The number of days ahead travelers purchase their airline tickets can be modeled by an exponential distribution with the average amount of time equal to 15 days. What is the probability that a traveler will purchase a ticket fewer than ten days in advance?

P(x<10) =0

Step-by-step explanation:

M =1/μ

M= 1/15 =0.066667

X- Exp (0.066667)

Where: X is the random variable for number of days

P (X<x) = 1 - e^-mx

P(X - x) = 1 - e^-0.06666 x10

P(x<10) = 0

let soccer = x and golf = y

part a

1st equation: x + y = 125

2nd equation: x = y +45

part b

replace x in the 1st equation with the 2nd one:

y + 45 + y = 125

simplify:

2y + 45 = 125

subtract 45 from each side:

2y = 80

divide both sides by 2:

y = 80/2

y = 40

she plays golf for 40 minutes.

part c

no because she plays soccer for 45 minutes longer than golf 40 +45 =   85 minutes, not 80 minutes.