3.5 Time, speed, distance. Suppose that ambulance attendants read the following data for four random ambulance responses:

pg164a.gif (13926 bytes)

a. Verify that the average distance per response dav = 5.1 miles, the average speed per response Sav = 12.5 miles/hr, and the average time per response tav = 19.1 minutes.

b. Intuitively, explain why (dav/Sav) x (60 minutes/hr) = 24.48 minutes is greater than tav = 19.1 minutes.

c. If we compute a weighted average speed Swt, where the weights sum to 1 and are proportional to the times driven at each speed, we find Swt = 16.03 miles/hr. We find that (dav/Swt x 60 minutes/hr) = tav = 19.1 minutes. Why is this a correct procedure?