In 1990 the life expectancy of males in a certain country was 70.9 years. In 1994, it was 73.5 years. Let E represent the life expectancy in year t and let t represent the number of years since 1990. The liner function E(t) that fits the data is E(t)= _t+_ . Round to the nearest 10th.Use the function to predict life expectancy in 2009. E(19) = round to the nearest 10th
To find the linear function E(t) that fits the given data, we need to determine the slope and y-intercept of the line.
We are given two data points: (1990, 70.9) and (1994, 73.5).
Let's find the slope first. The slope (m) is calculated using the formula:
m = (y2 - y1) / (x2 - x1)
Using the given data points:
m = (73.5 - 70.9) / (1994 - 1990)
= 2.6 / 4
= 0.65
Now, let's find the y-intercept (b). We can substitute the slope (m) and one of the data points into the equation:
y = mx + b
70.9 = 0.65 * 1990 + b
Simplifying:
70.9 = 1293.5 + b
b = 70.9 - 1293.5
b ≈ -1222.6
So, the linear function E(t) is given by:
E(t) = 0.65t - 1222.6
Now, to predict the life expectancy in 2009, we need to find E(19).
Plug t = 19 into the function:
E(19) = 0.65 * 19 - 1222.6
E(19) ≈ 12.35
Therefore, the predicted life expectancy in 2009 using the given linear function is approximately 12.35 years. Rounded to the nearest tenth, it is 12.4 years.
Life expectancy (E) increased (73.5-70.9)/4 = 0.65 years per year during that period.
If t=0 represents 1990, then
E = 70.9 + 0.65 t
Use that formula for E(19), and then do the rounding.
It is very unlikely that life expectancy will increase at that linear rate for 19 years, but that is what they are asking you to assume.