The relationship between the average rates is that ; The average rate of return trip is twice the average rate of front trip.
From the complete question seen online, we can say that;
Time taken on front trip = t
Time taken on return trip = 1/2 × t = t/2
Relationship between speed(Rate) and time is;
Rate ∝ 1/time
This means that Rate is inversely proportional to time.
Thus, for the front trip, the rate is; r = 1/t
For return trip, the rate is; R = 1/(t÷2)
R = 1 × 2/t
R = 2/t
Rate of front trip = 1/t
Rate of return trip = 2/t
Thus, we can say that the average rate of return trip is twice the average rate of front trip.
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