y=-5x-17
y=x-1
-5x-17=x-1
-17=6x-1
-16=6x
2 2/3=x
then you plug in the x for each of the equations to make sure that the answer you get is correct.
;)
x1:y1 = x2:y2 4:-2 = x2:5 x2 = (4*5)/-2 x2 = -10
It is linear. The highest power is 1 (x = x1, y = y1) so it is linear.
Suppose that you have simple two variable model: Y=b0+b1X1+e The least squares estimator for the slope coefficient, b1 can be obtained with b1=cov(X1,Y)/var(X1) the intercept term can be calculated from the means of X1 and Y b0=mean(Y)-b1*mean(X1) In a larger model, Y=b0+b1X1+b2X2+e the estimator for b1 can be found with b1=(cov(X1,Y)var(X2)-cov(X2,Y)cov(X1,X2))/(var(X1)var(X2)-cov(X1,X2)2) to find b2, simply swap the X1 and X2 terms in the above to get b2=(cov(X2,Y)var(X1)-cov(X1,Y)cov(X1,X2))/(var(X1)var(X2)-cov(X1,X2)2) Find the intercept with b0=mean(Y)-b1*mean(X1)-b2*mean(X2) Beyond two regressors, it just gets ugly.
x - y = xydifferentiating wrt x1 - (dy/dx) = x(dy/dx) + y(x + 1)(dy/dx) + y + 1 = 0
This question can only be answered if the probability distribution functions of X1, X2 and X3 are known. They are not and so the question cannot be answered.
x1:y1 = x2:y2 4:-2 = x2:5 x2 = (4*5)/-2 x2 = -10
use the formula y-y1=m(x-x1)
It is linear. The highest power is 1 (x = x1, y = y1) so it is linear.
Use the point-slope form: y - y1 = m(x - x1). From the givens, y1 = 2, x1 = 9, and m = -2. Thus, y - 2 = -2(x - 9) = -2x + 18, or y = -2x + 20.
Suppose that you have simple two variable model: Y=b0+b1X1+e The least squares estimator for the slope coefficient, b1 can be obtained with b1=cov(X1,Y)/var(X1) the intercept term can be calculated from the means of X1 and Y b0=mean(Y)-b1*mean(X1) In a larger model, Y=b0+b1X1+b2X2+e the estimator for b1 can be found with b1=(cov(X1,Y)var(X2)-cov(X2,Y)cov(X1,X2))/(var(X1)var(X2)-cov(X1,X2)2) to find b2, simply swap the X1 and X2 terms in the above to get b2=(cov(X2,Y)var(X1)-cov(X1,Y)cov(X1,X2))/(var(X1)var(X2)-cov(X1,X2)2) Find the intercept with b0=mean(Y)-b1*mean(X1)-b2*mean(X2) Beyond two regressors, it just gets ugly.
it equals x1 it equals x1
x - y = xydifferentiating wrt x1 - (dy/dx) = x(dy/dx) + y(x + 1)(dy/dx) + y + 1 = 0
85
(y-y1)=m(x-x1) OR we can write it y=m(x-x1)+y1
y - y1 = m(x - x1), where m is the slope of the line, and (x1, y1) is a point on the line.
This question can only be answered if the probability distribution functions of X1, X2 and X3 are known. They are not and so the question cannot be answered.
3