Irline passengers arrive randomly and independently at the passenger-screening facility at a major international airport. The mean arrival rate is 8 passengers per minute. a. Compute the probability of no arrivals in a one-minute period (to 6 decimals). b. Compute the probability that three or fewer passengers arrive in a one-minute period (to 4 decimals). c. Compute the probability of no arrivals in a -second period (to 4 decimals). d. Compute the probability of at least one arrival in a -second period (to 4 decimals).

A. 0.000045

B. 0.010336

C. 0.082085

D. 0.917915

Step-by-step explanation:

A. Compute the probability of no arrivals in a one-minute period.

P(x=o) = 0.000045

B. Compute the probability that three or fewer passengers arrive in a one-minute period.

p(x = 3) = 0.007567

p(x = 2) = 0.002270

p(x = 1) = 0.000454

p(x ≤ 3) = 0.000045 + 0.007567 + 0.002270 + 0.000454 = .010336

C. Compute the probability of no arrivals in a 15-second period.

p(x = 0) = 0.082085

D. Compute the probability of at least one arrival in a 15-second period.

p(x ≥ 1) = 1.0 - 0.082085 = 0.91791561

no one knows lol

step-by-step explanation:

10.416 m/s

step-by-step explantion:

distance=100m

time= 9.6s

average speed=100/9.6

=10.416 m/s