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2 years ago

What is 7 times the 6th power?

Mathematics

1 answer:

Sidana [21]2 years ago

5 0

7^6 = 117,649
7 x 7 x 7 x 7 x 7 x 7 = 117,649

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One leg of a right triangle is 10 units, and its hypotenuse is 12 units. What is the length of the other leg?

denis-greek [22]

Answer:

about 7 units

Step-by-step explanation:

10 $10^{2} + b^{2} = 12^{2} \\100+b^{2} = 144 \\144- 100= 44\\=\sqrt{44} \\= 6.6\\round \\=7$

3 0

3 years ago

The claim is that the white blood cell counts of adult females are normally distributed, with a standard deviation equal to 1.6

loris [4]

The population standard deviation, σ = 1.68, the sample standard deviation, s = 1.95, the sample mean =7.23, and the sample size, n = 39.The test statistics is given by
T=(n - 1) (s / σ)²
= (39 - 1) (1.95 / 1.68)²
= 38 (1.35)= 51.196

3 0

3 years ago

What percent of a yard is in a foot?

Norma-Jean [14]

There are 3 feet in 1 yard so each foot is 1/3

4 0

3 years ago

Read 2 more answers

Identify the segment bisector of MN Then find MN.<br> ????????? Helppp pleaseee

enot [183]

Given:

The figure.

To find:

The segment bisector of MN and value of MN.

Solution:

From the given figure it is clear that ray RP,i.e., $\overrightarrow {RP}$ is the segment bisector of MN because it divides segment MN in two equal parts.

Now,

$3\dfrac{5}{6}=\dfrac{3\times 6+5}{6}$

$3\dfrac{5}{6}=\dfrac{23}{6}$

Since, $\overrightarrow {RP}$ is the segment bisector of MN, therefore,

$MN=2\times \dfrac{23}{6}$

$MN=\dfrac{23}{3}$

$MN=7\dfrac{2}{3}$

Therefore, the length of MN is $7\dfrac{2}{3}$ .

6 0

3 years ago

A quality-conscious disk manufacturer wishes to know the fraction of disks his company makes which are defective. Step 2 of 2 :

const2013 [10]

Answer:

The 80% confidence interval for the population proportion of disks which are defective is (0.059, 0.079).

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of $\pi$ , and a confidence level of $1-\alpha$ , we have the following confidence interval of proportions.

$\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}$

In which

z is the zscore that has a pvalue of $1 - \frac{\alpha}{2}$ .

Suppose a sample of 1067 floppy disks is drawn. Of these disks, 74 were defective.

This means that $n = 1067, \pi = \frac{74}{1067} = 0.069$

80% confidence level

So $\alpha = 0.2$ , z is the value of Z that has a pvalue of $1 - \frac{0.2}{2} = 0.9$ , so $Z = 1.28$ .

The lower limit of this interval is:

$\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.069 - 1.28\sqrt{\frac{0.069*0.931}{1067}} = 0.059$

The upper limit of this interval is:

$\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.069 + 1.28\sqrt{\frac{0.069*0.931}{1067}} = 0.079$

The 80% confidence interval for the population proportion of disks which are defective is (0.059, 0.079).

7 0

2 years ago