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

9.6 in3
9.6 in3

12.8 in3
12.8 in3

22.4 in3
22.4 in3

35.2 in3

Answer 1

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

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

First, we need to find the radius for each ball:

- For the softball (diameter = 3.5 in):
\[
r_{\text{softball}} = \frac{3.5}{2} = 1.75 \text{ in}
\]

- For the baseball (diameter = 2.9 in):
\[
r_{\text{baseball}} = \frac{2.9}{2} = 1.45 \text{ in}
\]

Now, we will calculate the volumes of both the softball and the baseball using \( \pi \approx 3.14 \).

1. **Volume of the softball:**
\[
V_{\text{softball}} = \frac{4}{3} \times 3.14 \times (1.75)^3
\]
\[
(1.75)^3 = 5.359375
\]
\[
V_{\text{softball}} = \frac{4}{3} \times 3.14 \times 5.359375
\]
\[
V_{\text{softball}} \approx \frac{4}{3} \times 3.14 \times 5.359375 \approx 22.4 \text{ in}^3
\]

2. **Volume of the baseball:**
\[
V_{\text{baseball}} = \frac{4}{3} \times 3.14 \times (1.45)^3
\]
\[
(1.45)^3 = 3.052625
\]
\[
V_{\text{baseball}} = \frac{4}{3} \times 3.14 \times 3.052625
\]
\[
V_{\text{baseball}} \approx \frac{4}{3} \times 3.14 \times 3.052625 \approx 12.8 \text{ in}^3
\]

3. **Difference in volumes:**
\[
\text{Difference} = V_{\text{softball}} - V_{\text{baseball}}
\]
\[
\text{Difference} = 22.4 - 12.8 = 9.6 \text{ in}^3
\]

Therefore, the approximate difference in their volumes is **9.6 in³**.