in 1920, the record for a certain race was 45.8 sec. In 1980, it was 44.6 sec. Let R(t)=the record in the race and t=the number of years since 1920.

find the linear function that fits the data
use the function (a) to predict the record in 2003 and in 2006
find the year when the record will be 43.84 sec.

R(t)=?
what is the predicted record for 2003? sec

what is the predicted record for 2006? sec

In what year will the predicted record be 43.84 seconds?

Check my thinking on this.

1920 to 1980 = 60 years.
In 60 years the record decreased by 45.8-44.6 seconds = 1.2 seconds or 1.2/60 = 0.02 sec/year.
Therefore, R(t) = 45.8-years83((0.02).
For 2003, that is 2003-1920 = 83 years.
R(2003) = 45.8 - 83(0.02)= ??

2006 is done similarly.
For the last part,
43.84 = 45.8-years(0.02) and solve for years.
Again, check my thinking.

To find the linear function that fits the given data, we need to determine the rate at which the record changes over time. We can do this by calculating the slope of the line connecting the two data points.

First, let's find the slope (m) using the formula:

m = (change in y) / (change in x)

The change in y is the difference between the record times: 44.6 sec - 45.8 sec = -1.2 sec.
The change in x is the difference in years: 1980 - 1920 = 60 years.

Now, we can substitute the slope (m) and one of the data points (1920, 45.8 sec) into the point-slope form of a linear equation: y - y1 = m(x - x1).

Using (x1, y1) = (1920, 45.8) and m = -1.2 sec / 60 years, we get:

R(t) - 45.8 sec = (-1.2 sec / 60 years)(t - 1920)

Simplifying the equation gives:

R(t) = -0.02t + 83.8

So, the linear function that fits the data is R(t) = -0.02t + 83.8.

Now, let's use this function to predict the record in 2003 and 2006 by substituting the years (t) into the equation.

For 2003 (t = 2003 - 1920 = 83 years):

R(83) = -0.02(83) + 83.8
R(83) = 1.66 + 83.8
R(83) ≈ 85.46 sec

So, the predicted record for 2003 is approximately 85.46 seconds.

For 2006 (t = 2006 - 1920 = 86 years):

R(86) = -0.02(86) + 83.8
R(86) = 1.72 + 83.8
R(86) ≈ 85.52 sec

So, the predicted record for 2006 is approximately 85.52 seconds.

To find the year when the predicted record will be 43.84 seconds, we can set R(t) equal to 43.84 and solve for t:

43.84 = -0.02t + 83.8

Rearranging the equation gives:

-0.02t = 43.84 - 83.8
-0.02t = -39.96

Dividing both sides by -0.02 gives:

t = -39.96 / -0.02
t = 1998

So, the predicted record will be 43.84 seconds in the year 1998.