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3 years ago

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An article fills 1/2 of a magazine page. A corresponding photo takes up 3/8 of the article. How much of the page is taken up by

the photo

2 answers:

erastova [34]3 years ago

6 0

Answer:

$\frac{3}{16}$ part of a page.

Step-by-step explanation:

An article fills $\frac{1}{2}$ part of a magazine page.

Hence, if the total magazine page is x then $\frac{x}{2}$ of the magazine page is covered by an article.

Now, given that, a corresponding photo takes up $\frac{3}{8}$ part of the article.

Therefore, $\frac{x}{2}\times \frac{3}{8} =\frac{3x}{16}$ i.e. $\frac{3}{16}$ part of the total page is taken up by the photo. ( Answer )

tankabanditka [31]3 years ago

6 0

<u>Answer:</u>

An article fills 1/2 of a magazine page. The photo takes 3/16th of the page

<u>Solution:</u>

A corresponding photo takes up 3/8 of the article

We have to find out how much of the page is taken by photo

This can be calculated by finding out the space taken by photo in the total article space

Space taken by article = 1/2 page

Space taken by photo= 3/8th of article

3/8th of article means 3/8th of 1/2

$=\frac{3}{8} \times \frac{1}{2}=\frac{3}{16}$

Hence the corresponding photo takes 3/16 of the magazine page

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What is the solution to -3x<9

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dolphi86 [110]

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3 years ago

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3 years ago

In 2002, the mean age of an inmate on death row was 40.7 years with a standard deviation of 9.6 years according to the U.S. Depa

marissa [1.9K]

Answer:

The <em>95% confidence interval</em> for the current mean age of death-row inmates is between 42.23 years and 35.57 years.

Step-by-step explanation:

The <em>confidence interval</em> of the mean is given by the next formula:

$\\ \overline{x} \pm z_{1-\frac{\alpha}{2}}\frac{\sigma}{\sqrt{n}}$ [1]

We already know (according to the U.S. Department of Justice):

The (population) standard deviation for this case (mean age of an inmate on death row) has a standard deviation of 9.6 years ( $\\ \sigma = 9.6$ years).
The number of observations for the sample taken is $\\ n = 32$ .
The sample mean, $\\ \overline{x} = 38.9$ years.

For $\\ z_{1-\frac{\alpha}{2}}$ , we have that $\\ \alpha = 0.05$ . That is, the <em>level of significance</em> $\\ \alpha$ is 1 - 0.95 = 0.05. In this case, then, we have that the <em>z-score</em> corresponding to this case is:

$\\ z_{1-\frac{\alpha}{2}} = z_{1-\frac{0.05}{2}} = z_{1-0.025} = z_{0.975}$

Consulting a cumulative <em>standard normal table</em>, available on the Internet or in Statistics books, to find the z-score associated to the probability of, $\\ P(z$ , we have that $\\ z = 1.96$ .

Notice that we supposed that the sample is from a population that follows a <em>normal distribution</em>. However, we also have a value for n > 30, and we already know that for this result the sampling distribution for the sample means follows, approximately, a normal distribution with mean, $\\ \mu$ , and standard deviation, $\\ \sigma_{\overline{x}} = \frac{\sigma}{\sqrt{n}}$ .

Having all this information, we can proceed to answer the question.

Constructing the 95% confidence interval for the current mean age of death-row inmates

To construct the 95% confidence interval, we already know that this interval is given by [1]:

$\\ \overline{x} \pm z_{1-\frac{\alpha}{2}}\frac{\sigma}{\sqrt{n}}$

That is, we have:

$\\ \overline{x} = 38.9$ years.

$\\ z_{1-\frac{\alpha}{2}} = 1.96$

$\\ \sigma = 9.6$ years.

$\\ n = 32$

Then

$\\ 38.9 \pm 1.96*\frac{9.6}{\sqrt{32}}$

$\\ 38.9 \pm 1.96*\frac{9.6}{5.656854}$

$\\ 38.9 \pm 1.96*1.697056$

$\\ 38.9 \pm 3.326229$

Therefore, the Upper and Lower limits of the interval are:

Upper limit:

$\\ 38.9 + 3.326229$

$\\ 42.226229 \approx 42.23$ years.

Lower limit:

$\\ 38.9 - 3.326229$

$\\ 35.573771 \approx 35.57$ years.

In sum, the 95% confidence interval for the current mean age of death-row inmates is between 42.23 years and 35.57 years.

Notice that the "mean age of an inmate on death row was 40.7 years in 2002", and this value is between the limits of the 95% confidence interval obtained. So, according to the random sample under study, it seems that this mean age has not changed.

7 0

3 years ago