in 1920 the record for a certain race was 46.4 sec. In 1960 it was 45.6 sec.
Let R(t)=the record in the race and t= the number of years since 1920.
a. find a linear function that fits the data.
b. Use the function in (a) to predict the record in 2003 and in 2006.
c. Find the year when the record will be 44.52 sec
a)1920t=46.3r
1970t=45.8r
To find a linear function that fits the given data, we can use the slope-intercept form of a linear equation, which is given by:
y = mx + b
where y is the dependent variable (record time in this case), x is the independent variable (number of years since 1920), m is the slope, and b is the y-intercept.
a. We are given two data points: (1920, 46.4) and (1960, 45.6). Let's use these two points to find the slope (m) and the y-intercept (b).
The slope (m) can be calculated using the formula:
m = (y2 - y1) / (x2 - x1)
Substituting the values:
m = (45.6 - 46.4) / (1960 - 1920)
m = -0.8 / 40
m = -0.02
Now, let's substitute the slope (m) and one of the points (1920, 46.4) into the equation to find the y-intercept (b):
46.4 = -0.02 * 1920 + b
46.4 = -38.4 + b
b = 46.4 - (-38.4)
b = 46.4 + 38.4
b = 84.8
Therefore, the linear function that fits the given data is:
R(t) = -0.02t + 84.8
b. To predict the record in 2003 and 2006, substitute t as the number of years since 1920.
For 2003: t = 2003 - 1920 = 83
R(83) = -0.02 * 83 + 84.8
R(83) = 1.66 + 84.8
R(83) ≈ 86.46 seconds
For 2006: t = 2006 - 1920 = 86
R(86) = -0.02 * 86 + 84.8
R(86) = 1.72 + 84.8
R(86) ≈ 86.52 seconds
Therefore, the predicted records in 2003 and 2006 are approximately 86.46 seconds and 86.52 seconds, respectively.
c. To find the year when the record will be 44.52 seconds, substitute R(t) = 44.52 into the linear function and solve for t:
44.52 = -0.02t + 84.8
-0.02t = 44.52 - 84.8
-0.02t = -40.28
t = -40.28 / -0.02
t = 2014
Therefore, the year when the record is predicted to be 44.52 seconds is 2014.