2. CTfastrak bus waiting times are uniformly distributed from zero to 20 minutes. Find the probability that a randomly selected passenger will wait the following times for a CTfastrak bus. b. Between 5 and 10 minutes. c. Exactly 7.5922 minutes. d. Exactly 5 minutes. e. Between 15 and 25 minutes.

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:

b. the decimal could be two-tenths.

step-by-step explanation:

∠acb = 8°

step-by-step explanation:

∠xba lies on the circle and is half the measure of the intercepted arc ax

hence

measure of arc ax = 2 × 42° = 84°

∠acb is outside the circle and is

∠acb = (arcab - arcax ), hence

∠acb = 0.5 (100 - 84) = 0.5 × 16° = 8°