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)
.
Answer:
the radius is 7
Step-by-step explanation:
circumference is equal to pi(d)
you said it is 14pi
all you have to do is divide d by 2 to get r
A=1/2(b+b)•h
A=1/2(18+24)•8
A=1/2(42)•8
A=21•8
A=168ftsq
Just so you know, you can always use a calculator... or google which I know you have access to, considering you are on this site. Well, actually you could be using the brainly app but that's not the point. What I am trying to say is... you don't have to waste your points on questions like this! That's what Google is for :) Btw... your answer is 193
