Cars arrive at a toll booth according to a Poisson process with mean 90 cars per hour. Suppose the attendant makes a phone call. How long, in seconds, can the attendant's phone call last if the probability is at least 0.1 that no cars arrive during the call

92.12 seconds

Step-by-step explanation:

According to the poisson probability relation :

P(X =x) = (e^-λ * λ^x) / x!

For no calls to be reveived during the period, x = 0

P(X = 0) = (e^-λ * λ^0) / 0!

P(X = 0) = 0.1

0.1 = (e^-λ * λ^0) / 0!

0.1 = e^-λ

Take the In of both sides

In(0.1) = - λ

-2.303 = - λ

λ = 2.303

The length of call in second, l

l = λ / r ; r = arrival rate

r = 90 per hour ; this means ;

90 / 3600 = 0.025

l = 2.303 / 0.025

l = 92.12 seconds

step-by-step explanation:

x^2 -12x + 36 + y^2 = 36

x^2 + y^2 = 12x

r^2 = 12rcos(theta)

r = 12 cos(theta)

1,150 give brainliest

step-by-step explanation

2r= 2300

r =1150