Randomly take $51$ numbers from the set 1, 2, ..., 159. Find variance of their sum.
Variance of sum of $k$ randomly drawn numbers from $1,...,n$.
What causes the texture of bread to be open, ie with lots of large holes, or close, ie a regular, uniform tender crumb with no large holes? How can I achieve either? What is the fastest way to get from Courmayeur/Aosta to Milan? Where To start Learning With Salesforce Development Where should I start learning? There are various topics to learn like Objects, Classes, Apex, Visualforce, SOSL, SOQL, Triggers, etc. There are so many options out there to start learning. I couldn't decide what topic to learn first, and then step by step get proficiency in all aspects of Saleforce. In reading through the original GCM specification (), the composition of the 128 bit Initialization Vector as a concatenation of a 96b nonce and a 32b unsigned wrapping counter seems arbitrary and forces the scrambling pattern to repeat every 232 16-byte block.  Is the algorithm expected to be secure for significantly longer stream lengths if for example the IV were a 128b nonce XORed with a 64b or 128b counter, or are there known cryptanalysis issues that begin to arise? Spam iCloud Calendar Invitations What's the mathematics definition of the F value in Analysis of Variance Table? Suppose I have a linear regression model Y ~ b1*x1 + b2*x2 with the sample data:          x1      x2        y 0  -0.7251 -0.6033  10.2508 1  -1.3148  3.7067  10.3054 2  -1.2818 -5.8136   7.2010 3   0.6154  3.6336  11.2807 4   1.4887 -2.8337  10.6299 5  -2.0224 -0.3931   7.4502 6   0.5419 -1.7807  10.7339 7  -1.0745 -5.0296   5.8734 8   3.2116 -0.4681  11.7184 9   1.6767 -4.6040   9.0145 10 -1.3355 -2.6874   6.5014 11  0.8296  2.6793  11.8295 12 -2.8647  4.2916   7.8414 13  1.8445 -3.3924   9.6520 14 -0.1036  2.4273  11.9766 15  1.0886 -3.5636  10.6023 16 -3.3026 -6.9837   4.9962 17  1.4050 -3.8453  10.4630 18 -0.3111 -1.4278   9.2283 19 -3.5454  2.4960   7.3720   And using the following R code to get the Analysis of Variance Table:  &gt; df &lt;- read.csv('f:/data.csv', header=T, sep=",") &gt; model_All &lt;- lm(y ~ x1 + x2, data=df) &gt; anova(model_All) Analysis of Variance Table  Response: y           Df Sum Sq Mean Sq F value    Pr(&gt;F)     x1         1 44.440  44.440  58.277 6.865e-07 *** x2         1 26.640  26.640  34.934 1.711e-05 *** Residuals 17 12.964   0.763                       --- Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1   And I can calculate the F value of hypothesis test H0： b2=0 via:  &gt; model_Drop2 &lt;- lm(y ~ x1, data=df) &gt; SSR_All = sum(resid(model_All)^2) &gt; SSR_Dp2 = sum(resid(model_Drop2)^2) &gt; F_x2 = ((SSR_Dp2 - SSR_All) / 1) / (SSR_All / (20 - 3)) &gt; F_x2 [1] 34.93434   We can see that F_x2 is equal to F value of x2 in Analysis of Variance Table. However, the value F_x1 calculated as follow is not equal to F value of x1, I can't understand why:  &gt; model_Drop1 &lt;- lm(y ~ x2, data=df) &gt; SSR_Dp1 = sum(resid(model_Drop1)^2) &gt; F_x1 = ((SSR_Dp1 - SSR_All) / 1) / (SSR_All / (20 - 3)) &gt; F_x1 [1] 70.30392   So, what is the mathematics definition of the F value in table?  EDIT The result of the features reversed as commented by @Michael M:  &gt; model_All2 &lt;- lm(y ~ x2 + x1, data=df) &gt; anova(model_All2) Analysis of Variance Table  Response: y           Df Sum Sq Mean Sq F value    Pr(&gt;F)     x2         1 17.468  17.468  22.907 0.0001718 *** x1         1 53.612  53.612  70.304 1.914e-07 *** Residuals 17 12.964   0.763                       --- Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1 18.04: color emoji not showing up at all in Chrome, only partially in Firefox How do I prove $\sin x$ is uniformly continuous on $\mathbb R$ with delta and epsilon?   I proved geometrically that $\sin x&lt;x$ and thus, $$|f(x_1)-f(x_2)|=|\sin x_1 - \sin x_2|\le|\sin x_1|+|\sin x_2|&lt;|x_1|+|x_2|$$   But this doesn't help me much finding a delta...   Thanks for any help!  P.S. I'm only at the beginning of calculus so I can't use many theorems and derivation  (because they haven't been regorously proven). Is there a way to change the environment variables of another process in Unix? I've been trying to understand how the second order derivative "formula" works:  $$\lim_{h\to0} \frac{f(x+h) - 2f(x) + f(x-h)}{h^2}$$  So, the rate of change of the rate of change for an arbitrary continuous function. It basically feels right, since it samples "the after $x+h$ and the before $x-h$" and the $h^2$ is there (due to the expected /h/h -> /h*h), but I'm having trouble finding the equation on my own.  It's is basically a derivative of a derivative, right? Newtonian notation declares as $f''$ and Leibniz's as $\frac{\partial^2{y}}{\partial{x}^2}$ which dissolves into:  $$(f')'$$ and $$\frac{\partial{}}{\partial{x}}\frac{\partial{f}}{\partial{x}}$$  So, first derivation shows the rate of change of a function's value relative to input. The second derivative shows the rate of change of the actual rate of change, suggesting information relating to how frequenly it changes.  The original one is rather straightforward:  $$\frac{\Delta y}{\Delta x} = \lim_{h\to0} \frac{f(x+h) - f(x)}{x + h - x} = \lim_{h\to0} \frac{f(x+h) - f(x)}{h}$$  And can easily be shown that $f'(x) = nx^{n-1} + \dots$ is correct for the more forthcoming of polynomial functions. So, my logic suggests that to get the derivative of a derivative, one only needs to send the derivative function as input to finding the new derivative. I'll drop the $\lim_{h\to0}$ for simplicity:  $$f'(x) = \frac{f(x+h) - f(x)}{h}$$  So, the derivative of the derivative should be:  $$f''(x) = \lim_{h\to0} \frac{f'(x+h) - f'(x)}{h}$$  $$f''(x) = \lim_{h\to0} \frac{  \frac{ f(x+2h) - f(x+h)}{h} - \frac{ f(x+h) - f(x)}{h}  }{h}$$  $$f''(x) = \lim_{h\to0} \frac{  \frac{ f(x+2h) - f(x+h) - f(x+h) + f(x)}{h}  }{h}$$  $$f''(x) = \lim_{h\to0} \frac{  f(x+2h) - f(x+h) - f(x+h) + f(x)  }{h^2}$$  $$f''(x) = \lim_{h\to0} \frac{  f(x+2h) - 2f(x+h) + f(x)  }{h^2}$$  What am I doing wrong? Perhaps it is the mess of it all, but I just can't see it. Please help. Using Pigeonhole Principle to prove two numbers in a subset of $[2n]$ divide each other Let $n$ be greater or equal to $1$, and let $S$ be an $(n+1)$-subset of $[2n]$. Prove that there exist two numbers in $S$ such that one divides the other.   Any help is appreciated! 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