Abicyclist bikes the 56 mi to a city averaging a certain speed. the return trip is made at a speed that is 6 mph slower. total time for the round trip is 11 hr. find the bicyclist's average speed on each part of the trip.

Speed of the bicyclist when going to city =  14 miles per hour.

Speed while return trip = 8 miles per hour.

Explanation:

Let the speed of the bicyclist when going to city = x miles per hour.

Speed while return trip =  x - 6 miles per hour.

Total time taken = 11 hrs = Time for the trip to city + time taken for return trip.

Also, Time = Distance / Time.

So,

56 / x + 56 / ( x -6) = 11

11x² -178x + 336 = 0

Solving for x we get:

Acceptable x = 14 miles per hour.

Speed while return trip =  x - 6 miles per hour = 8 miles per hour.

it would be c. predictability