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
.
-2(2x-1)(x+3) = ((-2)2x - (-2)(-1))(x+3) = (-4x-2)(x+3)
3*2=6
6-x-2=0
4=x
The answer is 4
Answer:
Here is a link to help you. This is a very great video that will help you understand proportionalities .
Step-by-step explanation:
https://www.khanacademy.org/math/cc-seventh-grade-math/cc-7th-ratio-proportion/7th-constant-of-proportionality/v/constant-of-proportionality-from-equation