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Law Incorporation [45]
3 years ago
11

If parallelogram JKLM is rotated 270° counterclockwise around the origin, what are the coordinates of the endpoints of the side

congruent 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)

Mathematics
2 answers:
Sati [7]3 years ago
8 0

The <em>correct answer</em> is:

B) K’(6, 4); L’(3, 3)

Explanation:

A 270° clockwise rotation maps every point (x, y) to (y, -x). (It negates the x-coordinate and then switches it and the y-coordinate.)

This means the endpoints of the side congruent to KL, K' and L', would be mapped as:

K'(-4, 6)→(6, 4)

L'(-3, 3)→(3, 3)

Furkat [3]3 years ago
8 0

Answer:

the user above is correct

Step-by-step explanation:

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Step-by-step explanation:

Data given and notation  

n=1000 represent the random sample taken    

\hat p=0.52 estimated proportion of of U.S. employers were likely to require higher employee contributions for health care coverage

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Solution to the problem

The confidence interval would be given by this formula

\hat p \pm z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}

For the 95% confidence interval the value of \alpha=1-0.95=0.05 and \alpha/2=0.025, with that value we can find the quantile required for the interval in the normal standard distribution.

z_{\alpha/2}=1.96

The margin of error for this case is given by:

ME= z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}

And replacing we got:

ME = 1.64*\sqrt{\frac{0.52(1-0.52)}{1000}}=0.0259

And replacing into the confidence interval formula we got:

0.52 - 1.64*\sqrt{\frac{0.52(1-0.52)}{1000}}=0.4941

0.52 + 1.64*\sqrt{\frac{0.52(1-0.52)}{1000}}=0.5459

And the 95% confidence interval would be given (0.4941;0.5459).

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