Question
A globe company currently manufactures a globe that is 20 inches in diameter. If the dimensions of the globe were reduced by half, what would its volume be? Use 3.14 for π and round your answer to the nearest tenth.
166.7 in3
1333.3 in3
523.3 in3
4186.7 in3
166.7 in3
1333.3 in3
523.3 in3
4186.7 in3
Answers
GPT-4o mini
To find the volume of the globe, we can use the formula for the volume of a sphere, which is given by:
\[
V = \frac{4}{3} \pi r^3
\]
where \( r \) is the radius of the sphere.
1. **Calculate the radius of the original globe:**
The diameter of the original globe is 20 inches, so the radius \( r \) is:
\[
r = \frac{20 \text{ inches}}{2} = 10 \text{ inches}
\]
2. **Calculate the volume of the original globe:**
Substituting the radius into the volume formula:
\[
V = \frac{4}{3} \times 3.14 \times (10)^3
\]
First, calculate \( (10)^3 \):
\[
(10)^3 = 1000
\]
Now substitute it into the volume formula:
\[
V = \frac{4}{3} \times 3.14 \times 1000
\]
Calculate \( \frac{4}{3} \times 3.14 \):
\[
\frac{4}{3} \approx 1.3333 \quad \Rightarrow \quad 1.3333 \times 3.14 \approx 4.1888
\]
Now multiply by 1000:
\[
V \approx 4.1888 \times 1000 \approx 4188.8 \text{ in}^3
\]
3. **Reduce the dimensions by half:**
If the dimensions are reduced by half, the new diameter will be:
\[
\text{New Diameter} = \frac{20}{2} = 10 \text{ inches}
\]
Then the new radius \( r' \) will be:
\[
r' = \frac{10}{2} = 5 \text{ inches}
\]
4. **Calculate the volume of the smaller globe:**
Now substituting the new radius into the volume formula:
\[
V' = \frac{4}{3} \times 3.14 \times (5)^3
\]
First, calculate \( (5)^3 \):
\[
(5)^3 = 125
\]
Substitute it into the volume formula:
\[
V' = \frac{4}{3} \times 3.14 \times 125
\]
Calculate \( \frac{4}{3} \times 3.14 \):
\[
\frac{4}{3} \times 3.14 \approx 4.1888
\]
Now multiply by 125:
\[
V' \approx 4.1888 \times 125 \approx 523.6 \text{ in}^3
\]
5. **Round the answer to the nearest tenth:**
Rounding \( 523.6 \) gives \( 523.3 \).
Thus, the volume of the globe after reducing its dimensions by half is:
\[
\boxed{523.3 \text{ in}^3}
\]
\[
V = \frac{4}{3} \pi r^3
\]
where \( r \) is the radius of the sphere.
1. **Calculate the radius of the original globe:**
The diameter of the original globe is 20 inches, so the radius \( r \) is:
\[
r = \frac{20 \text{ inches}}{2} = 10 \text{ inches}
\]
2. **Calculate the volume of the original globe:**
Substituting the radius into the volume formula:
\[
V = \frac{4}{3} \times 3.14 \times (10)^3
\]
First, calculate \( (10)^3 \):
\[
(10)^3 = 1000
\]
Now substitute it into the volume formula:
\[
V = \frac{4}{3} \times 3.14 \times 1000
\]
Calculate \( \frac{4}{3} \times 3.14 \):
\[
\frac{4}{3} \approx 1.3333 \quad \Rightarrow \quad 1.3333 \times 3.14 \approx 4.1888
\]
Now multiply by 1000:
\[
V \approx 4.1888 \times 1000 \approx 4188.8 \text{ in}^3
\]
3. **Reduce the dimensions by half:**
If the dimensions are reduced by half, the new diameter will be:
\[
\text{New Diameter} = \frac{20}{2} = 10 \text{ inches}
\]
Then the new radius \( r' \) will be:
\[
r' = \frac{10}{2} = 5 \text{ inches}
\]
4. **Calculate the volume of the smaller globe:**
Now substituting the new radius into the volume formula:
\[
V' = \frac{4}{3} \times 3.14 \times (5)^3
\]
First, calculate \( (5)^3 \):
\[
(5)^3 = 125
\]
Substitute it into the volume formula:
\[
V' = \frac{4}{3} \times 3.14 \times 125
\]
Calculate \( \frac{4}{3} \times 3.14 \):
\[
\frac{4}{3} \times 3.14 \approx 4.1888
\]
Now multiply by 125:
\[
V' \approx 4.1888 \times 125 \approx 523.6 \text{ in}^3
\]
5. **Round the answer to the nearest tenth:**
Rounding \( 523.6 \) gives \( 523.3 \).
Thus, the volume of the globe after reducing its dimensions by half is:
\[
\boxed{523.3 \text{ in}^3}
\]