To solve this problem, we need to find a linear model for the population $P(x)$, where $P(x)$ is the population at time $x$ (in years). Then, we'll use the model to predict the population at year 83.

Step 1: Setting up the linear model

The general form of a linear model is:

$P(x) = mx + b$

Where:

• $m$ is the slope of the line (rate of change of population over time),
• $b$ is the y-intercept (population at year 0).

We can calculate $m$ using two points from the table. I'll select $(x_1, P(x_1)) = (0, 4)$ and $(x_2, P(x_2)) = (72, 81)$ to represent the data.

Step 2: Calculating the slope $m$

The slope $m$ is given by:

$m = \frac{P(x_2) - P(x_1)}{x_2 - x_1} = \frac{81 - 4}{72 - 0} = \frac{77}{72}$

So, the slope $m$ is approximately:

$m \approx 1.0694$

Step 3: Finding the y-intercept $b$

Since we know the population at $x = 0$ is 4, we can use the point $(0, 4)$ to find $b$.

$P(0) = m(0) + b = 4 \implies b = 4$

Step 4: Linear model

Thus, the linear model is:

$P(x) = 1.0694x + 4$

Step 5: Predicting the population at year 83

To find the population in year 83, substitute $x = 83$ into the model:

$P(83) = 1.0694(83) + 4$

Let's calculate the result.

The predicted population at year 83 is approximately 92.76. Since the problem asks us to round down, the final predicted population is:

$P(83) = 92$

Let me know if you need further details or clarifications.

Here are 5 related questions to expand your understanding:

1. How do you interpret the slope of the linear model in terms of population growth?
2. What would happen if you used different points to calculate the slope?
3. Could this model be improved by using a different type of regression (e.g., quadratic)?
4. How would the prediction change if the population grew at a non-linear rate?
5. Can this model be used reliably for very long-term predictions (e.g., year 150)?

Tip: When creating linear models from data, using points that are far apart often improves accuracy.