Question

If the diameter of a youth softball is 3.5 in.
and the diameter of an adult softball is 3.8 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

51.1 in3
51.1 in3

67.0 in3
67.0 in3

6.3 in3
6.3 in3

409.2 in3

Answer 1

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

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

Where \( r \) is the radius of the sphere. The radius is half of the diameter.

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

2. **Calculate the volume of the youth softball**:
\[
V_{\text{youth}} = \frac{4}{3} \pi (1.75)^3
\]
Calculating \( (1.75)^3 \):
\[
(1.75)^3 = 5.359375
\]

Now, substitute into the volume formula:
\[
V_{\text{youth}} = \frac{4}{3} \times 3.14 \times 5.359375
\]
\[
V_{\text{youth}} \approx \frac{4 \times 3.14 \times 5.359375}{3}
\]
\[
V_{\text{youth}} \approx \frac{67.1908225}{3} \approx 22.39694083 \text{ in}^3
\]

3. **Calculate the radius of the adult softball**:
\[
r_{\text{adult}} = \frac{3.8}{2} = 1.9 \text{ in}
\]

4. **Calculate the volume of the adult softball**:
\[
V_{\text{adult}} = \frac{4}{3} \pi (1.9)^3
\]
Calculating \( (1.9)^3 \):
\[
(1.9)^3 = 6.859
\]

Now, substitute into the volume formula:
\[
V_{\text{adult}} = \frac{4}{3} \times 3.14 \times 6.859
\]
\[
V_{\text{adult}} \approx \frac{4 \times 3.14 \times 6.859}{3}
\]
\[
V_{\text{adult}} \approx \frac{86.21352}{3} \approx 28.73784 \text{ in}^3
\]

5. **Calculate the difference in volumes**:
\[
\text{Difference} = V_{\text{adult}} - V_{\text{youth}}
\]
\[
\text{Difference} \approx 28.73784 - 22.39694083 \approx 6.34089917 \text{ in}^3
\]

Rounding this to the nearest tenth gives:
\[
\text{Difference} \approx 6.3 \text{ in}^3
\]

Thus, the approximate difference in their volumes is **6.3 in³.**