Multiply them back into each other... You would multiply your answer with the number you divided by (the number ouside the table if you used the table... ) and you should get the larger number you divided.
Answer:
j = 21 and n = 14
Step-by-step explanation:
we have the equations:
6j + n/3 = 134
j/3 + n = 31
54j + 3n = 1206
j + 3n = 93
53j = 1113
j = 21
(21)/3 + n = 31
7 + n = 31
n = 14
Let T = total cost
Let 8c = total amount of children tickets
Let 12a = total amount of adult tickets
T = 8c + 12a
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)
.
