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3 years ago

What are the coordinates of vertex J of the pre-image?

Mathematics

1 answer:

Dvinal [7]3 years ago

3 0

The figure is a rhombus J' K' L' M' in a coordinate plane, showing all its vertices.

The vertex J' is the point (0, -2).

And the statement indicates that the quadrilateral JKLM was transformed into the figure J'K'L'M' by the dilation (x,y) -> (x/2, y/2).

Then you have to find the preimag of J' which is J.

You know that x was multiplied by 1/2 to transform into 0, then x/2 = 0 => x =0

And y was mutilplied by 1/2 to transform into -2, then y/2 = -2 => y = -4.

And the pre-image is (0, -4)

Answer: (0, -4)

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The 98% confidence interval estimate of the proportion of adults who use social media is (0.56, 0.6034).

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of $\pi$ , and a confidence level of $1-\alpha$ , we have the following confidence interval of proportions.

$\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}$

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z is the zscore that has a pvalue of $1 - \frac{\alpha}{2}$ .

Of the 2809 people who responded to survey, 1634 stated that they currently use social media.

This means that $n = 2809, \pi = \frac{1634}{2809} = 0.5817$

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So $\alpha = 0.02$ , z is the value of Z that has a pvalue of $1 - \frac{0.02}{2} = 0.99$ , so $Z = 2.327$ .

The lower limit of this interval is:

$\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.5817 - 2.327\sqrt{\frac{0.5817*4183}{2809}} = 0.56$

The upper limit of this interval is:

$\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.5817 + 2.327\sqrt{\frac{0.5817*4183}{2809}} = 0.6034$

The 98% confidence interval estimate of the proportion of adults who use social media is (0.56, 0.6034).

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3 years ago