SOLUTION: In 1920, the record for a certain race was 45.7 sec. In 1990, it was 44.3 sec. Let R(t)= the record in the race and t= the number of years since 1920. So then what does R(t)= what

Question 410403: In 1920, the record for a certain race was 45.7 sec. In 1990, it was 44.3 sec. Let R(t)= the record in the race and t= the number of years since 1920. So then what does R(t)=
what is the predicted record for 2003?
what is the predicted record in 2006?
In what year will the predicted record be 43.74 seconds?
Answer by ankor@dixie-net.com(22740) (Show Source): You can put this solution on YOUR website!
In 1920, the record for a certain race was 45.7 sec.
In 1990, it was 44.3 sec.
Let R(t)= the record in the race and t= the number of years since 1920.
So then what does R(t)=
:
Find the slope of the equation Let yr 1920 = 0
t1=0; r1=45.7
t2=70; r2=44.3
:
m = = = -.02 is the slope
Find the equation
r - 45.7 = -.02(t - 0)
R(t) = -.02t + 45.7 is the equation
:
what is the predicted record for 2003?
t = 83
R(t) = -.02(83) + 45.7
R(t) = 44.04 sec in 2003
:
what is the predicted record in 2006?
t = 86
R(t) = -.02(86) + 45.7
R(t) = 43.98 sec in 2006
:
In what year will the predicted record be 43.74 seconds?
Solve for t
-.02t + 45.7 = 43.74
-.02t = 43.74 - 45.7
-.02t = -1.96
t =
t = +98
1920 + 98 = 2018, the predicted record of 43.74
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