The population size $P(t)$ of the species can be modeled by the given function:

$P(t) = \frac{1800}{1 + 6e^{-0.34t}}$

1. Finding the Initial Population Size $P(0)$

The initial population size corresponds to $t = 0$.

Substitute $t = 0$ into the equation:

$P(0) = \frac{1800}{1 + 6e^{-0.34(0)}}$

Since $e^0 = 1$:

$P(0) = \frac{1800}{1 + 6 \times 1} = \frac{1800}{1 + 6} = \frac{1800}{7} \approx 257.14$

So, the initial population size is approximately 257 fish.

2. Finding the Population After 9 Years $P(9)$

Substitute $t = 9$ into the equation:

$P(9) = \frac{1800}{1 + 6e^{-0.34(9)}}$

Now, calculate $-0.34 \times 9$ and the exponential:

$P(9) = \frac{1800}{1 + 6e^{-3.06}}$

Calculate $e^{-3.06}$ and substitute it back:

$P(9) = \frac{1800}{1 + 6 \times e^{-3.06}}$

$e^{-3.06} \approx 0.047$

$P(9) = \frac{1800}{1 + 6 \times 0.047} = \frac{1800}{1 + 0.282} = \frac{1800}{1.282} \approx 1403.43$

So, the population after 9 years is approximately 1403 fish.

Summary:

• Initial Population: Approximately 257 fish.
• Population After 9 Years: Approximately 1403 fish.

Would you like any further details or have any questions?

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Related Questions:

1. How does the population growth rate change over time?
2. What happens to the population as $t$ approaches infinity?
3. Can you derive the time at which the population will reach half of its carrying capacity?
4. What is the maximum population the lake can sustain based on this model?
5. How would changes in the growth rate constant affect the population over time?

Tip: Understanding the exponential function $e^x$ is crucial for solving differential equations and analyzing population dynamics.