$24.54 I hope this is the right answer
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
.
Ok so i just did 56/14 and got 4. So this means 14 goes into 56 4 times, so now i did 4 x 2 and got 8. So the answer is 8 free cones.
Answer:
3
Step-by-step explanation:
3 (1.5 - 0.5) = 3 (1) = 3 * 1 = 3. The value of the expression is 3.
Tell me if I'm wrong :)