In 1992, the life expectancy of males in a certain country was 68.3 years. In 1998, it was 71.1 years. Let E represent the life expectancy in year t and let t represent the number of years since 1992.

Question

The linear function E(t) that fits the data is... E(t)=_____t + _______ (round to nearest 10th)
Use the function to predict the life expectancy of males in 2006.
E(14) = ___________ (round to nearest 10th)

I come up with:
E(t)=6t + 68.3--- for the 1st answer and

E(14) = 74.9
i.e. 74.88 rounded to nearest 10th

Is it right?

Answer 1

The way to check your answer is to use the equation to find the first result, so

E(t)=6t+68.3

so for 1998 t=6 and
E(t)=6x6+68.3, which is not the correct answer so the equation is not correct.

From 1992 to 1998 the life expectancy has increased by (71.1-68.3) y = 2.8 y

So the rate of change (=gradient of the line, m) = 2.8 y/6 y

The starting point, c, is 68.3 y. So using the straight line equation of the form

y=mx+c

Thus at time t after 1992

E(t)=2.8/6(t) + 68.3 y

So at t=14 y

E(t)=2.8x14y/6 + 68.3 y