In 1992, the life expectancy of males in a certain country was 63.8 years. In 1999, it was 66.0 years. Let E represent the life expectancy in year t and let t represent the number of years since 1992. The lineahntthe nearest ter function E(t) that fits the date is? Use the function to predict the life expectancy of males in 2004. round both to

here is a variation of your question.

http://www.jiskha.com/display.cgi?id=1294619397

As you can see in the "related questions" below, this seems to be a popular type of question.

E = m y + b

y is year - 1992

63.8 = m(0) + b
b = 63.8
so
E = m y + 63.8
66.0 = m (1999-1992) + 63.8
2.2 = m (7)
m = .3143
so
E = .3143 y + 63.8
or
E = .3143(year-1992) + 63.8
part b:
in 2004 y = 2004-1992 = 12
E = .3143(12) + 63.8
= 67.6

Looks like I did it faster back then :)

To find the linear function that represents the data, we need to determine the slope and y-intercept using the given information.

Let's assign E(t) as the life expectancy in year t and t as the number of years since 1992.

Given data:
In 1992, E(0) = 63.8 years.
In 1999, E(7) = 66.0 years.

To find the slope:
slope (m) = (change in y) / (change in x) = (E(7) - E(0)) / (7 - 0) = (66.0 - 63.8) / 7 = 0.2 years per year.

To find the y-intercept:
We can use the point-slope equation y - y₁ = m(x - x₁), where (x₁, y₁) is any point on the line.
Choosing the point (0, 63.8):
y - 63.8 = 0.2(x - 0)
y - 63.8 = 0.2x
y = 0.2x + 63.8

Therefore, the linear function is E(t) = 0.2t + 63.8.

To predict the life expectancy of males in 2004 (12 years after 1992), substitute t = 12 into the equation:
E(12) = 0.2(12) + 63.8 = 2.4 + 63.8 = 66.2 years.

Rounding both to the nearest tenth, the predicted life expectancy of males in 2004 is approximately 66.2 years.