SOLUTION: I have asked for help on this story problem before however I have never heard back so I am still need of some help with solving this story problem. In 1991 the life expectancy o

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Question 331487: I have asked for help on this story problem before however I have never heard back so I am still need of some help with solving this story problem.
In 1991 the life expectancy of males in a certain country was 63.2 years. In 1996 it was 65.7 years. Let E represent the life expectancy in year t, and let + represent the number of years since 1991.
The linear function E(t) that fits the data is E)t)=_____t + _____.
Use the function to predict the life expectancy of males in 2006.
E (15)= _____ and round to the nearest tenth.
Thanks for your help :o)
Answer by vleith(2983) About Me  (Show Source):
You can put this solution on YOUR website!
Use the info provided to create a linear model of Life Expenctency.
You are given two points. With that data you have enough information to create the equation for a line.
First let's find the slope of the line
Slope = %28y%5B2%5D+-+y%5B1%5D%29+%2F+%28x%5B2%5D+-+x%5B1%5D%29
You are given the two data points (1991,63.2) and (1996, 65.7)
So the slope is %2865.7-63.2%29%2F%281996+-+1991%29 = 2.5%2F5+=+1%2F2
The problem says to model the Life Expectancy where t is the 'number of years since 1991"
So
E%28t%29+=+%281%2F2%29+t+%2B+63.2 that uses the slope we found above as the rate of change year to year. It also uses the expectency in the year 1991 as the starting point.
Check this now. In year 1991, t = 1991-1991 = 0. So E%280%29+=+0%2At+%2B+63.2+=+63.2 which is correct
IN 1996, t = 1996-1991 = 5. So E%285%29+=+%281%2F2%29%2A5+%2B+63.2+=+2.5%2B63.2+=+65.7 which is correct
You are then asked to use the equation to extrapolate the data to the year 2006
So, 2006-1991 = 15
E%2815%29+=+15%2F2+%2B+63.2+=+7.5+%2B+63.2+=+70.7
Again, this is just a model. There is nothing to suggest that Life expectency is linear. But that is all the data you have, so you have to go with that.
Make sense?