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:
Option (3). 120° will be the answer.
Step-by-step explanation:
By using the theorem of secants intersecting externally,
Measure of angle formed by two secants, from a point outside the circle is half the difference of the measures of the intercepted arcs.
m∠C = 
36° = 
72° =
°
= 72 + 48
= 120°
Therefore, Option (3) is the answer.
Answer:
x = 4
Step-by-step explanation:
10 = -1/4 x + 11
10 - 11 = -1/4 x
-1 = - 1/4 x
x = 4
Step-by-step explanation:
Price of one shirt =?
Six packages of three t-shirt = 6 x 3 = 18 shirts
18 shirts = 71.39
1 shirt = 71.39/18
1 shirt = 3.966 or 4