A Walmart employee has decided that they would like to completely fill the ball bin in the toy section with one type of ball. The ball has diameter of 9 inches. The bin is 20 inches tall and the opening at the top is 32 inches by 35 inches. How many balls should fit inside the bin? (Hint: Packina soheres in a rectanautar orism take ud \( 190 \% \) of their volume)

To determine how many balls can fit inside the bin, we need to follow these steps: 1. **Calculate the volume of one ball.** The formula for the volume \( V \) of a sphere is given by: \[ V = \frac{4}{3} \pi r^3 \] where \( r \) is the radius of the sphere. The diameter of the ball is 9 inches, so the radius \( r \) is: \[ r = \frac{9}{2} = 4.5 \text{ inches} \] 2. **Calculate the volume of the bin.** The volume \( V \) of a rectangular prism is given by: \[ V = \text{length} \times \text{width} \times \text{height} \] The dimensions of the bin are: - Height = 20 inches - Length = 32 inches - Width = 35 inches 3. **Adjust for packing efficiency.** Since the hint states that packing spheres in a rectangular prism takes up \( 190\% \) of their volume, we will multiply the volume of the balls by \( 1.9 \) to account for this packing efficiency. Now, let's perform the calculations step by step. ### Step 1: Calculate the volume of one ball \[ V_{\text{ball}} = \frac{4}{3} \pi (4.5)^3 \] ### Step 2: Calculate the volume of the bin \[ V_{\text{bin}} = 32 \times 35 \times 20 \] ### Step 3: Calculate the number of balls that fit in the bin The effective volume occupied by the balls will be: \[ V_{\text{effective}} = 1.9 \times V_{\text{ball}} \] The number of balls \( N \) that can fit in the bin is given by: \[ N = \frac{V_{\text{bin}}}{V_{\text{effective}}} \] Now, let's calculate these values. Calculate the value by following steps: - step0: Calculate: \(32\times 35\times 20\) - step1: Multiply the terms: \(1120\times 20\) - step2: Multiply the numbers: \(22400\) Calculate or simplify the expression \( (4/3) * \pi * (4.5)^3 \). Calculate the value by following steps: - step0: Calculate: \(\frac{4}{3}\pi \times 4.5^{3}\) - step1: Convert the expressions: \(\frac{4}{3}\pi \left(\frac{9}{2}\right)^{3}\) - step2: Multiply: \(\frac{4\pi }{3}\left(\frac{9}{2}\right)^{3}\) - step3: Evaluate the power: \(\frac{4\pi }{3}\times \frac{9^{3}}{2^{3}}\) - step4: Rewrite the expression: \(\frac{2^{2}\pi }{3}\times \frac{9^{3}}{2^{3}}\) - step5: Reduce the numbers: \(\frac{\pi }{3}\times \frac{9^{3}}{2}\) - step6: Rewrite the expression: \(\frac{\pi }{3}\times \frac{3^{6}}{2}\) - step7: Reduce the numbers: \(\pi \times \frac{3^{5}}{2}\) - step8: Multiply: \(\frac{\pi \times 3^{5}}{2}\) - step9: Multiply: \(\frac{243\pi }{2}\) Calculate or simplify the expression \( 1.9 * 381.703507 \). Calculate the value by following steps: - step0: Calculate: \(1.9\times 381.703507\) - step1: Multiply the numbers: \(725.2366633\) Calculate or simplify the expression \( 22400 / 725.2366633 \). Calculate the value by following steps: - step0: Calculate: \(\frac{22400}{725.2366633}\) - step1: Convert the expressions: \(\frac{22400}{\frac{7252366633}{10000000}}\) - step2: Multiply by the reciprocal: \(22400\times \frac{10000000}{7252366633}\) - step3: Multiply: \(\frac{22400\times 10000000}{7252366633}\) - step4: Multiply: \(\frac{224000000000}{7252366633}\) Now that we have all the necessary calculations, let's summarize the results: 1. **Volume of one ball:** \[ V_{\text{ball}} \approx 381.70 \text{ cubic inches} \] 2. **Volume of the bin:** \[ V_{\text{bin}} = 22400 \text{ cubic inches} \] 3. **Effective volume occupied by the balls (considering packing efficiency):** \[ V_{\text{effective}} \approx 725.24 \text{ cubic inches} \] 4. **Number of balls that can fit in the bin:** \[ N \approx \frac{22400}{725.24} \approx 30.89 \] Since we cannot have a fraction of a ball, we round down to the nearest whole number. Therefore, the number of balls that should fit inside the bin is: \[ \boxed{30} \]