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

a. Verify that the average distance per response d_av = 5.1 miles, the average speed per response S_av = 12.5 miles/hr, and the average time per response t_av = 19.1 minutes.

b. Intuitively, explain why (d_av/S_av) x (60 minutes/hr) = 24.48 minutes is greater than t_av = 19.1 minutes.

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