- Substitute the given height \(h = 8.5\) feet into the formula \(V = 0.95 h^{2.9}\).
- Calculate the power: \(8.5^{2.9} \approx 495.8095\).
- Multiply by the constant coefficient: \(0.95 \times 495.8095 \approx 471.019\).
- Round the result to the nearest hundredth to get \(\boxed{471.02 \text{ ft}^3}\).
Explanation
- Analysis of the Problem
The problem asks for the volume of water in a container, modeled by the equation \(V = 0.95 h^{2.9}\), where \(V\) is the volume in cubic feet and \(h\) is the height of the water in feet. We are given the height \(h = 8.5\) feet and need to calculate the corresponding volume \(V\), rounding the final result to the nearest hundredth. Our approach is to substitute \(h = 8.5\) into the given formula and perform the necessary calculations.
- Calculation of the Volume
To calculate the volume, we substitute \(h = 8.5\) into the expression \(0.95 \times h^{2.9}\):
\[V = 0.95 \times (8.5)^{2.9}\]
First, evaluate the exponential term \(8.5^{2.9}\). Using the calculation tool, we find:
\[(8.5)^{2.9} \approx 495.809546...\]
Next, multiply this result by \(0.95\):
\[V \approx 0.95 \times 495.809546 \approx 471.019069...\]
Rounding this value to the nearest hundredth, we get \(471.02\).
- Conclusion and Final Selection
Comparing the calculated value of \(471.02 \text{ ft}^3\) with the given options:
- \(434.29 \text{ ft}^3\)
- \(327.6 \text{ ft}^3\)
- \(385.97 \text{ ft}^3\)
- \(471.02 \text{ ft}^3\)
Our result matches the last option. Therefore, the volume of water in the container is \(\boxed{471.02 \text{ ft}^3}\).
Examples
This type of calculation is common in engineering and fluid mechanics, where the volume of oddly shaped tanks is determined by empirical models based on height. For instance, an environmental engineer might use such a formula to monitor water levels in a reservoir or a chemical processing tank to ensure safety and efficient resource management. Knowing how to manipulate exponential models and interpret them is a foundational skill in many scientific and technical fields.