SOLUTION: In 1920, the record for a certain race was 45.1 sec. In 1970, it was 44.1 sec. Let R(t)=the record in the race and t=the number of years since 1920. a)Find a linear function that

Algebra ->  Linear-equations -> SOLUTION: In 1920, the record for a certain race was 45.1 sec. In 1970, it was 44.1 sec. Let R(t)=the record in the race and t=the number of years since 1920. a)Find a linear function that       Log On


   

Question 275024: In 1920, the record for a certain race was 45.1 sec. In 1970, it was 44.1 sec. Let R(t)=the record in the race and t=the number of years since 1920.
a)Find a linear function that fits the data.
b)Use the function in (a) to predict the record in 2003 and in 2006.
c)Find the year when the record will be 43.20 sec.
Find a linear function that fits the data.
R(t)=
Answer by ankor@dixie-net.com(22740) About Me  (Show Source):
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In 1920, the record for a certain race was 45.1 sec. In 1970, it was 44.1 sec.
Let R(t)=the record in the race and t=the number of years since 1920.
:
a)Find a linear function that fits the data.
In 1920: x1=0; y1=45.1
In 1970: x2=50; y2=44.1
:
Find the slope:
m = %2844.1-45.1%29%2F%2850-0%29 = -1%2F50
write the equation using the form y - y1 = m(x - x1)
y - 45.1 = -1%2F50(x - 0)
y = -1%2F50x +45.1
or
R(t) = -1%2F50t + 45.1; is the linear function
:
:
b)Use the function in (a) to predict the record in 2003 and in 2006.
2003; t= 83
R(t) = -1%2F50(83) + 45.1
R(t) = 43.44 sec
2006; t=86
R(t) = -1%2F50(86) + 45.1
R(t) = 43.38 sec
:
:
c)Find the year when the record will be 43.20 sec.
-1%2F50t + 45.1 = 43.20
-1%2F50t = 43.20 - 45.1
-1%2F50t = -1.9
t = -50 * -1.9
t = 95, which is 1920 + 95 = 2015, the record should be 43.2


R(t)=