Solve the problem.
-A data-recording thermometer recorded the soil temperature in a field every 2 hours from noon to midnight, as shown in the following table. Use the Trapezoidal Rule to estimate the average temperature for the 12-hour period. Round your answer to the nearest hundredth if necessary.
$\begin{array}{l|l}
\text { Time } & \text { Temp }\left({ }^{\circ} \mathrm{F}\right) \
\hline \text { Noon } & 63 \
2 & 64 \
4 & 66 \
6 & 66 \
8 & 65 \
10 & 65 \
\text { Midnight } & 64
\end{array}$

A) $64.94 ^ { \circ } \mathrm { F }$
B) $77.9 ^ { \circ } \mathrm { F }$
C) $64.92 ^ { \circ } \mathrm { F }$
D) $75.5 ^ { \circ } \mathrm { F }$

Question

Solve the problem.
-A data-recording thermometer recorded the soil temperature in a field every 2 hours from noon to midnight, as shown in the following table. Use the Trapezoidal Rule to estimate the average temperature for the 12-hour period. Round your answer to the nearest hundredth if necessary.
$\begin{array}{l|l}
\text { Time } & \text { Temp }\left({ }^{\circ} \mathrm{F}\right) \
\hline \text { Noon } & 63 \
2 & 64 \
4 & 66 \
6 & 66 \
8 & 65 \
10 & 65 \
\text { Midnight } & 64
\end{array}$

A) $64.94 ^ { \circ } \mathrm { F }$
B) $77.9 ^ { \circ } \mathrm { F }$
C) $64.92 ^ { \circ } \mathrm { F }$
D) $75.5 ^ { \circ } \mathrm { F }$

Quizplus · Accepted Answer

Using the Trapezoidal Rule, the approximate area under the temperature vs. time graph can be found by finding the sum of the areas of trapezoids formed by adjacent temperature values at each time interval. Then, the average temperature can be estimated by dividing this area by the time interval (which is 12 hours in this case).

For example, the area of the trapezoid formed by the temperatures at noon (63°F) and 2pm (64°F) is:

$$\frac{1}{2}(2\text{ hours})(63^\circ\text{F}+64^\circ\text{F}) = 127^\circ \text{F}\cdot\text{hours}$$

Similarly, the areas of the other trapezoids can be found and added together:

$$\frac{1}{2}(2\text{ hours})(63^\circ\text{F}+64^\circ\text{F}) + \frac{1}{2}(2\text{ hours})(64^\circ\text{F}+66^\circ\text{F}) + \frac{1}{2}(2\text{ hours})(66^\circ\text{F}+66^\circ\text{F}) + \frac{1}{2}(2\text{ hours})(66^\circ\text{F}+65^\circ\text{F}) + \frac{1}{2}(2\text{ hours})(65^\circ\text{F}+65^\circ\text{F}) + \frac{1}{2}(2\text{ hours})(65^\circ\text{F}+64^\circ\text{F}) = 768^\circ\text{F}\cdot\text{hours}$$

Dividing by the time interval of 12 hours gives an average temperature of:

$$\frac{768^\circ\text{F}\cdot\text{hours}}{12\text{ hours}} = 64^\circ\text{F}$$

Therefore, the best choice is C: $64.92 ^ { \circ } \mathrm { F }$, since it is the closest answer to 64.00°F, which is the exact value obtained from the trapezoidal rule.

Solve the Problem A) 64.94∘F64.94 ^ { \circ } \mathrm { F }64.94∘F

Solve the Problem A) $64.94 ^ { \circ } \mathrm { F }$