The life expectancy of an American in 2000 was 76.1 years and the life expectancy in 2007 was 78.2 years. If this trend continues find the life expectancy in teh year 2015. (Assume the trend is linear)

Do i have to find the slope first? and if so do I use (76.1, 2000) and (78.2, 2007) or do I use one of the previous and (0,2015) to find slope?

Any help would be appreciated. I'm also not sure where to go from there.
Thanks!

2. Originally Posted by epetrik
The life expectancy of an American in 2000 was 76.1 years and the life expectancy in 2007 was 78.2 years. If this trend continues find the life expectancy in teh year 2015. (Assume the trend is linear)

Do i have to find the slope first? and if so do I use (76.1, 2000) and (78.2, 2007) or do I use one of the previous and (0,2015) to find slope?

Any help would be appreciated. I'm also not sure where to go from there.
Thanks!
There are several ways to generate an equation of a line. You are given two points, so from these two points you can find the slope of the line, then using one of these points and the slope you calculated, plug them into the point-slope formula and you will have your equation.

The choice of which is x and which is y is up to you, but normally you see "time" as the x-axis. If you use this convention, then "life-expectancy in years" would be your y-axis.

From these choices, P1=(2000, 76.1) and P2 = (2007, 78.2). From these two points you can calculate the slope m. Do you know how to do that?

Then when you have the slope, you use either of the above points and create the point-slope form of a linear equation,
$y - y_1 = m(x - x_1)$

3. Originally Posted by epetrik
The life expectancy of an American in 2000 was 76.1 years and the life expectancy in 2007 was 78.2 years. If this trend continues find the life expectancy in teh year 2015. (Assume the trend is linear)

Do i have to find the slope first? and if so do I use (76.1, 2000) and (78.2, 2007) or do I use one of the previous and (0,2015) to find slope?

Any help would be appreciated. I'm also not sure where to go from there.
Thanks!
to make the problem easier, let 2000 be year 0 ... 2007 is year 7 ... so, you make the prediction for year 15

(0,76.1)
(7,78.3)
(15, ?)

find the slope using the first two points and create a linear function

y = mx + b

y = life expectancy

m = slope

x = year, where x = 0 is 2000

b = y-intercept ... 76.1 in this case.

find y when x = 15.

4. I found slope but now i'm struggling with solving it in slope-intercept form. I have y-76.1=3/10(x-2000) which equals y-76.1=3/10x - 600. where do I go fromt there?

Thanks!

5. Originally Posted by epetrik
I found slope but now i'm struggling with solving it in slope-intercept form. I have y-76.1=3/10(x-2000) which equals y-76.1=3/10x - 600. where do I go fromt there?

Thanks!
Add 76.1 to both sides, to remove it from the left side. They you have the equation in the form of y = mx + b. Then you can solve for x = 2015 by substituting 2015 for x and evaluating this equation. Y will give you the expected life expectancy in 2015.

6. Thank you!