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It is a straight line equation.
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If y equals -2 when x equals 4 find x when y equals 5?
x1:y1 = x2:y2 4:-2 = x2:5 x2 = (4*5)/-2 x2 = -10
Is the equation x plus y equals 4 linear or not?
It is linear. The highest power is 1 (x = x1, y = y1) so it is linear.
How do you use a variance-covariance matrix to obtain least squares estimates?
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.
What is differential equation of x-y equals xy?
x - y = xydifferentiating wrt x1 - (dy/dx) = x(dy/dx) + y(x + 1)(dy/dx) + y + 1 = 0
How do find the probability of distribution Let Y equals X1 - X2 plus 2x3 Find Pr Y1.5?
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.
If y equals -2 when x equals 4 find x when y equals 5?
x1:y1 = x2:y2 4:-2 = x2:5 x2 = (4*5)/-2 x2 = -10
Determine if the point 3-4 is a solution for the equation y equals 8 -4x?
use the formula y-y1=m(x-x1)
Is the equation x plus y equals 4 linear or not?
It is linear. The highest power is 1 (x = x1, y = y1) so it is linear.
If f is a differentiable function such that f of 9 equals 2 and f' of 9 equals -2 then what is the linearization?
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.
X plus one over x squared plus two?
it equals x1 it equals x1
How do you use a variance-covariance matrix to obtain least squares estimates?
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.
What is differential equation of x-y equals xy?
x - y = xydifferentiating wrt x1 - (dy/dx) = x(dy/dx) + y(x + 1)(dy/dx) + y + 1 = 0
What is a point slope form equation?
(y-y1)=m(x-x1) OR we can write it y=m(x-x1)+y1
What does the point slope formula look like?
y - y1 = m(x - x1), where m is the slope of the line, and (x1, y1) is a point on the line.
Maxz equals 2x1 plus 2x2 stc 5x1 plus 3x2 equals 8 x1 plus 2x2 equals 4 x1 x2 equals 0 and integerssolve by ipp?
3
What is the y-intercept formula?
y-y1=m(x+x1)
How do find the probability of distribution Let Y equals X1 - X2 plus 2x3 Find Pr Y1.5?
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.
