Answer :
To solve this problem, we'll start by analysing the given function for the population size [tex]\( P(t) \)[/tex] of the species of fish in the lake:
[tex]\[ P(t) = \frac{1500}{1 + 9 e^{-0.42 t}} \][/tex]
We need to find the initial population size when [tex]\( t = 0 \)[/tex] and the population size after 9 years when [tex]\( t = 9 \)[/tex].
### Step 1: Initial Population Size ([tex]\( t = 0 \)[/tex])
Let's substitute [tex]\( t = 0 \)[/tex] into the function [tex]\( P(t) \)[/tex]:
[tex]\[ P(0) = \frac{1500}{1 + 9 e^{-0.42 \cdot 0}} \][/tex]
Since [tex]\( e^0 = 1 \)[/tex], this simplifies to:
[tex]\[ P(0) = \frac{1500}{1 + 9 \cdot 1} \][/tex]
[tex]\[ P(0) = \frac{1500}{1 + 9} \][/tex]
[tex]\[ P(0) = \frac{1500}{10} \][/tex]
[tex]\[ P(0) = 150 \][/tex]
Therefore, the initial population size of the species is [tex]\( 150 \)[/tex] fish.
### Step 2: Population Size After 9 Years ([tex]\( t = 9 \)[/tex])
Next, we substitute [tex]\( t = 9 \)[/tex] into the function [tex]\( P(t) \)[/tex]:
[tex]\[ P(9) = \frac{1500}{1 + 9 e^{-0.42 \cdot 9}} \][/tex]
First, calculate the exponent:
[tex]\[ -0.42 \cdot 9 = -3.78 \][/tex]
Now calculate [tex]\( e^{-3.78} \)[/tex]. This value is a bit intricate, but it's a small number. To ease the calculation, we know:
[tex]\[ e^{-3.78} \approx 0.0228 \][/tex]
Substitute this into the denominator:
[tex]\[ P(9) = \frac{1500}{1 + 9 \cdot 0.0228} \][/tex]
[tex]\[ P(9) = \frac{1500}{1 + 0.2052} \][/tex]
[tex]\[ P(9) = \frac{1500}{1.2052} \][/tex]
Finally, simplify the fraction:
[tex]\[ P(9) \approx 1244 \][/tex]
Therefore, the population size after 9 years is approximately [tex]\( 1244 \)[/tex] fish.
### Summary
- Initial population size: [tex]\( 150 \)[/tex] fish
- Population size after 9 years: [tex]\( 1244 \)[/tex] fish
[tex]\[ P(t) = \frac{1500}{1 + 9 e^{-0.42 t}} \][/tex]
We need to find the initial population size when [tex]\( t = 0 \)[/tex] and the population size after 9 years when [tex]\( t = 9 \)[/tex].
### Step 1: Initial Population Size ([tex]\( t = 0 \)[/tex])
Let's substitute [tex]\( t = 0 \)[/tex] into the function [tex]\( P(t) \)[/tex]:
[tex]\[ P(0) = \frac{1500}{1 + 9 e^{-0.42 \cdot 0}} \][/tex]
Since [tex]\( e^0 = 1 \)[/tex], this simplifies to:
[tex]\[ P(0) = \frac{1500}{1 + 9 \cdot 1} \][/tex]
[tex]\[ P(0) = \frac{1500}{1 + 9} \][/tex]
[tex]\[ P(0) = \frac{1500}{10} \][/tex]
[tex]\[ P(0) = 150 \][/tex]
Therefore, the initial population size of the species is [tex]\( 150 \)[/tex] fish.
### Step 2: Population Size After 9 Years ([tex]\( t = 9 \)[/tex])
Next, we substitute [tex]\( t = 9 \)[/tex] into the function [tex]\( P(t) \)[/tex]:
[tex]\[ P(9) = \frac{1500}{1 + 9 e^{-0.42 \cdot 9}} \][/tex]
First, calculate the exponent:
[tex]\[ -0.42 \cdot 9 = -3.78 \][/tex]
Now calculate [tex]\( e^{-3.78} \)[/tex]. This value is a bit intricate, but it's a small number. To ease the calculation, we know:
[tex]\[ e^{-3.78} \approx 0.0228 \][/tex]
Substitute this into the denominator:
[tex]\[ P(9) = \frac{1500}{1 + 9 \cdot 0.0228} \][/tex]
[tex]\[ P(9) = \frac{1500}{1 + 0.2052} \][/tex]
[tex]\[ P(9) = \frac{1500}{1.2052} \][/tex]
Finally, simplify the fraction:
[tex]\[ P(9) \approx 1244 \][/tex]
Therefore, the population size after 9 years is approximately [tex]\( 1244 \)[/tex] fish.
### Summary
- Initial population size: [tex]\( 150 \)[/tex] fish
- Population size after 9 years: [tex]\( 1244 \)[/tex] fish
