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.
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?
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
Yes, you are correct. The linear function that fits the given data is E(t) = 6t + 68.3, where t represents the number of years since 1992. This equation is derived from the given starting point in 1992 with a life expectancy of 68.3 and an increase of 2.8 years every 6 years until 1998.
To use this function to predict the life expectancy in 2006, you can substitute t = 14 into the equation:
E(14) = 6(14) + 68.3 = 84 + 68.3 = 152.3
Rounding this value to the nearest tenth gives approximately 152.3.
Therefore, you are correct that E(14) is approximately 152.3 or 152.4 when rounded to the nearest tenth.