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Question 288104: In 1920, the record for a certyain race was 46.4 seconds. In 1950, it wq 46.1 seconds. Let R(t)= the record in the race and t= the number of years since 1920.
a.)find a linear fucntion that fits the data
b.) Use the function in (a) to predict the record in 2003 and in 2006.
c.) find the year that the record will be 45.53 seconds.
Find a linear equation function that fits the data.
r(t)=______
ROUND TO THE NEAREST HUNDRETH
Answer by amnd(23)   (Show Source):
You can put this solution on YOUR website! a) The basic form of the linear equation would be: R(t) = At + 46.4
This is formed from the record for the year 1920 (46.4), and the fact that for the year 1920, t (which is the number of years since the same year, 1920) is zero. This can be proven mathematically as:
R(0) = At + 46.4 = 0 + 46.4 = 46.4
Now, to find the value of the constant for the variable t. The record in 1950 (with t = 1950 - 1920 = 30) would help. Insert it into the linear equation with the unknown constant A:
R(30)= A*30 +46.4
46.1 = 30A+46.4
30A = -0.3
A =-0.01
.
Knowing A, we can form the linear equation:
R(t) = -0.01t +46.4 or 46.4 - 0.01t
.
b) Simply input the value for t for both years (for 2003, t = 83, and for 2006, t = 86). This is easy enough math so I'm not going to jot down the whole thing, but these are the answers if you want to check yours:
For 2003: R(83)= 45.57
For 2006: R(86)= 45.54
.
c) R(t) = -0.01t + 46.4
45.53 = -0.01t + 46.4
0.01t = 46.4 - 45.53
t=0.87*100=87
The year would be 1920 + t = 1920 + 87 = 2007
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