To solve the given problem:

1. Determine an exponential equation to model the situation:

The exponential equation typically takes the form:

$A(t) = A_0 \cdot e^{-kt}$

Where:

• $A(t)$ is the amount of water remaining at time $t$,
• $A_0$ is the initial amount of water (500 gallons here),
• $k$ is the decay constant, and
• $t$ is time in hours.

From the table:

• At $t = 0$, $A(0) = 500$ (this confirms $A_0 = 500$).
• Use another data point, e.g., $t = 1$, $A(1) = 480$, to find $k$.

Substitute into the formula: $480 = 500 \cdot e^{-k(1)}$

Solve for $k$: $e^{-k} = \frac{480}{500} = 0.96$

Take the natural logarithm on both sides: $-k = \ln(0.96)$

$k \approx -\ln(0.96) \approx 0.04082$

Thus, the equation becomes: $A(t) = 500 \cdot e^{-0.04082t}$

2. Determine how much water remains after 10 hours:

Using the model $A(t) = 500 \cdot e^{-0.04082t}$, substitute $t = 10$:

$A(10) = 500 \cdot e^{-0.04082 \cdot 10}$

$A(10) = 500 \cdot e^{-0.4082} \approx 500 \cdot 0.6653 \approx 332.65 \, \text{gallons.}$

3. Find when the tank contains about half its volume:

Half of the initial volume is $\frac{500}{2} = 250$. Solve for $t$ when $A(t) = 250$:

$250 = 500 \cdot e^{-0.04082t}$

$e^{-0.04082t} = \frac{250}{500} = 0.5$

Take the natural logarithm on both sides: $-0.04082t = \ln(0.5)$

$t = \frac{\ln(0.5)}{-0.04082} \approx \frac{-0.6931}{-0.04082} \approx 16.98 \, \text{hours.}$

Final Answers:

1. The equation is $A(t) = 500 \cdot e^{-0.04082t}$.
2. After 10 hours, the tank contains approximately 332.65 gallons.
3. The tank will contain about half its volume after approximately 16.98 hours.

Would you like a more detailed explanation or have other questions? Here are 5 related questions you could explore:

1. How would the equation change if the evaporation rate doubled?
2. What is the physical significance of the constant $k$ in the equation?
3. Can we use logarithms to estimate the water level at any other time point?
4. How does the decay constant affect the speed of water reduction?
5. How could this model be adapted for non-constant evaporation rates?

Tip: Always check if the exponential model fits your data points closely by testing multiple values!