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
.
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
.
Answer:
the fourth answer
Step-by-step explanation:
this is because 6 x 10 is 60, and is you multiply the height by 3 it gives you 3(6) x 10 which can also be written as 18 x 10 which is equal to 180
divide 180 by 60 which will give you 3, this shows you that the growth from the original area to the new area is 3 times, so the ratio is 3:1
Answer:
X=2
Step-by-step explanation:
Answer:
9
Step-by-step explanation:
Given that :
Total Number of scorings (both Teams) = 18
Let :
x = number of 5 point score ; y = number of 2 point score
Total points scored = 34 + 29 = 63
x + y = 18 - - - (1)
5x + 2y = 63 - - (2)
From (1)
y = 18 - x
In (2) ; we have ;
5x + 2(18 - x) = 63
5x + 36 - 2x = 63
3x + 36 = 63
3x = 63 - 36
3x = 27
x = 9
9 5-point plays