5, 20, 80, 320, 1280, 5120, 20480, 81920, 327680, 1310720, 5242880, 20971520
The 12th term would be 20,971,520
Everything is being multiplied by 4 each time
Answer: ounces
Step-by-step explanation:
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
.
Area of circle = pi (d/2)^2
area of semi-circle =0.5* pi (d/2)^2
=0.5* pi (6/2)^2
=14.14
The graph is not proportional because it doesn’t go though the origin aka (0,0)