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
.
In order for us to answer this question, more information is needed.
Answer:
C. 10
Step-by-step explanation:
by deduction, 5 and 7 are prime numbers and 15 is not a multiple of 70, therefore the answer must be 10 as it is a non-prime number and is a factor of 70 (I.e. 10 and 7)
Answer:
no se podrías resolverlo tu
Answer:
The answer is y=8/9x
Step-by-step explanation: