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 35.2 in3 35.2 in3 22.4 in3 22.4 in3 12.8 in3 12.8 in3

Answer 1

To find the approximate difference in volumes between a softball and a baseball using the formula for the volume of a sphere \(V = \frac{4}{3}\pi r^3\), we will first determine the radii of the two balls and then calculate their respective volumes.

1. **Calculate the radius of the softball:**
\[
\text{Radius of softball} = \frac{3.5 \text{ in}}{2} = 1.75 \text{ in}
\]

2. **Calculate the radius of the baseball:**
\[
\text{Radius of baseball} = \frac{2.9 \text{ in}}{2} = 1.45 \text{ in}
\]

3. **Calculate the volume of the softball:**
\[
V_{\text{softball}} = \frac{4}{3} \pi (1.75)^3
\]
First, calculate \( (1.75)^3 \):
\[
(1.75)^3 = 1.75 \times 1.75 \times 1.75 = 5.359375
\]
Now, substitute that back into the volume formula:
\[
V_{\text{softball}} = \frac{4}{3} \times 3.14 \times 5.359375
\]
Calculate \( \frac{4}{3} \times 3.14 \):
\[
\frac{4}{3} \times 3.14 \approx 4.1867
\]
Now multiply:
\[
V_{\text{softball}} \approx 4.1867 \times 5.359375 \approx 22.4 \text{ in}^3
\]

4. **Calculate the volume of the baseball:**
\[
V_{\text{baseball}} = \frac{4}{3} \pi (1.45)^3
\]
First, calculate \( (1.45)^3 \):
\[
(1.45)^3 = 1.45 \times 1.45 \times 1.45 = 3.044625
\]
Now substitute that into the volume formula:
\[
V_{\text{baseball}} = \frac{4}{3} \times 3.14 \times 3.044625
\]
Now we calculate:
\[
V_{\text{baseball}} \approx 4.1867 \times 3.044625 \approx 12.8 \text{ in}^3
\]

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

The approximate difference in their volumes is **9.6 in³**.