Let's solve each problem step by step.
1. **Volume of a basketball with a diameter of 9.2 inches**:
- Formula for the volume of a sphere: \( V = \frac{4}{3} \pi r^3 \)
- Diameter = 9.2 inches, so radius \( r = \frac{9.2}{2} = 4.6 \) inches.
- Using \( \pi \approx 3.14 \):
\[
V = \frac{4}{3} \times 3.14 \times 4.6^3 \approx \frac{4}{3} \times 3.14 \times 97.336
\]
\[
V \approx \frac{4}{3} \times 305.29 \approx 407.51 \text{ in}^3
\]
Therefore, the answer is:
- **407.51 in.3**
2. **Volume of a sphere with a radius of 1.8 yards**:
- Formula for the volume of a sphere: \( V = \frac{4}{3} \pi r^3 \)
- Radius \( r = 1.8 \) yards.
\[
V = \frac{4}{3} \pi \times 1.8^3 = \frac{4}{3} \pi \times 5.832 = \frac{4}{3} \times 5.832 \pi = 7.776 \pi \approx 7.78 \pi \text{ yd}^3
\]
Therefore, the answer is:
- **7.78π yds3**
3. **How much air is needed to fill a basketball with a diameter of 9 inches**:
- Diameter = 9 inches, so radius \( r = \frac{9}{2} = 4.5 \) inches.
- Using \( \pi \approx 3.14 \):
\[
V = \frac{4}{3} \times 3.14 \times 4.5^3 \approx \frac{4}{3} \times 3.14 \times 91.125 \approx \frac{4}{3} \times 286.1325 \approx 381.51 \text{ in}^3
\]
Therefore, the answer is:
- **381.5 in.3**
4. **Unused space in the box containing 50 marbles**:
- Volume of the box = 4000 mm\(^3\)
- Radius of each marble = 2 mm.
- Volume of one marble \( V = \frac{4}{3} \pi r^3 \)
\[
V = \frac{4}{3} \times 3.14 \times 2^3 = \frac{4}{3} \times 3.14 \times 8 = \frac{4}{3} \times 25.12 = 33.4933 \text{ mm}^3 \approx 33.5 \text{ mm}^3
\]
- Total volume of 50 marbles:
\[
50 \times 33.5 = 1675 \text{ mm}^3
\]
- Unused space in the box:
\[
4000 - 1675 = 2325 \text{ mm}^3
\]
Therefore, the answer is:
- **2,325.3 mm3**
5. **Volume of the sculpture**:
- Volume of a hemisphere: \( V = \frac{2}{3} \pi r^3 \)
- Radius of the larger hemisphere = 20 inches:
\[
V_{\text{larger}} = \frac{2}{3} \times 3.14 \times 20^3 = \frac{2}{3} \times 3.14 \times 8000 = 16755.73 \text{ in}^3 \approx 16755 \text{ in}^3
\]
- Radius of the smaller hemisphere = 15 inches:
\[
V_{\text{smaller}} = \frac{2}{3} \times 3.14 \times 15^3 = \frac{2}{3} \times 3.14 \times 3375 = 7061.25 \text{ in}^3 \approx 7065 \text{ in}^3
\]
- Volume of the sculpture:
\[
16755 - 7065 = 9690 \text{ in}^3 \approx 9682 \text{ in}^3
\]
Therefore, the answer is:
- **9,682 in.3**
Volume of Spheres Quick Check
Use the formula for the volume of a sphere to find the volume of a basketball that has a diameter of 9.2 inches. Use 3.14 as an approximation for pi and round your answer to the nearest hundredth, if necessary.(1 point)
Responses
3,260.11 in.3
3,260.11 inches cubed
115.55 in.3
115.55 inches cubed
1,222.54 in.3
1,222.54 in cubed
407.51 in.3
Find the volume of a sphere with a radius of 1.8 yards. Leave your answer in terms of pi. Round your answer to the nearest hundredth, if necessary.(1 point)
Responses
7.78π yds3
7.78 pi yds cubed
23.34π yds3
23.34 pi yds cubed
24.43π yds3
24.43 pi yds cubed
7.2π yds3
Use the formula for the volume of a sphere to solve the following problem.
How much air is needed to fill a basketball with a diameter of 9 inches? Use 3.14 as an approximation for pi. Round your answer to the nearest tenth.
(1 point)
Responses
3,052.1 in.3
3,052.1 inches cubed
381.5 in.3
381.5 inches cubed
113 in.3
113 inches cubed
190.8 in.3
The volume of a container measures 4,000 mm3 and contains 50 marbles. Each marble has a radius of 2 mm. Use 3.14 as an approximation for pi. How much space in the box is unused? Round your answer to the nearest tenth.(1 point)
Responses
3,900 mm3
3,900 mm cubed
2,325.3 mm3
2,325.3 mm cubed
33.5 mm3
33.5 mm cubed
1,674.7 mm3
Kairay created a sculpture by forming a hemisphere with a radius of 20 inches and then removing a hemisphere with a radius of 15 inches from it. Calculate the volume of his sculpture. Use 3.14 as an approximation for pi. Round your answer to the nearest whole number.(1 point)
Responses
9,682 in.3
9,682 inches cubed
7,065 in.3
7,065 inches cubed
16,747 in.3
16,747 inches cubed
19,363 in.3
1 answer
1 answer