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 \( \pi \approx 3.14 \):
- 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:
\[ V_{softball} \approx 22.4 , \text{in}^3 \]
- 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:
\[ V_{baseball} \approx 12.8 , \text{in}^3 \]
- 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:
\[ \boxed{9.6 , \text{in}^3} \]