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Sagot :
To calculate the population of fish in the pond after one and two breeding seasons using the logistic growth model, we will follow these steps:
1. Determine the carrying capacity (K), which is 1200 fish.
2. Set the growth rate (r) at 180%, or as a decimal, 1.8.
3. Initial population (p0) of the pond is given as 400 fish.
Step-by-Step Solution:
### After One Breeding Season:
Using the logistic growth model, we calculate the population after one breeding season.
The logistic equation for one breeding season is:
[tex]\[ p_1 = p_0 + r \cdot p_0 \left(1 - \frac{p_0}{K}\right) \][/tex]
Plugging in the values:
[tex]\[ p_1 = 400 + 1.8 \cdot 400 \left(1 - \frac{400}{1200}\right) \][/tex]
First, simplify the fraction:
[tex]\[ 1 - \frac{400}{1200} = 1 - \frac{1}{3} = \frac{2}{3} \][/tex]
Now, multiply and calculate:
[tex]\[ 1.8 \cdot 400 \cdot \frac{2}{3} = 1.8 \cdot 400 \cdot 0.6667 \approx 480 \][/tex]
Add this to the initial population:
[tex]\[ p_1 = 400 + 480 = 880 \][/tex]
So, the population after one breeding season [tex]\( p_1 \)[/tex] is:
[tex]\[ p_1 = 880 \][/tex]
### After Two Breeding Seasons:
We use the new population [tex]\( p_1 \)[/tex] to find [tex]\( p_2 \)[/tex]:
The logistic equation for the next breeding season is:
[tex]\[ p_2 = p_1 + r \cdot p_1 \left(1 - \frac{p_1}{K}\right) \][/tex]
Plugging in the values:
[tex]\[ p_2 = 880 + 1.8 \cdot 880 \left(1 - \frac{880}{1200}\right) \][/tex]
First, simplify the fraction:
[tex]\[ 1 - \frac{880}{1200} = 1 - \frac{22}{30} = \frac{8}{30} = \frac{4}{15} \][/tex]
Now, multiply and calculate:
[tex]\[ 1.8 \cdot 880 \cdot \frac{4}{15} = 1.8 \cdot 880 \cdot 0.2667 \approx 422.4 \][/tex]
Add this to the population after one breeding season:
[tex]\[ p_2 = 880 + 422.4 = 1302.4 \][/tex]
So, the population after two breeding seasons [tex]\( p_2 \)[/tex] is:
[tex]\[ p_2 = 1302.4 \][/tex]
Therefore, the population of the pond after one breeding season is:
[tex]\[ p_1 = 880 \][/tex]
And after two breeding seasons the population is:
[tex]\[ p_2 = 1302.4 \][/tex]
1. Determine the carrying capacity (K), which is 1200 fish.
2. Set the growth rate (r) at 180%, or as a decimal, 1.8.
3. Initial population (p0) of the pond is given as 400 fish.
Step-by-Step Solution:
### After One Breeding Season:
Using the logistic growth model, we calculate the population after one breeding season.
The logistic equation for one breeding season is:
[tex]\[ p_1 = p_0 + r \cdot p_0 \left(1 - \frac{p_0}{K}\right) \][/tex]
Plugging in the values:
[tex]\[ p_1 = 400 + 1.8 \cdot 400 \left(1 - \frac{400}{1200}\right) \][/tex]
First, simplify the fraction:
[tex]\[ 1 - \frac{400}{1200} = 1 - \frac{1}{3} = \frac{2}{3} \][/tex]
Now, multiply and calculate:
[tex]\[ 1.8 \cdot 400 \cdot \frac{2}{3} = 1.8 \cdot 400 \cdot 0.6667 \approx 480 \][/tex]
Add this to the initial population:
[tex]\[ p_1 = 400 + 480 = 880 \][/tex]
So, the population after one breeding season [tex]\( p_1 \)[/tex] is:
[tex]\[ p_1 = 880 \][/tex]
### After Two Breeding Seasons:
We use the new population [tex]\( p_1 \)[/tex] to find [tex]\( p_2 \)[/tex]:
The logistic equation for the next breeding season is:
[tex]\[ p_2 = p_1 + r \cdot p_1 \left(1 - \frac{p_1}{K}\right) \][/tex]
Plugging in the values:
[tex]\[ p_2 = 880 + 1.8 \cdot 880 \left(1 - \frac{880}{1200}\right) \][/tex]
First, simplify the fraction:
[tex]\[ 1 - \frac{880}{1200} = 1 - \frac{22}{30} = \frac{8}{30} = \frac{4}{15} \][/tex]
Now, multiply and calculate:
[tex]\[ 1.8 \cdot 880 \cdot \frac{4}{15} = 1.8 \cdot 880 \cdot 0.2667 \approx 422.4 \][/tex]
Add this to the population after one breeding season:
[tex]\[ p_2 = 880 + 422.4 = 1302.4 \][/tex]
So, the population after two breeding seasons [tex]\( p_2 \)[/tex] is:
[tex]\[ p_2 = 1302.4 \][/tex]
Therefore, the population of the pond after one breeding season is:
[tex]\[ p_1 = 880 \][/tex]
And after two breeding seasons the population is:
[tex]\[ p_2 = 1302.4 \][/tex]
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