To find $P(15)$ using the given population model $P(t) = \frac{600}{1 + 7e^{-0.125t}}$, follow these steps:

1. Substitute $t = 15$ into the equation: $P(15) = \frac{600}{1 + 7e^{-0.125 \times 15}}$

2. Calculate the exponent: $-0.125 \times 15 = -1.875$

3. Compute $e^{-1.875}$: $e^{-1.875} \approx 0.153$

4. Substitute this value back into the equation: $P(15) = \frac{600}{1 + 7 \times 0.153}$ $P(15) = \frac{600}{1 + 1.071}$ $P(15) = \frac{600}{2.071}$

5. Divide to find $P(15)$: $P(15) \approx 289.7$

Rounding to the nearest whole number: $P(15) \approx 290$

The population at $t = 15$ is approximately 290.

Would you like more details or have any questions? Here are some questions you might consider:

1. How do we compute exponential values like $e^{-1.875}$?
2. Why is it important to round the final result?
3. Can this model be used to predict long-term population behavior?
4. What happens to the population as $t$ increases indefinitely?
5. How does the initial population size affect the model?

Tip: When rounding numbers, if the digit after the rounding place is 5 or more, round up. Otherwise, round down.