All categories

3 years ago

What is the prime factorization for 42

Mathematics

2 answers:

Semenov [28]3 years ago

7 0

Answer:

2*3*7

Step-by-step explanation:

Cerrena [4.2K]3 years ago

4 0

Answer:

2x3x7

Step-by-step explanation:

You might be interested in

The distance between points (1, 1) and (4, 7) is 6.71. is it true or false

svet-max [94.6K]

To determine whether the statement is true or false, we need to calculate for the distance of the two points. To do this, we can use the pythagorean theorem as follows:

d = √[(y2-y1)^2 + (x2-x1)^2]
d = √[(7-1)^2 + (4-1)^2]
d = √[(6)^2 + (3)^2]
d = 2√13 = 7.21

Therefore, the statement is false.

3 0

4 years ago

An environmentalist wants to find out the fraction of oil tankers that have spills each month. Step 2 of 2 : Suppose a sample of

rewona [7]

Answer:

The 80% confidence interval for the population proportion of oil tankers that have spills each month is (0.199, 0.257).

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 333 tankers is drawn. Of these ships, 257 did not have spills.

333 - 257 = 76 have spills.

This means that $n = 333, \pi = \frac{76}{333} = 0.228$

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:

$\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.228 - 1.28\sqrt{\frac{0.228*0.772}{333}} = 0.199$

The upper limit of this interval is:

$\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.228 + 1.28\sqrt{\frac{0.228*0.772}{333}} = 0.257$

The 80% confidence interval for the population proportion of oil tankers that have spills each month is (0.199, 0.257).

6 0

3 years ago

In order to set rates, an insurance company is trying to estimate the number of sick days that full time workers at an auto rep

son4ous [18]

Answer:

D. 31

Step-by-step explanation:

We have that to find our $\alpha$ level, that is the subtraction of 1 by the confidence interval divided by 2. So:

$\alpha = \frac{1 - 0.95}{2} = 0.025$

Now, we have to find z in the Ztable as such z has a pvalue of $1 - \alpha$ .

That is z with a pvalue of $1 - 0.025 = 0.975$ , so Z = 1.96.

Now, find the margin of error M as such

$M = z\frac{\sigma}{\sqrt{n}}$

In which $\sigma$ is the standard deviation of the population and n is the size of the sample.

A previous study indicated that the population standard deviation is 2.8 days.

This means that $\sigma = 2.8$

How large a sample must be selected if the company wants to be 95% confident that the true mean differs from the sample mean by no more than 1 day?

This is n for which M = 1. So

$M = z\frac{\sigma}{\sqrt{n}}$

$1 = 1.96\frac{2.8}{\sqrt{n}}$