Telephone calls arrive at the Global Airline reservation office in Louisville according to a Poisson distribution with a mean of 5.4 calls per minute. What is the probability of receiving exactly 3 call during a one-minute interval?

0.119

Step-by-step explanation:

Given that :

Mean (m) = 5.4

Probability of receiving exactly 3 calls ; X = 3

P(X, μ) = [(e^-μ) * (μ^x)] / x!

P(3, 5.4) = [(e^-5.4) * (5.4^3)] / 3!

P(3, 5.4) = (0.0045165 * 157.464) / (3*2*1)

P(3 ; 5.4) = 0.711186156 / 6

P(3 ; 5.4) = 0.118531026

= 0.119

b) well because you get well

y < -3

step-by-step explanation:

-3y - 2 > 7

-3y -2 + 2 > 7+2

-3y > 9

-3y/-3 < 9/-3   (change the inequality sign when multiplying or dividing by a negative number)

y < -3