in 1992, the life expectancy of males in a certain country was 69.8 years. in 1999, it was 72.4 years. Let E represent the life expectency 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+_
Use the function to predict the life expecteancy of males in 2006.
E(14)=
The question below differs from yours only in the numbers.
follow its method.
http://www.jiskha.com/display.cgi?id=1269785566
To find the linear function that fits the given data, we need to determine the slope and y-intercept.
Let's first find the slope (m) using the given data points (1992, 69.8) and (1999, 72.4):
m = (72.4 - 69.8) / (1999 - 1992)
= 2.6 / 7
= 0.3714 (rounded to four decimal places)
Now, let's find the y-intercept (b) using one of the data points. We'll use (1992, 69.8):
69.8 = 0.3714 * 1992 + b
69.8 = 739.7088 + b
b = 69.8 - 739.7088
b = - 669.9088 (rounded to four decimal places)
Therefore, the linear function E(t) that fits the data is:
E(t) = 0.3714t - 669.9088
Now, we can use this function to predict the life expectancy of males in 2006 by substituting t = 14 into the equation:
E(14) = 0.3714 * 14 - 669.9088
E(14) = 5.1984 - 669.9088
E(14) = -664.7104
Therefore, the predicted life expectancy of males in 2006 is approximately -664.7104 years.
Please note that this negative value is only obtained because the linear function is extrapolated beyond the range of the given data, which may not accurately represent the actual life expectancy in 2006.