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)
Look at any of those examples down below. for example:
http://www.jiskha.com/display.cgi?id=1294619397
To find the linear function that fits the data, we can use the formula for a straight line:
E(t) = mt + b
Here, m represents the slope of the line, and b represents the y-intercept.
To find the slope (m), we can use the formula:
m = (E2 - E1) / (t2 - t1)
Where E1 and E2 are the life expectancy values for years 1992 and 1997 respectively, and t1 and t2 are the corresponding number of years since 1992.
E1 = 71.1 years
E2 = 73.4 years
t1 = 0 years (Since it is the starting year)
t2 = 1997 - 1992 = 5 years
m = (73.4 - 71.1) / (5 - 0)
= 2.3 / 5
= 0.46 (rounded to two decimal places)
Now, to find the y-intercept (b), we can substitute the values of E and t into the equation and solve for b. Using the point (t1, E1):
E(t1) = m * t1 + b
71.1 = 0.46 * 0 + b
71.1 = b
Therefore, b = 71.1
So, the linear function E(t) that fits the data is:
E(t) = 0.46t + 71.1
Now, to predict the life expectancy in 2003 (11 years since 1992), we substitute t = 11 into the equation:
E(11) = 0.46 * 11 + 71.1
= 5.06 + 71.1
= 76.16 (rounded to the nearest tenth)
Therefore, the predicted life expectancy of males in 2003 is 76.2 years.