Question 311333
In 1990, the life expectancy of males in a certain country was 61.1 years.
 In 1996, it was 63.5 years.
 Let E represent the life expectancy in year t and let t represent the number of years since 1990.
 The linear function E(t) that fits the data is E(t)= ?t + ?(round to the nearest tenth).
 Then use the fuunction to predict the life expectancy of males in 2009.
 E(19)=? (round to the nearest tenth).
:
Find the slope
In 1990; x1=0 and y1=61.1
In 1996; x2=6 and y2=63.5
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Find the slope (m) using the slope equation: m = {{{(y2-y1)/(x2-x1)}}}
m = {{{(63.5-61.1)/(6 - 0))}}} = {{{(2.4)/(6)}}} = .4
:
Use the point/slope formula to write the equation; y - y1 = m(x - x1)
y - 61.1 = .4(x - 0)
y = .4x + 61.1
:
therefore:
The Linear function E(t) that fits the data is.
E(t)= .4t + 61.1
:
Use the Function to predict the life expectancy of males in 2009.
This means t=19; substitute 19 for t in the equation
E(19)= .4(19) + 61.1
E(19)= 7.6 + 61.1
E(19)= 68.7 yrs is the life expectancy in 2009
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Did this method make some sense to you?