54:36 is your answer.
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It should help
Step-by-step explanation:
(x-2)^3 = x^3-6x^2+12x-8
-(x^3-6x^2+12x-8)=
-x^3+6x^2-12x+8
-x^3+6x^2-12x+8 -5=
y=-x^3+6x^2-12x+3
I attached the picture
So the exchange rate from one US dollar to a Canadian dollar is equal to 1.30 Canadian dollars. So in this current situation if you swapped all of your US dollars for Canadian dollars you would have $13000 Canadian dollars
The answer to the question is boneless
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