Answer:
9,240
Step-by-step explanation:
For the first number, there are 22 numbers available to choose from. For the second number, there are 21 numbers available, since the first number cannot be the same as the second number. And for the last number, there are 20 remaining numbers to choose from. So, there are 22*21*20=9240 possible sequences.
What's the rest of the question?
Answer:
b. 0.25
c. 0.05
d. 0.05
e. 0.25
Step-by-step explanation:
if the waiting time x follows a uniformly distribution from zero to 20, the probability that a passenger waits exactly x minutes P(x) can be calculated as:

Where a and b are the limits of the distribution and x is a value between a and b. Additionally the probability that a passenger waits x minutes or less P(X<x) is equal to:

Then, the probability that a randomly selected passenger will wait:
b. Between 5 and 10 minutes.

c. Exactly 7.5922 minutes

d. Exactly 5 minutes

e. Between 15 and 25 minutes, taking into account that 25 is bigger than 20, the probability that a passenger will wait between 15 and 25 minutes is equal to the probability that a passenger will wait between 15 and 20 minutes. So:

Answer:
3r^2 /2
Step-by-step explanation:
r^3 cancels out r in the denominator remaining with r^2 in the numerator.
so 3r^2/2
= 3r^2 /2