SOLUTION: Hello! I need a help ASAP!!! Problem 3. In 1920, the record for a certain race was 46.4 sec. In 1990, it was 45.0 sec. Let R(t)= the record in the race and t= the number of the

Algebra ->  Graphs -> SOLUTION: Hello! I need a help ASAP!!! Problem 3. In 1920, the record for a certain race was 46.4 sec. In 1990, it was 45.0 sec. Let R(t)= the record in the race and t= the number of the      Log On


   

Question 201400: Hello! I need a help ASAP!!!
Problem 3.
In 1920, the record for a certain race was 46.4 sec. In 1990, it was 45.0 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 44.56 sec
Find a linear function that fits the data.
R(t)= ??
What is the predicted record for 2003?
What is the predicted record for 2006?
In what year will the predicted record be 44.56 sec??
Thank you so much for your help!!!

Found 2 solutions by RAY100, jsmallt9:
Answer by RAY100(1637) About Me  (Show Source):
You can put this solution on YOUR website!
As a linear relationship, it follows the form, y=mx +b.
.
In this case,,,R=mt +b,,,with points (0,46.4),,and (70,45)
.
m= (y2-y1) / (x2-x1) = - (46.4 -45) /(1990 - 1920) = -1.4 / 70 = -.02
.
R = (-.02)t +b,,,,,@(0,46.4),,,46.4 = 0 +b,,,b=46.4
.
Base eqn ,,,,R = -.02t +46.4
.
@ 1990, t=1990-1920=70,,,R=-.02(70) +46.4 = 45,,,,,good check
.
@ 2003, t=2003-1920=83,,,R=-.02(83)+46.4 = 44.74
.
@ 2006, t=2006-1920=86,,,R=-.02(86)+46.4 = 44.68
.
Have a great day

Answer by jsmallt9(3758) About Me  (Show Source):
You can put this solution on YOUR website!
In general when you want to find the equation of a line, you need the slope of the line and at least one point on the line. To start with you have two points:
Year   t (Year-1920)   R(t)   Point
1920   0               46.4   (0, 46.4)
1990   70              45.0   (70, 45.0)

We have the point(s) we need but we do not have the slope. With the two points we can calculate the slope using the slope formula:
m+=+%28y%5B2%5D-y%5B1%5D%29%2F%28x%5B2%5D-x%5B1%5D%29
Substituting the coordinates of the two points above in the formula:
m+=+%2846.4+-+45.0%29%2F%280+-+70%29
Subtracting we get
m+=+%281.4%29%2F%28-70%29
Dividing 1.4 by -70 we get
m+=+-0.02
Now we have the slope and at least one point. There are a couple of ways to find the equation of the line:
  • Using the Point-Slope formula: y+-+y%5B1%5D+=+m%28x+-+x%5B1%5D%29
  • Using the Slope-Intercept form: y+=+mx+%2B+b
Since most students seem to the prefer the second method that is how we will proceed.
First we substitute the slope and the coordinates of a point (it doesn't matter which point as long as it is on the line) into the Slope-Intercept form.
y+=+mx+%2B+b
Substituting the coordinates of (0, 46.4) in for the x and the y and substituting -0.02 in for the m we get
%2846.4%29+=+%28-0.02%29%2A%280%29+%2B+b
We can now solve for b. Since -0.02 * 0 is zero and since 0 + b is b we get
46.4+=+b
Now that we have m (-0.02) and b (46.4) we can write the equation of our line. In doing so we will use t instead of x and R(t) instead of y.
R%28t%29+=+-0.02t+%2B+46.4
which is the answer to part (a).

For part (b), to calculate the record in 2003 we will use 83 for t (since 2003-1920 = 83) and figure out R(83). To calculate the record in 2006 we will use t = 86 and calculate R(86).
R%2883%29+=+-0.02%2883%29+%2B+46.4
Multiplying -0.02 by 83 we get
R%2883%29+=+-1.66+%2B+46.4
Adding we get
R%2883%29+=+44.74
Similarly
R%2886%29+=+-0.02%2886%29+%2B+46.4
Simplifying like above
R%2886%29+=+44.68

For part (c) we will use R(t) = 44.56 and solve for t:
44.56+=+-0.02t+%2B+46.4
Subtracting 46.4 from both sides we get
-1.84+=+-0.02t
Dividing both sides by -0.02 we get
92+=+t
Since t = 92 the year must be 92+1920 or 2012