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: ]1, 2[
Step-by-step explanation:
![A \cap B=[0; 2[\\\\(A \cap B)-C=]1, 2[](https://tex.z-dn.net/?f=A%20%5Ccap%20B%3D%5B0%3B%202%5B%5C%5C%5C%5C%28A%20%5Ccap%20B%29-C%3D%5D1%2C%202%5B)
Answer:
angle R
Step-by-step explanation:
To solve this we will use cosine rule
Cosine rule
cos(A) = 
suppose,
q = 6.25
s = 11.04
r = 13.19
angleQ = 
= 30.58
angleR = 
= 93.82
angleS = 
= 56.63
Answer:
FALSE
Step-by-step explanation:
:)
Derek ran less than 6 miles. If the race is 6 miles, and he only ran 3/8, he ran less than a half of the race. If he didn't finish the race of 6 miles, he couldn't possibly have run 6 miles.
