Mathematics, 02.12.2020 17:10 shambrayperkins

Fewer young people are driving. In , of people under years old who were eligible had a driver's license. Bloomberg reported that percentage had dropped to in . Suppose these results are based on a random sample of people under years old who were eligible to have a driver's license in and again in . a. At confidence, what is the margin of error and the interval estimate of the number of eligible people under years old who had a driver's license in ?

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Answer from: perfect32

Complete Question

Fewer young people are driving. In 1995, 63.9% of people under years 20 old who were eligible had a driver's license. Bloomberg reported that percentage had dropped to 41.7% in 2016. Suppose these results are based on a random sample 1,200 of people under 20 years old who were eligible to have a driver's license in 1995 and again in 2016.

a. At 95% confidence, what is the margin of error and the interval estimate of the number of eligible people under 20 years old who had a driver's license in 1995?

Margin of error(to four decimal places)

Interval estimate (to four decimal places)

b. At 95% confidence, what is the margin of error and the interval estimate of the number of eligible people under 20 years old who had a driver's license in 2016?

Margin of error(to four decimal places)

Interval estimate to (to four decimal places)

0.6120 < p < 0.639 + 0.6670

0.3900 < p < 0.4440

Step-by-step explanation:

Considering question a

The sample proportion is 1995 is $\^ p_1 = 0.639$

The sample size is n = 1200

From the question we are told the confidence level is 95% , hence the level of significance is

$\alpha = (100 - 95 ) \%$

=> $\alpha = 0.05$

Generally from the normal distribution table the critical value of $\frac{\alpha }{2}$ is

$Z_{\frac{\alpha }{2} } = 1.96$

Generally the margin of error is mathematically represented as

$E = Z_{\frac{\alpha }{2} } * \sqrt{\frac{\^ p (1- \^ p)}{n} }$

=> $E = 1.96 * \sqrt{\frac{0.639 (1- 0.639)}{1200} }$

=> E = 0.027

Generally 95% Interval estimate is mathematically represented as

$\^ p -E < p < \^ p +E$

=> 0.639 -0.027 < p < 0.639 + 0.027

=> 0.6120 < p < 0.639 + 0.6670

Considering question b

The sample proportion is 1995 is $\^ p_2 = 0.417$

The sample size is n = 1200

From the question we are told the confidence level is 95% , hence the level of significance is

$\alpha = (100 - 95 ) \%$

=> $\alpha = 0.05$

Generally from the normal distribution table the critical value of $\frac{\alpha }{2}$ is

$Z_{\frac{\alpha }{2} } = 1.96$

Generally the margin of error is mathematically represented as

$E = Z_{\frac{\alpha }{2} } * \sqrt{\frac{\^ p (1- \^ p)}{n} }$

=> $E = 1.96 * \sqrt{\frac{0.417 (1- 0.417)}{1200} }$

=> E = 0.027

Generally 95% Interval estimate is mathematically represented as

$\^ p -E < p < \^ p +E$

=> 0.417 -0.027 < p < 0.417 + 0.027

=> 0.3900 < p < 0.4440

Answer from: Quest

i think the correct option is the second option.

Which function is shown on the graph below

Answer from: Quest

when rotating a figure 180 degrees, imagine that you are able to take the figure and turn it clockwise or counterclockwise around a center point. to rotate a figure 180 degrees, you will need to apply the rule (x, y) → (-x, -y). start by using a coordinate grid with coordinates for each vertex of the figure. the center point of the coordinate grid is located at (0, 0), which is what you will rotate the figure around. write down the original coordinates of the shape you are going to rotate. then, apply the rule. for example, coordinate (1, 2) becomes (-1, -2). another example would be, coordinate (-4, -2) becoming (4, 2). after you change each original coordinate to the rotated coordinate, you will draw your new figure.

step-by-step explanation:

Identify the transformation that maps the figure with center (7, 1) onto itself. a) rotate 180° cloc

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