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Consider the following table and use the formulas that are given for computing the correlation coefficient. (Give your answer correct to two decimal places.)

x 1 1 0 0 0
y 7 3 5 6 6


r =
n


xy







x







y






n


x2








x



2


n


y2








y



2


(1)


r =
SS(xy)

SS(x) ยท SS(y)

(2)

This is what I done (2)(27)/5 =10.8 for ss(xy) then 2^/5=.8 for SS(x), ten (27^)/5=145.8 for SS(y), then I put 10.8 over sqrt.8 X 145.8 = 1.00

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