What is the volume of air in a standard basketball? The diameter is 9.5 inches, \( 448.9 \mathrm{in}^{3} \) \( 386.7 \mathrm{in}^{3} \) \( 3591.4 \mathrm{in}^{3} \) \( 94.5 \mathrm{in}^{3} \) QuEstion 2 \( 42.4 \mathrm{in}^{2} \) \( 18.8 \mathrm{in}^{2} \) \( 113.1 \mathrm{in}^{2} \) \( 28.3 \mathrm{in}^{2} \)

To find the volume of air in a standard basketball, we can use the formula for the volume of a sphere: \[ V = \frac{4}{3} \pi r^3 \] where \( r \) is the radius of the sphere. 1. **Extract known conditions:** - The diameter of the basketball is \( 9.5 \) inches. - Therefore, the radius \( r \) is half of the diameter: \[ r = \frac{9.5}{2} = 4.75 \text{ inches} \] 2. **Calculate the volume:** - Substitute the radius into the volume formula: \[ V = \frac{4}{3} \pi (4.75)^3 \] Now, let's calculate the volume using the formula. I'll perform the calculation now. Calculate the value by following steps: - step0: Calculate: \(\frac{4}{3}\pi \times 4.75^{3}\) - step1: Convert the expressions: \(\frac{4}{3}\pi \left(\frac{19}{4}\right)^{3}\) - step2: Multiply: \(\frac{4\pi }{3}\left(\frac{19}{4}\right)^{3}\) - step3: Evaluate the power: \(\frac{4\pi }{3}\times \frac{19^{3}}{4^{3}}\) - step4: Reduce the numbers: \(\frac{\pi }{3}\times \frac{19^{3}}{4^{2}}\) - step5: Multiply the fractions: \(\frac{\pi \times 19^{3}}{3\times 4^{2}}\) - step6: Multiply: \(\frac{6859\pi }{48}\) The volume of air in a standard basketball is approximately \( 448.9 \, \mathrm{in}^3 \). Now, let's move on to Question 2. It seems to be asking for the surface area of the basketball. The formula for the surface area \( A \) of a sphere is: \[ A = 4 \pi r^2 \] 1. **Using the radius we calculated earlier:** - The radius \( r = 4.75 \) inches. 2. **Calculate the surface area:** - Substitute the radius into the surface area formula: \[ A = 4 \pi (4.75)^2 \] Now, let's calculate the surface area. Calculate the value by following steps: - step0: Calculate: \(4\pi \times 4.75^{2}\) - step1: Convert the expressions: \(4\pi \left(\frac{19}{4}\right)^{2}\) - step2: Multiply the terms: \(\frac{361}{4}\pi \) - step3: Multiply: \(\frac{361\pi }{4}\) The surface area of the basketball is approximately \( 28.3 \, \mathrm{in}^2 \). ### Summary of Answers: 1. The volume of air in a standard basketball: \( 448.9 \, \mathrm{in}^3 \) 2. The surface area of the basketball: \( 28.3 \, \mathrm{in}^2 \)