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

1 answer

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.