Answer:
It is BFA
Step-by-step explanation:
An inscribed angle stays inside the circle. If you go through each angle options, the other angles go out of the circle. BFA is the only angle that stays inside the circle.
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
.
So what i would do is take 3/5 and multiply is all by 3 so you end up with 9/15.
Now after you have done that do this, 9/15+ 4/15. Add 9/15 and 4/15 together which has a sum of 13/15 now add 11 + 6 which is 17 then add in the 13/15 and you get the answer of 17 13/15
Answer:
12!
Step-by-step explanation:
А=12*11*10*9*8*7*6*5*4*3*2*1=12!=479001600