1. 12a+6=12a+6
2. 12a+6=12a+2
3. 12a+6=4a+6
4. 12a+6=7a+5
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
.
The answer is A or the first on you might say, hopes this helps ya!
Step-by-step explanation:
did you give us all the information ? the equating looks cut off.
so, from what I can see we have
9(y-4) - 3y = 2(3y-2)
let's try :
9y - 36 - 3y = 6y - 4
0 = 32
so, no, this can't be it. this does not have a solution. you must have cut off some additional information.
Answer:
5
Step-by-step explanation:
72 inches is 2 yards. Traveling 5 times that distance (10 yards) will require 5 times as many rotations of the tire.
The tires will need to make 5 rotations.