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 ![E[L]=1.375n](https://tex.z-dn.net/?f=E%5BL%5D%3D1.375n)
.
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 ![\mathrm{Var}[L]=1.859n](https://tex.z-dn.net/?f=%5Cmathrm%7BVar%7D%5BL%5D%3D1.859n)
.
Answer:
11 and 16
Step-by-step explanation:
A good way to solve these is to work backwards.
X= larger and Y= smaller
if X is 5 more than 2Y than subtract 5 from the total
38 - 5 = 33
divide by 3
33/3 = 11
11 = Y and double that is 22 so add the five back into X and you get 16
Answer:
x=1.4 or 21/15
Step-by-step explanation:
Answer:
1
Step-by-step explanation:
We have the graph of f(x).
In order to find f(0), we simply need to find the y-coordinate of the coordinate when x is 0.
From the graph, we can see that there is a point at (0,1).
In other words, when x is 0, f(x) is 1.
Therefore:
Mr. Cecere scored 24 points. Mr. Neason scored 12 points. Mr. Rudolph scored 34 points.
