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: The answer is B. 34
Step-by-step explanation:
3(12−9)+5^2
=(3)(3)+5^2
=9+5^2
=9+25
=34
150+3p
p=30
150+3(30)
150+90
240
When p=30 in 150+3p, the final value is 240.
Answer:

Step-by-step explanation:
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Answer:
B and C
Step-by-step explanation:
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