im stuck on precalculus hw

Question

In 1992, the moose population in a park was measured to be 4880. By 1999, the population was measured again to be 5650. If the population continues to change linearly:Find a formula for the moose population, , in terms of , the years since 1990.   What does your model predict the moose population to be in 2003?

Answer 1

This is a point slope formula question. You are given two data points and the question is asking you to assume that the relationship between moose population and years passed is linear.

We know that slope intercept form is: y = m * x + b. In this equation, where m = ((y₂-y₁)/(x₂-x₁)).

We will write our data as coordinates. (x1,y1) and (x2, y2). We will make our x-axis the time axis and our y-axis the population axis. This is because the dependent variable always goes on the y-axis. Because the population of moose depends on the year, the year is our independent variable.

Don't forget, t is the number of years since 1990!

In 1992, moose population was 4880 and 1992 is 2 years past 1990. So (2, 4880) is our first data point.

In 1999, moose population was 5650 and 1999 is 9 years past 1990. So (9, 5650) is our second data point.

We can plug this into the slope equation.

m = ((y₂-y₁)/(x₂-x₁)) = ((5650 - 4880) / (9 - 2)) = 770/7 = 110.

This means the population of moose increases by 110 every year. Now we can plug this into our slope intercept form with either data point and solve for b.

y = m * x + b

P(t) = 110 * t + b

I have inserted P(t) for y and t for x since this is what the problem asks and input our slope.

We know from the first data point that P(2) = 4880 = 110 * (2) + b.

From this equation, b = 4660.

To verify, we can use the second data point to recalculate b.

P(9) = 5650 = 110 * (9) + b.

b = 4660.

Our equation becomes: P(t) = 110 * t + 4660.

2003 is 13 years after 1990 so we set t equal to 13 and solve for the population.

P(13) = 110 * (13) + 4660

P(13) = 6090

We predict the moose population to be 6090 moose in 2003.

Answer 2