Suppose you walk to your friend's house in $15 \mathrm{~min}$, then to the store which takes $45 \mathrm{~min}$, and then to school which takes $25 \mathrm{~min}$. If the distances are $A=707 \mathrm{~m}, \mathrm{~B}=1423 \mathrm{~m}$, and $\mathrm{C}=606 \mathrm{~m}$ what was the magnitude of your average velocity in kilometers per hour?

Step 1 ：Given that the distances to your friend's house, the store, and the school are 707 m, 1423 m, and 606 m respectively, and the time taken to walk to each location is 15 min, 45 min, and 25 min respectively.

Step 2 ：First, convert the distances from meters to kilometers and the time from minutes to hours. This gives us 0.707 km, 1.423 km, and 0.606 km for the distances, and 0.25 hr, 0.75 hr, and 0.4166666666666667 hr for the times.

Step 3 ：Next, calculate the total displacement by adding up the distances. This gives us \(0.707 + 1.423 + 0.606 = 2.736 \) km.

Step 4 ：Then, calculate the total time by adding up the times. This gives us \(0.25 + 0.75 + 0.4166666666666667 = 1.4166666666666667 \) hr.

Step 5 ：Finally, calculate the magnitude of the average velocity by dividing the total displacement by the total time. This gives us \(\frac{2.736}{1.4166666666666667} = 1.9312941176470586 \) km/hr.

Step 6 ：So, the magnitude of your average velocity was approximately \(\boxed{1.93 \mathrm{~km/hr}}\).