I'm a bit stuck trying to figure out what this problem wants, can anyone help?
In 1991, the life expectancy of males in a certain country was 62.3 years. In 1996, it was 65.1 years. Let E represent the life expectancy in years and t represent the number of years since 1991.
a. The linear function E(2) that fits the data is
E(t)=___t + ___ (round to the nearest tenth).
b. Use the function to predict the life expectancy of males in 2007. E(18)=____.
The slope is deltaE/deltaTime=2.8/5
E=2.8/5 * t + 62.3
where t is in years greater than 1991
Thank you, I finally got the answer I needed after playing around with the original figures before writing the actual equation. I got E(t)=.56t+62.3 for the equation and E(18)=.56(18)+62.3=72.38 or 72.4....sound about right?
To solve part a of the problem, we need to find the equation of a linear function that fits the given data. The equation will have the form E(t) = mt + b, where m represents the slope and b represents the y-intercept.
Step 1: Determine the change in life expectancy over the given time period.
The change in life expectancy is calculated by subtracting the life expectancy in 1991 from the life expectancy in 1996:
ΔE = (65.1 years) - (62.3 years) = 2.8 years
Step 2: Determine the rate of change.
To find the rate of change, divide the change in life expectancy by the number of years:
Rate of change, m = ΔE / Δt (where Δt represents the change in time)
m = 2.8 years / 5 years = 0.56 years per year
Step 3: Determine the y-intercept.
To find the y-intercept, substitute one of the given points into the equation and solve for b. Let's use the point (1991, 62.3).
62.3 = 0.56 * 0 + b
b = 62.3
Therefore, the linear function E(t) that fits the data is:
E(t) = 0.56t + 62.3 (rounded to the nearest tenth).
For part b of the problem, we need to use the linear function we found to predict the life expectancy of males in 2007, which is 18 years after 1991.
E(t) = 0.56 * 18 + 62.3
E(18) = 10.08 + 62.3
E(18) = 72.38
Therefore, the predicted life expectancy of males in 2007 is 72.4 years (rounded to the nearest tenth).