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Ok here are a few more

1. collect like terms

2. translate to an algebraic expression
The product of 46% and some number
A: y=0.46x

3. In 1993, the life expectancy of males in a certain country was 61.3 yrs. In 1999, it was 64.7yrs. Let E represent the life expectancy in year t and let t represent the number of years since 1993.

The linear function E(t) that fits the data is
A: E(t)=-0.03t+61.3
Use the function to predict the life expectancy of males in 2003.
A: E(10)=61.6

4. solve by substitution
A: (-2,9)

5. solve by elimination

  • m -

    They are all correct except #3
    For #4 and #5 it is very simple to try your answer in both equations.
    Why did you not do that?

    for #3, you should define t as the number of years since 1993
    I got a "slope" of (64.7-61.3)/6 = .566667

    so E(t) = .566667t + 61.3
    check for the second point
    LS = 64.7
    RS = .566667(6) + 61.3 = 64.7

    then E(10) = .566667(10) + 61.3 = 66.966
    or 67.0

    You should have realized that your answer of 61.6 for 2003 could not be right, since it was less than the 1999 figure.

  • m -

    Thank you very much.

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