The population of Austin, Texas in 1980 was 346,000 people. The population in 1990 was 472,000 people. What is the rate of change of the population with respect to the year for this function?

Answer:

12.6 thousand a year

Step-by-step explanation:

take the 346 add 126 thousand to it equaling a good 472 thousand divide 126 thousand by 10 cause it is a ten year span that equals you get 12.6 thousand.

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We want to find the rate of change of the population of Austin, Texas in the years between 1980 and 1990.

The answer is: r = 12,600 people/year.

We will assume that this increase in population is modeled with a linear equation, remember that a general linear equation is:

y = a*x + b

Where the rate of change, or slope, is a.

We know that if a line passes through two points (x₁, y₁) and (x₂, y₂) the slope is given by the formula:

[tex]a = \frac{y_2 - y_1}{x_2 - x_1}[/tex]

For this case we have two points, these are:

(1980, 346,000)

(1990, 472,000)

Then the slope or rate of change is just:

[tex]a = \frac{472,000 - 346,000}{1990 - 1980} = 12,600[/tex]

To be complete we need to add the correspondent units, which are people per year.

Then the rate of change is:

r = 12,600 people/year.

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