SOLUTION: A school record in a certain race in 1960 was 3.8 minutes. In 1980 the school record in the same event was 3.65 minutes.Choose the linear equation for these data points.(y=-.0075x+

Question 1042615: A school record in a certain race in 1960 was 3.8 minutes. In 1980 the school record in the same event was 3.65 minutes.Choose the linear equation for these data points.(y=-.0075x+18.5),(y=.0075x-18.5), (y=.0075x 0.18.5)
Answer by Boreal(15235) (Show Source): You can put this solution on YOUR website!
The year is the x-value (fixed, known) and the time for the race the y-value.
slope is
change in y/change in x.
It is 3.65-3.80/1980-1960
That is -0.15/20= -0.0075
Now use the point slope formula
y-y1=m(x-x1)
y-3.8=-0.0075(x-1960), using the first point (the second will give the same answer)
y=-.0075x-(-0.0075*1960)+3.8
y=-0.0075x+14.7+3.8
y=-0.0075x+18.5, the first ANSWER
Check using second point.
y-3.65=-0.0075(x-1980)
y-3.65=-0.0075x+14.85
y=-0.0075x+18.5
RELATED QUESTIONS
In 1920, the record for a certain race was 46.8 sec. In 1980, the record was 45.6 sec.... (answered by mangopeeler07)

In 1920, the record for a certain race was 45.7 sec. In 1980, it was 43.9 sec. Let R(t)=... (answered by richwmiller)

Please help. In 1920, the record for a certain race was 45.4 sec. In 1960, it was 44.6... (answered by Mathtut)

Solve problem using an absolute value equation or inequality In 1985 Steve Cram set a (answered by bucky)

Lost again, this one seems like it is not even in the book for how to solve it. Any help (answered by stanbon)

In 1920, the record for a certain race was 45.7 sec. In 1990, it was 44.3 sec. Let R(t)=... (answered by ankor@dixie-net.com)

Need some help please. Question : The world record for the mile was 4 minutes and 10... (answered by wuwei96815)

In 1920, the record for a certain race was 46.7 seconds in 1940, it was 46.1 sec. Let... (answered by ankor@dixie-net.com)

In 1920 the record for a certain race was 45.1 sec. In, 1980, it was 4303 sec. Let R(t)= (answered by ewatrrr)