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?
1 answer:
Answer: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
You might be interested in
Here is the answer with an explanation!
Answer:
whats the question though?
3-y/2=1
-3 -3
-y/2=-2
x2 x2
-y=-4
/-1 /-1
Y=4
2. an=20+an−1 and a1=145
she starts with 145 so a1=145 and then adds 20 every month: an=20+an−1
3,190,000 is 319 x 10^5 in scientific notation
