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Question 209683: Story problems always confuse me. I have tried and tried to get this one, and I can't do it. Can some one please help me?
The 1990 life expectancy of males in a certain country was 71.3 years. In 1995, it was 74.5 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)
Use the function to predict the life expectancy in males in 2006
E(16) = ____ (round to the nearest tenth)
Found 2 solutions by Alan3354, stanbon:
Answer by Alan3354(69443)   (Show Source):
You can put this solution on YOUR website! The 1990 life expectancy of males in a certain country was 71.3 years. In 1995, it was 74.5 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)
Use the function to predict the life expectancy in males in 2006
E(16) = ____ (round to the nearest tenth)
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1990 is the starting point, t = 0 years
In 1990, t=0 and E = 71.3
In 1995, t=5 and E = 74.5
The change in E is 74.5-71.3 = 3.2 years
The change per year is 3.2/5 = 0.64 increase per year.
E(t) = 71.3 + 0.64t
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For 2006: t is 2006 - 1990 = 16
E(16) = 71.3 + 0.64*16 = 71.3 + 10.24 = 81.54
--> 81.5 years in 2006
Answer by stanbon(75887)  (Show Source):
You can put this solution on YOUR website! The 1990 life expectancy of males in a certain country was 71.3 years. In 1995, it was 74.5 years. Let E represent the life expectancy in year t and let t represent the number of years since 1990.
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You have two points relating year and life expectancy: (0,71.3)and(5,74.5)
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slope = (74.5-71.3)/(5-0) = 3.2/5 = 0.64
intercept: when t = 0, expected life = 71.3
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The linear function E(t) that fits the data is
E(t) = 0.64t + 71.3 (round to the nearest tenth)
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Use the function to predict the life expectancy in males in 2006
E(16) = 0.64*16 + 71.3 (round to the nearest tenth)
E(16) = 81.54 yrs.
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Cheers,
Stan H.
Reply to stanbon@comcast.net
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