Answer:
I think it's 4y
Step-by-step explanation:
I think it's 4y is because if y = 0 they all work but 4y which is 0 while all the others are 4
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)
.
cannot be simplified any further :
5=5x1
206=2*103
Answer:
$3.25
Step-by-step explanation:
Given that:
Mean, λ = 1.4
Strike within next minute = $3 won
Strike between one and 2 minutes = $5
Strike more than 2 minutes = $1
Probability that next strike occurs within the next minute :
Using poisson :
P(x < 1) = 1 - e^-(λx) ;
P(x < 1) = 1 - e^-(1.4*1) = 1 - e^-1.4
P(x < 1) = 1 - 0.2465969
P(x < 1) = 0.7534030
Next strike occurs between 1 and 2 minutes :
(1 < x < 2) :
P(x < 2) - P(x < 1)
P(x < 2) = 1 - e^-(λx) ;
P(x < 2) = 1 - e^-(1.4*2) = 1 - e^-2.8
P(x < 2) = 1 - 0.0608100
P(x < 2) = 0.9391899
P(x < 2) - P(x < 1)
0.9391899 - 0.7534030 = 0.1857869
P(striking after 2 minutes)
P(x > 2) = e^-(λx) ;
P(x > 2) = e^-(1.4*2) = e^-2.8
P(x > 2) = 0.0608100
Amount charged :
(0.7534030 * 3) + (0.1857869 * 5) + (0.06081 * 1)
= 3.2499
= $3.25
Answer:
16π/9
Step-by-step explanation:
40°/360° = 1/9
40° is 1/9 of a circle
Arc length = 1/9 of circumference = (1/9)2πr = 16π/9
