36! goodluck hope it good
SO SORRY I ANSWERED ON THE OTHER ONE! The answer is C. This line has a negative slope, and the ending number tells you the y-intercept.
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:
y=-6
Step-by-step explanation:
y+6=0(x-5)
Solve
y=-6
The image is not very much enlarged, and is slanted, so answers might be just approximate.
(a) median is 39 or 40, your choice.
(b) 3rd quartile of y looks 47 or 48
(c) variable y has more disperson, since both the extremes and the quartiles are spread out longer that those of variable x.
(d) since both halfs of the blue box (of x) and both lines outside of the box are basically equal, variable x's distribution is very much symmetric.
Sorry, cannot read question part (e) completely.