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Question 201381: Hello! I need a help with ASAP!!!
Problem 1.
In 1920, the record for a certain race was 45.4 sec. In 1990, it was 44.0 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 43.56 sec
Find a linear function that fits the data.
R(t)= ?? -002-45.4???
What is the predicted record for 2003? 43.74??
What is the predicted record for 2006? 43.68??
In what year will the predicted record be 43.56 sec??
I appreciate this.
Thank you!
Answer by nerdybill(7384)   (Show Source):
You can put this solution on YOUR website! In 1920, the record for a certain race was 45.4 sec. In 1990, it was 44.0 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
Think of the x-axis representing the year and the y-axis representing the "record-time" in seconds.
The problem, then, really gives you two data points: (remember, since 1920)
1920 we get: (0,45.4)
1990 we get: (70,44)
Slope then is
(y2-y1)/(x2-x1) = (44-45.4)/(70-0) = -1.4/70 = -0.02
.
Using the slope and one (70,44) of the two points substitute it into the "point-slope" form:
y-y1 = m(x-x1)
y-44 = -0.02(x-70)
y = -0.02(x-70) + 44
y = -0.02x + 1.4+44
y = -0.02x + 45.4
Replacing y with R(t) and x with t we get:
R(t) = -0.02t + 45.4
.
b) Use the function in (a) to predict the record in 2003 and in 2006.
For 2003:
t = 2003-1920 = 83
R(t) = -0.02t + 45.4
R(t) = -0.02(83) + 45.4
R(t) = -1.66 + 45.4
R(t) = -1.66 + 45.4
R(t) = 43.74 seconds
.
For 2006:
t = 2006-1920 = 86
R(t) = -0.02t + 45.4
R(t) = -0.02(86) + 45.4
R(t) = -1.66 + 45.4
R(t) = -1.72 + 45.4
R(t) = 43.68 seconds
.
c) Find the year when the record will be 43.56 sec
R(t) = -0.02t + 45.4
43.56 = -0.02t + 45.4
43.56-45.4 = -0.02t
-1.84 = -0.02t
92 = t
Year then = 1920+92 = 2012
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