Are we assuming that the increase is linear ?
If so, then
treat you given data as two ordered pairs
(0,61.6) and (5,63.8)
slope = (63.8-61.6)/5
= .44
then E(t) = .44t + b , the standard form of a linear equation
plug in (0,61.6) which is like the y-intercept, so b= 61.6
E(t) = .44t + 61.6
test it for t=5 , the year 1999
E(5) = .44(5) + 61.6 = 63.8
as expected
now when year is 2007, t = 13
E(13) = .44(13) + 61.6 = 67.3
In 1994 the life expectancy of males in a certain country was 61.6 years. In 1999, it was 63.8 years. Let E?
represent the life expectancy in year t and let t represent the number of years since 1994.
1) The linear function E(t) that fits the data is: E(t)=___t+___
2) Use the function to predict the life expectancy of males in 2007.
E(13)=______
2 answers
Thank you so much, I was having problems with this question.