If parallelogram JKLM is rotated 270° clockwise around the origin, what are the coordinates of the endpoints of the side congrue
nt to side KL in the image parallelogram?
A. K’(-4, -6); L’(3, -3)
B. K’(-6, -4); L’(-3, -3)
C. K’(6, 4); L’(3, 3)
D. K’(-4, 6); L’(-3, -3)
A. K’(-4, -6); L’(3, -3)
B. K’(-6, -4); L’(-3, -3)
C. K’(6, 4); L’(3, 3)
D. K’(-4, 6); L’(-3, -3)
2 answers:
Answer: The answer is (B). K'(-6,-4) and L(-3,-3).
Step-by-step explanation: In the figure, the parallelogram JKLM is drawn on a co-ordinate plane. Here, the co-ordinates of points K and L are (-4,6) and (-3,3) respectively.
Now, when we rotate the parallelogram about the origin through 270° , then we have to make a jump of three consecutive quadrants,.ie., presently we are in the fourth quadrant, so we will the resulting co-ordinates in the third quadrant and so both x and y co-ordinates of the new points will be negative. Also, the x-co-ordinate and y-co-ordinates of both the points K and L will get reversed. Please see the attached figure.
Thus, the new co-ordinates of K and L are K'(-6,-4) and L(-3,-3).
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Answer:
Step-by-step explanation:
We are given the following in the question:
Sample size, n = 67
Variance = 3.85
We have to find 80% confidence interval for the population variance of the weights.
Degree of freedom = 67 - 1 = 66
Level of significance = 0.2
Chi square critical value for lower tail =
Chi square critical value for upper tail =
80% confidence interval:
Putting values, we get,
Thus, (3.13,4.91) is the required 80% confidence interval for the population variance of the weights.