In 1920, the record for a certain race was 46.6 sec. In 1990, it was 45.9 sec. Let R(t)=therecord 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 45.7 sec.
Find a linear function that fits the data.
R(t)=
just like the one about salaries
m = change in seconds per year = (45.9-46.6)/(1990-1920) = -.01 seconds / year
so
R = -.01 *t + 46.6
check when t = 70 (that is 1990)
R = -.01(70) +46.6 = 45.9 check
b) t = 2003 - 1920 = 83
R = -.01(83) + 46.6 = 45.77
c) 45.7 = -.01 t + 46.6
t = 90 years
90 + 1920 = 2010
To find a linear function that fits the data, we can use the formula for the equation of a straight line:
y = mx + b
In this case, we can let y represent the record time (R(t)) and x represent the number of years since 1920 (t). The slope of the line (m) will give us the rate at which the record time changes over time, and the y-intercept (b) will give us the initial record time in 1920.
To find the slope, we can use the formula:
m = (y2 - y1) / (x2 - x1)
Let's use the data points (1920, 46.6) and (1990, 45.9) to find the slope:
m = (45.9 - 46.6) / (1990 - 1920)
= -0.7 / 70
= -0.01
So, the slope of the line is -0.01.
To find the y-intercept, we can substitute the values of one of the data points into the equation and solve for b.
Using the point (1920, 46.6):
46.6 = -0.01 * 1920 + b
Simplifying the equation:
46.6 = -19.2 + b
Adding 19.2 to both sides:
b = 46.6 + 19.2
= 65.8
So, the y-intercept is 65.8.
Putting it all together, the linear function that fits the data is:
R(t) = -0.01t + 65.8
Now, let's use this function to answer the remaining questions.
b) To predict the record in 2003 and 2006, we need to find the values of R(t) for t = 2003 - 1920 = 83 and t = 2006 - 1920 = 86, respectively.
R(83) = -0.01 * 83 + 65.8
= 64.97 (approximately)
R(86) = -0.01 * 86 + 65.8
= 64.94 (approximately)
Therefore, the predicted record for 2003 is approximately 64.97 seconds, and for 2006, it is approximately 64.94 seconds.
c) To find the year when the record will be 45.7 seconds, we can substitute R(t) = 45.7 into the linear function and solve for t:
45.7 = -0.01t + 65.8
Rearranging the equation:
-0.01t = 45.7 - 65.8
= -20.1
Dividing by -0.01:
t = (-20.1) / (-0.01)
= 2010
Therefore, the record will be 45.7 seconds in the year 2010.