Trains headed for destination A arrive at the train station at 15-minute intervals tarting at 7am, where as trains headed for destination B arrive at 15-minute intervals tarting at 7:05am. (a) If a certain passenger arrives at the station at a time uniformly distributed between 7 and 8am, and then gets on the first train that arrives, what proportion of time does he or she go to destination A? (b) What if the passenger arrives at a time uniformly distributed between 7:10am and 8:10am?

a.) 2/3 ; b.) 7/12

Step-by-step explanation:

Given the following details :

Destination A trains arrives at station at 15 minutes interval starting from 7am

Destination B trains arrives at station at 15 minutes interval starting from 7.05am

A.) if a passenger arrives at station uniformly between 7am and 8am and gets on first train. What proportion of time does the passenger leave from destination A?

To leave for destination A. The passenger has to arrive before train to A departs and after train to B leaves.

Train to B leaves, starting at at 7.05 a.m and at interval of 15 minutes

Train to A leaves at 7.00 am and at interval of 15 minutes.

Therefore,

(Train B departs - train A arrives) (between 7 to 8am)

(7.05 - 7.15) + (7.20 - 7.30) + (7.35 - 7.45) + (7.50 - 8.00) = 40 minutes

Total time = 7.00 - 8.00 = 60 minutes

(40 ÷ 60) = 2/3 or 0.667

B.) If passenger arrives uniformly between 7.10am - 8:10am

(7.10 - 7.15) + (7.20 - 7.30) + (7.35 - 7.45) + (7.50 - 8)

5 + 10 + 10 + 10 = 35

35/60 = 7/12 or 0.583

mx+89b(-98)nb

step-by-step explanation:

b. y = 4x - 2

step-by-step explanation:

to find a parallel equation to an equation, make sure both equations have the same slope but different y-intercepts.

y = 4x + 1

4 = slope

1 = y-intercept

y = 4x - 2

4 = slope

-2 = y-intercept

they both have the same slope but different y-intercepts, so, b. is the correct answer.

and to check your work, you can graph both equations. : )