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2 years ago

Pythagoras' Theorem:

Mathematics

2 answers:

sdas [7]2 years ago

6 0

Yes, it’s the exact same thing

kvv77 [185]2 years ago

3 0

Answer:

Yes.

Step-by-step explanation:

Yes

c^2 = a^2 + b^2

We take the square root of both sides to give:

√c^2 = √(a^2 + b^2)

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Is 68 prime because i am doing homework and im stuck on this and it is very hard so help me please

BlackZzzverrR [31]

If 68 is a prime number, then the only factors it has are 1 and 68.
If it has any other factors besides 1 and 68, then it's NOT prime.

Right away, without any higher math, you can look at just the last digit
in 68 . The last digit is '8'. That tells you that '68' is an even number,
and THAT tells you that '2' must be one of its factors. So '68' is not a
prime number.

The factors of 68 are 1, 2, 4, 17, 34, and 68 .

68 has four more factors besides 1 and 68, so it's not a prime number.

8 0

2 years ago

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Do the following setm of lengths form a right triangle 36,77, 85

xxTIMURxx [149]

Answer:

They can make a right traingle because both sides of the equation are equal these lengths do form a right triangle.

3 0

2 years ago

Read 2 more answers

43. What is the translation rule to map AHIJ to AH ! J'? 11 H

lutik1710 [3]

Left 4. Or x-4 because the way it moves

3 0

2 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

Kelsey entered the triathlon at camp Meadowlark. The total distance was 15 11/16 miles. The bike segment was 12 1/4 miles, and t

Novay_Z [31]

Answer:

3/8 was the swimming segment!

Step-by-step explanation:

We only need to subtract the bike and running segment from the total distance:

15 11/16 - 12 1/4 - 3 1/16

15.6875 - 12.25 - 3.0625 = 0.375 = 3/8

5 0

2 years ago