In 1920, the record for a certain race was 46.6 sec. In 1980, it was 45.4 sec. Let R(t)= the record in the race and t = the number of years snce 1920.
a) Find a linear function that fits the data R(t) =
b) Use the function in (a) to predict the record in 2003 and in 2006.
2003 =
2006 =
c
consider 1920 --- > t = 0
then your given data can be represented by two ordered pairs
(0,46.6) and (60,45.4)
slope = (45.4-46.6)/(60-0) = - 0.02
R(t) = - 0.02t + b
sub in (0,46.6)
b = 46.6
R(t) = -0.02t + 46.6
now just plug in values of t corresponding to the years,
e.g. 2006 ---> t = 86
R(86) = -.02(86) + 46.6 = 44.88
To find a linear function that fits the given data, we need to determine the equation of the line that passes through the two given points: (1920, 46.6) and (1980, 45.4).
Let's use the point-slope form of a linear equation:
y - y1 = m(x - x1)
First, let's find the slope (m) using the formula:
m = (y2 - y1) / (x2 - x1)
Substituting the given values:
m = (45.4 - 46.6) / (1980 - 1920)
= -1.2 / 60
= -0.02
Now, let's choose one of the points to plug into the equation. Let's choose (1920, 46.6).
Using (x1, y1) = (1920, 46.6) and m = -0.02, the equation becomes:
y - 46.6 = -0.02(x - 1920)
Expanding the equation:
y - 46.6 = -0.02x + 38.4
Simplifying the equation:
y = -0.02x + 85
Therefore, the linear function that fits the data is:
R(t) = -0.02t + 85
Now, let's use this function to predict the record in 2003 and 2006.
For 2003, the number of years since 1920 would be 2003 - 1920 = 83.
Plugging this value into the equation:
R(83) = -0.02(83) + 85
= -1.66 + 85
= 83.34
Therefore, the predicted record in 2003 would be approximately 83.34 seconds.
For 2006, the number of years since 1920 would be 2006 - 1920 = 86.
Plugging this value into the equation:
R(86) = -0.02(86) + 85
= -1.72 + 85
= 83.28
Therefore, the predicted record in 2006 would be approximately 83.28 seconds.
Now, let's move to part (c) of the question.