Question

If the diameter of a softball is 3.5 in.
and the diameter of a baseball is 2.9 in.
, what is the approximate difference in their volumes?

Use 3.14 for π
and round your answer to the nearest tenth.

V=43πr3
(1 point)
Responses

12.8 in3
12.8 in3

9.6 in3
9.6 in3

35.2 in3
35.2 in3

22.4 in3

Answer 1

To find the difference in volumes between a softball and a baseball, we can use the formula for the volume of a sphere:

\[
V = \frac{4}{3} \pi r^3
\]

First, we need to calculate the radius for both the softball and the baseball.

### Step 1: Calculate the radii
- Radius of the softball:
\[
r_s = \frac{3.5 \, \text{in}}{2} = 1.75 \, \text{in}
\]
- Radius of the baseball:
\[
r_b = \frac{2.9 \, \text{in}}{2} = 1.45 \, \text{in}
\]

### Step 2: Calculate the volumes
- Volume of the softball:
\[
V_s = \frac{4}{3} \pi r_s^3 = \frac{4}{3} \times 3.14 \times (1.75)^3
\]
Calculating \( (1.75)^3 \):
\[
(1.75)^3 = 5.359375
\]
Now substituting back:
\[
V_s = \frac{4}{3} \times 3.14 \times 5.359375 \approx \frac{4 \times 3.14 \times 5.359375}{3}
\]
Calculating \( 4 \times 3.14 \):
\[
4 \times 3.14 = 12.56
\]
Now multiply:
\[
12.56 \times 5.359375 \approx 67.3
\]
Then divide by 3:
\[
V_s \approx \frac{67.3}{3} \approx 22.43 \, \text{in}^3
\]

Now rounding:
\[
V_s \approx 22.4 \, \text{in}^3
\]

- Volume of the baseball:
\[
V_b = \frac{4}{3} \pi r_b^3 = \frac{4}{3} \times 3.14 \times (1.45)^3
\]
Calculating \( (1.45)^3 \):
\[
(1.45)^3 = 3.049625
\]
Now substituting back:
\[
V_b = \frac{4}{3} \times 3.14 \times 3.049625 \approx \frac{4 \times 3.14 \times 3.049625}{3}
\]
Calculating:
\[
4 \times 3.14 = 12.56
\]
Now multiply:
\[
12.56 \times 3.049625 \approx 38.34
\]
Then divide by 3:
\[
V_b \approx \frac{38.34}{3} \approx 12.78 \, \text{in}^3
\]

### Step 3: Calculate the difference
Now, find the difference in volumes:
\[
\text{Difference} = V_s - V_b \approx 22.4 - 12.8 \approx 9.6 \, \text{in}^3
\]

### Final Answer
Thus, the approximate difference in their volumes is:

**9.6 in³**