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March 29, 2017

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in 1920 the record for a certain race was 46.9 sec. In 1990 it was 45.5 sec. let R(t) = the record in the race and t = the number of years since 1920

1.)find a linear function that fits this data?

2.) use a function to predict the record in 2003 and 2006.

3.) find the year when the record will be 44.98 sec.


I think number 1 is 46.9

  • math (am i correct) - ,

    Record = m (year -1920) + b
    where year -1920 = t so
    R = m t + b
    46.9 = m (0) + b
    so b = 46.9
    in 1990, t = 1990 - 1920 = 70
    so
    45.5 = m (70) + 46.9
    m = -.02
    so
    R = -.02 t + 46.9

    for 2003 for example
    t = 2003 - 1920 = 83
    so
    R = -.02(83)+46.9
    R = 45.24

  • math (am i correct) - ,

    1 is not 46.9.
    Question 1 asks you to find a linear function that fits the data.

    Find a function f(t) where t is the number of years since 1920. There are two points given.
    1920, 46.9s
    1990, 45.5s
    Or, for t expressed as years since 1920...
    0 years, 46.9s
    70 years, 45.5s
    for a linear function, y = mx + b
    Where m is the slope and b is the y axis intercept.
    The slope, m, is (y2 - y1)/(x2 - x1)
    or
    (45.5 - 46.9)/(70 - 0)
    Calculate this value for the slope.
    The y intercept is at x=0 which is already given as one of the points (0, 46.9)

  • math (am i correct) - ,

    ok so what is the predicted record for 2006?


    in what year will the predicted record be 44.98 seconds?

  • math (am i correct) - ,

    Scooby, we are not going to do everything for you.
    From the way I did 2003 you should be able to do the rest.

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