Posted by **deedra** on Thursday, January 8, 2009 at 3:43pm.

Ok here are a few more

1. collect like terms

12r+6s-3r-9s

A:9r-3s

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=52-6y

A: (-2,9)

5. solve by elimination

2r-7s=-31

7r+2s=24

A:(2,5)

- m -
**Reiny**, Thursday, January 8, 2009 at 4:00pm
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 -
**deedra**, Thursday, January 8, 2009 at 4:06pm
Thank you very much.

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