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:
x = 3
Step-by-step explanation:
Given the linear equation :
1/4(x-5)+4=1/3(2x+7)-5/6
(x-5)/4 + 4 = (2x+7)/3 - 5/6
Take the lcm and sum
(x-5+16)/4 = (4x+14-5)/6
(x+11)/4 = (4x+9)/6
Cross multiply
6(x+11) = 4(4x+9)
6x + 66 = 16x + 36
Collect like terms
6x - 16x = 36 - 66
-10x = - 30
x = 30/10
x = 3
Answer:
38
Step-by-step explanation:
if there are bars around it its automatically positive even if it was -38 the absolute would still be 38.
The cost of the burger went up 2%/ and you can also subtract 6 from8 and get the answer. Thank you have a nice day
It should have 2 solutions
