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:
8x -16y = 24
Step-by-step explanation:
8x - 16y= -14
A parallel line (i.e. same slope) will be: 8x - 16y= c
We have to find the value of "c".
If it passes through (x, y) = (3, 0) ⇒ 24 - 0 = c,
so equation is 8x -16y = 24
Answer: C
Step-by-step explanation:
Simplify the expression to 2^1/4
Now transform the expression using a^m/n = n root a raised to a power of m.
And that's how you get your answer.
Radii? Please explain what that means
So the person is out of pocket of $10.00
