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3 years ago

(-5⁵)²as a single power

Mathematics

2 answers:

lord [1]3 years ago

7 0

Answer:-9765625

Step-by-step explanation: multiply the exponents first and then include the base. So, you would have 5x2=10, then se -5 to the 10th power (use a calculator).

goldenfox [79]3 years ago

6 0

Answer:-5^7

Step-by-step explanation:

as you have -5 to the power of 5 and u need to square that u have to add the powers therefore making it -5 to the power of

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The difference of two numbers is 5, and their sum is 17. What are the two numbers?

inysia [295]

Answer:

11 and 6

Step-by-step explanation:

11-6=5 <--- difference

11+6=17 <--- sum

3 0

2 years ago

A recipe uses 1 cup of sugar to make 6 cookies. If you want to make 30 cookies, how much sugar will you need? Use cups as your l

sergij07 [2.7K]

Answer:

the answer is 5 cups

Step-by-step explanation:

let x rep a cup of sugar

and y rep a cookie

and k be constant

x/kx=6y/30y

k=5

8 0

2 years ago

Someone PLEASE HELP IF YOUR GOOD AT MATH

Zina [86]

Answer:

your answer is 33

Step-by-step explanation:

33×6=198

8 0

3 years ago

Please someone help me !!!!!

Sveta_85 [38]

X² = 16
x = ±√16
x = ±4

The roots of this equation are -4 and 4
Сorrect answer is third from above

6 0

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