SOLUTION: this question has been driving me insane for days. x and y are connected by y=a/1-bx^2. a and b are constants. i have been given a table of experimental values for x and y which.

Question 868974: this question has been driving me insane for days.
x and y are connected by y=a/1-bx^2. a and b are constants. i have been given a table of experimental values for x and y which. i have been asked to plot a suitable graph to find a and b. i am not allowed to use logs and it will produce a straight line graph.
i have tried re arranging to get it to the form of y=mx+b. finding it difficult due to the 2 constants not being grouped.

any help would be greately appreciated
Answer by rothauserc(4718) (Show Source): You can put this solution on YOUR website!
y = a / (1 - bx^2)
x (6, 8, 10, 11, 12), y (5.50, 6.76, 9.10, 11.60, 16.67)
using the values for x and y we can calculate a regression line, y = mx + b
first calculate the mean of the x's = summation of x's / number of x's
mean for x = (6+8+10+11+12)/5 = 9.4
mean for y = (5.50 + 6.76 + 9.10 + 11.60 + 16.67) / 5 = 9.926
m is the slope = [ summation of (XiYi) - N*mean of x*mean of y ] / (summation of (Xi)^2 - N * mean of x)
m = [ 6*5.50 + 8*6.76 + 10*9.10 + 11*11.60 + 12*16.67) - 5*9.4*9.926 ] / ( 6^2 + 8^2 + 10^2 + 11^2 + 12^2) - 5*9.4^2) = -8.802/23.2 = -0.38
b = mean of y - m*mean of x = 9.926 - (-0.38*9.4) = 13.50
regression line is y = -0.38x + 13.50

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