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2 years ago

Is 0.25 the decimal of 4/16

Mathematics

2 answers:

Bumek [7]2 years ago

6 0

Yes 0.25 is the decimal of 4/16. Because 1/4 divided by 4/16 is 1/4. And 100/4 is 25. So 1/4 is 0.25 or 4/16.

frosja888 [35]2 years ago

4 0

Yes.
A fraction means that you divide the top number (numerator) by the bottom number (denominator). So, if you divide 4 by 16, then the answer is 0.25
Another way to get this answer is if you know your basic fraction-decimal conversions. If you reduce the fraction by dividing both the numerator and the denominator by the same number (in this case, 4 works), then you get the fraction 1/4, which is also equal to 0.25
Good luck with your math, and I hope this helps :)

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3 years ago

If the greatest possible error of a measurement is 1/32 inches, what is the actual precision of this measurement?

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2 years ago

Very important

Veronika [31]

Using the z-distribution, we have that:

For a 99% confidence level, a sample size of 127 is needed.

For a 95% confidence level, a sample size of 74 is needed, meaning that a decrease in the confidence level decreases the needed sample size, as M and n are inverse proportional.

<h3>What is a z-distribution confidence interval?</h3>

The confidence interval is:

$\overline{x} \pm z\frac{\sigma}{\sqrt{n}}$

The margin of error is:

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

In which:

$\overline{x}$ is the sample mean.

z is the critical value.

n is the sample size.

$\sigma$ is the standard deviation for the population.

For a 99% confidence interval, $\alpha = 0.99$ , hence z is the value of Z that has a p-value of $\frac{1+0.99}{2} = 0.995$ , so the critical value is z = 2.575.

The margin of error and population standard deviation are:

$M = 3, \sigma = 13.1$

Hence we have to solve for n to find the needed sample size, as follows:

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

$3 = 2.575\frac{13.1}{\sqrt{n}}$

$3\sqrt{n} = 2.575 \times 13.1$

$\sqrt{n} = \frac{2.575 \times 13.1}{3}$

$(\sqrt{n})^2 = \left(\frac{2.575 \times 13.1}{3}\right)^2$

n = 126.4.

Rounding up, for a 99% confidence level, a sample size of 127 is needed.

For the 95% confidence interval, we have that z = 1.96, hence:

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

$3 = 1.96\frac{13.1}{\sqrt{n}}$

$3\sqrt{n} = 1.96 \times 13.1$

$\sqrt{n} = \frac{1.96 \times 13.1}{3}$

$(\sqrt{n})^2 = \left(\frac{1.96 \times 13.1}{3}\right)^2$

n = 73.3.

Rounding up, for a 95% confidence level, a sample size of 74 is needed, meaning that a decrease in the confidence level decreases the needed sample size, as M and n are inverse proportional.

More can be learned about the z-distribution at brainly.com/question/25890103

#SPJ1

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1 year ago