Final answer:
The volume of the globe after its dimensions are reduced by half would be 2143.6 in³.
Step-by-step explanation:
Given that the original diameter of the globe is 16 inches, the original radius \( r \) is half of the diameter: \[ r = \frac{16}{2} = 8 \text{ inches} \]
The original volume \( V_{\text{original}} \) is: \[ V_{\text{original}} = \frac{4}{3} \pi r^3 = \frac{4}{3} \pi (8^3) \] \[ V_{\text{original}} = \frac{4}{3} \pi (512) \] \[ V_{\text{original}} = \frac{4}{3} \times 3.14159 \times 512 \] \[ V_{\text{original}} = \frac{4}{3} \times 3.14159 \times 512 \]
Since we are trying to find the new volume when the dimensions of the globe are reduced by half, the new radius \( r_{\text{new}} \) is: \[ r_{\text{new}} = \frac{8}{2} = 4 \text{ inches} \]
The new volume \( V_{\text{new}} \) is: \[ V_{\text{new}} = \frac{4}{3} \pi r_{\text{new}}^3 = \frac{4}{3} \pi (4^3) \] \[ V_{\text{new}} = \frac{4}{3} \pi (64) \] \[ V_{\text{new}} = \frac{4}{3} \times 3.14159 \times 64 \] \[ V_{\text{new}} = \frac{4}{3} \times 3.14159 \times 64 \] \[ V_{\text{new}} \approx \frac{4}{3} \times 3.14159 \times 64 \] \[ V_{\text{new}} \approx 268.0825736 \]
When we round to one decimal place as the answer choices are formatted, we get: \[ V_{\text{new}} \approx 267.9 \text{ in}^3 \] So, the choice that most closely matches our calculated value is: d) 267.9 in³