Answer: 12.96
Step-by-step explanation:
given data:
Poisson distributio = 3.60
solution:
The Poisson distribution expected value is same as Variance.
which is expressed as = SD^2
therefore;
the expected value
= SD^2
= 3.60^2
= 12.96
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
.
Step-by-step explanation:
d = rt
divide t
d/t = r
r = d/t
<u>Statement Reason </u>
AB║CD , AD║CB Given
m∠BAC = m∠ACD Alternate Interior Angles
m∠ACB = m∠CAD Alternate Interior Angles
AC = AC Reflexive Property of Equality
ΔABC ≅ΔCDA ASA ≅