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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?
3
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
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
What is in a flaming dark light deck?
This deck contains. Monsters Flamvell firedog x2 flamvell magician x3 Flamvell guard x2 masked dragon x3 dread dragon x2 majestic dragon x2 worm caratos x2 worm dimikles x2 worm gulse x1 mystic tomato x3 changer synchron x2 reptilianne naga x2 drill barnicle x1 spells pot of avarice x1 pot of benevolence x2 monster reborn x1 cards of consonence x1 scapegoat x1 stardust shimmer x2 dark hole x1 heavy storm x1 burden of the mighty x1 crashbug road x2 swords of revealing light x1 spirit burner x1 Traps parallel selection x2 bottomless trap hole x2 trap hole x2 threating roar x2 call of the huanted x1 shadow spell x1 starlight road x2 urgent tuning x2 magic cylinder x1
