Question 430403: In 1920, the record for a certain race was 45.1 seconds. In 1950, it was 44.8 seconds. Let R(t)= the record in the race and t= the number of years since 1920. Find a linear function that fits the data. Find the year when the record will be 44.12 sec.
Answer by nerdybill(7384)   (Show Source):
You can put this solution on YOUR website! In 1920, the record for a certain race was 45.1 seconds. In 1950, it was 44.8 seconds. Let R(t)= the record in the race and t= the number of years since 1920. Find a linear function that fits the data. Find the year when the record will be 44.12 sec.
.
year| x-value | y-value
1920| 0 | 45.1
1950| 30 | 44.8
.
two points provided then is:
(0, 45.1)
(30,44.8)
.
Slope:
(y2-y1)/(x2-x1)
(44.8-45.1)/(30-0)
(-0.3)/(30)
-0.01
.
using one point (0,45.1) and the slope -.01
plug into the "point-slope" form
y - y1 = m(x -x1)
y - 45.1 = -.01(x -0)
y - 45.1 = -.01x
y = -.01x+45.1
.
R(t)= -.01t+45.1
.
Now, to find when record will be 44.12 -- set it in place of R(t) and solve for t:
44.12 = -.01t+45.1
-0.98 = -.01t
98 = t (years after 1920)
answer then is
1920+98 = 2018
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