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March 31, 2015

March 31, 2015

Posted by **scooby9132002** on Sunday, June 14, 2009 at 6:33pm.

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) -
**Damon**, Sunday, June 14, 2009 at 6:43pmRecord = 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) -
**Quidditch**, Sunday, June 14, 2009 at 6:49pm1 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) -
**scooby9132002**, Sunday, June 14, 2009 at 6:55pmok so what is the predicted record for 2006?

in what year will the predicted record be 44.98 seconds?

- math (am i correct) -
**Damon**, Sunday, June 14, 2009 at 7:08pmScooby, 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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