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

6.3 in3
6.3 in3

67.0 in3
67.0 in3

409.2 in3
409.2 in3

1 answer

To find the difference in volumes between the youth and adult softballs, we can use the formula for the volume of a sphere:

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

First, we need to find the radius of each softball:

  1. Youth softball:

    • Diameter = 3.5 in
    • Radius \( r = \frac{3.5}{2} = 1.75 \) in
  2. Adult softball:

    • Diameter = 3.8 in
    • Radius \( r = \frac{3.8}{2} = 1.9 \) in

Next, we can calculate the volume of each softball using \( \pi \approx 3.14 \).

Volume of youth softball:

\[ V_{youth} = \frac{4}{3} \times 3.14 \times (1.75)^3 \] Calculating \( (1.75)^3 \): \[ (1.75)^3 = 1.75 \times 1.75 \times 1.75 \approx 5.359375 \] Now substitute this value into the volume formula: \[ V_{youth} \approx \frac{4}{3} \times 3.14 \times 5.359375 \approx \frac{4}{3} \times 16.82781 \approx 22.43708 \] Thus, \[ V_{youth} \approx 22.4 \text{ in}^3 \]

Volume of adult softball:

\[ V_{adult} = \frac{4}{3} \times 3.14 \times (1.9)^3 \] Calculating \( (1.9)^3 \): \[ (1.9)^3 = 1.9 \times 1.9 \times 1.9 \approx 6.859 \] Substituting this into the volume formula: \[ V_{adult} \approx \frac{4}{3} \times 3.14 \times 6.859 \approx \frac{4}{3} \times 21.57726 \approx 28.76968 \] Thus, \[ V_{adult} \approx 28.8 \text{ in}^3 \]

Now find the difference in volume: \[ \text{Difference} = V_{adult} - V_{youth} \approx 28.8 - 22.4 \approx 6.4 \text{ in}^3 \]

Rounding to the nearest tenth, the approximate difference in their volumes is:

\[ \text{Difference} \approx 6.3 \text{ in}^3 \]

So the correct response is:

6.3 in³