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3 years ago

Is 9/31 an irrational number

Mathematics

2 answers:

elena55 [62]3 years ago

5 0

Answer:No

Step-by-step explanation:

tatyana61 [14]3 years ago

5 0

Answer:

No, 9/31 is not an irrational number.

Step-by-step explanation:

A real number is called rational number when it can be expressed in the form of $\frac{p}{q}$ ,

Where, both p and q are integers and q ≠ 0,

Otherwise, it is called irrational number.

Here, the given number is,

$\frac{9}{31}$

Since,both 9 and 31 are integers,

Also, 31 ≠ 0,

⇒ 9/31 is a rational number,

Hence, It is not an irrational number.

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Answer:

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And rounded up we have that n=1068

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

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Solution to the problem

In order to find the critical value we need to take in count that we are finding the interval for a proportion, so on this case we need to use the z distribution. Since our interval is at 95% of confidence, our significance level would be given by $\alpha=1-0.95=0.05$ and $\alpha/2 =0.025$ . And the critical value would be given by:

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The margin of error for the proportion interval is given by this formula:

$ME=z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}$ (a)

And on this case we have that $ME =\pm 0.03$ and we are interested in order to find the value of n, if we solve n from equation (a) we got:

$n=\frac{\hat p (1-\hat p)}{(\frac{ME}{z})^2}$ (b)

We don't have a prior estimation for the proportion $\hat p$ so we can use 0.5 as an approximation for this case

And replacing into equation (b) the values from part a we got:

$n=\frac{0.5(1-0.5)}{(\frac{0.03}{1.96})^2}=1067.11$

And rounded up we have that n=1068

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