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i think its pretty much the same thing because matrix X1 X2 IS ACTUALLY X1 X2
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∙ 12y ago
Anonymous
Econometrics
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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
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.
M is the midpoint of pq verify that this is the midpoint by using the distance formula to show tha pm equals mq?
Let P(x1, y1), Q(x2, y2), and M(x3, y3).If M is the midpoint of PQ, then,(x3, y3) = [(x1 + x2)/2, (y1 + y2)/2]We need to verify that,√[[(x1 + x2)/2 - x1]^2 + [(y1 + y2)/2 - y1]^2] = √[[x2 - (x1 + x2)/2]^2 + [y2 - (y1 + y2)/2]^2]]Let's work separately in both sides. Left side:√[[(x1 + x2)/2 - x1]^2 + [(y1 + y2)/2 - y1]^2]= √[[(x1/2 + x2/2)]^2 - (2)(x1)[(x1/2 + x2/2)) + x1^2] + [(y1/2 + y2/2)]^2 - (2)(y1)[(y1/2 + y2/2)] + y1^2]]= √[[(x1)^2]/4 + [(x1)(x2)]/2 + [(x2)^2]/4 - (x1)^2 - (x1)(x2) + (x1)^2 +[(y1)^2]/4 + [(y1)(y2)]/2 + [(y2)^2]/4 - (y1)^2 - (y1)(y2) + (y1)^2]]= √[[(x1)^2]/4 - [(x1)(x2)]/2 + [(x2)^2]/4 + [(y1)^2]/4 - [(y1)(y2)]/2 + [(y2)^2]/4]]Right side:√[[x2 - (x1 + x2)/2]^2 + [y2 - (y1 + y2)/2]^2]]= √[[(x2)^2 - (2)(x2)[(x1/2 + x2/2)] + [(x1/2 + x2/2)]^2 + [(y2)^2 - (2)(y2)[(y1/2 + y2/2)] + [(y1/2 + y2/2)]^2]]= √[[(x2)^2 - (x1)(x2) - (x2)^2 + [(x1)^2]/4 + [(x1)(x2)]/2 + [(x2)^2]/4 + (y2)^2 - (y1)[(y2) - (y2)^2 + [(y1)^2]/4) + [(y1)(y2)]/2 + [(y2)^2]/4]]= √[[(x1)^2]/4 - [(x1)(x2)]/2 + [(x2)^2]/4 + [(y1)^2]/4 - [(y1)(y2)]/2 + [(y2)^2]/4]]Since the left and right sides are equals, the identity is true. Thus, the length of PM equals the length of MQ. As the result, M is the midpoint of PQ
How do you do equations and graphs for slope?
The equation for the slope between the points A = (x1, y1) and B = (x2, y2) = (y2 - y1)/(x2 - x1), provided x1 is different from x2. If x1 and x2 are the same then the slope is not defined.
How would you graph y equals 5x-12?
Select any tow values of x, say x1 and x2.Calculate y1 = 5*x1 - 12 and y2 = 5*x2 - 12. Mark the points (x1, y1) and (x2, y2). Join them with a straight line and extend in both directions.
Matrix inverse method ------ x1 x2 x33000 -x1 5x2-x30 2x1-3x30?
inverse matrix. x1+x2+x3=3,000 -x1+5x2=0 2x1-3x3=0.
What is the formula to calculate the slope?
It's m = y2 - y1/ x2- x1 It's m equals y2 minus y1 over x2 minus x1
Let A denote the event that the sum of 2 dice equals 11 Let x1 x2 denote the outcome first die equals x1 second die equals x2 Then?
A={(6,5),(5,6)} And The probability P(A) equals1/8
How do you swap two numbers without third variables?
If the variables are x1 & x2 the solution is : 1) x1=x1+x2; 2) x2=x1-x2; 3) x1=x1-x2; EX: x1=1 , x2=6; 1) x1= 1+6 = 7 2) x2= 7-6 =1 3 x1=7-1 =6 ============================================
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
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.
Average Rate of Change Fx equals x2 plus 3x plus 5.1 from x1 equals -5 to x2 equals 6?
f(x1) = (-5)2 + 3*(-5) + 5.1 = 25 - 15 + 5.1 = 15.1 f(x2) = (6)2 + 3*(6) + 5.1 = 36 + 18 + 5.1 = 59.1 f(x2)-f(x1) = 59.1 - 15.1 = 44 x2 - x1 = 6 - (-5) = 11 So average rate of change = 44/11 = 4
Maxz equals 2x1 plus 2x2 stc 5x1 plus 3x2 equals 8 x1 plus 2x2 equals 4 x1 x2 equals 0 and integerssolve by ipp?
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M is the midpoint of pq verify that this is the midpoint by using the distance formula to show tha pm equals mq?
Let P(x1, y1), Q(x2, y2), and M(x3, y3).If M is the midpoint of PQ, then,(x3, y3) = [(x1 + x2)/2, (y1 + y2)/2]We need to verify that,√[[(x1 + x2)/2 - x1]^2 + [(y1 + y2)/2 - y1]^2] = √[[x2 - (x1 + x2)/2]^2 + [y2 - (y1 + y2)/2]^2]]Let's work separately in both sides. Left side:√[[(x1 + x2)/2 - x1]^2 + [(y1 + y2)/2 - y1]^2]= √[[(x1/2 + x2/2)]^2 - (2)(x1)[(x1/2 + x2/2)) + x1^2] + [(y1/2 + y2/2)]^2 - (2)(y1)[(y1/2 + y2/2)] + y1^2]]= √[[(x1)^2]/4 + [(x1)(x2)]/2 + [(x2)^2]/4 - (x1)^2 - (x1)(x2) + (x1)^2 +[(y1)^2]/4 + [(y1)(y2)]/2 + [(y2)^2]/4 - (y1)^2 - (y1)(y2) + (y1)^2]]= √[[(x1)^2]/4 - [(x1)(x2)]/2 + [(x2)^2]/4 + [(y1)^2]/4 - [(y1)(y2)]/2 + [(y2)^2]/4]]Right side:√[[x2 - (x1 + x2)/2]^2 + [y2 - (y1 + y2)/2]^2]]= √[[(x2)^2 - (2)(x2)[(x1/2 + x2/2)] + [(x1/2 + x2/2)]^2 + [(y2)^2 - (2)(y2)[(y1/2 + y2/2)] + [(y1/2 + y2/2)]^2]]= √[[(x2)^2 - (x1)(x2) - (x2)^2 + [(x1)^2]/4 + [(x1)(x2)]/2 + [(x2)^2]/4 + (y2)^2 - (y1)[(y2) - (y2)^2 + [(y1)^2]/4) + [(y1)(y2)]/2 + [(y2)^2]/4]]= √[[(x1)^2]/4 - [(x1)(x2)]/2 + [(x2)^2]/4 + [(y1)^2]/4 - [(y1)(y2)]/2 + [(y2)^2]/4]]Since the left and right sides are equals, the identity is true. Thus, the length of PM equals the length of MQ. As the result, M is the midpoint of PQ
X2 - x - 12 equals 0?
x²-x-12=0 x1=-(-1/2) - Square root of ((-1/2)²+12) x1=0.5 - 3.5 x1=-3 x2=-(-1/2) + Square root of ((-1/2)²+12) x2=0.5 + 3.5 x2=4
Show magnitude in a graph?
the absolute value of displacement which is delta equals x2-x1
X2 plus 2x - 15 equals 0?
x²+2x-15=0 x1=-2/2 - Square root of ((2/2)²+15) x1=-1-4 x1=-5 x2=-2/2 + Square root of ((2/2)²+15) x2=-1+4 x2=3
