To determine the fish population after 2 years using the given model [tex]\( P(t) = \frac{1300}{1 + 9e^{-0.9t}} \)[/tex]:
1. Identify the Constants and the Exponent:
- The given equation is [tex]\( P(t) = \frac{1300}{1 + 9e^{-0.9t}} \)[/tex].
- We need to calculate the population at [tex]\( t = 2 \)[/tex] years.
2. Substitute 2 for t:
- Replace [tex]\( t \)[/tex] with 2 in the equation [tex]\( P(t) \)[/tex]:
[tex]\[
P(2) = \frac{1300}{1 + 9 e^{-0.9 \cdot 2}}
\][/tex]
3. Simplify the Exponent:
- Calculate the exponent value:
[tex]\[
-0.9 \cdot 2 = -1.8
\][/tex]
- So the expression becomes:
[tex]\[
P(2) = \frac{1300}{1 + 9 e^{-1.8}}
\][/tex]
4. Evaluate [tex]\(e^{-1.8}\)[/tex]:
- Find the numerical value of [tex]\( e^{-1.8} \)[/tex]:
[tex]\[
e^{-1.8} \approx 0.165298
\][/tex]
5. Simplify the Denominator:
- Compute the denominator:
[tex]\[
1 + 9 \cdot 0.165298 \approx 1 + 1.487682 = 2.487682
\][/tex]
6. Apply the Denominator in the Population Formula:
- Substitute the calculated denominator back into the equation:
[tex]\[
P(2) = \frac{1300}{2.487682}
\][/tex]
7. Compute the Population:
- Divide 1300 by the calculated denominator:
[tex]\[
P(2) \approx 522.573151
\][/tex]
8. Round to the Nearest Whole Number:
- Round the result to the nearest whole number:
[tex]\[
P(2) \approx 523
\][/tex]
Therefore, to the nearest whole number, the fish population after 2 years will be approximately 523 fish.
