Question

If the diameter of a softball is%C2%A0 3.5%C2%A0in. %C2%A0and the diameter of a baseball is%C2%A0 2.9%C2%A0in. , what is the approximate difference in their volumes? Use 3.14 for%C2%A0 π %C2%A0and round your answer to the nearest tenth. V=43πr3 (1 point) Responses 9.6%C2%A0in3 9.6%C2%A0in3 35.2%C2%A0in3 35.2%C2%A0in3 12.8%C2%A0in3 12.8%C2%A0in3 22.4%C2%A0in3 22.4%C2%A0in3

Answer 1

To find the approximate 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 each ball:

- **Softball:** Diameter = 3.5 in; Radius \( r = \frac{3.5}{2} = 1.75 \) in
- **Baseball:** Diameter = 2.9 in; Radius \( r = \frac{2.9}{2} = 1.45 \) in

Now, we can calculate the volumes for each ball using \( \pi \approx 3.14 \):

1. **Volume of the softball:**

\[
V_{softball} = \frac{4}{3} \pi (1.75)^3
\]

Calculating \( (1.75)^3 \):

\[
(1.75)^3 = 1.75 \times 1.75 \times 1.75 = 5.359375
\]

Now, substituting the value into the volume formula:

\[
V_{softball} = \frac{4}{3} \times 3.14 \times 5.359375 \approx \frac{4 \times 3.14 \times 5.359375}{3}
\]

Calculating \( 4 \times 3.14 \):

\[
4 \times 3.14 = 12.56
\]

Now calculate:

\[
V_{softball} \approx \frac{12.56 \times 5.359375}{3} \approx \frac{67.3271}{3} \approx 22.44237
\]

Rounding to the nearest tenth, we get:

\[
V_{softball} \approx 22.4 \, \text{in}^3
\]

2. **Volume of the baseball:**

\[
V_{baseball} = \frac{4}{3} \pi (1.45)^3
\]

Calculating \( (1.45)^3 \):

\[
(1.45)^3 = 1.45 \times 1.45 \times 1.45 = 3.043125
\]

Substituting the value into the volume formula:

\[
V_{baseball} = \frac{4}{3} \times 3.14 \times 3.043125 \approx \frac{4 \times 3.14 \times 3.043125}{3}
\]

Calculate \( 4 \times 3.14 \):

\[
4 \times 3.14 = 12.56
\]

Now calculate:

\[
V_{baseball} \approx \frac{12.56 \times 3.043125}{3} \approx \frac{38.2648}{3} \approx 12.75493
\]

Rounding to the nearest tenth, we get:

\[
V_{baseball} \approx 12.8 \, \text{in}^3
\]

3. **Calculate the difference in volumes:**

\[
\text{Difference} = V_{softball} - V_{baseball} \approx 22.4 - 12.8 = 9.6 \, \text{in}^3
\]

Thus, the approximate difference in their volumes is:

\[
\boxed{9.6 \, \text{in}^3}
\]