Sure, let's solve this step-by-step.
1. Calculating the Volume of the Initial Globe:
- The initial diameter of the globe is 16 inches.
- The radius is half of the diameter, so the initial radius is [tex]\( \frac{16}{2} = 8 \)[/tex] inches.
- The formula for the volume of a sphere is [tex]\( V = \frac{4}{3} \pi r^3 \)[/tex].
- Plugging in the initial radius and [tex]\( \pi = 3.14 \)[/tex],
[tex]\[
V_{\text{initial}} = \frac{4}{3} \times 3.14 \times (8)^3 = \frac{4}{3} \times 3.14 \times 512.
\][/tex]
- Simplifying this:
[tex]\[
V_{\text{initial}} = \frac{4}{3} \times 3.14 \times 512 = 2143.5733333333333 \text{ cubic inches}.
\][/tex]
2. Calculating the Volume of the Reduced Globe:
- If the dimensions are reduced by half, the new diameter is [tex]\( \frac{16}{2} = 8 \)[/tex] inches.
- The new radius is half of the new diameter, so the new radius is [tex]\( \frac{8}{2} = 4 \)[/tex] inches.
- Using the volume formula again for the reduced globe:
[tex]\[
V_{\text{reduced}} = \frac{4}{3} \times 3.14 \times (4)^3 = \frac{4}{3} \times 3.14 \times 64.
\][/tex]
- Simplifying this:
[tex]\[
V_{\text{reduced}} = \frac{4}{3} \times 3.14 \times 64 = 267.94666666666666 \text{ cubic inches}.
\][/tex]
3. Rounding to the Nearest Tenth:
- The volume of the reduced globe after simplifying is [tex]\( 267.94666666666666 \)[/tex].
- Rounding this to the nearest tenth:
[tex]\[
V_{\text{reduced rounded}} = 267.9 \text{ cubic inches}.
\][/tex]
So, if the dimensions of the globe were reduced by half, the volume of the new globe would be 267.9 cubic inches.
Therefore, the answer is:
[tex]\( 267.9 \, \text{in}^3 \)[/tex]
