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:
The second box
Step-by-step explanation:
The volume of a pyramid can be given by the formula: V=31Ah. Where. A: Area of the V=(1/3)(B)(h)
multiply both sides by 3
3V=Bh
divide both sides by B
3V/B=h
Therefore, If it is wider it has more space to carry more items such as macaroni.
I hope this helps! (:
Answer:
y=1/2x-3
Step-by-step explanation:
m=(y2-y1)/(x2-x1)
m=(-7-(-1))/(-8-4)
m=(-7+1)/-12
m=-6/-12
m=1/2
y-y1=m(x-x1)
y-(-1)=1/2(x-4)
y+1=1/2x-4/2
y=1/2x-2-1
y=1/2x-3
Answer:
(-3, 4) is a solution
Step-by-step explanation:
The point (-3, 4) is inside the shaded area of the graph, so is a solution.
You can check in the inequality
y > -2x -3
4 > -2(-3) -3 . . . . substitute for x and y
4 > 3 . . . . . . . true; the given point is a solution
Answer:
Step-by-step explanation:
5%×g= 5g/100
g-5g/100
