The length of time for one individual to be served at a cafeteria is a random variable having an exponential distribution with a mean of 77 minutes. what is the probability that a person is served in less than 22 minutes on at least 55 of the next 77 days?

0.01265

Step-by-step explanation:

Since, if the time to be served has an exponential distribution with a mean of 7, then

Chance to be served in under 2 minutes:

Let A represents the number of days when a person is served in less than 2 minutes,

Hence,

the probability that a person is served in less than 2 minutes on at least 5 of the next 7 days ( using binomial distribution )

= P(A=5) + P(A=6) + P(A=7)

≈ 0.01265

10a^3+277a^2-132a-792

step-by-step explanation:

(2a+11)(5a-12)(foil)

=10a^2-24a+55a-132 =10a^2+31a-132

(10a^2+31a-132)(a+6)

=10a^3+60a^2+31a^2+186a^2-132a-792 =10a^3+277a^2-132a-792

i hope this is what you are looking for : )

50*10 is greater

step-by-step explanation:

if you add the zero from 10 to 50, it will give you a bigger number.