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X1 plus y2?
4
Is the smallest LCM of a number is the number itself?
Umm, you're question seems to have been a bit garbled. If the question was meant to be: Is the LCM of a single number the number itself? In that case the answer is...N/A... LCM (Least Common Multiple) MUST be at least two numbers. Otherwise "common" has no meaning in the title and the least multiple of any number would be 1 x0 =1, x1 =x, etc...However, if the question was meant to be something like: If a number is a multiple of another number, is their LCM the larger number? In that case, yes.P.S. "Is the smallest lcm" is redundant. Smallest and least are identical in this situation.
The distance between points x1 y1 and 5 -5 is the square root of x1 plus 5 2 plus y1 - 52?
false
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 the answer to Evaluate x1 - x-1 plus x0 for x equals 2?
shut up and do your hw
What is the least common multiple of 3x4 and 12x2?
Not sure this question is stated correctly. Answering logically, the LCM of 12 (3x4) and 24 (12x2) is 24. This is because 12(x2=24) and 24(x1=24) are both factors of 24. Altering the question slightly, the least common multiple of 2, 3, 4, and 12 is 12. This is because 2(x6=12), 3(x4=12), 4(x3=12), and 12(x1=12) are all factors of 12.
X plus one over x squared plus two?
it equals x1 it equals x1
X1 plus y2?
4
Is the smallest LCM of a number is the number itself?
Umm, you're question seems to have been a bit garbled. If the question was meant to be: Is the LCM of a single number the number itself? In that case the answer is...N/A... LCM (Least Common Multiple) MUST be at least two numbers. Otherwise "common" has no meaning in the title and the least multiple of any number would be 1 x0 =1, x1 =x, etc...However, if the question was meant to be something like: If a number is a multiple of another number, is their LCM the larger number? In that case, yes.P.S. "Is the smallest lcm" is redundant. Smallest and least are identical in this situation.
The distance between points x1 y1 and 5 -5 is the square root of x1 plus 5 2 plus y1 - 52?
false
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 X3 plus X2 plus X plus 1 divided by X 1?
2x2+7/x1
When does an LPP possess a multiple optimal solutions?
minimize Z=-X1+2X2=
What size does a GUI have to be on roblox?
A GUI size has to be at least X1 Y1
X1 plus x2 divided by 2 y1 plus y2 divided by 2?
Oh honey, you've got yourself a classic case of finding the average of two points in a coordinate plane. All you need to do is add the x-coordinates (x1 + x2) and divide by 2 to get the x-coordinate of the midpoint. Then do the same for the y-coordinates (y1 + y2), divide by 2, and voila, you've got the y-coordinate of the midpoint. Easy peasy lemon squeezy!
The distance between points (x1 y1) and (4 8) is the square root of (x1 - 8)2 plus (y1 - 4)2.?
It is false-apex
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.
