B=30. all you have to do is minus 9 from both sides then multiply 6 to each side so all thats left is the variable
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:
Last choice. d.
Step-by-step explanation:
We are given the first term and the sixth term.
The first term is
.
The sixth them is
.
Let's solve for the common ratio, r.

Divide both sides by -9:

Take the fifth root of both sides:


So the common ratio is 4.






Answer:
1.125 pages per minute
Step-by-step explanation:
Mika read a 405 page book in 6 hours.
6 hours is equivalent to
6 × 60 minutes = 360 minutes
Number of pages read per minute = 405/360
= 9/8 = 1.125 pages
Mika read 1.125 pages per minute.