Answered

Join IDNLearn.com and start exploring the answers to your most pressing questions. Get the information you need from our community of experts who provide accurate and comprehensive answers to all your questions.

A species of fish was added to a lake. The population size [tex]P(t)[/tex] of this species can be modeled by the following function, where [tex]t[/tex] is the number of years from the time the species was added to the lake.
[tex]
P(t) = \frac{1500}{1 + 9 e^{-0.42 t}}
[/tex]

Find the initial population size of the species and the population size after 9 years. Round your answers to the nearest whole number as necessary.

Initial population size: [tex]\square[/tex] fish

Population size after 9 years: [tex]\square[/tex] fish

Sagot :

GinnyAnsweridn
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
We value your presence here. Keep sharing knowledge and helping others find the answers they need. This community is the perfect place to learn together. For clear and precise answers, choose IDNLearn.com. Thanks for stopping by, and come back soon for more valuable insights.