SOLUTION: Lost again, this one seems like it is not even in the book for how to solve it. Any help would be appreciated. In 1920, the record for a certain race was 45.4 sec. In 1960, it

Algebra ->  College  -> Linear Algebra -> SOLUTION: Lost again, this one seems like it is not even in the book for how to solve it. Any help would be appreciated. In 1920, the record for a certain race was 45.4 sec. In 1960, it      Log On


   



Question 144527: Lost again, this one seems like it is not even in the book for how to solve it. Any help would be appreciated.
In 1920, the record for a certain race was 45.4 sec. In 1960, it was 44.6. Let R(t)= the record in the race and t= the number of years since 1920.
a) Find a linear frunction 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 43.6 sec.

Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
In 1920, the record for a certain race was 45.4 sec.
In 1960, it was 44.6.
Let R(t)= the record in the race and t= the number of years since 1920.
---------------------
a) Find a linear frunction that fits the data.
You have two points relating years since 1920 and race records:
The points are (0,45.4) and (40,44.6)
slope = (44.6-45.4)/(40-0) = -0.02
intercept = 45.4
EQUATION: record = (-0.02)(years since 1920) + 45.4
r = -0.02t + 45.4
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b) Use the function in (a) to predict the record in 2003 and in 2006.
r(2003) = -0.02(2003-1920) + 45.4
r(2003) = -0.02(83) + 45.4 = 43.74
----------------------------------------
r(2006) = ?????
I'll let you do this following the same pattern as above.
------------------------------
c) Find the year when the record will be 43.6 sec.
43.6 = -0.02t + 45.4
-0.02t = -1.8
t = 90
year = 1920+90 = 2010
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Cheers,
Stan H.