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Let's break down the problem step-by-step.

### Given:
The population of deer in an area is modeled by the equation:
\[ P(t) = \frac{1600}{1 + 19 \cdot e^{-0.6t}} \]
where:
- \( P(t) \) represents the number of deer at time \( t \) (years after the start of 2020).
- \( t \) is the number of years after the start of 2020.
  
### Questions to Answer:
#### (a) How many deer are there in the area at the start of 2020?
To find the number of deer at the start of 2020, we need to calculate \( P(t) \) when \( t = 0 \).

\[ P(0) = \frac{1600}{1 + 19 \cdot e^{-0.6 \cdot 0}} \]
Simplify the exponent:
\[ e^{-0.6 \cdot 0} = e^0 = 1 \]

Therefore, the equation becomes:
\[ P(0) = \frac{1600}{1 + 19 \cdot 1} = \frac{1600}{1 + 19} = \frac{1600}{20} = 80 \]

**Answer:** There are 80 deer in the area at the start of 2020.

#### (b) Determine graphically and by calculation when the population is 1000 deer according to the model.
We need to find the value of \( t \) such that \( P(t) = 1000 \).

Set \( P(t) = 1000 \) in the model equation:
\[ 1000 = \frac{1600}{1 + 19 \cdot e^{-0.6t}} \]

Cross-multiply to get rid of the fraction:
\[ 1000 \cdot (1 + 19 \cdot e^{-0.6t}) = 1600 \]
\[ 1000 + 19000 \cdot e^{-0.6t} = 1600 \]

Isolate the exponential term:
\[ 19000 \cdot e^{-0.6t} = 1600 - 1000 \]
\[ 19000 \cdot e^{-0.6t} = 600 \]

Divide by 19000:
\[ e^{-0.6t} = \frac{600}{19000} \]
\[ e^{-0.6t} = 0.03158 \]

Take the natural logarithm (ln) of both sides to solve for \( t \):
\[ -0.6t = \ln(0.03158) \]
Calculate \( \ln(0.03158) \):
\[ \ln(0.03158) \approx -3.454 \]

Now solve for \( t \):
\[ t = \frac{-3.454}{-0.6} \]
\[ t \approx 5.76 \]

**Answer:** The population will be 1000 deer approximately 5.76 years after the start of 2020, which is around the middle of 2025.

#### (c) Use the model to estimate how big the population can get.
To find the maximum population, we need to consider the behavior of the function as \( t \) approaches infinity. The exponential term \( e^{-0.6t} \) will approach zero as \( t \) increases.

Thus, the function \( P(t) \) approaches:
\[ P(t) \approx \frac{1600}{1 + 0} = 1600 \]

**Answer:** The maximum population the model predicts is 1600 deer.

### Do you want more details or have any questions about this? Here are some follow-up questions:

1. How does the exponential decay factor \( -0.6 \) affect the population growth over time?
2. What would happen if the initial population was different, for example, 200 deer instead of 80?
3. How can we verify these results using a graphing tool?
4. What real-world factors might limit the deer population, despite the model's prediction?
5. How does the logistic model compare to exponential growth models?

**Tip:** Always double-check calculations, especially when dealing with exponential and logarithmic functions, to ensure precision.

The population of deer in an area is expected to follow the model: P(t) = 1600 / (1 + 19 * e^(-0.6t)), where P(t) is the number of animals, and t is the number of years after the start of 2020. a) How many deer are there in the area at the start of 2020? b) Determine graphically and by calculation when the population is 1000 deer. c) Use the model to estimate how big the population can get.

Explore how to calculate deer population growth using the logistic model P(t) = 1600 / (1 + 19 * e^(-0.6t)). This problem involves solving for the population at the start of 2020, determining when it reaches 1000 deer, and estimating the maximum population size. Suitable for students in Grades 10-12.

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