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.

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

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?

To find the linear function that fits the given data, we can use the slope-intercept form of a linear equation: y = mx + b, where y represents the life expectancy (E) and x represents the number of years since 1993 (t).

We have two data points:
- In 1993 (t = 0), the life expectancy was 68.6 years (E = 68.6).
- In 2000 (t = 7), the life expectancy was 71.1 years (E = 71.1).

Knowing these two data points, we can find the slope (m) and the y-intercept (b) of the linear equation.

First, let's find the slope (m):
m = (E2 - E1) / (t2 - t1)
= (71.1 - 68.6) / (7 - 0)
= 2.5 / 7
≈ 0.3571 (rounded to four decimal places)

Now, let's find the y-intercept (b) by substituting one of the given data points into the equation and solving for b:
68.6 = 0.3571(0) + b
68.6 = b
b = 68.6

Therefore, the linear equation for life expectancy (E) based on the number of years since 1993 (t) is:
E(t) = 0.3571t + 68.6 (rounded to the nearest tenth)

To predict the life expectancy in 2003 (t = 2003 - 1993 = 10), we can substitute t = 10 into the equation:
E(10) = 0.3571(10) + 68.6
E(10) ≈ 3.571 + 68.6
E(10) ≈ 72.171

So, the predicted life expectancy in 2003 is approximately 72.171 years.