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:
violates the rules or bedmas
Step-by-step explanation:
When doing questions like these, bedmas comes into play. althought multiplication and division are the same in terms of priority, you should do the one that comes first when reading the equation/formula left to right.
this would mean you would do 2 ÷ 3 and the answer would be multiplied by 12.
0.12 = 1 ounce. 12 goes into 342, 28.5 times. Your answer would be 28.5.
(3.42 divided by 0.12)