SOLUTION: Problem 2. In 1920, the record for a certain race was 45.6 sec. In 1990, it was 44.8 sec. Let R(t)= the record in the race and t= the number of the year since 1920. a) Find a

Algebra ->  Graphs -> SOLUTION: Problem 2. In 1920, the record for a certain race was 45.6 sec. In 1990, it was 44.8 sec. Let R(t)= the record in the race and t= the number of the year since 1920. a) Find a      Log On


   

Question 201382: Problem 2.
In 1920, the record for a certain race was 45.6 sec. In 1990, it was 44.8 sec.
Let R(t)= the record in the race and t= the number of the year since 1920.

a) Find a linear function that the date
b) Use the function in (a) to predict the record in 2003 and in 2006.
c) Find the year when the record will be 43.86 sec

Find a linear function that fits the data.
R(t)= ?? -0.02t-45.6???
What is the predicted record for 2003?
What is the predicted record for 2006?
In what year will the predicted record be 43.86 sec??

I appreciate this.
Thank you!

Answer by user_dude2008(1862) About Me  (Show Source):
You can put this solution on YOUR website!
a)

Function is R(t) = -0.04t + 45.6 (close, but no cigar)

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b)

2003 ---> t=83
--> R(83) = -0.04(83) + 45.6 = 42.28

In 2003, time is approx 42.28 sec.


2006 ---> t=86
--> R(86) = -0.04(86) + 45.6 = 42.16

In 2006, time is approx 42.16 sec.


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c)

43.86 = -0.04t + 45.6

43.86 - 45.6 = -0.04t

-1.74= -0.04t

t = 43.5

Year will be 1920+43 = 1963 (mid year)