posted by deedra on .
Ok here are a few more
1. collect like terms
2. translate to an algebraic expression
The product of 46% and some number
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
Use the function to predict the life expectancy of males in 2003.
4. solve by substitution
5. solve by elimination
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
You should have realized that your answer of 61.6 for 2003 could not be right, since it was less than the 1999 figure.
Thank you very much.