Question

In 1920, the record for a certain race was 46.1 sec. In 1960, it was 44.9 sec. Let R(t) = the record in the race and t = the nbr of years since 1920.
a) Find the 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 43.19 sec.

The linear function that fits the data:
R(t) = ?
Predict record in 2003 and 2006
The year when the record will be 43.19 sec.

Answer 1

\[R(t) = 46.1 + 0.03 \times t\]

Answer 2

I've given you the function now you can solve your question Good Luck !

Answer 3

dumbunny did it help ?

Answer 4

Not yet...still working on it. LOL

Answer 5

The answer you gave me for R(t) isn't right.

Answer 6

R(t)= -0.03t+46.1

Answer 7

shadow, that's not right either.

Answer 8

oh yeah sorry it would be minus -0.03 t

Answer 9

shadwofax is right

Answer 10

physopholy it's 19998 now

Answer 11

and 5% if we account only maths online students lol

Answer 12

okay, I didn't see the -

Answer 13

since t is years-1920, you know the second data point has 40 for the t. Since the difference between them (the R(t)s) is 1.2, the rate of change is -1.2/40 or -.03 seconds/year. This is your slope. Since at t=0 your R(t) is 46.1, it's R(t)=46.1-.03t. That answers a. For part B you must plug in 83 and 86 for the t. For c you do 43.19=46.1-.03t and solve for t. On an unrelated note, it's pretty sad how this problem sets the human race up to get slower over time. The average getting lower makes sense, but with steroids and stuff like that our maximum should get higher over time.....

Answer 14

I am still not understanding how you get part b and c...

Answer 15

That's the easiest part. t is years past 1920, so you just take 2003-1920 and 2006-1920 to get your t values to plug into your function that you get for answer A. C is the same thing but plugging in the R(t) (the record speed) for an unknown number of years, solving for t, and then adding it onto 1920 to get the year

Answer 16

physopholy, the way I am reading it, we are getting faster not slower.

Answer 17

You make it sound easy, but I am confused as hell.

Answer 18

You got the A part, correct? R(t)=46.1-.03t. The t here means "years past 1920". So if t=1 it's 1921. The R(t) is the new record time for that year. See how when t=0 the R(t)=your given R(t) for 1920? That's how we know I'm right

Answer 19

Yea, I got part A and I figured there are 83 years from 1920 to 2003 and 86 years from 1920 to 2006. That is as far as I am getting.

Answer 20

I can't figure this out..

Answer 21

Plug in the 83 into part A for the t on the right side beside the -.03. Tell me what you get.

Answer 22

R(83) = 0.03(83) + 46.1

Answer 23

no...R(83) = -0.03(83) + 46.1

Answer 24

r(t) = 46.1 - .03 x 83
yeah you got that right just solve it using calculator

Answer 25

that is the record for 1920+83, or 2003. Use 86 instead of 83 to get the value for 2006 afterwards

Answer 26

I don't have an algebraic calculator so trying to multiply by a negative number is about impossible.

Answer 27

I got 38.2

Answer 28

That can't be right.

Answer 29

Then 39.6

Answer 30

lol, just multiply by a positive and then stick a negative in front of it. 83*-.03=-2.49. For 2003 it's 46.1-2.49=43.61

Answer 31

another three years will lower it by another 3*.03 so the answer then is 43.52

Answer 32

Thanks...