SOLUTION: In 1920, the record for a certain race was 46.7 seconds in 1940, it was 46.1 sec. Let R(T)= the record in the record in the race and t=the number of yes since 1920 A. find a lin

Algebra ->  Graphs -> SOLUTION: In 1920, the record for a certain race was 46.7 seconds in 1940, it was 46.1 sec. Let R(T)= the record in the record in the race and t=the number of yes since 1920 A. find a lin      Log On


   

Question 330600: In 1920, the record for a certain race was 46.7 seconds in 1940, it was 46.1 sec. Let R(T)= the record in the record in the race and t=the number of yes since 1920
A. find a linear function that fits the data
b. Use the function in (a) to predict the record in 2003
c Find the year when the record will be 43.88 seconds
Answer by ankor@dixie-net.com(22740) About Me  (Show Source):
You can put this solution on YOUR website!
In 1920, the record for a certain race was 46.7 seconds in 1940, it was 46.1 sec.
Let R(T)= the record in the record in the race and t=the number of yes since 1920
:
A. find a linear function that fits the data
Find the slope, assign the given value as follows:
x1=0, y1=46.7; (1920)
x2=20,y2=46.1; (1940)
slope(m) = %2846.1-46.7%29%2F%2820-0%29 = -.6%2F20 = -.03
:
the point/slope equation; y - y1 = m(x - x1)
y - 46.7 = -.03(x - 0)
y = -.03x + 46.7
;
R(t) = -.03t + 46.7, is the function
:
:
b. Use the function in (a) to predict the record in 2003
2003 - 1920 = 83; t=83
R(83) = -.03(83) + 46.7
R(83) = -2.49 + 46.7
R(83) = 44.21 sec in 2003
:
c Find the year when the record will be 43.88 seconds
-.03t + 46.7 = 43.88
-.03t = 43.88 - 46.7
-.03t = -2.82
t = %28-2.82%29%2F%28-.03%29
t = +94 yrs
1920 + 94 = 2014 the yr it should be 43.88 sec