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:
coincident
Step-by-step explanation:
The first equation is 3 times the second equation, so describes exactly the same line. The lines are "coincident".
None of the answers are correct because it would be 4, 10, 12, 17, then 41
Answer:
sixty-two plus 7 times a number h
Step-by-step explanation:
"Joint variation<span>" means "directly, but with two or more variables". Thus, we write as follows:
j </span>α gv
<span>
We insert proportionality constant k, to change it to equality.
j = kgv
We solve k,
1 = k (4)(3)
k =1/12
j = gv/12
j = 8(11)/12
j = 22/3</span>
