Question

In 2006, a town's population was 590. By 2011, the population had grown to 740.

- a. Create a linear equation for the town's population "n" years from 2006.
Pn=
b. Suppose you want to use your equation to predict the population of the town in 2021. What value would you use for n? __
What does your equation predict for 2021? __
c. How many years will it be after 2006 for the town to triple its 2006 population? Round to at least 3 decimal places if necessary. __

Answer 1

The linear equation for the town's population n years from 2006 is Pn = 590 + 30n. For 2021, predicted population using n=15 is 1,040. To triple the 2006 population, it takes approximately 39.333 years.

Step-by-step explanation:

To create a linear equation for the town's population n years from 2006, we first find the rate of change in population between 2006 and 2011. The population increased from 590 to 740 in 5 years, which is an increase of 740 - 590 = 150 over 5 years. Therefore, the yearly rate of increase is 150/5 = 30 people per year.

Next, we use the point-slope form of a linear equation, which is y - y1 = m(x - x1), where (x1, y1) is a point on the line (in this case, x1=0 for the year 2006, and y1=590 for the population in that year), and m is the slope (rate of change per year, which we have found to be 30). Therefore, the linear equation for the population Pn is:

Pn = 590 + 30n

To predict the population of the town in 2021, we would use n = 15, since 2021 is 15 years after 2006. Substituting n=15 into our equation gives us Pn = 590 + 30(15), resulting in Pn = 590 + 450, which equals Pn = 1040. So, the predicted population for 2021 is 1,040.

To find how many years after 2006 the town will triple its 2006 population, we set the population to 3 times the population in 2006, which is 3(590) = 1770. We solve for n in the equation 1770 = 590 + 30n. Subtracting 590 from both sides gives 1180 = 30n, and dividing by 30 gives n ≈ 39.333, rounding to at least three decimal places.