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3 years ago

Please help me figure out these two problems!! Thanks

Mathematics

1 answer:

erastovalidia [21]3 years ago

4 0

Answer:

A) eight legs per one spider (56/7)

B) one pretzel is 2.25 (6.75/3)

Step-by-step explanation:

56 divided by 7 is 8

6.75 divided by 3 is 2.25

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How do you solve this absolute value equatuon? 4|x - 1| - 7 = -3

MatroZZZ [7]

4|x - 1| - 7 = -3 |add 7 to both sides

4|x - 1| = 4 |divide both sides by 4

|x - 1| = 1 ⇔ x - 1 = 1 or x - 1 = -1 |add 1 to both sides

x = 2 or x = 0

Answer: B.

3 0

4 years ago

During the election of 1824, Jackson won the popular vote, but no canididate won a mojority of the electoral votes. How was this

Reil [10]

The cabinet got do deside I think sorry if I’m wrong

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

Step-by-step explanation:

The confidence interval 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 level of significance $\\ \alpha$ is 1 - 0.95 = 0.05. In this case, then, we have that the z-score 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 standard normal table, 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 normal distribution. 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

Find the probability of this event. Enter the answer as a fraction in simplest form, as a decimal, and as a percent.

olasank [31]

Answer:

12/60, 0.2, 20%

hope i was helpful :)

5 0

3 years ago

Noah wants to put a varitey of fish in his new tank his tank is large enough to hold 70 fish find how many fish of each breed he

Kryger [21]

Answer:

14 Rainbow, 28 Swordtail, and 21 Molly. He can still have 7 more fish.

Step-by-step explanation:

6 0

3 years ago