m
posted by deedra .
Ok here are a few more
1. collect like terms
12r+6s3r9s
A:9r3s
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
3x+2y=12
x=526y
A: (2,9)
5. solve by elimination
2r7s=31
7r+2s=24
A:(2,5)

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 (64.761.3)/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. 
Thank you very much.
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