[b]Q6.[/b] At $3:00$ AM, the temperature was $13^o$ below zero. By none it has risen to $32^o$. What is the average hourly increase in temperature ?
1. Identify the initial temperature at 3:00 AM, which is \( -13^\circ \) (13 degrees below zero).
2. Identify the final temperature at noon, which is \( 32^\circ \).
3. Calculate the total change in temperature:
   \[
   \Delta T = 32^\circ - (-13^\circ) = 32^\circ + 13^\circ = 45^\circ
   \]
4. Determine the time interval over which this change occurred. From 3:00 AM to noon is 9 hours.
5. Calculate the average hourly increase in temperature by dividing the total change in temperature by the number of hours:
   \[
   \text{Average hourly increase} = \frac{\Delta T}{\text{Time interval}} = \frac{45^\circ}{9 \text{ hours}} = 5 \text{ deg/hr}
   \]

The final answer is \(\boxed{5 \text{ deg/hr}}\).