In 1994 the life expectancy of males in a certain country was 61.6 years. In 1999, it was 63.8 years. Let E?

represent the life expectancy in year t and let t represent the number of years since 1994.

1) The linear function E(t) that fits the data is: E(t)=___t+___

2) Use the function to predict the life expectancy of males in 2007.
E(13)=______

Are we assuming that the increase is linear ?

If so, then
treat you given data as two ordered pairs
(0,61.6) and (5,63.8)

slope = (63.8-61.6)/5
= .44
then E(t) = .44t + b , the standard form of a linear equation
plug in (0,61.6) which is like the y-intercept, so b= 61.6

E(t) = .44t + 61.6

test it for t=5 , the year 1999
E(5) = .44(5) + 61.6 = 63.8
as expected

now when year is 2007, t = 13
E(13) = .44(13) + 61.6 = 67.3

Thank you so much, I was having problems with this question.

To find the linear function that fits the data, we need to determine the slope and the y-intercept.

To find the slope, we can use the formula:

slope = (change in y-values) / (change in x-values)

Given that in 1994 the life expectancy was 61.6 years (E(0) = 61.6) and in 1999 it was 63.8 years (E(5) = 63.8), we can calculate the change in y-values:

(change in y-values) = E(5) - E(0) = 63.8 - 61.6 = 2.2

And since the time interval between these two years is 5 years, we have:

(change in x-values) = 5

Now we can calculate the slope:

slope = (change in y-values) / (change in x-values) = 2.2 / 5 = 0.44

Next, we need to determine the y-intercept. We can use the formula:

y-intercept = E(0) - slope * t

Since we already know that E(0) = 61.6 and the slope is 0.44, we can substitute these values into the formula:

y-intercept = 61.6 - 0.44 * 0 = 61.6

Therefore, the linear function that fits the data is:

E(t) = 0.44t + 61.6

Now, to predict the life expectancy of males in 2007 (t = 2007 - 1994 = 13), we can substitute t = 13 into the function:

E(13) = 0.44 * 13 + 61.6 = 5.72 + 61.6 = 67.32

Therefore, the predicted life expectancy of males in 2007 is 67.32 years.