Should be About 25 feet as your answer
31. There are C(5, 2) = 10 ways to choose 2 colors from a group of 10 colors if you don't care about the order. (Here, we treat blue background with violet letters as being indistinguishable from violet background with blue letters.)
Only one of those 10 pairs is "B and V", so the probability is 1/10.
a) P(B and V) = 10%
32. The 12 inch dimension on the figure is 0 inches for the cross section. The remaining dimensions of the cross section are
c) 5 in. × 4 in.
_____
C(n, k) = n!/(k!·(n-k)!)
C(5, 2) = 5!/(2!·3!) = 5·4/(2·1) = 10
The expected length of code for one encoded symbol is
where 
is the probability of picking the letter 
, and 
is the length of code needed to encode . is given to us, and we have
so that we expect a contribution of
bits to the code per encoded letter. For a string of length 
, we would then expect ![E[L]=1.375n](https://tex.z-dn.net/?f=E%5BL%5D%3D1.375n)
.
By definition of variance, we have
![\mathrm{Var}[L]=E\left[(L-E[L])^2\right]=E[L^2]-E[L]^2](https://tex.z-dn.net/?f=%5Cmathrm%7BVar%7D%5BL%5D%3DE%5Cleft%5B%28L-E%5BL%5D%29%5E2%5Cright%5D%3DE%5BL%5E2%5D-E%5BL%5D%5E2)
For a string consisting of one letter, we have
so that the variance for the length such a string is
"squared" bits per encoded letter. For a string of length , we would get ![\mathrm{Var}[L]=1.859n](https://tex.z-dn.net/?f=%5Cmathrm%7BVar%7D%5BL%5D%3D1.859n)
.
Graph 2, Option 4 would be your answer because:
If you substitute 1 into the equation (x), you would get about 212
If you substitute 2, you would get approximately 225
If you substitute 10, you would get approximately 364.
Looking at the graph, your answer would be going up and to the right, the population is getting bigger while the years are increasing as well.
If we substitute 15 into x in the equation, we would get approximately 492.
Your Answer Is approximately 0.143
