SOLUTION: In 1920, the record for a certain race was 45.4 sec. In 1990, it was 44.7 sec. Let R(t) = the record race and t = number of years since 1920.
I have found the linear function that
Algebra ->
Linear-equations
-> SOLUTION: In 1920, the record for a certain race was 45.4 sec. In 1990, it was 44.7 sec. Let R(t) = the record race and t = number of years since 1920.
I have found the linear function that
Log On
Question 257686: In 1920, the record for a certain race was 45.4 sec. In 1990, it was 44.7 sec. Let R(t) = the record race and t = number of years since 1920.
I have found the linear function that fits the dats which is R(t) = -0.0175 + 45.4 and found the predicted record for 2003 which is 43.95 and the predicted record for 2006 which is 43.9.
The problem I am having is finding the year that 44.50 is the predicted record. Could someone please help me with this on how to figure In what year will the predicted record be 44.50?
Thank you so much
Freda 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 45.4 sec. In 1990, it was 44.7 sec. Let R(t) = the record race and t = number of years since 1920.
In what year will the predicted record be 44.50?
:
I think there is a slight inaccuracy in the equation you came up with
:
The slope: m = = = -.01
The equation should be:
f(t)= -.01t + 45.4; where t is the years from 1920
then
2003: t=83
f(t) = -.01(83) + 45.4
f(t) = 44.57
and the same way:
2006: t=86
f(t) = 44.54
:
Find the year when it's 44.50; find t when f(t) = 44.5
-.01t + 45.4 = 44.5
-.01t = 44.5 - 45.4
-.01t = -.9
t =
t = +90 yrs
:
1920 + 90 = 2010 the year for a record of 44.5 seconds
