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Problem:
The car has a standard transmission, so it had to be put in gear by shifting three times during the 15 seonds in which the speed was being measured. The accelerator was being depressed at a constant rate during the 15-second period and then released at the same rate until the pedal was no longer being depressed.

Table, time is x, speed is y, in each an ordered pair:
(x,y),
(1,6),
(2,14),
(3,20),
(4,30)
(5,35.5)
(6,40),
(7,48),
(8,52),
(9,54),
(10, 57),
(11, 61),
(12, 65.5),
(13, 66.5),
(14, 67),
(15, 68.2),
(16, 65),

I'm supposed to graph it using a scatterplot and determine the equation of best fit. I graphed it, but I don't know how to find the equation of best fit, if you could help, (AND EXPLAIN IT), not just give me the answer, that'd be really great, thanks!

2007-02-15 08:37:57 · 3 answers · asked by laura.ispurple 1 in Education & Reference Homework Help

3 answers

Let the equation of line be y = a + bx ..now our aim is to find the value of constant a and b ...

so we form to equation, to find a and b .

sum of y cordinate = n a + b (sum of x terms)
sum of (x * y ) cordinate = a ( sum of x terms) + b (sum of x^2 )

sum of y - cordinate = 749.7
sum of x cordinate = 136
sum of xy cordinate = 7762
and sum of x ^2 = 1496
number of x , y cordinate n =16 ..

plug in these values in the equation we get
749.7 = 16a + 136b
7762 = 136a + 1496b

solve these equation for a and b
we get a = 12.11
and b = 4.08
so best line of fit is y = 12.11 + 4.08x ...

Note : please cross check the calculation
Hope this will help you ..:)

2007-02-15 08:53:30 · answer #1 · answered by RAKESHtutor 3 · 0 0

Best fit is calculated as the gradient dy/dx of the slope of the line you have forced through your group of coordinates as listed.

Pick two distant points on this line and drop a perpendicular from each; also draw lined parallel to x axis and intercept with y axis.

Difference in y values divided by differences in x values provides m in y = mx + c. c is the point at which the line crosses the y axis. This is ultimately your equation.

2007-02-15 08:45:46 · answer #2 · answered by Modern Major General 7 · 0 0

Ah, good old geomeotry days. Man I miss them.

2007-02-15 08:42:08 · answer #3 · answered by Edgar 3 · 0 0

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