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
.
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
.
Answer: (1, 2) (The third option)
Step-by-step explanation: A solution has to be on the line. (1, 2) is the only point that isn't on the line.
Answer: d. 6
Explanation: you multiply each number by 6 every time
Answer:
d=4.14
Step-by-step explanation:
use the equation d=2r
d=C/π=13/π≈4.13803
4.13803≈4.14