Let X be the number of burglaries in a week. X follows Poisson distribution with mean of 1.9
We have to find the probability that in a randomly selected week the number of burglaries is at least three.
P(X ≥ 3 ) = P(X =3) + P(X=4) + P(X=5) + ........
= 1 - P(X < 3)
= 1 - [ P(X=2) + P(X=1) + P(X=0)]
The Poisson probability at X=k is given by
P(X=k) =
Using this formula probability of X=2,1,0 with mean = 1.9 is
P(X=2) =
P(X=2) =
P(X=2) = 0.2698
P(X=1) =
P(X=1) =
P(X=1) = 0.2841
P(X=0) =
P(X=0) =
P(X=0) = 0.1495
The probability that at least three will become
P(X ≥ 3 ) = 1 - [ P(X=2) + P(X=1) + P(X=0)]
= 1 - [0.2698 + 0.2841 + 0.1495]
= 1 - 0.7034
P(X ≥ 3 ) = 0.2966
The probability that in a randomly selected week the number of burglaries is at least three is 0.2966
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Answer:
B
Step-by-step explanation:
Answer:
Given
f(x) = 2x+7
g(x) = x^2-4
h(x) = 5x
a. 4h(x)
= 4 * 5x
= 20x
b. f(x) - g(x)
f(x) - g(x) = 2x + 7 - (x^2 - 4)
= 2x+7-x^2+4
=-x^2+2x+7+4
=-x^2+2x+11
c. f(g(x)) = 2(g(x))+7
=2(x^2-4) +7
=2x^2-8+7
=2x^2-1
d. g(x)h(x) = (x^2-4)(5x)
= 5x^3 - 20x
e. g(x) / f(x) = x2 - 4/ 2x + 7