# Thread: Linear Function

1. ## Linear Function

In 1995, the life expectancy of males in a certain county was 69.5 years. In 1999, it ws 72.1 years. Let E represent the life expectancy in year t and let t represent the number of years since 1995.

The Linear function E(t) that fits the data is.
E(t)=______ t + ______

Use the Function to predict the life expectancy of males in 2005.

E(10)=

Round to the nearest tenth.

2. Let's say that at t = 0, it is the year 1995, so we know that:

$E(0) = 69.5$

So, at t = 4, we get:

$E(4) = 72.1$

We know that:

$E(x) = at + b$

Where b is the y-intercept, which is $E(0)$, that gives us:

$E(t) = at + 69.5$

Now, all we need to to is set it equal to $E(4)$ to solve for a:

$E(4) = a(4) + 69.5 = 72.1$

$4a = 2.6$

$a = 0.65$

So, we now have an equation:

$E(t) = 0.65t + 69.5$

And, now we find t = 10:

$E(10) = 0.65(10) + 69.5 = 6.5 + 69.5 = 76$

And there you go.

3. ## ok...

Originally Posted by Aryth
Let's say that at t = 0, it is the year 1995, so we know that:

$E(0) = 69.5$

So, at t = 4, we get:

$E(4) = 72.1$

We know that:

$E(x) = at + b$

Where b is the y-intercept, which is $E(0)$, that gives us:

$E(t) = at + 69.5$

Now, all we need to to is set it equal to $E(4)$ to solve for a:

$E(4) = a(4) + 69.5 = 72.1$

$4a = 2.6$

$a = 0.65$

So, we now have an equation:

$E(t) = 0.65t + 69.5$

And, now we find t = 10:

$E(10) = 0.65(10) + 69.5 = 6.5 + 69.5 = 76$

And there you go.
Great job and easy reading.