SOLUTION: (Can someone please help me with this question?) In 1992, the life expectancy of males in a certain country was 71.1 years. In 1997, it was 73.4 years. Let E represent the life ex

Click here to see ALL problems on Linear-equations

Question 841582: (Can someone please help me with this question?)
In 1992, the life expectancy of males in a certain country was 71.1 years. In 1997, it was 73.4 years. Let E represent the life expectancy in year t and let t represent the number of years since 1992.
The linear function E(t) that fits the data is E(t)=___?t+__?(Round to the nearest tenth)
Use the function to also predict the life expectancy of males in 2003.
E(11)=___?(Round to the nearest tenth)
Found 2 solutions by ewatrrr, josmiceli:
Answer by ewatrrr(24785) (Show Source):
You can put this solution on YOUR website!

Hi,
E(t)=___?t+__? |
find slope:
(1997,73.4)
(1992, 71.1) m = = 2.3/5 = .46
Use 2nd point to find 'b' where t = 0
E(t)= .46t+ 71.1
E(11) = .46(11) + 71.1

Answer by josmiceli(19441) (Show Source):
You can put this solution on YOUR website!
What they want you to do is find the slope of
a straight line that passes through the points
( 71.1, 0 ) and ( 73.4, 5 )
-----------------------
is plotted on the vertical axis
is plotted on the horizontal axis
I call the year 1992
Then 1997 becomes ( 5 years later )
------------------------------------------
Use the general point-slope formula

Multiply both sides by

-----------------------
Check the answer:
( 73.4, 5 )

OK
-----------------------
2003 - 1992 = 11 years

----------------------
also
----------------------
Here's a plot of the line: