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3 years ago

26. A positive whole number is called stable if at least one of its digits has the same value

Mathematics

1 answer:

Kisachek [45]3 years ago

5 0

Answer:

Step-by-step explanation:

Given the value 78247 a s a stable number because at least one of its digits has the same value as its position in the number. The 4th number in the value is 4, this makes the number a stable number.

The following are the 3-digits stable numbers that appears in 78247

The first number is 824. This digits are stable numbers because 2 as a number is situated in the same place as the number (2nd position).

Hence, there are only 1 stable 3-digit numbers in the value 78247 since only a value exists as 2 in the value and there is no 1 and 3 in the value.

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6.7-3^2.9+4^3 evaluate the expression. show your work on a piece of paper. 50+ points~!

vovangra [49]

Answer:

Step-by-step explanation:

$6.7 - {3}^{2} .9 + {4}^{3} \\ \\ = 42 - 9.9 + 64 \\ \\ = 42 - 81 + 64 \\ \\ = 106 - 81 \\ \\ = 25$

7 0

3 years ago

Read 2 more answers

Find the geometric mean of 14 and 20.

Tom [10]

Answer:

16.73 to nearest hundredth.

Step-by-step explanation:

The geometric mean of x and y is √xy.

So out geometric mean is √(14*20)

= √280

= 16.7332

7 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

I can type 19 words in 38 seconds. How many can I type in 60 seconds?

mel-nik [20]

Hi!

We can solve this by setting up a proportion

$\frac{19}{38} \frac{x}{60}$

Solve for x by cross multiplying.

19 x 60 = 1140
1140/38 = 30

You can type 30 words in 60 seconds

Hope this helps! :)

6 0

3 years ago

Read 2 more answers

What is the volume of the cylinder below

Alika [10]

Answer:

Option A, 80π

Step-by-step explanation:

4²×5π

= 80π

7 0

3 years ago