Volume of Spheres Quick Check

1. 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

1,222.54 in.3
1,222.54 in cubed

115.55 in.3
115.55 inches cubed

407.51 in.3
407.51 inches cubed

3,260.11 in.3

2. 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

7.2π yds3
7.2 pi yds cubed

24.43π yds3
24.43 pi yds cubed

23.34π yds3
23.34 pi yds cubed

7.78π yds3

3. 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

113 in.3
113 inches cubed

3,052.1 in.3
3,052.1 inches cubed

190.8 in.3
190.8 inches cubed

381.5 in.3

4. 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

2,325.3 mm3
2,325.3 mm cubed

33.5 mm3
33.5 mm cubed

3,900 mm3
3,900 mm cubed

1,674.7 mm3

5. 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

2,325.3 mm3
2,325.3 mm cubed

33.5 mm3
33.5 mm cubed

3,900 mm3
3,900 mm cubed

1,674.7 mm3

User Icon for Bot GPT-4o Bot GPT-4o answered
4 months ago

Let's solve the problems step by step.

1. **Volume of a basketball with diameter 9.2 inches:**

The formula for the volume of a sphere is:
\[
V = \frac{4}{3} \pi r^3
\]
First, find the radius (r), which is half of the diameter:
\[
r = \frac{9.2}{2} = 4.6 \text{ inches}
\]
Plug in the values and use 3.14 for \(\pi\):
\[
V = \frac{4}{3} \times 3.14 \times (4.6)^3
\]
Computing the volume:
\[
V = \frac{4}{3} \times 3.14 \times 97.336
\]
\[
V \approx 407.51 \text{ in}^3
\]
So, the correct answer is:
\[
407.51 \text{ inches cubed}
\]

2. **Volume of a sphere with radius 1.8 yards:**

Using the formula \( V = \frac{4}{3} \pi r^3 \):
\[
V = \frac{4}{3} \pi (1.8)^3
\]
Let’s compute \( (1.8)^3 \):
\[
(1.8)^3 = 5.832
\]
So,
\[
V = \frac{4}{3} \pi \times 5.832
\]
\[
V \approx 7.776 \pi \, \text{yds}^3
\]
Rounded to the nearest hundredth:
\[
V \approx 7.78 \pi \, \text{yds}^3
\]
The correct answer is:
\[
7.78 \pi \text{ yds}^3
\]

3. **Air needed to fill a basketball with diameter 9 inches:**

Radius \( r = \frac{9}{2} = 4.5 \text{ inches} \):
\[
V = \frac{4}{3} \times 3.14 \times (4.5)^3
\]
Compute \( (4.5)^3 \):
\[
(4.5)^3 = 91.125
\]
\[
V = \frac{4}{3} \times 3.14 \times 91.125
\]
\[
V \approx 381.51 \text{ in}^3
\]
Rounded to the nearest tenth:
\[
V \approx 381.5 \text{ in}^3
\]
The correct answer is:
\[
381.5 \text{ inches cubed}
\]

4 and 5. **Unused space in the container with total volume 4000 mm³ and 50 marbles with radius 2 mm:**
Calculate the volume of one marble:
\[
r = 2 \text{ mm}
\]
Using the volume formula for a sphere:
\[
V = \frac{4}{3} \times 3.14 \times (2)^3
\]
Compute:
\[
V = \frac{4}{3} \times 3.14 \times 8
\]
\[
V = \frac{4}{3} \times 25.12
\]
\[
V \approx 33.49 \text{ mm}^3
\]
Volume of 50 marbles:
\[
50 \times 33.49 \approx 1674.5 \text{ mm}^3
\]
Unused space can be calculated as:
\[
4000 \text{ mm}^3 - 1674.5 \text{ mm}^3 = 2325.5 \text{ mm}^3
\]
Round to the nearest tenth:
\[
2325.5 \approx 2325.3
\]
The correct answer is:
\[
2,325.3 \text{ mm}^3
\]

So, the correct answers are:
1. 407.51 inches cubed
2. 7.78 π yds cubed
3. 381.5 inches cubed
4. 2,325.3 mm cubed
5. 2,325.3 mm cubed