Answer: 1.5
-1.5
Since any 2 points determine 1 line, take 2 of the points and find the equation of the line drawn thru these 2 points. Substitute the x and y of the either point into the equation and find the y-intercept (b) Then, substitute the x and y of the 3rd point into the equation and see if the both sides of the equation are =. (y2-y1) ÷ (x2 - x1) = slope y = slope * x + b Point # 1 = (6, 5) Point # 2 = (10, 25) Point # 3 = (12, 30) Point # 4 = (12, 35) (y2 - y1) ÷ (x2 - x1) = slope (25 - 5) ÷ (10 - 6) = slope (20) ÷ (4) = slope Slope = 5 y = m * x + b y = 5 * x + b Substitute the x and y of the point (6, 5) into the equation and find the y-intercept (b) y = 5 * x + b 5 = 5 * 6 + b 5 = 30 + b b = -25 y = 5 * x - 25 . Check your points Point # 1 = (6, 5) 5 = 5 * 6 - 25 5 = 30 - 25 OK . Point # 2 = (10, 25) 25 = 5 * 10 - 25 25 = 5 * 10 - 25 OK . Then, substitute the x and y of the 3rd point into the equation and see if the both sides of the equation are Point # 3 = (12, 30) . y = 5 * x - 25 30 = 5 * 12 - 25 30 = 60 - 25 = 35 Point # 3 = (12, 30) is not on the line . . Point # 4 = (12, 35) 35 = 5 * 12 - 25 35 = 60 - 25 =35 Point # 4 = (12, 35) is on the line
If the quadrilateral physics of the number can be squared be point 12 of a dove, then it is possible.
Since there are 12 inches in one foot, you divide 68 by 12 and you get 5 feet 6 point 66 inches.
Answer: 1.5
The residual for a particular point in a regression is negative if the estimated or fitted value at that point is greater than the observed value.
-1.5
A residual is defined in the context of some "expected" value. There is no information in the question regarding expected values.
Residual point
sin = -12/13 cos = 5/12 tan = -5/12 cosec = -13/12 sec = 12/5 cotan = -12/5
Turning Point - 2011 The Happiness 5-12 was released on: USA: 13 March 2013
If a data point has a residual of zero, it means that the observed value of the data point matches the value predicted by the regression model. In other words, there is no difference between the actual value and the predicted value for that data point.
The point lies 1 unit below the regression line.
If you mean a slope of 2/5 and the point (-15, 12) then equation is 5y = 2x+90
No.
The point lies one unit below the regression line.