Question
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.
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.
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.
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.
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.
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.
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.
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.
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.
Answers
GPT-4o mini
To calculate the volumes requested using the formula for the volume of a sphere, we will use the formula:
\[
V = \frac{4}{3} \pi r^3
\]
where \( r \) is the radius of the sphere.
### 1. Volume of a basketball with a diameter of 9.2 inches
First, we find the radius:
\[
r = \frac{diameter}{2} = \frac{9.2}{2} = 4.6 \text{ inches}
\]
Now we use the volume formula:
\[
V = \frac{4}{3} \pi (4.6)^3
\]
Calculating \( (4.6)^3 \):
\[
(4.6)^3 = 97.336
\]
Now substituting back into the volume formula:
\[
V = \frac{4}{3} \times 3.14 \times 97.336
\]
Calculating \( \frac{4}{3} \):
\[
\frac{4}{3} \approx 1.3333
\]
Now calculating the volume:
\[
V \approx 1.3333 \times 3.14 \times 97.336 \approx 406.193
\]
Rounding to the nearest hundredth:
\[
\text{Volume} \approx 406.19 \text{ cubic inches}
\]
### 2. Volume of a sphere with a radius of 1.8 yards
Using the volume formula:
\[
V = \frac{4}{3} \pi (1.8)^3
\]
Calculating \( (1.8)^3 \):
\[
(1.8)^3 = 5.832
\]
So, the volume is:
\[
V = \frac{4}{3} \pi (5.832) = \frac{4 \cdot 5.832}{3} \pi
\]
Calculating:
\[
\frac{4 \cdot 5.832}{3} = 7.776
\]
Thus, the volume in terms of \( \pi \) is:
\[
V \approx 7.776\pi \text{ cubic yards}
\]
### 3. How much air is needed to fill a basketball with a diameter of 9 inches?
First, we find the radius:
\[
r = \frac{9}{2} = 4.5 \text{ inches}
\]
Now apply the volume formula:
\[
V = \frac{4}{3} \pi (4.5)^3
\]
Calculating \( (4.5)^3 \):
\[
(4.5)^3 = 91.125
\]
Thus, the volume is:
\[
V = \frac{4}{3} \times 3.14 \times 91.125 \approx 381.696
\]
Rounding to the nearest tenth:
\[
\text{Volume} \approx 381.7 \text{ cubic inches}
\]
### 4. Unused space in a box that measures 4,000 mm³ containing 50 marbles with a radius of 2 mm
First, find the volume of one marble:
Using the volume formula:
\[
V = \frac{4}{3} \pi (2)^3
\]
Calculating \( (2)^3 = 8 \):
\[
V = \frac{4}{3} \pi (8) = \frac{32}{3} \pi \approx 33.51 \text{ mm}^3
\]
Now for 50 marbles:
\[
50 \cdot \frac{32}{3} \approx 50 \cdot 33.51 \approx 1675.5 \text{ mm}^3
\]
Calculating unused space:
\[
\text{Unused space} = 4000 - 1675.5 \approx 2324.5
\]
Rounding to the nearest tenth:
\[
\text{Unused space} \approx 2324.5 \text{ mm}^3
\]
### 5. Volume of Kairay’s sculpture
Calculating the volume of the large hemisphere with a radius of 20 inches:
\[
V_{large} = \frac{2}{3} \pi (20)^3
\]
Calculating \( (20)^3 = 8000 \):
\[
V_{large} = \frac{2}{3} \times 3.14 \times 8000 \approx 16758.67 \text{ cubic inches}
\]
Now for the small hemisphere with a radius of 15 inches:
\[
V_{small} = \frac{2}{3} \pi (15)^3
\]
Calculating \( (15)^3 = 3375 \):
\[
V_{small} = \frac{2}{3} \times 3.14 \times 3375 \approx 7068.75 \text{ cubic inches}
\]
Now finding the volume of the sculpture:
\[
V_{sculpture} = V_{large} - V_{small} \approx 16758.67 - 7068.75 \approx 9689.92
\]
Rounding to the nearest whole number:
\[
\text{Volume of sculpture} \approx 9690 \text{ cubic inches}
\]
In summary:
1. Volume of the basketball: 406.19 cubic inches
2. Volume of sphere in terms of \( \pi \): \( 7.776\pi \) cubic yards
3. Air needed for filling basketball: 381.7 cubic inches
4. Unused space in box: 2324.5 mm³
5. Volume of Kairay's sculpture: 9690 cubic inches
\[
V = \frac{4}{3} \pi r^3
\]
where \( r \) is the radius of the sphere.
### 1. Volume of a basketball with a diameter of 9.2 inches
First, we find the radius:
\[
r = \frac{diameter}{2} = \frac{9.2}{2} = 4.6 \text{ inches}
\]
Now we use the volume formula:
\[
V = \frac{4}{3} \pi (4.6)^3
\]
Calculating \( (4.6)^3 \):
\[
(4.6)^3 = 97.336
\]
Now substituting back into the volume formula:
\[
V = \frac{4}{3} \times 3.14 \times 97.336
\]
Calculating \( \frac{4}{3} \):
\[
\frac{4}{3} \approx 1.3333
\]
Now calculating the volume:
\[
V \approx 1.3333 \times 3.14 \times 97.336 \approx 406.193
\]
Rounding to the nearest hundredth:
\[
\text{Volume} \approx 406.19 \text{ cubic inches}
\]
### 2. Volume of a sphere with a radius of 1.8 yards
Using the volume formula:
\[
V = \frac{4}{3} \pi (1.8)^3
\]
Calculating \( (1.8)^3 \):
\[
(1.8)^3 = 5.832
\]
So, the volume is:
\[
V = \frac{4}{3} \pi (5.832) = \frac{4 \cdot 5.832}{3} \pi
\]
Calculating:
\[
\frac{4 \cdot 5.832}{3} = 7.776
\]
Thus, the volume in terms of \( \pi \) is:
\[
V \approx 7.776\pi \text{ cubic yards}
\]
### 3. How much air is needed to fill a basketball with a diameter of 9 inches?
First, we find the radius:
\[
r = \frac{9}{2} = 4.5 \text{ inches}
\]
Now apply the volume formula:
\[
V = \frac{4}{3} \pi (4.5)^3
\]
Calculating \( (4.5)^3 \):
\[
(4.5)^3 = 91.125
\]
Thus, the volume is:
\[
V = \frac{4}{3} \times 3.14 \times 91.125 \approx 381.696
\]
Rounding to the nearest tenth:
\[
\text{Volume} \approx 381.7 \text{ cubic inches}
\]
### 4. Unused space in a box that measures 4,000 mm³ containing 50 marbles with a radius of 2 mm
First, find the volume of one marble:
Using the volume formula:
\[
V = \frac{4}{3} \pi (2)^3
\]
Calculating \( (2)^3 = 8 \):
\[
V = \frac{4}{3} \pi (8) = \frac{32}{3} \pi \approx 33.51 \text{ mm}^3
\]
Now for 50 marbles:
\[
50 \cdot \frac{32}{3} \approx 50 \cdot 33.51 \approx 1675.5 \text{ mm}^3
\]
Calculating unused space:
\[
\text{Unused space} = 4000 - 1675.5 \approx 2324.5
\]
Rounding to the nearest tenth:
\[
\text{Unused space} \approx 2324.5 \text{ mm}^3
\]
### 5. Volume of Kairay’s sculpture
Calculating the volume of the large hemisphere with a radius of 20 inches:
\[
V_{large} = \frac{2}{3} \pi (20)^3
\]
Calculating \( (20)^3 = 8000 \):
\[
V_{large} = \frac{2}{3} \times 3.14 \times 8000 \approx 16758.67 \text{ cubic inches}
\]
Now for the small hemisphere with a radius of 15 inches:
\[
V_{small} = \frac{2}{3} \pi (15)^3
\]
Calculating \( (15)^3 = 3375 \):
\[
V_{small} = \frac{2}{3} \times 3.14 \times 3375 \approx 7068.75 \text{ cubic inches}
\]
Now finding the volume of the sculpture:
\[
V_{sculpture} = V_{large} - V_{small} \approx 16758.67 - 7068.75 \approx 9689.92
\]
Rounding to the nearest whole number:
\[
\text{Volume of sculpture} \approx 9690 \text{ cubic inches}
\]
In summary:
1. Volume of the basketball: 406.19 cubic inches
2. Volume of sphere in terms of \( \pi \): \( 7.776\pi \) cubic yards
3. Air needed for filling basketball: 381.7 cubic inches
4. Unused space in box: 2324.5 mm³
5. Volume of Kairay's sculpture: 9690 cubic inches