Question 252643: In 1920, the record for a certain race was 46.6 sec. In 1940 it was 46.2 sec.
Let R(t)= the record in the race and t= the number of years since 1920.
Find the linear function that fits the data
R(t)=
What is the predicted record for 2003? 2006?
In what year will the record be 44.62 sec?
Found 2 solutions by drk, richwmiller:
Answer by drk(1908)   (Show Source):
You can put this solution on YOUR website! We have coordinates: (0,46.6) and (20,46.2). WE can find the slope of these using:
So, we get
m = (46.2 - 46.6) / (20 - 0)
m = -.02
Now, we put our slope and 1 point into y = mx + b to find b.
46.6 = -.02(0 + b.
b = 46.6
We now have a linear equation to model our data:
Y = -.02X + 46.6
2003 is 83 years after 1920, so x = 83.
Y = -.02(83) + 46.6 = 44.94
2006 is 86 years after 1920, so x = 86.
Y = -.02(86) + 46.6 = 44.88.
When will it be 44.62. This is the Y value and we want to find x. So,
44.62 = -.02X + 46.6
-1.98 = -.02X
X = 99 years, or 2019.
Answer by richwmiller(17219)  (Show Source):
You can put this solution on YOUR website! set t=year and R(t)=race time use r(t) as the y and t as the x
find the find the slope difference in y's over difference in x's
m is slope
m=(y1-y2)/(x1-x2)
then plug one point data as y and x
y=m(x)+b
and solve for b
then substitute r(t) for y and t for x
and then substitute 2003 and later 2006 for t
and find r(t)
for the final question, find t when r(t)=44.62
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