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3 years ago

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A rectangular field is three times as long as it is wide. If the perimeter of the field is 400 feet, what are the dimensions of

the field

1 answer:

ArbitrLikvidat [17]3 years ago

8 0

Answer:

L=150 W=50

Step-by-step explanation:

150 is 3 times 50, 150+150+50+50 is 400

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Solve for x <br>please I need the answer the picture shows the question

ozzi

Answer:

15inch

Step-by-step explanation:

X^2=8^2+7^2

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2 years ago

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Which expression is equivalent to 7 3/4 - 1/2 Choose one answer A: 1/2 + 7 3/4 B: 7 3/4 + (-1/2) C: 1/2 + (-7 3/4) D: -7 3/4 + (

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3 years ago

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The length of a rectangle is eight more than twice its width. The perimeter is 96 feet. Find the dimensions of the rectangle

blagie [28]

Step-by-step explanation:

Length = L

Width = W

L=8 more than 2W

L = 8 * 2W

L = 16W

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perimeter is the lengths of all sides added together

so

2L + 2W = 96

we have

L = 16W

2L + 2W = 96

let's do substitution

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2(16W) + 2W = 96

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2 years ago

Which are compostie number 49, 35, 32, 41, 47

tensa zangetsu [6.8K]

Answer: 49, 35, and 32

Step-by-step explanation:

7 * 7 = 49

7 * 5 = 35

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3 0

4 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