• August 14th 2009, 02:06 PM
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!
• August 14th 2009, 02:19 PM
QM deFuturo
Quote:

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,
$y - y_1 = m(x - x_1)$
• August 14th 2009, 02:22 PM
skeeter
Quote:

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.
• August 14th 2009, 02:35 PM
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!
• August 14th 2009, 03:16 PM
QM deFuturo
Quote:

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?