In 1992, the life expectancy of males in a certain country was 71.1 years. In 1997, it was 73.4 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 the nearest tenth)
Use the function to also predict the life expectancy of males in 2003.
E(11)=___?(Round to the nearest tenth)
here is the method. Put your numbers in:
http://www.jiskha.com/display.cgi?id=1294619397
To find the linear function that fits the data, we need to use the given information about the life expectancy in 1992 and 1997.
Let's start by determining the slope of the linear equation. The slope represents the rate of change in life expectancy per year. We can find the slope using the formula:
Slope (m) = (change in life expectancy)/(change in years)
(change in life expectancy) = 73.4 years - 71.1 years = 2.3 years
(change in years) = 1997 - 1992 = 5 years
Substituting the values into the formula:
Slope (m) = 2.3 years / 5 years = 0.46
Now that we have the slope, we can use the point-slope form of the linear equation, which is:
y - y1 = m(x - x1)
where (x1, y1) represents a point on the line.
Since we are given the life expectancy in 1992, we can use that as our point (x1, y1) = (0, 71.1). Substituting this into the equation:
E - 71.1 = 0.46(t - 0)
Simplifying:
E - 71.1 = 0.46t
To isolate E, we add 71.1 to both sides:
E = 0.46t + 71.1
Therefore, the linear function E(t) is E(t) = 0.46t + 71.1. (rounded to the nearest tenth)
To predict the life expectancy in 2003 (t = 11), we substitute the value of 11 into the equation:
E(11) = 0.46(11) + 71.1
Calculating this:
E(11) = 5.06 + 71.1 ≈ 76.2
Therefore, the predicted life expectancy of males in 2003 is approximately 76.2 years. (rounded to the nearest tenth)