1. **State the problem:** We are given average monthly high temperatures for certain months and a sinusoidal model predicting temperature as a function of month number $t$. We need to find the predicted temperature for June ($t=6$). 2. **Model form:** A sinusoidal model for temperature can be written as: $$T(t) = A \sin(B(t - C)) + D$$ where: - $A$ is the amplitude (half the difference between max and min temperatures), - $B = \frac{2\pi}{12} = \frac{\pi}{6}$ since the period is 12 months, - $C$ is the horizontal shift (phase shift), - $D$ is the vertical shift (average of max and min temperatures). 3. **Find amplitude $A$ and vertical shift $D$:** - Max temperature from data is about 67.4°F, - Min temperature from data is about 42.0°F. Calculate: $$A = \frac{67.4 - 42.0}{2} = \frac{25.4}{2} = 12.7$$ $$D = \frac{67.4 + 42.0}{2} = \frac{109.4}{2} = 54.7$$ 4. **Find phase shift $C$:** The maximum temperature occurs near $t=8$ (August), so the sine function reaches its maximum at $t=8$. Since sine reaches max at $\frac{\pi}{2}$, solve: $$B(t - C) = \frac{\pi}{2}$$ $$\frac{\pi}{6}(8 - C) = \frac{\pi}{2}$$ Divide both sides by $\frac{\pi}{6}$: $$8 - C = \frac{\frac{\pi}{2}}{\frac{\pi}{6}} = \frac{\pi}{2} \times \frac{6}{\pi} = 3$$ $$C = 8 - 3 = 5$$ 5. **Write the model:** $$T(t) = 12.7 \sin\left(\frac{\pi}{6}(t - 5)\right) + 54.7$$ 6. **Calculate temperature for June ($t=6$):** $$T(6) = 12.7 \sin\left(\frac{\pi}{6}(6 - 5)\right) + 54.7 = 12.7 \sin\left(\frac{\pi}{6}\right) + 54.7$$ Since $\sin\left(\frac{\pi}{6}\right) = \frac{1}{2}$: $$T(6) = 12.7 \times \frac{1}{2} + 54.7 = 6.35 + 54.7 = 61.05$$ 7. **Compare with answer choices:** Closest to 61.05°F is (B) 61.8°F. **Final answer:** (B) 61.8°F