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:
B. 4m
Step-by-step explanation:
What you do is you just take the 8 and put it into the m.
m-8, 8-8=0 so A isn't your answer.
4m, 4(8)=32 so B is your answer.
B. 4m is your answer.
Answer:
Leena consumed 1,500 calories at dinner
Step-by-step explanation:
The reason the answer is Leena consumed 1,500 calories at dinner is because she consumes 2/3 of her daily calories at dinner.
U/72 = 2/4
means u/72 = 0.5
therefore u = 72*0.5
which equals 36
The answer for this problem is 12
i hope this help you bro
