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:
Yes
Step-by-step explanation:
4(1)-3(3)=-5
4-9=-5
-5=-5
Answer:(0,6)
Step-by-step explanation:
go to the left 3 points from T
then you go up 2 points from the number 4 which you should end up on after going over 3 spots.
Answer:
104
Step-by-step explanation:
women = 8/8
men = 5/8
all workers = 8/8 + 5/8 = 13/8
extra women workers = 8/8 - 5/8 = 3/8
so...
3/8 = 24
13/8 = x
( cross multiplication method)
3/8 x = 39
x = 104
Hence, there are 104 workers
Answer:
Its Ok! I'm here to help! The answer would be 0.01152 If I'm correct!
Step-by-step explanation:
I really hope this help!
