Question 9
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The air temperature of a mountain drops by $4^{\circ} \mathrm{C}$ for every $1000 \mathrm{~m}$ of elevation.
If the air temperature is $25^{\circ} \mathrm{C}$ at the base of the mountain, at what elevation will the air temperature be $17^{\circ} \mathrm{C}$ ?
$1000 \mathrm{~m} \quad 2000 \mathrm{~m}$
$4000 \mathrm{~m} \quad 8000 \mathrm{~m}$

Step 1 ：First, we need to find the temperature difference between the base of the mountain and the desired elevation. The temperature at the base is \(25^\circ\mathrm{C}\) and the desired temperature is \(17^\circ\mathrm{C}\). So, the temperature difference is \(25^\circ\mathrm{C} - 17^\circ\mathrm{C} = 8^\circ\mathrm{C}\).

Step 2 ：Next, we know that the temperature drops by \(4^\circ\mathrm{C}\) for every \(1000\mathrm{m}\) of elevation. So, we can set up a proportion to find the elevation: \(\frac{8^\circ\mathrm{C}}{4^\circ\mathrm{C}/1000\mathrm{m}}\).

Step 3 ：Now, we can solve the proportion: \(\frac{8^\circ\mathrm{C}}{4^\circ\mathrm{C}/1000\mathrm{m}} = \frac{8^\circ\mathrm{C}}{1^\circ\mathrm{C}} * 1000\mathrm{m} = 8 * 1000\mathrm{m} = 8000\mathrm{m}\).

Step 4 ：So, the elevation at which the air temperature will be \(17^\circ\mathrm{C}\) is \(\boxed{8000\mathrm{m}}\).