A winter storm is approaching Morristown, and the temperature is dropping at a rate of $$ 3^{\circ} \mathrm{F} $$ per hour. How many hours, $$ h $$, will it take for the temperature to drop $$ 18^{\circ} \mathrm{F} $$ ?

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To determine how many hours it will take for the temperature to drop by $$ 18^{\circ} \mathrm{F} $$ at a rate of $$ 3^{\circ} \mathrm{F} $$ per hour, you can set up a simple equation based on the relationship between rate, time, and temperature change.

**Step 1: Understand the Relationship**

The temperature drop ($$ \Delta T $$) is related to the rate of temperature change ($$ R $$) and the time ($$ h $$) by the equation:

$$
\Delta T = R \times h
$$

Where:
- $$ \Delta T $$ is the total temperature drop ($$ 18^{\circ} \mathrm{F} $$)
- $$ R $$ is the rate of temperature drop ($$ 3^{\circ} \mathrm{F} $$ per hour)
- $$ h $$ is the time in hours

**Step 2: Plug in the Known Values**

$$
18^{\circ} \mathrm{F} = 3^{\circ} \mathrm{F}/\text{hour} \times h
$$

**Step 3: Solve for $$ h $$**

$$
h = \frac{18^{\circ} \mathrm{F}}{3^{\circ} \mathrm{F}/\text{hour}} = 6 \text{ hours}
$$

**Conclusion**

It will take **6 hours** for the temperature in Morristown to drop by $$ 18^{\circ} \mathrm{F} $$ at the given rate.
It will take 6 hours for the temperature to drop by $$ 18^{\circ} \mathrm{F} $$.

A winter storm is approaching Morristown, and the temperature is dropping at a rate of \( 3^{\circ} \mathrm{F} \) per hour. How many hours, \( h \), will it take for the temperature to drop \( 18^{\circ} \mathrm{F} \) ?

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