In 1920, the record for a race was 46.9 seconds. In 1940 it was 46.5 seconds. Let R(t)= the record in the race and t= the number of years since 1920.
a) Find a linear function that fits data.
b)Predict the records for 2003 and 2006.
c) Year when record will be 44.90 seconds.
To find a linear function that fits the given data, we can use the general form of a linear function, y = mx + b, where m is the slope and b is the y-intercept.
a) We are given two points: (0, 46.9) and (20, 46.5). The first point represents the year 1920 (since it is 0 years since 1920) and the record time is 46.9 seconds. The second point represents the year 1940 (20 years since 1920) and the record time is 46.5 seconds.
To find the slope, we can use the formula:
m = (y2 - y1) / (x2 - x1)
m = (46.5 - 46.9) / (20 - 0)
m = -0.4 / 20
m = -0.02
Now, substitute one of the points into the linear function equation to find the y-intercept:
46.9 = (-0.02)(0) + b
b = 46.9
Therefore, the linear function that fits the data is:
R(t) = -0.02t + 46.9
b) To predict the records for 2003 and 2006, we need to find the number of years since 1920 for each of those years and then substitute those values into the linear function.
For 2003, t = 2003 - 1920 = 83
R(83) = -0.02(83) + 46.9
R(83) ≈ 45.44 seconds (rounded to two decimal places)
For 2006, t = 2006 - 1920 = 86
R(86) = -0.02(86) + 46.9
R(86) ≈ 45.38 seconds (rounded to two decimal places)
Therefore, the predicted record for 2003 is approximately 45.44 seconds and for 2006 is approximately 45.38 seconds.
c) To determine the year when the record will be 44.90 seconds, we can set up the equation and solve for t in the linear function equation:
44.90 = -0.02t + 46.9
Rearranging the equation to solve for t:
-0.02t = 44.90 - 46.9
-0.02t = -2
t = -2 / (-0.02)
t = 100
Therefore, the record will be 44.90 seconds in the year 1920 + 100 = 2020.