digraph.DiGraph
Failed example:
Expected:
Traceback (most recent call last):
...
RuntimeError: The string (IRAaDCIIOEOKcPWAo) seems corrupt: for n = 10, the string is too short.
Got:
Traceback (most recent call last):
...
RuntimeError: the string (IRAaDCIIOEOKcPWAo) seems corrupt: for n = 10, the string is too short
```
and similar in `graph.py`. Once fixed and the patchbot comes back green, then it will be a positive review.
### comment:11 Changed 2 years ago by git
• Commit changed from 93ea7bdd587ab946b3c68ebd9ef52d3407704ae0 to 62403cd23042e6f8697dba2800648a2946a0919b
Branch pushed to git repo; I updated commit sha1. New commits:
62403cd `trac #26478: fix doctests`
### comment:12 Changed 2 years ago by dcoudert
I fixed the doctests. Now waiting for the patchbot.
### comment:13 Changed 2 years ago by tscrim
• Reviewers set to Travis Scrimshaw
• Status changed from needs_review to positive_review
Thanks.
### comment:14 Changed 2 years ago by vbraun
• Status changed from positive_review to needs_work
Merge conflict
### comment:15 Changed 2 years ago by git
• Commit changed from 62403cd23042e6f8697dba2800648a2946a0919b to 2cf89df2339e0a1cb2e98fb0af9bc3ddeb0a993f
Branch pushed to git repo; I updated commit sha1. New commits:
2cf89df `trac #26478: fix merge conflict with 8.5.beta0`
### comment:16 Changed 2 years ago by dcoudert
• Status changed from needs_work to needs_review
I fixed the merge conflict with 8.5.beta0
### comment:17 Changed 2 years ago by tscrim
• Status changed from needs_review to positive_review
### comment:18 Changed 2 years ago by vbraun
• Branch changed from public/26478_cleaning to 2cf89df2339e0a1cb2e98fb0af9bc3ddeb0a993f
• Resolution set to fixed
• Status changed from positive_review to closed
Note: See TracTickets for help on using tickets. | 1,688 | 5,939 | {"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} | 2.875 | 3 | CC-MAIN-2021-17 | latest | en | 0.860744 |
https://delanceyplace.com/view-archives.php?p=4208 | 1,603,224,760,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107874135.2/warc/CC-MAIN-20201020192039-20201020222039-00147.warc.gz | 294,466,747 | 10,379 | ## how many stars are there? -- 10/14/20
Today's selection from Space at the Speed of Light by Dr. Rebecca Smethurst. How many stars are there in the universe?
"It is nigh on physically impossible to count the number of galaxies in the universe because a) there are so many and b) how can we be sure we've found them all?
"But, to put some sort of lower limit on this, we can use an image that the famous Hubble Space Telescope (HST) took about a decade ago. Astronomers decided to use HST to stare at the darkest patch of sky that we know, in the constellation Fornax, in the Southern Hemisphere night sky, to see what they could find. They took an image that was a 2 x 2 arcminute square patch of the sky. An arcminute is a funny unit -- it is a sixtieth (1/60) of a degree, and an arcsecond is a sixtieth (1/60) of an arcminute. Given that the whole sky is 360 degrees around, it's a very small patch. It's an image that is 5 percent of the size of the Full Moon in the sky. Astronomers didn't really know what to expect to find in this tiny dark patch of sky, but the latest count on the number of stars found in the image of that patch is four. And the latest count of the number of galaxies is about five thousand -- everything from beautiful nearby spiral galaxies to distant galaxies that we detect as just a single pixel.
Hubble eXtreme Deep Field (HXDF) taken in 2012
"If we take that number and apply it to the rest of the area of the sky, we can estimate that there are at least 100 billion galaxies in the universe. Remember, this image was taken of the darkest patch of the sky, so in other regions we should see even more galaxies (plus all the galaxies that are still too faint for us to see). It's more likely that there are another couple of zeros on the end of that number. So, let's make our estimate a round trillion galaxies. Not only that, let's say that each galaxy contains, roughly, 100 billion stars. So, perhaps, we can estimate that there are at least 100 sextillion stars. That's 100,000,000,000,000,000,000,000 stars in the universe. So, if one in a quintillion stars might develop life, and there are at least a hundred sextillion stars in the universe, then perhaps there are a hundred thousand planets out there in the vastness of space that might have the right conditions to develop intelligent life!
"People often ask me, how -- as an astrophysicist knowing and thinking about these numbers all the time -- I don't get overwhelmed by it all. How do I stare at the sky without being entirely crippled by anxiety at the sheer scale of the whole thing and our own insignificance? Firstly, in day-to-day life -- whether you're at your desk crunching data, in an office, at home, or on a train -- there isn't time to stop and think about it. But, when I look at the majesty of the night sky, with the Milky Way stretching out overhead in a huge arc of stars, I don't feel anxious. I feel limitless. Like there are infinite possibilities out there and I could be part of any one of them. The scale doesn't scare me; it thrills me. Like the protagonist at the beginning of a good adventure novel yearning to see the world and get out of their small town. When I look up at the sky and think about the sheer number of stars out there, I can't help but get excited about drawing the conclusion that we can't be the only planet whose cards came up right in the game of life."
#### author:
Dr. Rebecca Smethurst
#### title:
Space at the Speed of Light
Ten Speed Press | 836 | 3,510 | {"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 | 0.95435 |
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# At a college party, 70% of the women are wearing red t-shirts. If 60%
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At a college party, 70% of the women are wearing red t-shirts. If 60% [#permalink]
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18 Jul 2018, 22:42
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At a college party, 70% of the women are wearing red t-shirts. If 60% of the people at the party are wearing red t-shirts, what is the ratio of women to men at the party?
(1) Forty percent of the men at the party are wearing red t-shirts.
(2) There are 120 people at the party.
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Re: At a college party, 70% of the women are wearing red t-shirts. If 60% [#permalink]
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18 Jul 2018, 23:53
Bunuel wrote:
At a college party, 70% of the women are wearing red t-shirts. If 60% of the people at the party are wearing red t-shirts, what is the ratio of women to men at the party?
(1) Forty percent of the men at the party are wearing red t-shirts.
(2) There are 120 people at the party.
I'm poor with Percentages. Please let me know if the below answer is wrong.
Given:
Women with Red T-Shirts, W = 70%
Total people with Red T shirts, P = 60%
Total People at Party = P
Total Women at Party = W
Total Men at Party = M
Find - W/M
Statement 1:
Men with Red T-Shirts, M = 40%
Insufficient as we don't know the total number of people, Total percentage of Women/Men.
Statement 2:
Total People at Party = 120
We can find the number of people wearing Red T shirts at the party i.e., 60% of 120 = 72.
But we don't know the number/percentage of Women/Men at the party
Hence insufficient
Statement 1 & 2:
Insufficient. We still don't get the percentage of Women/Men at the party to find the ratio.
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Re: At a college party, 70% of the women are wearing red t-shirts. If 60% [#permalink]
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19 Jul 2018, 00:09
2
Akash720 wrote:
Bunuel wrote:
At a college party, 70% of the women are wearing red t-shirts. If 60% of the people at the party are wearing red t-shirts, what is the ratio of women to men at the party?
(1) Forty percent of the men at the party are wearing red t-shirts.
(2) There are 120 people at the party.
I'm poor with Percentages. Please let me know if the below answer is wrong.
Given:
Women with Red T-Shirts, W = 70%
Total people with Red T shirts, P = 60%
Total People at Party = P
Total Women at Party = W
Total Men at Party = M
Find - W/M
Statement 1:
Men with Red T-Shirts, M = 40%
Insufficient as we don't know the total number of people, Total percentage of Women/Men.
Statement 2:
Total People at Party = 120
We can find the number of people wearing Red T shirts at the party i.e., 60% of 120 = 72.
But we don't know the number/percentage of Women/Men at the party
Hence insufficient
Statement 1 & 2:
Insufficient. We still don't get the percentage of Women/Men at the party to find the ratio.
Hi Akash720
Lets consider total people as 100X
so 60X wearing tshirt.
in which 70% is women so 42X is female
Therefore 18X is male.
From A it is given 40% male so 40% of M =18X
So total M=45X ,Then total Female will be 55X
so ratio will be 45/55.
We can get sol from A. Total number is not required.
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At a college party, 70% of the women are wearing red t-shirts. If 60% [#permalink]
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19 Jul 2018, 00:20
kunalcvrce wrote:
Akash720 wrote:
Bunuel wrote:
At a college party, 70% of the women are wearing red t-shirts. If 60% of the people at the party are wearing red t-shirts, what is the ratio of women to men at the party?
(1) Forty percent of the men at the party are wearing red t-shirts.
(2) There are 120 people at the party.
I'm poor with Percentages. Please let me know if the below answer is wrong.
Given:
Women with Red T-Shirts, W = 70%
Total people with Red T shirts, P = 60%
Total People at Party = P
Total Women at Party = W
Total Men at Party = M
Find - W/M
Statement 1:
Men with Red T-Shirts, M = 40%
Insufficient as we don't know the total number of people, Total percentage of Women/Men.
Statement 2:
Total People at Party = 120
We can find the number of people wearing Red T shirts at the party i.e., 60% of 120 = 72.
But we don't know the number/percentage of Women/Men at the party
Hence insufficient
Statement 1 & 2:
Insufficient. We still don't get the percentage of Women/Men at the party to find the ratio.
Hi Akash720
Lets consider total people as 100X
so 60X wearing tshirt.
in which 70% is women so 42X is female
Therefore 18X is male.
From A it is given 40% male so 40% of M =18X
So total M=45X ,Then total Female will be 55X
so ratio will be 45/55.
We can get sol from A. Total number is not required.
Hi Kunal
Given data is 70% of women are wearing Red t shirts. So I assume this means there are x women and 70% of x are wearing Red t-shirts. So we really don't have the value of x women.
I hope my concern is clear. Correct me if I'm wrong
Posted from my mobile device
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Re: At a college party, 70% of the women are wearing red t-shirts. If 60% [#permalink]
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19 Jul 2018, 00:51
Akash720 wrote:
Bunuel wrote:
At a college party, 70% of the women are wearing red t-shirts. If 60% of the people at the party are wearing red t-shirts, what is the ratio of women to men at the party?
(1) Forty percent of the men at the party are wearing red t-shirts.
(2) There are 120 people at the party.
I'm poor with Percentages. Please let me know if the below answer is wrong.
Given:
Women with Red T-Shirts, W = 70%
Total people with Red T shirts, P = 60%
Total People at Party = P
Total Women at Party = W
Total Men at Party = M
Find - W/M
Statement 1:
Men with Red T-Shirts, M = 40%
Insufficient as we don't know the total number of people, Total percentage of Women/Men.
Statement 2:
Total People at Party = 120
We can find the number of people wearing Red T shirts at the party i.e., 60% of 120 = 72.
But we don't know the number/percentage of Women/Men at the party
Hence insufficient
Statement 1 & 2:
Insufficient. We still don't get the percentage of Women/Men at the party to find the ratio.
Hi Akash720
You are right i misread the question.
Intern
Joined: 18 Apr 2018
Posts: 3
Re: At a college party, 70% of the women are wearing red t-shirts. If 60% [#permalink]
### Show Tags
19 Jul 2018, 01:41
1
Hi,
I will go with A.
At a college party, 70% of the women are wearing red t-shirts. If 60% of the people at the party are wearing red t-shirts, what is the ratio of women to men at the party?
(1) Forty percent of the men at the party are wearing red t-shirts.
(2) There are 120 people at the party.
Let Men be "x" and Women be "y". It says that 70% of women are wearing red t-shirts i.e. 7y/10. 60% of the people are wearing red t-shirts i.e. 3(x+y)/5.
1.2x/5 men are wearing red t-shirt. Hence, 2x/5 + 7y/10 = 3(x+y)/5. you get y/x= 2. Sufficient.
2. Insufficient. It just tells (x+y)=120. Doesn't tell anything about either percentage or number of men wearing red t-shirt.
Correct me if i am wrong in my approach.
ISB, NUS, NTU Moderator
Joined: 11 Aug 2016
Posts: 312
At a college party, 70% of the women are wearing red t-shirts. If 60% [#permalink]
### Show Tags
19 Jul 2018, 04:02
Akash720 wrote:
Bunuel wrote:
At a college party, 70% of the women are wearing red t-shirts. If 60% of the people at the party are wearing red t-shirts, what is the ratio of women to men at the party?
(1) Forty percent of the men at the party are wearing red t-shirts.
(2) There are 120 people at the party.
I'm poor with Percentages. Please let me know if the below answer is wrong.
Given:
Women with Red T-Shirts, W = 70%
Total people with Red T shirts, P = 60%
Total People at Party = P
Total Women at Party = W
Total Men at Party = M
Find - W/M
Statement 1:
Men with Red T-Shirts, M = 40%
Insufficient as we don't know the total number of people, Total percentage of Women/Men.
Statement 2:
Total People at Party = 120
We can find the number of people wearing Red T shirts at the party i.e., 60% of 120 = 72.
But we don't know the number/percentage of Women/Men at the party
Hence insufficient
Statement 1 & 2:
Insufficient. We still don't get the percentage of Women/Men at the party to find the ratio.
No of women at party: W
No of men at party: M
No of women wearing Red t shirt: Wr
No of men wearing Red t shirt: Mr
Total No of people at party:W+M
Now,
Given:
70% of the women are wearing red t-shirts : $$Wr=\frac{70W}{100}$$ .....................................(1)
60% of the people at the party are wearing red t-shirts : $$(Wr+Mr)=\frac{60(W+M)}{100}$$ ......(2)
To Calculate:
$$\frac{W}{M}$$
Statement1: Forty percent of the men at the party are wearing red t-shirts.
Therefore, $$Mr=\frac{40M}{100}$$
(Wr+Mr) = $$\frac{70W}{100}+\frac{40M}{100}$$
=$$\frac{(7W+4M)}{10}$$
Equating the above equation with (2)
We get, $$(7W+4M)=6(W+M)$$
This implies, W=2M
or, $$\frac{W}{M}=2$$
Sufficient.
Statement1: There are 120 people at the party
Therefore, M+W=120
We need either M or W to solve this equation. We have neither.
Hence, InSufficient.
_________________
~R.
If my post was of any help to you, You can thank me in the form of Kudos!!
At a college party, 70% of the women are wearing red t-shirts. If 60% &nbs [#permalink] 19 Jul 2018, 04:02
Display posts from previous: Sort by | 3,266 | 11,310 | {"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} | 4.0625 | 4 | CC-MAIN-2018-47 | latest | en | 0.890497 |
https://encyclopedia2.thefreedictionary.com/planar+graph | 1,568,903,913,000,000,000 | text/html | crawl-data/CC-MAIN-2019-39/segments/1568514573533.49/warc/CC-MAIN-20190919142838-20190919164838-00034.warc.gz | 479,512,935 | 11,952 | # planar graph
Also found in: Wikipedia.
## planar graph
[′plā·nər ‚graf]
(mathematics)
A graph that can be drawn in a plane without any lines crossing.
References in periodicals archive ?
The current best algorithm for 4-coloring an m-vertex planar graph runs in O([m.
In particular, one well-known result of Schnyder states that every planar graph has arboricity at most 3, see [5].
Corresponding to every connected link diagram we can find a connected signed planar graph and vice versa.
In a maximal planar graph G = ((V (G), E(G)) with [absolute value of V (G)]=n and [absolute value of E (G)]=m, we have m = 3n - 6.
There's an interactive planar graph as well that can be rotated with your mouse.
This graph is still too complex to be displayed with reasonably true distances and without crossing lines, so I removed the link between difficult and heavy to obtain at least an approximately planar graph.
Yannakakis has investigated the book embedding problem for planar graphs, showing that four pages are sufficient to book embed a planar graph [12].
In this paper it is proved that this is not the case (not even for every 5-connected planar graph [G.
However, the number is easily applied to most polyhedra, for by looking 'through a face' of the polyhedron we can obtain a planar graph.
An example of orthogonal embedding of a planar graph is illustrated in Fig.
Observation 2 -- Considering a particular planar graph representation: This observation illustrates how students over emphasize the visual aspect of Graph Theory, and base their answers on a specific planar representation of a graph's vertices and arcs.
For example, the crossing number of a planar graph is 0 and the crossing number of [K.
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https://schoolhousereviewcrew.com/i-see-cards/ | 1,722,967,347,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640497907.29/warc/CC-MAIN-20240806161854-20240806191854-00881.warc.gz | 412,017,871 | 31,827 | # I See Cards Reviews
FractazmicTM is one of the fun educational card games of the makers of I See CardsTM. The aim of this game is to help your child/student to learn and practice fractions in a fun way at their individual level.
It is impossible to put an age limit on FUN, so these cards are for All Ages – although it is specifically aimed at 1st-8th grade students.
The FractazmicTM set of cards is made from sturdy material and is bright and colorful.
## Fractazmic Decks Include:
• 60 Math Learning Cards
• Each card clearly numbered
• Each card clearly labeled with the fraction and reinforcing graphic.
The deck of cards are divided into three different sets. These sets are:
• The Twelfths suit – Blue
• The Tenths suit – Green
• The Sixteenths suit – Red
### What makes Fractazmic Decks Special:
• Quickly learn to convert and add fractions.
• Reinforces relationship between fractions and measurements.
• Shows real world use of fractions.
The Rules:
The cards can be used in various ways to fit best with where your child / student is at from using the individual sets of cards to first familiarize the child with the different fractions to playing the different games. FractazmicTM can be played by 2-4 players, but also be used individually by the student.
The aim of the game is to combine the different cards in a suit to make 1(a hand). The player with the most ‘hands’ wins.
Your first stop might be this webpage HERE where you can actually play the speed game online as well as WIN A FREE SET OF CARDS if your time is the fastest.
Here are some links to video clips that explain how to play Fractazmic Rummy and Trap:
FractazmicTM Rummy
Trap
You can also download a booklet HERE explaining how to play the different games.
In this booklet you will also find games for the other card games from I See CardsTM.
Other games from I See CardsTM are:
PyramathTM (which the crew reviewed last year)
Fractazmic’s amazing creators can be contacted for support or advice through their contact page HERE.
Follow the linky below to see what the Crew thought of this product. | 490 | 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} | 2.515625 | 3 | CC-MAIN-2024-33 | latest | en | 0.936698 |
https://www.aqua-calc.com/convert/density/gram-per-us-teaspoon | 1,627,943,017,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046154385.24/warc/CC-MAIN-20210802203434-20210802233434-00250.warc.gz | 639,303,927 | 16,472 | # Select a table to convert grams per US teaspoon
## grams/US teaspoon [g/tsp] of density conversion tables
Select units whose names contain any of the following checked keywords, where ∗ is a wildcard for partial matches:
Conversion tableCorrelationFormula
gram ⁄ US teaspoon ➔ microgram ⁄ centimeter³1 g/tsp = 202 884.136 µg/cm³202 884.136 × ρg/tsp = ρµg/cm³
gram ⁄ US teaspoon ➔ microgram ⁄ decimeter³1 g/tsp = 202 884 136 µg/dm³202 884 136 × ρg/tsp = ρµg/dm³
gram ⁄ US teaspoon ➔ microgram ⁄ US fluid ounce1 g/tsp = 6 000 000 µg/fl.oz6 000 000 × ρg/tsp = ρµg/fl.oz
gram ⁄ US teaspoon ➔ microgram ⁄ foot³1 g/tsp = 5 745 038 957 µg/ft³5 745 038 957 × ρg/tsp = ρµg/ft³
gram ⁄ US teaspoon ➔ microgram ⁄ inch³1 g/tsp = 3 324 675.32 µg/in³3 324 675.32 × ρg/tsp = ρµg/in³
gram ⁄ US teaspoon ➔ microgram ⁄ liter1 g/tsp = 202 884 136 µg/l202 884 136 × ρg/tsp = ρµg/l
gram ⁄ US teaspoon ➔ microgram ⁄ meter³1 g/tsp = 202 884 136 000 µg/m³202 884 136 000 × ρg/tsp = ρµg/m³
gram ⁄ US teaspoon ➔ microgram ⁄ metric cup1 g/tsp = 50 721 034 µg/metric c50 721 034 × ρg/tsp = ρµg/metric c
gram ⁄ US teaspoon ➔ microgram ⁄ metric tablespoon1 g/tsp = 3 043 262.04 µg/metric tbsp3 043 262.04 × ρg/tsp = ρµg/metric tbsp
gram ⁄ US teaspoon ➔ microgram ⁄ metric teaspoon1 g/tsp = 1 014 420.68 µg/metric tsp1 014 420.68 × ρg/tsp = ρµg/metric tsp
gram ⁄ US teaspoon ➔ microgram ⁄ milliliter1 g/tsp = 202 884.136 µg/ml202 884.136 × ρg/tsp = ρµg/ml
gram ⁄ US teaspoon ➔ microgram ⁄ millimeter³1 g/tsp = 202.884136 µg/mm³202.884136 × ρg/tsp = ρµg/mm³
gram ⁄ US teaspoon ➔ microgram ⁄ US pint1 g/tsp = 95 999 999.9 µg/pt95 999 999.9 × ρg/tsp = ρµg/pt
gram ⁄ US teaspoon ➔ microgram ⁄ US tablespoon1 g/tsp = 3 000 000 µg/tbsp3 000 000 × ρg/tsp = ρµg/tbsp
gram ⁄ US teaspoon ➔ microgram ⁄ US teaspoon1 g/tsp = 999 999.998 µg/tsp999 999.998 × ρg/tsp = ρµg/tsp
gram ⁄ US teaspoon ➔ microgram ⁄ US cup1 g/tsp = 48 000 000.1 µg/US c48 000 000.1 × ρg/tsp = ρµg/US c
gram ⁄ US teaspoon ➔ microgram ⁄ US gallon1 g/tsp = 767 999 998 µg/US gal767 999 998 × ρg/tsp = ρµg/US gal
gram ⁄ US teaspoon ➔ microgram ⁄ US quart1 g/tsp = 192 000 000 µg/US qt192 000 000 × ρg/tsp = ρµg/US qt
gram ⁄ US teaspoon ➔ microgram ⁄ yard³1 g/tsp = 155 116 051 790 µg/yd³155 116 051 790 × ρg/tsp = ρµg/yd³
gram ⁄ US teaspoon ➔ pennyweight ⁄ centimeter³1 g/tsp = 0.1304575288 dwt/cm³0.1304575288 × ρg/tsp = ρdwt/cm³
gram ⁄ US teaspoon ➔ pennyweight ⁄ decimeter³1 g/tsp = 130.457529 dwt/dm³130.457529 × ρg/tsp = ρdwt/dm³
gram ⁄ US teaspoon ➔ pennyweight ⁄ US fluid ounce1 g/tsp = 3.85808959 dwt/fl.oz3.85808959 × ρg/tsp = ρdwt/fl.oz
gram ⁄ US teaspoon ➔ pennyweight ⁄ foot³1 g/tsp = 3 694.14583 dwt/ft³3 694.14583 × ρg/tsp = ρdwt/ft³
gram ⁄ US teaspoon ➔ pennyweight ⁄ inch³1 g/tsp = 2.13781587 dwt/in³2.13781587 × ρg/tsp = ρdwt/in³
gram ⁄ US teaspoon ➔ pennyweight ⁄ liter1 g/tsp = 130.457529 dwt/l130.457529 × ρg/tsp = ρdwt/l
gram ⁄ US teaspoon ➔ pennyweight ⁄ meter³1 g/tsp = 130 457.529 dwt/m³130 457.529 × ρg/tsp = ρdwt/m³
gram ⁄ US teaspoon ➔ pennyweight ⁄ metric cup1 g/tsp = 32.6143822 dwt/metric c32.6143822 × ρg/tsp = ρdwt/metric c
gram ⁄ US teaspoon ➔ pennyweight ⁄ metric tablespoon1 g/tsp = 1.95686293 dwt/metric tbsp1.95686293 × ρg/tsp = ρdwt/metric tbsp
gram ⁄ US teaspoon ➔ pennyweight ⁄ metric teaspoon1 g/tsp = 0.6522876458 dwt/metric tsp0.6522876458 × ρg/tsp = ρdwt/metric tsp
gram ⁄ US teaspoon ➔ pennyweight ⁄ milliliter1 g/tsp = 0.1304575288 dwt/ml0.1304575288 × ρg/tsp = ρdwt/ml
gram ⁄ US teaspoon ➔ pennyweight ⁄ millimeter³1 g/tsp = 0.0001304575288 dwt/mm³0.0001304575288 × ρg/tsp = ρdwt/mm³
gram ⁄ US teaspoon ➔ pennyweight ⁄ US pint1 g/tsp = 61.7294333 dwt/pt61.7294333 × ρg/tsp = ρdwt/pt
gram ⁄ US teaspoon ➔ pennyweight ⁄ US cup1 g/tsp = 30.8647167 dwt/US c30.8647167 × ρg/tsp = ρdwt/US c
gram ⁄ US teaspoon ➔ pennyweight ⁄ US gallon1 g/tsp = 493.835467 dwt/US gal493.835467 × ρg/tsp = ρdwt/US gal
gram ⁄ US teaspoon ➔ pennyweight ⁄ US quart1 g/tsp = 123.458867 dwt/US qt123.458867 × ρg/tsp = ρdwt/US qt
gram ⁄ US teaspoon ➔ pennyweight ⁄ US tablespoon1 g/tsp = 1.9290448 dwt/US tbsp1.9290448 × ρg/tsp = ρdwt/US tbsp
gram ⁄ US teaspoon ➔ pennyweight ⁄ US teaspoon1 g/tsp = 0.6430149333 dwt/US tsp0.6430149333 × ρg/tsp = ρdwt/US tsp
gram ⁄ US teaspoon ➔ pennyweight ⁄ yard³1 g/tsp = 99 741.9375 dwt/yd³99 741.9375 × ρg/tsp = ρdwt/yd³
gram ⁄ US teaspoon ➔ gram ⁄ centimeter³1 g/tsp = 0.202884136 g/cm³0.202884136 × ρg/tsp = ρg/cm³
gram ⁄ US teaspoon ➔ gram ⁄ decimeter³1 g/tsp = 202.884136 g/dm³202.884136 × ρg/tsp = ρg/dm³
gram ⁄ US teaspoon ➔ gram ⁄ US fluid ounce1 g/tsp = 6 g/fl.oz6 × ρg/tsp = ρg/fl.oz
gram ⁄ US teaspoon ➔ gram ⁄ foot³1 g/tsp = 5 745.03896 g/ft³5 745.03896 × ρg/tsp = ρg/ft³
gram ⁄ US teaspoon ➔ gram ⁄ inch³1 g/tsp = 3.32467532 g/in³3.32467532 × ρg/tsp = ρg/in³
gram ⁄ US teaspoon ➔ gram ⁄ liter1 g/tsp = 202.884136 g/l202.884136 × ρg/tsp = ρg/l
gram ⁄ US teaspoon ➔ gram ⁄ meter³1 g/tsp = 202 884.136 g/m³202 884.136 × ρg/tsp = ρg/m³
gram ⁄ US teaspoon ➔ gram ⁄ metric cup1 g/tsp = 50.721034 g/metric c50.721034 × ρg/tsp = ρg/metric c
gram ⁄ US teaspoon ➔ gram ⁄ metric tablespoon1 g/tsp = 3.04326204 g/metric tbsp3.04326204 × ρg/tsp = ρg/metric tbsp
gram ⁄ US teaspoon ➔ gram ⁄ metric teaspoon1 g/tsp = 1.01442068 g/metric tsp1.01442068 × ρg/tsp = ρg/metric tsp
gram ⁄ US teaspoon ➔ gram ⁄ milliliter1 g/tsp = 0.202884136 g/ml0.202884136 × ρg/tsp = ρg/ml
gram ⁄ US teaspoon ➔ gram ⁄ millimeter³1 g/tsp = 0.000202884136 g/mm³0.000202884136 × ρg/tsp = ρg/mm³
gram ⁄ US teaspoon ➔ gram ⁄ US pint1 g/tsp = 96 g/pt96 × ρg/tsp = ρg/pt
gram ⁄ US teaspoon ➔ gram ⁄ US tablespoon1 g/tsp = 3 g/tbsp3 × ρg/tsp = ρg/tbsp
gram ⁄ US teaspoon ➔ gram ⁄ US cup1 g/tsp = 48 g/US c48 × ρg/tsp = ρg/US c
gram ⁄ US teaspoon ➔ gram ⁄ US gallon1 g/tsp = 768 g/US gal768 × ρg/tsp = ρg/US gal
gram ⁄ US teaspoon ➔ gram ⁄ US quart1 g/tsp = 192 g/US qt192 × ρg/tsp = ρg/US qt
gram ⁄ US teaspoon ➔ gram ⁄ yard³1 g/tsp = 155 116.052 g/yd³155 116.052 × ρg/tsp = ρg/yd³
gram ⁄ US teaspoon ➔ grain ⁄ centimeter³1 g/tsp = 3.13098069 gr/cm³3.13098069 × ρg/tsp = ρgr/cm³
gram ⁄ US teaspoon ➔ grain ⁄ decimeter³1 g/tsp = 3 130.98069 gr/dm³3 130.98069 × ρg/tsp = ρgr/dm³
gram ⁄ US teaspoon ➔ grain ⁄ US fluid ounce1 g/tsp = 92.5941501 gr/fl.oz92.5941501 × ρg/tsp = ρgr/fl.oz
gram ⁄ US teaspoon ➔ grain ⁄ foot³1 g/tsp = 88 659.5 gr/ft³88 659.5 × ρg/tsp = ρgr/ft³
gram ⁄ US teaspoon ➔ grain ⁄ inch³1 g/tsp = 51.3075809 gr/in³51.3075809 × ρg/tsp = ρgr/in³
gram ⁄ US teaspoon ➔ grain ⁄ liter1 g/tsp = 3 130.98069 gr/l3 130.98069 × ρg/tsp = ρgr/l
gram ⁄ US teaspoon ➔ grain ⁄ meter³1 g/tsp = 3 130 980.69 gr/m³3 130 980.69 × ρg/tsp = ρgr/m³
gram ⁄ US teaspoon ➔ grain ⁄ metric cup1 g/tsp = 782.745173 gr/metric c782.745173 × ρg/tsp = ρgr/metric c
gram ⁄ US teaspoon ➔ grain ⁄ metric tablespoon1 g/tsp = 46.9647104 gr/metric tbsp46.9647104 × ρg/tsp = ρgr/metric tbsp
gram ⁄ US teaspoon ➔ grain ⁄ metric teaspoon1 g/tsp = 15.6549035 gr/metric tsp15.6549035 × ρg/tsp = ρgr/metric tsp
gram ⁄ US teaspoon ➔ grain ⁄ milliliter1 g/tsp = 3.13098069 gr/ml3.13098069 × ρg/tsp = ρgr/ml
gram ⁄ US teaspoon ➔ grain ⁄ millimeter³1 g/tsp = 0.003130980691 gr/mm³0.003130980691 × ρg/tsp = ρgr/mm³
gram ⁄ US teaspoon ➔ grain ⁄ US pint1 g/tsp = 1 481.5064 gr/pt1 481.5064 × ρg/tsp = ρgr/pt
gram ⁄ US teaspoon ➔ grain ⁄ US cup1 g/tsp = 740.753201 gr/US c740.753201 × ρg/tsp = ρgr/US c
gram ⁄ US teaspoon ➔ grain ⁄ US gallon1 g/tsp = 11 852.0512 gr/US gal11 852.0512 × ρg/tsp = ρgr/US gal
gram ⁄ US teaspoon ➔ grain ⁄ US quart1 g/tsp = 2 963.0128 gr/US qt2 963.0128 × ρg/tsp = ρgr/US qt
gram ⁄ US teaspoon ➔ grain ⁄ US tablespoon1 g/tsp = 46.2970751 gr/US tbsp46.2970751 × ρg/tsp = ρgr/US tbsp
gram ⁄ US teaspoon ➔ grain ⁄ US teaspoon1 g/tsp = 15.4323584 gr/US tsp15.4323584 × ρg/tsp = ρgr/US tsp
gram ⁄ US teaspoon ➔ grain ⁄ yard³1 g/tsp = 2 393 806.5 gr/yd³2 393 806.5 × ρg/tsp = ρgr/yd³
gram ⁄ US teaspoon ➔ kilogram ⁄ centimeter³1 g/tsp = 0.000202884136 kg/cm³0.000202884136 × ρg/tsp = ρkg/cm³
gram ⁄ US teaspoon ➔ kilogram ⁄ decimeter³1 g/tsp = 0.202884136 kg/dm³0.202884136 × ρg/tsp = ρkg/dm³
gram ⁄ US teaspoon ➔ kilogram ⁄ US fluid ounce1 g/tsp = 0.006 kg/fl.oz0.006 × ρg/tsp = ρkg/fl.oz
gram ⁄ US teaspoon ➔ kilogram ⁄ foot³1 g/tsp = 5.74503895 kg/ft³5.74503895 × ρg/tsp = ρkg/ft³
gram ⁄ US teaspoon ➔ kilogram ⁄ inch³1 g/tsp = 0.003324675321 kg/in³0.003324675321 × ρg/tsp = ρkg/in³
gram ⁄ US teaspoon ➔ kilogram ⁄ liter1 g/tsp = 0.202884136 kg/l0.202884136 × ρg/tsp = ρkg/l
gram ⁄ US teaspoon ➔ kilogram ⁄ meter³1 g/tsp = 202.884136 kg/m³202.884136 × ρg/tsp = ρkg/m³
gram ⁄ US teaspoon ➔ kilogram ⁄ metric cup1 g/tsp = 0.050721034 kg/metric c0.050721034 × ρg/tsp = ρkg/metric c
gram ⁄ US teaspoon ➔ kilogram ⁄ metric tablespoon1 g/tsp = 0.00304326204 kg/metric tbsp0.00304326204 × ρg/tsp = ρkg/metric tbsp
gram ⁄ US teaspoon ➔ kilogram ⁄ metric teaspoon1 g/tsp = 0.00101442068 kg/metric tsp0.00101442068 × ρg/tsp = ρkg/metric tsp
gram ⁄ US teaspoon ➔ kilogram ⁄ milliliter1 g/tsp = 0.000202884136 kg/ml0.000202884136 × ρg/tsp = ρkg/ml
gram ⁄ US teaspoon ➔ kilogram ⁄ millimeter³1 g/tsp = 2.02884136 × 10-7 kg/mm³2.02884136 × 10-7 × ρg/tsp = ρkg/mm³
gram ⁄ US teaspoon ➔ kilogram ⁄ US pint1 g/tsp = 0.096 kg/pt0.096 × ρg/tsp = ρkg/pt
gram ⁄ US teaspoon ➔ kilogram ⁄ US tablespoon1 g/tsp = 0.003 kg/tbsp0.003 × ρg/tsp = ρkg/tbsp
gram ⁄ US teaspoon ➔ kilogram ⁄ US teaspoon1 g/tsp = 0.001 kg/tsp0.001 × ρg/tsp = ρkg/tsp
gram ⁄ US teaspoon ➔ kilogram ⁄ US cup1 g/tsp = 0.048 kg/US c0.048 × ρg/tsp = ρkg/US c
gram ⁄ US teaspoon ➔ kilogram ⁄ US gallon1 g/tsp = 0.7680000002 kg/US gal0.7680000002 × ρg/tsp = ρkg/US gal
gram ⁄ US teaspoon ➔ kilogram ⁄ US quart1 g/tsp = 0.1920000001 kg/US qt0.1920000001 × ρg/tsp = ρkg/US qt
gram ⁄ US teaspoon ➔ kilogram ⁄ yard³1 g/tsp = 155.116052 kg/yd³155.116052 × ρg/tsp = ρkg/yd³
gram ⁄ US teaspoon ➔ pound ⁄ centimeter³1 g/tsp = 0.0004472829558 lb/cm³0.0004472829558 × ρg/tsp = ρlb/cm³
gram ⁄ US teaspoon ➔ pound ⁄ decimeter³1 g/tsp = 0.4472829558 lb/dm³0.4472829558 × ρg/tsp = ρlb/dm³
gram ⁄ US teaspoon ➔ pound ⁄ US fluid ounce1 g/tsp = 0.01322773573 lb/fl.oz0.01322773573 × ρg/tsp = ρlb/fl.oz
gram ⁄ US teaspoon ➔ pound ⁄ foot³1 g/tsp = 12.6656429 lb/ft³12.6656429 × ρg/tsp = ρlb/ft³
gram ⁄ US teaspoon ➔ pound ⁄ inch³1 g/tsp = 0.00732965443 lb/in³0.00732965443 × ρg/tsp = ρlb/in³
gram ⁄ US teaspoon ➔ pound ⁄ liter1 g/tsp = 0.4472829558 lb/l0.4472829558 × ρg/tsp = ρlb/l
gram ⁄ US teaspoon ➔ pound ⁄ meter³1 g/tsp = 447.282956 lb/m³447.282956 × ρg/tsp = ρlb/m³
gram ⁄ US teaspoon ➔ pound ⁄ metric cup1 g/tsp = 0.111820739 lb/metric c0.111820739 × ρg/tsp = ρlb/metric c
gram ⁄ US teaspoon ➔ pound ⁄ metric tablespoon1 g/tsp = 0.006709244337 lb/metric tbsp0.006709244337 × ρg/tsp = ρlb/metric tbsp
gram ⁄ US teaspoon ➔ pound ⁄ metric teaspoon1 g/tsp = 0.002236414779 lb/metric tsp0.002236414779 × ρg/tsp = ρlb/metric tsp
gram ⁄ US teaspoon ➔ pound ⁄ milliliter1 g/tsp = 0.0004472829558 lb/ml0.0004472829558 × ρg/tsp = ρlb/ml
gram ⁄ US teaspoon ➔ pound ⁄ millimeter³1 g/tsp = 4.472829558 × 10-7 lb/mm³4.472829558 × 10-7 × ρg/tsp = ρlb/mm³
gram ⁄ US teaspoon ➔ pound ⁄ US pint1 g/tsp = 0.211643772 lb/pt0.211643772 × ρg/tsp = ρlb/pt
gram ⁄ US teaspoon ➔ pound ⁄ US tablespoon1 g/tsp = 0.00661386787 lb/tbsp0.00661386787 × ρg/tsp = ρlb/tbsp
gram ⁄ US teaspoon ➔ pound ⁄ US teaspoon1 g/tsp = 0.00220462262 lb/tsp0.00220462262 × ρg/tsp = ρlb/tsp
gram ⁄ US teaspoon ➔ pound ⁄ US cup1 g/tsp = 0.105821886 lb/US c0.105821886 × ρg/tsp = ρlb/US c
gram ⁄ US teaspoon ➔ pound ⁄ US gallon1 g/tsp = 1.69315017 lb/US gal1.69315017 × ρg/tsp = ρlb/US gal
gram ⁄ US teaspoon ➔ pound ⁄ US quart1 g/tsp = 0.4232875434 lb/US qt0.4232875434 × ρg/tsp = ρlb/US qt
gram ⁄ US teaspoon ➔ pound ⁄ yard³1 g/tsp = 341.972358 lb/yd³341.972358 × ρg/tsp = ρlb/yd³
gram ⁄ US teaspoon ➔ long ton ⁄ centimeter³1 g/tsp = 1.99679891 × 10-7 long tn/cm³1.99679891 × 10-7 × ρg/tsp = ρlong tn/cm³
gram ⁄ US teaspoon ➔ long ton ⁄ decimeter³1 g/tsp = 0.000199679891 long tn/dm³0.000199679891 × ρg/tsp = ρlong tn/dm³
gram ⁄ US teaspoon ➔ long ton ⁄ US fluid ounce1 g/tsp = 5.905239165 × 10-6 long tn/fl.oz5.905239165 × 10-6 × ρg/tsp = ρlong tn/fl.oz
gram ⁄ US teaspoon ➔ long ton ⁄ foot³1 g/tsp = 0.005654304866 long tn/ft³0.005654304866 × ρg/tsp = ρlong tn/ft³
gram ⁄ US teaspoon ➔ long ton ⁄ inch³1 g/tsp = 3.272167156 × 10-6 long tn/in³3.272167156 × 10-6 × ρg/tsp = ρlong tn/in³
gram ⁄ US teaspoon ➔ long ton ⁄ liter1 g/tsp = 0.000199679891 long tn/l0.000199679891 × ρg/tsp = ρlong tn/l
gram ⁄ US teaspoon ➔ long ton ⁄ meter³1 g/tsp = 0.1996798911 long tn/m³0.1996798911 × ρg/tsp = ρlong tn/m³
gram ⁄ US teaspoon ➔ long ton ⁄ metric cup1 g/tsp = 4.991997277 × 10-5 long tn/metric c4.991997277 × 10-5 × ρg/tsp = ρlong tn/metric c
gram ⁄ US teaspoon ➔ long ton ⁄ metric tablespoon1 g/tsp = 2.995198365 × 10-6 long tn/metric tbsp2.995198365 × 10-6 × ρg/tsp = ρlong tn/metric tbsp
gram ⁄ US teaspoon ➔ long ton ⁄ metric teaspoon1 g/tsp = 9.983994549 × 10-7 long tn/metric tsp9.983994549 × 10-7 × ρg/tsp = ρlong tn/metric tsp
gram ⁄ US teaspoon ➔ long ton ⁄ milliliter1 g/tsp = 1.99679891 × 10-7 long tn/ml1.99679891 × 10-7 × ρg/tsp = ρlong tn/ml
gram ⁄ US teaspoon ➔ long ton ⁄ millimeter³1 g/tsp = 1.99679890982 × 10-10 long tn/mm³1.99679890982 × 10-10 × ρg/tsp = ρlong tn/mm³
gram ⁄ US teaspoon ➔ long ton ⁄ US pint1 g/tsp = 9.448382679 × 10-5 long tn/pt9.448382679 × 10-5 × ρg/tsp = ρlong tn/pt
gram ⁄ US teaspoon ➔ long ton ⁄ US cup1 g/tsp = 4.724191339 × 10-5 long tn/US c4.724191339 × 10-5 × ρg/tsp = ρlong tn/US c
gram ⁄ US teaspoon ➔ long ton ⁄ US gallon1 g/tsp = 0.0007558706116 long tn/US gal0.0007558706116 × ρg/tsp = ρlong tn/US gal
gram ⁄ US teaspoon ➔ long ton ⁄ US quart1 g/tsp = 0.0001889676533 long tn/US qt0.0001889676533 × ρg/tsp = ρlong tn/US qt
gram ⁄ US teaspoon ➔ long ton ⁄ US tablespoon1 g/tsp = 2.952619585 × 10-6 long tn/US tbsp2.952619585 × 10-6 × ρg/tsp = ρlong tn/US tbsp
gram ⁄ US teaspoon ➔ long ton ⁄ US teaspoon1 g/tsp = 9.842065268 × 10-7 long tn/US tsp9.842065268 × 10-7 × ρg/tsp = ρlong tn/US tsp
gram ⁄ US teaspoon ➔ long ton ⁄ yard³1 g/tsp = 0.1526662313 long tn/yd³0.1526662313 × ρg/tsp = ρlong tn/yd³
gram ⁄ US teaspoon ➔ milligram ⁄ centimeter³1 g/tsp = 202.884136 mg/cm³202.884136 × ρg/tsp = ρmg/cm³
gram ⁄ US teaspoon ➔ milligram ⁄ decimeter³1 g/tsp = 202 884.136 mg/dm³202 884.136 × ρg/tsp = ρmg/dm³
gram ⁄ US teaspoon ➔ milligram ⁄ US fluid ounce1 g/tsp = 6 000 mg/fl.oz6 000 × ρg/tsp = ρmg/fl.oz
gram ⁄ US teaspoon ➔ milligram ⁄ foot³1 g/tsp = 5 745 038.96 mg/ft³5 745 038.96 × ρg/tsp = ρmg/ft³
gram ⁄ US teaspoon ➔ milligram ⁄ inch³1 g/tsp = 3 324.67532 mg/in³3 324.67532 × ρg/tsp = ρmg/in³
gram ⁄ US teaspoon ➔ milligram ⁄ liter1 g/tsp = 202 884.136 mg/l202 884.136 × ρg/tsp = ρmg/l
gram ⁄ US teaspoon ➔ milligram ⁄ meter³1 g/tsp = 202 884 136 mg/m³202 884 136 × ρg/tsp = ρmg/m³
gram ⁄ US teaspoon ➔ milligram ⁄ metric cup1 g/tsp = 50 721.034 mg/metric c50 721.034 × ρg/tsp = ρmg/metric c
gram ⁄ US teaspoon ➔ milligram ⁄ metric tablespoon1 g/tsp = 3 043.26204 mg/metric tbsp3 043.26204 × ρg/tsp = ρmg/metric tbsp
gram ⁄ US teaspoon ➔ milligram ⁄ metric teaspoon1 g/tsp = 1 014.42068 mg/metric tsp1 014.42068 × ρg/tsp = ρmg/metric tsp
gram ⁄ US teaspoon ➔ milligram ⁄ milliliter1 g/tsp = 202.884136 mg/ml202.884136 × ρg/tsp = ρmg/ml
gram ⁄ US teaspoon ➔ milligram ⁄ millimeter³1 g/tsp = 0.202884136 mg/mm³0.202884136 × ρg/tsp = ρmg/mm³
gram ⁄ US teaspoon ➔ milligram ⁄ US pint1 g/tsp = 96 000 mg/pt96 000 × ρg/tsp = ρmg/pt
gram ⁄ US teaspoon ➔ milligram ⁄ US tablespoon1 g/tsp = 3 000 mg/tbsp3 000 × ρg/tsp = ρmg/tbsp
gram ⁄ US teaspoon ➔ milligram ⁄ US teaspoon1 g/tsp = 1 000 mg/tsp1 000 × ρg/tsp = ρmg/tsp
gram ⁄ US teaspoon ➔ milligram ⁄ US cup1 g/tsp = 48 000 mg/US c48 000 × ρg/tsp = ρmg/US c
gram ⁄ US teaspoon ➔ milligram ⁄ US gallon1 g/tsp = 768 000 mg/US gal768 000 × ρg/tsp = ρmg/US gal
gram ⁄ US teaspoon ➔ milligram ⁄ US quart1 g/tsp = 192 000 mg/US qt192 000 × ρg/tsp = ρmg/US qt
gram ⁄ US teaspoon ➔ milligram ⁄ yard³1 g/tsp = 155 116 052 mg/yd³155 116 052 × ρg/tsp = ρmg/yd³
gram ⁄ US teaspoon ➔ ounce ⁄ centimeter³1 g/tsp = 0.0071565273 oz/cm³0.0071565273 × ρg/tsp = ρoz/cm³
gram ⁄ US teaspoon ➔ ounce ⁄ decimeter³1 g/tsp = 7.1565273 oz/dm³7.1565273 × ρg/tsp = ρoz/dm³
gram ⁄ US teaspoon ➔ ounce ⁄ US fluid ounce1 g/tsp = 0.2116437717 oz/fl.oz0.2116437717 × ρg/tsp = ρoz/fl.oz
gram ⁄ US teaspoon ➔ ounce ⁄ foot³1 g/tsp = 202.650286 oz/ft³202.650286 × ρg/tsp = ρoz/ft³
gram ⁄ US teaspoon ➔ ounce ⁄ inch³1 g/tsp = 0.1172744709 oz/in³0.1172744709 × ρg/tsp = ρoz/in³
gram ⁄ US teaspoon ➔ ounce ⁄ liter1 g/tsp = 7.1565273 oz/l7.1565273 × ρg/tsp = ρoz/l
gram ⁄ US teaspoon ➔ ounce ⁄ meter³1 g/tsp = 7 156.5273 oz/m³7 156.5273 × ρg/tsp = ρoz/m³
gram ⁄ US teaspoon ➔ ounce ⁄ metric cup1 g/tsp = 1.78913183 oz/metric c1.78913183 × ρg/tsp = ρoz/metric c
gram ⁄ US teaspoon ➔ ounce ⁄ metric tablespoon1 g/tsp = 0.1073479095 oz/metric tbsp0.1073479095 × ρg/tsp = ρoz/metric tbsp
gram ⁄ US teaspoon ➔ ounce ⁄ metric teaspoon1 g/tsp = 0.0357826365 oz/metric tsp0.0357826365 × ρg/tsp = ρoz/metric tsp
gram ⁄ US teaspoon ➔ ounce ⁄ milliliter1 g/tsp = 0.0071565273 oz/ml0.0071565273 × ρg/tsp = ρoz/ml
gram ⁄ US teaspoon ➔ ounce ⁄ millimeter³1 g/tsp = 7.1565273 × 10-6 oz/mm³7.1565273 × 10-6 × ρg/tsp = ρoz/mm³
gram ⁄ US teaspoon ➔ ounce ⁄ US pint1 g/tsp = 3.38630035 oz/pt3.38630035 × ρg/tsp = ρoz/pt
gram ⁄ US teaspoon ➔ ounce ⁄ US tablespoon1 g/tsp = 0.1058218859 oz/tbsp0.1058218859 × ρg/tsp = ρoz/tbsp
gram ⁄ US teaspoon ➔ ounce ⁄ US teaspoon1 g/tsp = 0.03527396192 oz/tsp0.03527396192 × ρg/tsp = ρoz/tsp
gram ⁄ US teaspoon ➔ ounce ⁄ US cup1 g/tsp = 1.69315018 oz/US c1.69315018 × ρg/tsp = ρoz/US c
gram ⁄ US teaspoon ➔ ounce ⁄ US gallon1 g/tsp = 27.0904027 oz/US gal27.0904027 × ρg/tsp = ρoz/US gal
gram ⁄ US teaspoon ➔ ounce ⁄ US quart1 g/tsp = 6.77260069 oz/US qt6.77260069 × ρg/tsp = ρoz/US qt
gram ⁄ US teaspoon ➔ ounce ⁄ yard³1 g/tsp = 5 471.55772 oz/yd³5 471.55772 × ρg/tsp = ρoz/yd³
gram ⁄ US teaspoon ➔ troy ounce ⁄ centimeter³1 g/tsp = 0.006522876438 oz t/cm³0.006522876438 × ρg/tsp = ρoz t/cm³
gram ⁄ US teaspoon ➔ troy ounce ⁄ decimeter³1 g/tsp = 6.52287644 oz t/dm³6.52287644 × ρg/tsp = ρoz t/dm³
gram ⁄ US teaspoon ➔ troy ounce ⁄ US fluid ounce1 g/tsp = 0.1929044794 oz t/fl.oz0.1929044794 × ρg/tsp = ρoz t/fl.oz
gram ⁄ US teaspoon ➔ troy ounce ⁄ foot³1 g/tsp = 184.707292 oz t/ft³184.707292 × ρg/tsp = ρoz t/ft³
gram ⁄ US teaspoon ➔ troy ounce ⁄ inch³1 g/tsp = 0.1068907935 oz t/in³0.1068907935 × ρg/tsp = ρoz t/in³
gram ⁄ US teaspoon ➔ troy ounce ⁄ liter1 g/tsp = 6.52287644 oz t/l6.52287644 × ρg/tsp = ρoz t/l
gram ⁄ US teaspoon ➔ troy ounce ⁄ meter³1 g/tsp = 6 522.87644 oz t/m³6 522.87644 × ρg/tsp = ρoz t/m³
gram ⁄ US teaspoon ➔ troy ounce ⁄ metric cup1 g/tsp = 1.63071911 oz t/metric c1.63071911 × ρg/tsp = ρoz t/metric c
gram ⁄ US teaspoon ➔ troy ounce ⁄ metric tablespoon1 g/tsp = 0.09784314667 oz t/metric tbsp0.09784314667 × ρg/tsp = ρoz t/metric tbsp
gram ⁄ US teaspoon ➔ troy ounce ⁄ metric teaspoon1 g/tsp = 0.03261438229 oz t/metric tsp0.03261438229 × ρg/tsp = ρoz t/metric tsp
gram ⁄ US teaspoon ➔ troy ounce ⁄ milliliter1 g/tsp = 0.006522876438 oz t/ml0.006522876438 × ρg/tsp = ρoz t/ml
gram ⁄ US teaspoon ➔ troy ounce ⁄ millimeter³1 g/tsp = 6.52287644 × 10-6 oz t/mm³6.52287644 × 10-6 × ρg/tsp = ρoz t/mm³
gram ⁄ US teaspoon ➔ troy ounce ⁄ US pint1 g/tsp = 3.08647167 oz t/pt3.08647167 × ρg/tsp = ρoz t/pt
gram ⁄ US teaspoon ➔ troy ounce ⁄ US cup1 g/tsp = 1.54323584 oz t/US c1.54323584 × ρg/tsp = ρoz t/US c
gram ⁄ US teaspoon ➔ troy ounce ⁄ US gallon1 g/tsp = 24.6917733 oz t/US gal24.6917733 × ρg/tsp = ρoz t/US gal
gram ⁄ US teaspoon ➔ troy ounce ⁄ US quart1 g/tsp = 6.17294333 oz t/US qt6.17294333 × ρg/tsp = ρoz t/US qt
gram ⁄ US teaspoon ➔ troy ounce ⁄ US tablespoon1 g/tsp = 0.09645223979 oz t/US tbsp0.09645223979 × ρg/tsp = ρoz t/US tbsp
gram ⁄ US teaspoon ➔ troy ounce ⁄ US teaspoon1 g/tsp = 0.03215074667 oz t/US tsp0.03215074667 × ρg/tsp = ρoz t/US tsp
gram ⁄ US teaspoon ➔ troy ounce ⁄ yard³1 g/tsp = 4 987.09688 oz t/yd³4 987.09688 × ρg/tsp = ρoz t/yd³
gram ⁄ US teaspoon ➔ short ton ⁄ centimeter³1 g/tsp = 2.236414779 × 10-7 short tn/cm³2.236414779 × 10-7 × ρg/tsp = ρshort tn/cm³
gram ⁄ US teaspoon ➔ short ton ⁄ decimeter³1 g/tsp = 0.0002236414779 short tn/dm³0.0002236414779 × ρg/tsp = ρshort tn/dm³
gram ⁄ US teaspoon ➔ short ton ⁄ US fluid ounce1 g/tsp = 6.613867865 × 10-6 short tn/fl.oz6.613867865 × 10-6 × ρg/tsp = ρshort tn/fl.oz
gram ⁄ US teaspoon ➔ short ton ⁄ foot³1 g/tsp = 0.00633282145 short tn/ft³0.00633282145 × ρg/tsp = ρshort tn/ft³
gram ⁄ US teaspoon ➔ short ton ⁄ inch³1 g/tsp = 3.664827215 × 10-6 short tn/in³3.664827215 × 10-6 × ρg/tsp = ρshort tn/in³
gram ⁄ US teaspoon ➔ short ton ⁄ liter1 g/tsp = 0.0002236414779 short tn/l0.0002236414779 × ρg/tsp = ρshort tn/l
gram ⁄ US teaspoon ➔ short ton ⁄ meter³1 g/tsp = 0.223641478 short tn/m³0.223641478 × ρg/tsp = ρshort tn/m³
gram ⁄ US teaspoon ➔ short ton ⁄ metric cup1 g/tsp = 5.59103695 × 10-5 short tn/metric c5.59103695 × 10-5 × ρg/tsp = ρshort tn/metric c
gram ⁄ US teaspoon ➔ short ton ⁄ metric tablespoon1 g/tsp = 3.354622168 × 10-6 short tn/metric tbsp3.354622168 × 10-6 × ρg/tsp = ρshort tn/metric tbsp
gram ⁄ US teaspoon ➔ short ton ⁄ metric teaspoon1 g/tsp = 1.11820739 × 10-6 short tn/metric tsp1.11820739 × 10-6 × ρg/tsp = ρshort tn/metric tsp
gram ⁄ US teaspoon ➔ short ton ⁄ milliliter1 g/tsp = 2.236414779 × 10-7 short tn/ml2.236414779 × 10-7 × ρg/tsp = ρshort tn/ml
gram ⁄ US teaspoon ➔ short ton ⁄ millimeter³1 g/tsp = 2.236414779 × 10-10 short tn/mm³2.236414779 × 10-10 × ρg/tsp = ρshort tn/mm³
gram ⁄ US teaspoon ➔ short ton ⁄ US pint1 g/tsp = 0.000105821886 short tn/pt0.000105821886 × ρg/tsp = ρshort tn/pt
gram ⁄ US teaspoon ➔ short ton ⁄ US cup1 g/tsp = 5.2910943 × 10-5 short tn/US c5.2910943 × 10-5 × ρg/tsp = ρshort tn/US c
gram ⁄ US teaspoon ➔ short ton ⁄ US gallon1 g/tsp = 0.000846575085 short tn/US gal0.000846575085 × ρg/tsp = ρshort tn/US gal
gram ⁄ US teaspoon ➔ short ton ⁄ US quart1 g/tsp = 0.0002116437717 short tn/US qt0.0002116437717 × ρg/tsp = ρshort tn/US qt
gram ⁄ US teaspoon ➔ short ton ⁄ US tablespoon1 g/tsp = 3.306933935 × 10-6 short tn/US tbsp3.306933935 × 10-6 × ρg/tsp = ρshort tn/US tbsp
gram ⁄ US teaspoon ➔ short ton ⁄ US teaspoon1 g/tsp = 1.10231131 × 10-6 short tn/US tsp1.10231131 × 10-6 × ρg/tsp = ρshort tn/US tsp
gram ⁄ US teaspoon ➔ short ton ⁄ yard³1 g/tsp = 0.170986179 short tn/yd³0.170986179 × ρg/tsp = ρshort tn/yd³
gram ⁄ US teaspoon ➔ slug ⁄ centimeter³1 g/tsp = 1.390197927 × 10-5 sl/cm³1.390197927 × 10-5 × ρg/tsp = ρsl/cm³
gram ⁄ US teaspoon ➔ slug ⁄ decimeter³1 g/tsp = 0.01390197927 sl/dm³0.01390197927 × ρg/tsp = ρsl/dm³
gram ⁄ US teaspoon ➔ slug ⁄ US fluid ounce1 g/tsp = 0.0004111305962 sl/fl.oz0.0004111305962 × ρg/tsp = ρsl/fl.oz
gram ⁄ US teaspoon ➔ slug ⁄ foot³1 g/tsp = 0.393660215 sl/ft³0.393660215 × ρg/tsp = ρsl/ft³
gram ⁄ US teaspoon ➔ slug ⁄ inch³1 g/tsp = 0.000227812624 sl/in³0.000227812624 × ρg/tsp = ρsl/in³
gram ⁄ US teaspoon ➔ slug ⁄ liter1 g/tsp = 0.01390197927 sl/l0.01390197927 × ρg/tsp = ρsl/l
gram ⁄ US teaspoon ➔ slug ⁄ meter³1 g/tsp = 13.9019793 sl/m³13.9019793 × ρg/tsp = ρsl/m³
gram ⁄ US teaspoon ➔ slug ⁄ metric cup1 g/tsp = 0.003475494818 sl/metric c0.003475494818 × ρg/tsp = ρsl/metric c
gram ⁄ US teaspoon ➔ slug ⁄ metric tablespoon1 g/tsp = 0.0002085296891 sl/metric tbsp0.0002085296891 × ρg/tsp = ρsl/metric tbsp
gram ⁄ US teaspoon ➔ slug ⁄ metric teaspoon1 g/tsp = 6.950989635 × 10-5 sl/metric tsp6.950989635 × 10-5 × ρg/tsp = ρsl/metric tsp
gram ⁄ US teaspoon ➔ slug ⁄ milliliter1 g/tsp = 1.390197927 × 10-5 sl/ml1.390197927 × 10-5 × ρg/tsp = ρsl/ml
gram ⁄ US teaspoon ➔ slug ⁄ millimeter³1 g/tsp = 1.390197927 × 10-8 sl/mm³1.390197927 × 10-8 × ρg/tsp = ρsl/mm³
gram ⁄ US teaspoon ➔ slug ⁄ US pint1 g/tsp = 0.00657808952 sl/pt0.00657808952 × ρg/tsp = ρsl/pt
gram ⁄ US teaspoon ➔ slug ⁄ US tablespoon1 g/tsp = 0.000205565298 sl/tbsp0.000205565298 × ρg/tsp = ρsl/tbsp
gram ⁄ US teaspoon ➔ slug ⁄ US teaspoon1 g/tsp = 6.85217659 × 10-5 sl/tsp6.85217659 × 10-5 × ρg/tsp = ρsl/tsp
gram ⁄ US teaspoon ➔ slug ⁄ US cup1 g/tsp = 0.00328904476 sl/US c0.00328904476 × ρg/tsp = ρsl/US c
gram ⁄ US teaspoon ➔ slug ⁄ US gallon1 g/tsp = 0.05262471631 sl/US gal0.05262471631 × ρg/tsp = ρsl/US gal
gram ⁄ US teaspoon ➔ slug ⁄ US quart1 g/tsp = 0.01315617908 sl/US qt0.01315617908 × ρg/tsp = ρsl/US qt
gram ⁄ US teaspoon ➔ slug ⁄ yard³1 g/tsp = 10.6288258 sl/yd³10.6288258 × ρg/tsp = ρsl/yd³
gram ⁄ US teaspoon ➔ stone ⁄ centimeter³1 g/tsp = 3.194878256 × 10-5 st/cm³3.194878256 × 10-5 × ρg/tsp = ρst/cm³
gram ⁄ US teaspoon ➔ stone ⁄ decimeter³1 g/tsp = 0.03194878256 st/dm³0.03194878256 × ρg/tsp = ρst/dm³
gram ⁄ US teaspoon ➔ stone ⁄ US fluid ounce1 g/tsp = 0.0009448382664 st/fl.oz0.0009448382664 × ρg/tsp = ρst/fl.oz
gram ⁄ US teaspoon ➔ stone ⁄ foot³1 g/tsp = 0.9046887786 st/ft³0.9046887786 × ρg/tsp = ρst/ft³
gram ⁄ US teaspoon ➔ stone ⁄ inch³1 g/tsp = 0.000523546745 st/in³0.000523546745 × ρg/tsp = ρst/in³
gram ⁄ US teaspoon ➔ stone ⁄ liter1 g/tsp = 0.03194878256 st/l0.03194878256 × ρg/tsp = ρst/l
gram ⁄ US teaspoon ➔ stone ⁄ meter³1 g/tsp = 31.9487826 st/m³31.9487826 × ρg/tsp = ρst/m³
gram ⁄ US teaspoon ➔ stone ⁄ metric cup1 g/tsp = 0.007987195643 st/metric c0.007987195643 × ρg/tsp = ρst/metric c
gram ⁄ US teaspoon ➔ stone ⁄ metric tablespoon1 g/tsp = 0.0004792317384 st/metric tbsp0.0004792317384 × ρg/tsp = ρst/metric tbsp
gram ⁄ US teaspoon ➔ stone ⁄ metric teaspoon1 g/tsp = 0.0001597439128 st/metric tsp0.0001597439128 × ρg/tsp = ρst/metric tsp
gram ⁄ US teaspoon ➔ stone ⁄ milliliter1 g/tsp = 3.194878256 × 10-5 st/ml3.194878256 × 10-5 × ρg/tsp = ρst/ml
gram ⁄ US teaspoon ➔ stone ⁄ millimeter³1 g/tsp = 3.194878256 × 10-8 st/mm³3.194878256 × 10-8 × ρg/tsp = ρst/mm³
gram ⁄ US teaspoon ➔ stone ⁄ US pint1 g/tsp = 0.01511741229 st/pt0.01511741229 × ρg/tsp = ρst/pt
gram ⁄ US teaspoon ➔ stone ⁄ US cup1 g/tsp = 0.007558706143 st/US c0.007558706143 × ρg/tsp = ρst/US c
gram ⁄ US teaspoon ➔ stone ⁄ US gallon1 g/tsp = 0.1209392979 st/US gal0.1209392979 × ρg/tsp = ρst/US gal
gram ⁄ US teaspoon ➔ stone ⁄ US quart1 g/tsp = 0.03023482453 st/US qt0.03023482453 × ρg/tsp = ρst/US qt
gram ⁄ US teaspoon ➔ stone ⁄ US tablespoon1 g/tsp = 0.0004724191336 st/US tbsp0.0004724191336 × ρg/tsp = ρst/US tbsp
gram ⁄ US teaspoon ➔ stone ⁄ US teaspoon1 g/tsp = 0.0001574730443 st/US tsp0.0001574730443 × ρg/tsp = ρst/US tsp
gram ⁄ US teaspoon ➔ stone ⁄ yard³1 g/tsp = 24.426597 st/yd³24.426597 × ρg/tsp = ρst/yd³
gram ⁄ US teaspoon ➔ tonne ⁄ centimeter³1 g/tsp = 2.02884136 × 10-7 t/cm³2.02884136 × 10-7 × ρg/tsp = ρt/cm³
gram ⁄ US teaspoon ➔ tonne ⁄ decimeter³1 g/tsp = 0.000202884136 t/dm³0.000202884136 × ρg/tsp = ρt/dm³
gram ⁄ US teaspoon ➔ tonne ⁄ US fluid ounce1 g/tsp = 6.0 × 10-6 t/fl.oz6.0 × 10-6 × ρg/tsp = ρt/fl.oz
gram ⁄ US teaspoon ➔ tonne ⁄ foot³1 g/tsp = 0.00574503895 t/ft³0.00574503895 × ρg/tsp = ρt/ft³
gram ⁄ US teaspoon ➔ tonne ⁄ inch³1 g/tsp = 3.324675321 × 10-6 t/in³3.324675321 × 10-6 × ρg/tsp = ρt/in³
gram ⁄ US teaspoon ➔ tonne ⁄ liter1 g/tsp = 0.000202884136 t/l0.000202884136 × ρg/tsp = ρt/l
gram ⁄ US teaspoon ➔ tonne ⁄ meter³1 g/tsp = 0.202884136 t/m³0.202884136 × ρg/tsp = ρt/m³
gram ⁄ US teaspoon ➔ tonne ⁄ metric cup1 g/tsp = 5.0721034 × 10-5 t/metric c5.0721034 × 10-5 × ρg/tsp = ρt/metric c
gram ⁄ US teaspoon ➔ tonne ⁄ metric tablespoon1 g/tsp = 3.04326204 × 10-6 t/metric tbsp3.04326204 × 10-6 × ρg/tsp = ρt/metric tbsp
gram ⁄ US teaspoon ➔ tonne ⁄ metric teaspoon1 g/tsp = 1.01442068 × 10-6 t/metric tsp1.01442068 × 10-6 × ρg/tsp = ρt/metric tsp
gram ⁄ US teaspoon ➔ tonne ⁄ milliliter1 g/tsp = 2.02884136 × 10-7 t/ml2.02884136 × 10-7 × ρg/tsp = ρt/ml
gram ⁄ US teaspoon ➔ tonne ⁄ millimeter³1 g/tsp = 2.02884136 × 10-10 t/mm³2.02884136 × 10-10 × ρg/tsp = ρt/mm³
gram ⁄ US teaspoon ➔ tonne ⁄ US pint1 g/tsp = 9.6 × 10-5 t/pt9.6 × 10-5 × ρg/tsp = ρt/pt
gram ⁄ US teaspoon ➔ tonne ⁄ US tablespoon1 g/tsp = 3.0 × 10-6 t/tbsp3.0 × 10-6 × ρg/tsp = ρt/tbsp
gram ⁄ US teaspoon ➔ tonne ⁄ US teaspoon1 g/tsp = 1.0 × 10-6 t/tsp1.0 × 10-6 × ρg/tsp = ρt/tsp
gram ⁄ US teaspoon ➔ tonne ⁄ US cup1 g/tsp = 4.8 × 10-5 t/US c4.8 × 10-5 × ρg/tsp = ρt/US c
gram ⁄ US teaspoon ➔ tonne ⁄ US gallon1 g/tsp = 0.0007680000002 t/US gal0.0007680000002 × ρg/tsp = ρt/US gal
gram ⁄ US teaspoon ➔ tonne ⁄ US quart1 g/tsp = 0.0001920000001 t/US qt0.0001920000001 × ρg/tsp = ρt/US qt
gram ⁄ US teaspoon ➔ tonne ⁄ yard³1 g/tsp = 0.1551160517 t/yd³0.1551160517 × ρg/tsp = ρt/yd³
gram ⁄ US teaspoon ➔ troy pound ⁄ centimeter³1 g/tsp = 0.0005435730365 troy/cm³0.0005435730365 × ρg/tsp = ρtroy/cm³
gram ⁄ US teaspoon ➔ troy pound ⁄ decimeter³1 g/tsp = 0.5435730365 troy/dm³0.5435730365 × ρg/tsp = ρtroy/dm³
gram ⁄ US teaspoon ➔ troy pound ⁄ US fluid ounce1 g/tsp = 0.01607537328 troy/fl.oz0.01607537328 × ρg/tsp = ρtroy/fl.oz
gram ⁄ US teaspoon ➔ troy pound ⁄ foot³1 g/tsp = 15.3922743 troy/ft³15.3922743 × ρg/tsp = ρtroy/ft³
gram ⁄ US teaspoon ➔ troy pound ⁄ inch³1 g/tsp = 0.008907566128 troy/in³0.008907566128 × ρg/tsp = ρtroy/in³
gram ⁄ US teaspoon ➔ troy pound ⁄ liter1 g/tsp = 0.5435730365 troy/l0.5435730365 × ρg/tsp = ρtroy/l
gram ⁄ US teaspoon ➔ troy pound ⁄ meter³1 g/tsp = 543.573036 troy/m³543.573036 × ρg/tsp = ρtroy/m³
gram ⁄ US teaspoon ➔ troy pound ⁄ metric cup1 g/tsp = 0.1358932592 troy/metric c0.1358932592 × ρg/tsp = ρtroy/metric c
gram ⁄ US teaspoon ➔ troy pound ⁄ metric tablespoon1 g/tsp = 0.008153595556 troy/metric tbsp0.008153595556 × ρg/tsp = ρtroy/metric tbsp
gram ⁄ US teaspoon ➔ troy pound ⁄ metric teaspoon1 g/tsp = 0.002717865191 troy/metric tsp0.002717865191 × ρg/tsp = ρtroy/metric tsp
gram ⁄ US teaspoon ➔ troy pound ⁄ milliliter1 g/tsp = 0.0005435730365 troy/ml0.0005435730365 × ρg/tsp = ρtroy/ml
gram ⁄ US teaspoon ➔ troy pound ⁄ millimeter³1 g/tsp = 5.435730366 × 10-7 troy/mm³5.435730366 × 10-7 × ρg/tsp = ρtroy/mm³
gram ⁄ US teaspoon ➔ troy pound ⁄ US pint1 g/tsp = 0.2572059722 troy/pt0.2572059722 × ρg/tsp = ρtroy/pt
gram ⁄ US teaspoon ➔ troy pound ⁄ US cup1 g/tsp = 0.1286029863 troy/US c0.1286029863 × ρg/tsp = ρtroy/US c
gram ⁄ US teaspoon ➔ troy pound ⁄ US gallon1 g/tsp = 2.05764778 troy/US gal2.05764778 × ρg/tsp = ρtroy/US gal
gram ⁄ US teaspoon ➔ troy pound ⁄ US quart1 g/tsp = 0.5144119444 troy/US qt0.5144119444 × ρg/tsp = ρtroy/US qt
gram ⁄ US teaspoon ➔ troy pound ⁄ US tablespoon1 g/tsp = 0.008037686649 troy/US tbsp0.008037686649 × ρg/tsp = ρtroy/US tbsp
gram ⁄ US teaspoon ➔ troy pound ⁄ US teaspoon1 g/tsp = 0.002679228889 troy/US tsp0.002679228889 × ρg/tsp = ρtroy/US tsp
gram ⁄ US teaspoon ➔ troy pound ⁄ yard³1 g/tsp = 415.591406 troy/yd³415.591406 × ρg/tsp = ρtroy/yd³
What is a gram per US teaspoon (g/tsp)?
#### Foods, Nutrients and Calories
LARGE COOKED SHRIMP, UPC: 078742258447 contain(s) 71 calories per 100 grams (≈3.53 ounces) [ price ]
83 foods that contain Stigmasterol. List of these foods starting with the highest contents of Stigmasterol and the lowest contents of Stigmasterol
#### Gravels, Substances and Oils
CaribSea, Freshwater, Eco-Complete Cichlid, Ivory Coast weighs 1 281.48 kg/m³ (80.00018 lb/ft³) with specific gravity of 1.28148 relative to pure water. Calculate how much of this gravel is required to attain a specific depth in a cylindricalquarter cylindrical or in a rectangular shaped aquarium or pond [ weight to volume | volume to weight | price ]
Cobalt octacarbonyl [Co2(CO)8] weighs 1 780 kg/m³ (111.12177 lb/ft³) [ weight to volume | volume to weight | price | mole to volume and weight | mass and molar concentration | density ]
Volume to weightweight to volume and cost conversions for Peanut oil with temperature in the range of 10°C (50°F) to 140°C (284°F)
#### Weights and Measurements
The millisievert beta-radiation [mSv β] is a derived unit of ionizing radiation dose in the International System of Units (SI) and is a measure of the effective biological damage of low levels of ionizing radiation on the human body caused by exposure to beta (β) radiation in thousandths of a sievert.
Acceleration (a) of an object measures the object's change in velocity (v) per unit of time (t): a = v / t.
centner/ha to short tn/pm² conversion table, centner/ha to short tn/pm² unit converter or convert between all units of surface density measurement.
#### Calculators
Online Food Calories and Nutrients Calculator | 16,263 | 32,227 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.5625 | 3 | CC-MAIN-2021-31 | longest | en | 0.287535 |
https://techwhiff.com/learn/brief-exercise-18-06-determine-the-missing/438912 | 1,669,849,395,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710777.20/warc/CC-MAIN-20221130225142-20221201015142-00542.warc.gz | 581,857,975 | 13,141 | # Brief Exercise 18-06 Determine the missing amounts. Unit Selling Price Unit Variable Costs Unit Contribution Margin...
###### Question:
Brief Exercise 18-06
Determine the missing amounts.
##### How do you factor the expression 2a ^5 - 32ab ^8?
How do you factor the expression 2a ^5 - 32ab ^8?...
##### A R1 B 2kn 10 V 2kn R2 2k R3 Figure 2 in Fiqure 2 above,...
A R1 B 2kn 10 V 2kn R2 2k R3 Figure 2 in Fiqure 2 above, calculate the following, Show work 34. Total Current 35. VAB Je 2A 36. VBc= R1 75 VIN 10 V R, R2 Vy 470 VoUT Figure 4 Figure 3 37. Figure 3 is an example of a C. voltage multiplier D. voltage divider A. semiconductor B. current source 38-39. R...
##### Uestion 13. S-I +-:neMJ4-1. :ne N) Find: (i Isolated points of S: (ii) Boundary points of...
uestion 13. S-I +-:neMJ4-1. :ne N) Find: (i Isolated points of S: (ii) Boundary points of S: (iii) Accumulation points of S: (iv) Interior points of S:...
##### How does the law of diminishing marginal utility affect quantity demanded?
How does the law of diminishing marginal utility affect quantity demanded?...
##### Question 31 3 pts Proportion Problem. Each year Delta Dental does a survey of parents concerning...
Question 31 3 pts Proportion Problem. Each year Delta Dental does a survey of parents concerning the tradition of the Tooth Fairy. A random sample of 1,058 parents were asked about how much money they leave for a tooth and questions concerning how their child responded to the concept of the Tooth Fa...
##### Can u tell me plzz 10) Provide four (4) examples of how a settlement may grow....
can u tell me plzz 10) Provide four (4) examples of how a settlement may grow. You may choose to illustrate your answer with a diagram....
##### 15 and 11 11. ssm A bicyclist makes a trip that consists of three parts, each...
15 and 11 11. ssm A bicyclist makes a trip that consists of three parts, each in the same direction (due north) along a straight road. During the first part, she rides for 22 minutes at an average speed of 7.2 m/s. During the second part, she rides for 36 minutes at an average speed of 5.1 m/s.... | 585 | 2,108 | {"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} | 3.015625 | 3 | CC-MAIN-2022-49 | latest | en | 0.836542 |
https://www.accountingtools.com/articles/what-is-sales-mix.html | 1,547,692,278,000,000,000 | text/html | crawl-data/CC-MAIN-2019-04/segments/1547583658681.7/warc/CC-MAIN-20190117020806-20190117042806-00173.warc.gz | 700,703,299 | 15,029 | # Sales mix
Sales mix is the proportion of different products and services that comprise the total sales of a company. Sales mix is a key issue in businesses that sell products with differing profit levels, since a change in the mix of product sold can trigger a change in net profits, even when total sales remain approximately the same from period to period.
In many cases, each product or service that a company provides has a different profit, so changes in sales mix (even if sales levels remain the same) usually result in differing amounts of profit from period to period. Thus, if a company introduces a new product that has a low profit, and which it sells aggressively, it is quite possible that profits will decline even as total sales increase. Conversely, if a company elects to drop a low-profit product line and instead push sales of a higher-profit product line, total profits can actually increase even as total sales decline.
One of the best ways for a company to improve its profits in a low-growth market where increases in market share are difficult to obtain is to use its marketing and sales activities to alter the sales mix in favor of those products having the largest amount of profit associated with them.
When adjusting the sales mix, it is of considerable importance to understand the impact on the company constraint. Some products require more bottleneck time than others, and so may leave little room for the production of additional units.
Sales managers have to be aware of sales mix when they devise commission plans for the sales staff, since the intent should be to incentivize the sales staff to sell high-profit items. Otherwise, a poorly-constructed commission plan could push the sales staff in the direction of selling the wrong products, which alters the sales mix and results in lower profits.
A cost accounting variance called sales mix variance is used to measure the difference in unit volumes in the actual sales mix from the planned sales mix. Follow these steps to calculate it at the individual product level:
1. Subtract budgeted unit volume from actual unit volume and multiply by the standard contribution margin
2. Do the same for each of the products sold
3. Aggregate this information to arrive at the sales mix variance for the company
The formula is:
(Actual unit sales - Budgeted unit sales) x Budgeted contribution margin
Sales Mix Variance Example
ABC International expects to sell 100 blue widgets, which have a contribution margin of \$12 per unit, but actually sells only 80 units. Also, ABC expects to sell 400 green widgets, which have a contribution margin of \$6, but actually sells 500 units. The sales mix variance is:
Blue widget: (80 actual units - 100 budgeted units) x \$12 contribution margin = -\$240
Green widget: (500 actual units - 400 budgeted units) x \$6 contribution margin = \$600
Thus, the aggregate sales mix variance is \$360, which reflects a large increase in the sales volume of a product having a lower contribution margin, combined with a decline in sales for a product that has a higher contribution margin.
Related Courses | 620 | 3,133 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.671875 | 3 | CC-MAIN-2019-04 | latest | en | 0.950089 |
http://mathhelpforum.com/algebra/39488-ages-problem.html | 1,527,046,810,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794865411.56/warc/CC-MAIN-20180523024534-20180523044534-00056.warc.gz | 175,172,367 | 9,957 | 1. ages problem
The ration of Jonathan's age to David's is 6:5 in seven years it will be 7:6. Whar are their ages now?
This is my homework, and although I tried many times over was not able to solve it. Can anybody explain? Thanks!!!!
2. You can set up two systems and solve.
Let j = Jonathans age and d = davids age.
NOW: 6j=5d
7 years from now: 7(j+7)=6(d+7)
Solve for j and d.
3. Hello, AlexOrme!
Another approach . . .
The ratio of Jonathan's age to David's is 6:5
Iin seven years it will be 7:6.
Whar are their ages now?
We are told that: .$\displaystyle \frac{\text{Jonathan}}{\text{David}} \:=\;\frac{6a}{5a}$ . for some integer $\displaystyle a.$
In seven years, they are both seven years older
. . and the ratio is 7-to-6: . $\displaystyle \frac{6a+7}{5a+7} \:=\:\frac{7}{6}$
Solve for $\displaystyle a$ . . . and then their ages. | 277 | 852 | {"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} | 4.1875 | 4 | CC-MAIN-2018-22 | latest | en | 0.926021 |
http://math.stackexchange.com/questions/tagged/function-fields | 1,469,455,513,000,000,000 | text/html | crawl-data/CC-MAIN-2016-30/segments/1469257824230.71/warc/CC-MAIN-20160723071024-00290-ip-10-185-27-174.ec2.internal.warc.gz | 144,182,051 | 25,712 | # Tagged Questions
The tag has no usage guidance.
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### Is $\mathbb{C}(x,y)$ a rational function field?
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### Compute $\ell(W+P)$ and $\ell(W-P)$ for a place $P\in\mathbb{P}(F/K)$ of degree $1$ and any $W$ canonical divisor.
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### profinite completion of a ring of integers in a global field
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### Cokernel of map, function field.
Let $F$ be a function field in one variable with total constant field $k$, let $X$ be the set of all places of $F$, and let $S$ be a nonempty finite subset of $X$. We are interested in the dimension ... | 986 | 3,466 | {"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} | 3.328125 | 3 | CC-MAIN-2016-30 | latest | en | 0.758415 |
theoryinpracticebyprati.wordpress.com | 1,526,858,038,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794863811.3/warc/CC-MAIN-20180520224904-20180521004904-00402.warc.gz | 657,995,300 | 18,186 | # The Magic of Bayesian Methods for Missing Data
Missing data – it is almost inevitable in any dataset, and a perennial concern to anyone using that data to analyze trends or make decisions. The frequent “head in the sand” approach, where one just deletes any observations with missing data, is convenient, but most likely not very realistic. When data is missing, it is often missing for a reason.
Part of the problem with missing data is that, at least within the frequentist framework, there is no satisfactory way to address the missing-ness. One common solution is to impute each missing data with the mean or median as calculated from the observations that are missing. This could work, but should be used with extreme caution. Take, for example, a study on the efficacy of a new drug for headaches. Someone for whom the drug is less effective might be less likely to respond to follow up surveys about the drug, and therefore would be missing data. If one were to assign to this person the average treatment effect of the drug calculated from the people for whom the drug was effective, one would over-estimate the efficacy of the drug. Therefore, imputing with the average (or even the median) is not always a sound solution, but oftentimes it is the only solution available so it may be better than nothing.
Considering missing data within the Bayesian framework provides an analyst with much more appealing options. In simplified terms, the Bayesian framework considers data to be a particular realization from a distribution containing all possible data, and this underlying distribution can inform the analyst as to possible values for the missing data. If the particular data seems to have sparser realizations in the part of the distribution for less drug efficacy, then Bayesian methods impute values to fill in that sparser portion of the distribution, giving a more realistic picture of the actual treatment effect of the drug. This frankly magical method is only possible with this underlying distribution, a concept unique to Bayesianism.
Let me illustrate with an example, using R and the Bayesian analysis software Stan and the 12-month follow up survey for the Oregon Health Insurance Experiment (OHIE) data (WordPress does not recognize Rmd or Stan files, so the specific code is available upon request – comment below). The OHIE data was collected in 2008 under an expansion of Medicaid, where 30,000 low-income adults were randomly selected via lottery from a 90,000-person waiting list for the opportunity to apply for 10,000 additional spots in the Medicaid program. In particular, the 12-month follow up survey was sent to all 30,000 persons from the lottery to study whether having insurance improved health. Specifically for this exercise, I looked at whether having insurance decreased the number of days in the last month someone had mental health problems. Since it was a pseudo-experiment, in theory I should not have to include other covariates. The results indicate that, when missing data is simply deleted, having insurance is associated with a strong positive effect in the number of days with mental health problems, i.e. having insurance actually makes someone’s mental health worse off – a rather implausible scenario. When the missing data is imputed from the underlying distribution, having insurance is associated with an effect that spans 0, i.e. having insurance has no effect on someone’s mental health – a result which, considering that some of the population may have had no mental health problems at all, minor mental health problems that could be treated with some medication, and severe mental health problems that would take years of medication and/or therapy to properly treat, seems much more plausible.
Note: these results are part of a larger problem set I did for a Bayesian Statistics course at Columbia University. | 788 | 3,897 | {"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-2018-22 | latest | en | 0.938789 |
https://forum.vectorworks.net/index.php?/topic/104753-question-easy-way-to-calculate-percentage-in-worksheets/ | 1,674,960,644,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764499697.75/warc/CC-MAIN-20230129012420-20230129042420-00477.warc.gz | 284,716,040 | 17,986 | # Question: Easy way to calculate percentage in Worksheets
## Recommended Posts
Hi and greetings.
At the moment, I try to find a more easy way to calculate a relative percentage
in a Worksheet without of using criteria in a cell.
Attached is a small file with a use case.
B3 list the area of the listed rectangles.
B4 = formula with criteria
In C3 I want to have the percentage of the area of one rectangle, relative to the summarized area of all rectangles.
I only got my percentage, if I set criteria in B4 and calculate with it.
Otherwise, it won't work.
For this relative easy calculation, this solution works "okay".
For more complex calculations (complex if-formulas), this could be a tricky one.
Another solution would be to use two worksheets and reference
some cells with each other.
I want to know if there is more easy way without, of a use of criteria or a use of two worksheets?
like:
C3 = B3 / B4
B4 should list the summarized area in a way, that it is possible to calculate with it in C3
without a criteria.
If not possible, I will write a Wish in Jira.
Greetings, and thanks for your help
Tobi
I think there is a bigger bug and I think I have reported it previously. It appears that 2023 (through at least SP2) can not handle a database calculation that references a different part of the database.
You should be able to do something like:
Set Cell B1 = B3 to grab the total from the sum at the top of the database and then use a formula of =B3/B1 to get your percentage.
When you try and do this, B1 displays the proper value when first entered, but when you add the B3/B1 formula, the value of B1 resets to zero. 😞
This appears to happen anytime you use a referenced cell back into the database.
It looks like your using criteria is the only current work around.
You can simplify your criteria. you don't need the Database and in most cases you don't need the DLVP criteria. So your original of:
=AREA(DATABASE(NOTINDLVP & (NOTINREFDLVP) & (T=RECT)))
Can become just
=AREA(((T=RECT)))
While they are getting better, worksheets still need some more love.
And I did file this before:
Hi Pat,
greetings and thnx for your help and reporting VB-191196.
Hope we get a fix soon!
Greetings from Germany and have a good start in the week.
Tobi
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• Create New... | 655 | 2,754 | {"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-2023-06 | latest | en | 0.912967 |
https://www.convert-measurement-units.com/convert+1+min+to+Decahertz.php | 1,627,660,720,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046153971.20/warc/CC-MAIN-20210730154005-20210730184005-00544.warc.gz | 736,811,353 | 17,007 | Convert 1/min to daHz (1/min to Decahertz)
## 1/min into Decahertz
numbers in scientific notation
https://www.convert-measurement-units.com/convert+1+min+to+Decahertz.php
## How many Decahertz make 1 1/min?
1 1/min = 0.001 666 666 666 666 7 Decahertz [daHz] - Measurement calculator that can be used to convert 1/min to Decahertz, among others.
# Convert 1/min to Decahertz (1/min to daHz):
1. Choose the right category from the selection list, in this case 'Frequency'.
2. Next enter the value you want to convert. The basic operations of arithmetic: addition (+), subtraction (-), multiplication (*, x), division (/, :, ÷), exponent (^), brackets and π (pi) are all permitted at this point.
3. From the selection list, choose the unit that corresponds to the value you want to convert, in this case '1/min'.
4. Finally choose the unit you want the value to be converted to, in this case 'Decahertz [daHz]'.
5. Then, when the result appears, there is still the possibility of rounding it to a specific number of decimal places, whenever it makes sense to do so.
With this calculator, it is possible to enter the value to be converted together with the original measurement unit; for example, '499 1/min'. In so doing, either the full name of the unit or its abbreviation can be used. Then, the calculator determines the category of the measurement unit of measure that is to be converted, in this case 'Frequency'. After that, it converts the entered value into all of the appropriate units known to it. In the resulting list, you will be sure also to find the conversion you originally sought. Alternatively, the value to be converted can be entered as follows: '25 1/min to daHz' or '16 1/min into daHz' or '72 1/min -> Decahertz' or '74 1/min = daHz' or '52 1/min to Decahertz' or '19 1/min into Decahertz'. For this alternative, the calculator also figures out immediately into which unit the original value is specifically to be converted. Regardless which of these possibilities one uses, it saves one the cumbersome search for the appropriate listing in long selection lists with myriad categories and countless supported units. All of that is taken over for us by the calculator and it gets the job done in a fraction of a second.
Furthermore, the calculator makes it possible to use mathematical expressions. As a result, not only can numbers be reckoned with one another, such as, for example, '(88 * 67) 1/min'. But different units of measurement can also be coupled with one another directly in the conversion. That could, for example, look like this: '499 1/min + 1497 Decahertz' or '61mm x 94cm x 23dm = ? cm^3'. The units of measure combined in this way naturally have to fit together and make sense in the combination in question.
If a check mark has been placed next to 'Numbers in scientific notation', the answer will appear as an exponential. For example, 1.975 308 624 ×1020. For this form of presentation, the number will be segmented into an exponent, here 20, and the actual number, here 1.975 308 624. For devices on which the possibilities for displaying numbers are limited, such as for example, pocket calculators, one also finds the way of writing numbers as 1.975 308 624 E+20. In particular, this makes very large and very small numbers easier to read. If a check mark has not been placed at this spot, then the result is given in the customary way of writing numbers. For the above example, it would then look like this: 197 530 862 400 000 000 000. Independent of the presentation of the results, the maximum precision of this calculator is 14 places. That should be precise enough for most applications. | 876 | 3,650 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5625 | 4 | CC-MAIN-2021-31 | longest | en | 0.834072 |
http://ken.duisenberg.com/potw/archive/arch98/980415sol.html | 1,637,982,093,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964358078.2/warc/CC-MAIN-20211127013935-20211127043935-00325.warc.gz | 40,933,448 | 1,613 | ## Couples in a Candy Store
1. Three couples bought candy at a candy store. Each person paid as many cents per candy as pieces of candy he or she bought, and everyone bought a different number of pieces (no one was empty handed.) When they compared their bills, they found that the difference between the amounts paid by each husband and wife was the same. What is the smallest number of candies purchased? Show the amounts for the difference and for each person/couple.
2. What if there were two couples?
3. Four couples?
4. Five couples?
5. Can a method be extended for M couples?
6. If the difference between each husband's and wife's amount is less than 500 cents, what was the largest number of couples possible in the store?
Source: 1. Somewhere I can't remember (if you find it, let me know...), 2-6: Original.
Solutions were received from MANY people.
1. For 3 couples: (1,7) (4,8) (11,13). Difference is \$0.48, total candies = 44.
2. For 2 couples: (1,4) (7,8). Difference is \$0.15. Total candies is 20.
3. For 4 couples: (5,13) (9,15) (16,20) (35,37). Difference is \$1.44. Total candies is 150.
4. For 5 couples: (1,17) (6,18) (14,22) (21,27) (34,38). Difference it \$2.88. Total candies is 198.
5. In general for M couples, find a difference D with M pairs of factors: (B>A, A*B=D, A and B both odd or both even.) Then for each pair of factors A and B, the couple's values can be found as (B+A)/2 and (B-A)/2.
6. For a difference less than 500 [cents], you can get 7 couples: (2,22) (7,23) (14,26) (19,29) (37,43) (58,62) (119,121). Difference is \$4.80.
Solutions can be found for some of the answers above with a smaller difference, but some numbers of candies are then duplicated.
Mail to Ken | 500 | 1,712 | {"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-2021-49 | latest | en | 0.949406 |
http://databasefaq.com/index.php/tag/quantile | 1,553,360,155,000,000,000 | text/html | crawl-data/CC-MAIN-2019-13/segments/1552912202889.30/warc/CC-MAIN-20190323161556-20190323183556-00489.warc.gz | 62,986,482 | 4,667 | FAQ Database Discussion Community
## R function with functions as arguments, each with variable arguments
r,function,arguments,simulation,quantile
In answer to a question on Cross Validated, I wrote a simple function that used arbitrary quantile functions as its arguments etacor=function(rho=0,nsim=1e4,fx=qnorm,fy=qnorm){ #generate a bivariate correlated normal sample x1=rnorm(nsim);x2=rnorm(nsim) if (length(rho)==1){ y=pnorm(cbind(x1,rho*x1+sqrt((1-rho^2))*x2)) return(cor(fx(y[,1]),fy(y[,2]))) } coeur=rho rho2=sqrt(1-rho^2) for (t in 1:length(rho)){ y=pnorm(cbind(x1,rho[t]*x1+rho2[t]*x2)) coeur[t]=cor(fx(y[,1]),fy(y[,2]))} return(coeur) } However, both fx and fy may...
## Plot quantiles of distribution in ggplot2 with facets
r,ggplot2,quantile,density-plot
I'm currently plotting a number of different distributions of first differences from a number of regression models in ggplot. To facilitate interpretation of the differences, I want to mark the 2.5% and the 97.5% percentile of each distribution. Since I will be doing quite a few plots, and because the...
## Applying a function to each quantile of an R dataframe
r,data.frame,quantile
I have an R dataframe and I want to apply an estimation function for each of its quantiles. Here's an example with lm(): df <- data.frame(Y = sample(100), X1 = sample(100), X2 = sample(100)) estFun <- function(df){lm(Y ~ X1 + X2, data = df)} If I split that in two...
## Quantile per group in ddply
r,classification,plyr,quantile
I tried to classify my grouped data into quartiles, therefore adding a column "diam_quart" to the dataframe z assigning each row one of the four classes 1, 2, 3, or 4: quart = ddply(z, .(Code), transform, diam_quart = ifelse(Diameter <= quantile(Diameter , 0.25), 1, ifelse(Diameter <= quantile(Diameter , 0.5), 2,...
## R qqplot argument “y” is missing error
r,plot,quantile
I am relatively new to R and I am struggling with a error messages related to qqplot. Some sample data are at the bottom. I am trying to do a qqplot on some azimuth data, i.e. like compass directions. I've looked around here and the ?qqplot R documentation, but I...
## r get value only from quantile() function
r,quantile
I'm sorry for what may be a silly question. When I do: > quantile(df\$column, .75) #get 3rd quartile I get something like 75% 1234.5 Is there a way to just get the value (1234.5) without the descriptive "75%" string? Thank you very much....
## create quantile category variables using defined cut-points in Stata
category,stata,quantile
I am trying to create indicator variables using different quantile levels. I am creating a variable that contains categories corresponding to quantiles. For one variable, the code I am using is xtile PH_scale = PH, nq(4) tab PH_scale, gen(PH_scale_) Also, I know that if I want to use my own...
## Quantize Integers into discrete buckets
d3.js,charts,quantile,quantization
I have a list of ~7500 items which all have a similar signature: { revenue: integer, title: string, sector: string } The revenue will range from 0 to ~1 Billion. I'd like to construct a scale such that, given a particular company's revenue..it returns its position relative to the following...
## Quantile-Quantile plot using two vectors with ggplot
r,ggplot2,quantile
I am looking for a quantile-quantile plot using ggplot2. I have these data: model <- c(539.4573, 555.9882, 594.6838, 597.5712, 623.6659, 626.7169, 626.8539, 627.9992, 629.1139, 634.7405, 636.3438, 646.4692, 654.3024, 663.0181, 670.0985, 672.8391, 680.5557, 683.2600, 683.5159, 692.0328, 695.7185, 698.9505, 702.3676, 707.4271, 726.6507, 726.8524, 732.1197, 741.6183, 750.3617, 752.5978, 757.1609, 762.2874, 767.0678, 776.9476, 779.2352,...
## Assigning Percentile Based Groups to Dataframe in R
r,quantile
I am having trouble figuring out how to take on this particular problem. Suppose I have the following data frame: set.seed(123) Factors <- sample(LETTERS[1:26],50,replace=TRUE) Values <- sample(c(5,10,15,20,25,30),50,replace=TRUE) df <- data.frame(Factors,Values) df Factors Values 1 H 5 2 U 15 3 K 25 4 W 5 5 Y 20 6 B...
## Find top deciles from dataframe by group
r,data.frame,rank,quantile,split-apply-combine
I am attempting to create new variables using a function and lapply rather than working right in the data with loops. I used to use Stata and would have solved this problem with a method similar to that discussed here. Since naming variables programmatically is so difficult or at least...
## Matlab Calculating mean of distribution quantile in a for-loop
matlab,loops,for-loop,distribution,quantile
I am trying to calculate portfolio cVaR (conditional value at risk) levels from my simulated data for various portfolios. I am able to do that for one single portfolio using the following code: % Without a for-loop for series 1 test2 = test(:,1) VaR_Calib_EVT = 100 * quantile(test2, VarLevel_Calib); help1...
## How can I get a percentile value for each dataframe row considering a subset of the data?
r,data.frame,quantile
I have a dataframe obs with 145 rowns and more than 1000 columns. For each row I would like to extract the value of the 95th percentile but calculated only on the data greater or equal to 1. I managed calculating a value for each row, considering all data, as... | 1,532 | 5,294 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3 | 3 | CC-MAIN-2019-13 | latest | en | 0.729696 |
https://vdocuments.site/slope-intercept-gameboard.html | 1,585,561,909,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585370496901.28/warc/CC-MAIN-20200330085157-20200330115157-00078.warc.gz | 754,884,824 | 18,663 | # slope intercept gameboard
Post on 02-Aug-2015
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## Documents
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Students roll 2 dice to determine which problem to work. The problems are listed in a hotlinked table in the powerpoint presentation. Students identify the slope and/or the y-intercept of a linear function. The game moves quickly until all squares have been selected.
TRANSCRIPT
Slope Intercept Race
Slope Intercept RaceMaterials: Powerpoint Gameboard Poster of Gameboard or Draw on Whiteboard to keep score Number Cube (1 6) Letter Cube (A F)
Directions1. Divide the students into teams (2 to 4 teams will work or have students play against the teacher.) 2. Choose a team to go first. That team rolls both dice, which designates a square on the problem grid. 3. Randomly choose 1 student from that team to answer the question. 4. If the answer is correct, the team owns that square, and it is marked off the grid. If not, the other team (or next team if you have more than 2) has an opportunity to guess correctly and gain ownership. 5. Play rotates until all squares are taken or time expires. 6. If a team rolls a square that is already owned, they lose their turn and play goes to the next team. 7. The team with the most squares wins.
SLOPE-INTERCEPT RACE
SLOPE/INTERCEPT GAMEBOARD1 A A1 2 A2 3 A3 4 A4 5 A5 6 A6
BC D E F
B1C1 D1 E1 F1
B2C2 D2 E2 F2
B3C3 D3 E3 F3
B4C4 D4 E4 F4
B5C5 D5 E5 F5
B6C6 D6 E6 F6
A1: FIND THE X-INTERCEPT
B1: FIND THE SLOPE
C1: FIND THE Y-INTERCEPT
A2: FIND THE Y-INTERCEPT
A3: FIND THE X-INTERCEPT
A4: FIND THE SLOPE
D1: FIND THE SLOPE
E1: FIND THE Y-INTERCEPT
F1: FIND THE X-INTERCEPT
A5: FIND THE SLOPE
A6: FIND BOTH: THE X-INTERCEPT & THE Y-INTERCEPT
E2: FIND THE SLOPE
F2: FIND THE X-INTERCEPT
F5: FIND THE Y-INTERCEPT
B3: FIND THE SLOPE
B4: FIND THE X-INTERCEPT & THE Y-INTERCEPT
B2: FIND THE X-INTERCEPT
C2: FIND THE Y-INTERCEPT
D2: FIND THE SLOPE
D5: FIND BOTH THE X-INTERCEPT & THE Y-INTERCEPT
D6: FIND THE SLOPE
D4: FIND THE X-INTERCEPT
E4: FIND THE Y-INTERCEPT
F4: FIND THE SLOPE
D3: FIND THE SLOPE
E3: FIND THE X-INTERCEPT
F3: FIND THE Y-INTERCEPT
E5: FIND THE SLOPE
C5: FIND THE X-INTERCEPT
C6: FIND THE Y-INTERCEPT
C3: SURPRISE! FREE POINT
C4: IS THE GRAPH A NEGATIVE POSITIVE - NO OR UNDEFINED SLOPE?
B5: IS THE GRAPH A NEGATIVE POSITIVE - NO OR UNDEFINED SLOPE?
B6: IS THE GRAPH A NEGATIVE POSITIVE - NO OR UNDEFINED SLOPE?
E6: IS THE GRAPH A NEGATIVE POSITIVE - NO OR UNDEFINED SLOPE?
F6: IS THE GRAPH A NEGATIVE POSITIVE - NO OR UNDEFINED SLOPE?
Stay Connected!
My Blog: http://algebrasfriend.blogspot.com/My Email Address: educator.ch@gmail.com
CHickman (Algebras Friend), 2013 | 867 | 2,697 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5 | 4 | CC-MAIN-2020-16 | latest | en | 0.810542 |
https://pfeiffertheface.com/how-do-i-prepare-for-the-ged-math-test/ | 1,695,587,230,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233506669.30/warc/CC-MAIN-20230924191454-20230924221454-00165.warc.gz | 506,982,275 | 10,265 | Pfeiffertheface.com
Discover the world with our lifehacks
# How do I prepare for the GED math test?
## How do I prepare for the GED math test?
10 Tips for the Math GED Test
Table of Contents
1. Take and Use Notes.
2. Study on Your Mobile Device.
3. Set Goals.
4. Learn from Wrong Answers.
5. Study Every Day.
6. Read Questions Carefully.
7. Eliminate Wrong Answers.
8. Skip Hard Questions.
What kind of math is on the GED test?
Topics on the GED Math are: number operations and number sense (about 20% to 30% of the test), measurement & geometry (approximately 20% to 30% of the test), data analysis and statistics (about 20% to 30% ), and Algebra (approximately 25% to 30%of the test).
Is the GED math test hard to pass?
Depending on which test you’re taking, you only need to answer 40-50% of the questions correctly to get a passing score. Passing the test can be fast and easy, with a little preparation. If you don’t study for the test, it might be too hard. Very few people can pass the test without preparing for it first.
### How many questions is the GED math practice test?
The GED math test consists of 46 questions.
Can you pass the math GED without studying?
I want to pass the GED test without studying. Is this possible? This is one of the frequently asked questions we receive regularly. The short answer is yes but keep reading.
How many questions can you miss on the GED math test to pass?
To pass, you need at least 21 correct answers in the GED Mathematical Reasoning section, and you should have no more than 24 wrong answers.
## Can you pass GED without studying?
Is there a GED math practice test for 2019?
GED Math Practice Test 1 This is the first of our free GED Math practice tests, and it has been fully updated to reflect the latest 2019 GED revisions. To prepare for your GED Math test you will want to work through as many practice questions as possible.
To prepare for your GED Math test you will want to work through as many practice questions as possible. After answering each of these questions, the correct answer will be provided along with a detailed explanation. Get started on your test prep right now with this GED Math practice test.
What is a passing score on the GED math test?
The GED Math test is scored via a scale from 100 to 200 points. To pass the GED Math test, you must earn a score of at least 145. 2. FREQUENTLY ASKED QUESTIONS (FAQs) ABOUT GED MATH TEST 2.1. Who is eligible to take the GED Math Test? 2.2. How much is the GED Math Test? In most states, the GED Math test costs \$30 or less.
### What is on the GED test?
The GED, or General Education Development test, is a high school equivalency exam. Broken into four main subject areas – math, science, social studies, and reasoning through language arts – a passing score grants the test taker the equivalent of a high school diploma. | 700 | 2,868 | {"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-2023-40 | longest | en | 0.930938 |
https://www.techscience.com/cmc/v71n2/45759 | 1,716,543,118,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058709.9/warc/CC-MAIN-20240524091115-20240524121115-00656.warc.gz | 885,458,329 | 20,193 | Open Access
ARTICLE
# Structure Preserving Algorithm for Fractional Order Mathematical Model of COVID-19
Zafar Iqbal1,2, Muhammad Aziz-ur Rehman1, Nauman Ahmed1,2, Ali Raza3,4, Muhammad Rafiq5, Ilyas Khan6,*, Kottakkaran Sooppy Nisar7
1 Department of Mathematics, University of Management and Technology, Lahore, Pakistan
2 Department of Mathematics and Statistics, The University of Lahore, Lahore, Pakistan
3 Stochastic Analysis & Optimization Research Group, Department of Mathematics, Air University, Islamabad, 44000, Pakistan
4 Department of Mathematics, National College of Business Administration and Economics, Lahore, Pakistan
5 Department of Mathematics, Faculty of Sciences, University of Central Punjab, Lahore, 54500, Pakistan
6 Faculty of Mathematics and Statistics, Ton Duc Thang University, Ho Chi Minh City, 72915, Vietnam
7 Department of Mathematics, College of Arts and Science at Wadi Aldawaser, Prince Sattam Bin Abdulaziz University, Alkharj, 11991, Kingdom of Saudi Arabia
* Corresponding Author: Ilyas Khan. Email:
(This article belongs to the Special Issue: Mathematical aspects of the Coronavirus Disease 2019 (COVID-19): Analysis and Control)
Computers, Materials & Continua 2022, 71(2), 2141-2157. https://doi.org/10.32604/cmc.2022.013906
## Abstract
In this article, a brief biological structure and some basic properties of COVID-19 are described. A classical integer order model is modified and converted into a fractional order model with as order of the fractional derivative. Moreover, a valued structure preserving the numerical design, coined as Grunwald–Letnikov non-standard finite difference scheme, is developed for the fractional COVID-19 model. Taking into account the importance of the positivity and boundedness of the state variables, some productive results have been proved to ensure these essential features. Stability of the model at a corona free and a corona existing equilibrium points is investigated on the basis of Eigen values. The Routh–Hurwitz criterion is applied for the local stability analysis. An appropriate example with fitted and estimated set of parametric values is presented for the simulations. Graphical solutions are displayed for the chosen values of (fractional order of the derivatives). The role of quarantined policy is also determined gradually to highlight its significance and relevancy in controlling infectious diseases. In the end, outcomes of the study are presented.
## Keywords
APA Style
Iqbal, Z., Rehman, M.A., Ahmed, N., Raza, A., Rafiq, M. et al. (2022). Structure preserving algorithm for fractional order mathematical model of COVID-19. Computers, Materials & Continua, 71(2), 2141-2157. https://doi.org/10.32604/cmc.2022.013906
Vancouver Style
Iqbal Z, Rehman MA, Ahmed N, Raza A, Rafiq M, Khan I, et al. Structure preserving algorithm for fractional order mathematical model of COVID-19. Comput Mater Contin. 2022;71(2):2141-2157 https://doi.org/10.32604/cmc.2022.013906
IEEE Style
Z. Iqbal et al., "Structure Preserving Algorithm for Fractional Order Mathematical Model of COVID-19," Comput. Mater. Contin., vol. 71, no. 2, pp. 2141-2157. 2022. https://doi.org/10.32604/cmc.2022.013906
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## Example Questions
### Example Question #1 : Sequences
The sequence is defined by:
What is ?
Explanation:
Begin by interpreting the general definition:
This means that every number in the sequence is five greater than the element preceding it. For instance:
It is easiest to count upwards:
### Example Question #106 : Arithmetic
The sequence is defined by:
What is the value of ?
Explanation:
For this problem, you definitely do not want to "count upwards" to the full value of the sequence. Therefore, the best approach is to consider the general pattern that arises from the general definition:
This means that for every element in the list, each one is greater than the one preceding it. For instance:
Now, notice that the first element is:
The second is:
The third could be represented as:
And so forth...
Now, notice that for the third element, there are only two instances of . We could rewrite our sequence:
This value will always "lag behind" by one. Therefore, for the st element, you will have:
### Example Question #107 : Arithmetic
The sequence is defined by:
What is the value of ?
Explanation:
For this problem, you definitely do not want to "count upwards" to the full value of the sequence. Therefore, the best approach is to consider the general pattern that arises from the general definition:
This means that for every element in the list, each one is less than the one preceding it. For instance:
Now, notice that the first element is:
The second is:
The third could be represented as:
And so forth...
Now, notice that for the third element, there are only two instances of . We could rewrite our sequence:
This value will always "lag behind" by one. Therefore, for the th element, you will have:
Tired of practice problems?
Try live online GRE prep today. | 432 | 1,891 | {"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} | 4.21875 | 4 | CC-MAIN-2024-33 | latest | en | 0.896283 |
https://www.saranextgen.com/homeworkhelp/doubts.php?id=50068 | 1,720,909,233,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514517.94/warc/CC-MAIN-20240713212202-20240714002202-00204.warc.gz | 860,721,073 | 5,643 | # ABCD is a rectangle and ABE is a triangle whose vertex E lies on CD. If AB = 5 cm and the area of the triangle is 10 sq. cm, then the perimeter of the rectangle is (a) 14 cm (b) 15 cm (c) 18 cm (d) 20 cm
## Question ID - 50068 | SaraNextGen Top Answer ABCD is a rectangle and ABE is a triangle whose vertex E lies on CD. If AB = 5 cm and the area of the triangle is 10 sq. cm, then the perimeter of the rectangle is (a) 14 cm (b) 15 cm (c) 18 cm (d) 20 cm
Answer Key / Explanation : (c) -
Area of ABE =
Perimeter = =18 cm. | 176 | 529 | {"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-2024-30 | latest | en | 0.758083 |
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