Look at my reply to Kiki's question
It is the same as yours, just the numbers have been changed.
http://www.jiskha.com/display.cgi?id=1269785566
iN 1993, THE LIFE EXPECTANCY of males was 68.6 years. In 2000, it was 71.1 years. Let E represent the life expectancy in year t and t represent the number of years since 1993.
The linear function E(t) that fits this data is: E(t)=_t + _ (round to the nearest tenth)
Use the function to predict the life expectancy in 2003.
E(10)=
_= blank spaces to fill in. I'm not sure how to do this. I've done these problems, but usually the formaula is given.
3 answers
Where did you get .55? I don't need the slope I don't think. This is a pretty straightforward Q, I don't need to graph it.
I was hoping you would follow the example.
in E(t) = _t + _
they want you to express it in the standard form of a straigh line equation
f(x) = mx + b, where m is the slope
so the first thing I did was find the slope, which turned out to be .55
In your case, the two point would be
(0,68.6) and (7,71.1)
so your slope = (71.1-68.6)/(7-0) = .3571
so E(t) = .3571t + b
put in (0,68.6)
68.6 = 0 + b
b = 68.6
so E(t) = .3571 + 68.6
then for 2003, t = 10
E(10) = .3571(10) + 68.6 = 72.17
Don't you see that your equation is identical is style and setup to hers?
in E(t) = _t + _
they want you to express it in the standard form of a straigh line equation
f(x) = mx + b, where m is the slope
so the first thing I did was find the slope, which turned out to be .55
In your case, the two point would be
(0,68.6) and (7,71.1)
so your slope = (71.1-68.6)/(7-0) = .3571
so E(t) = .3571t + b
put in (0,68.6)
68.6 = 0 + b
b = 68.6
so E(t) = .3571 + 68.6
then for 2003, t = 10
E(10) = .3571(10) + 68.6 = 72.17
Don't you see that your equation is identical is style and setup to hers?
0 answers