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Aleonysh [2.5K]
3 years ago
6

You are given a fraction in simplest form. the numerator is not zero. when you write the fraction as a decimal,it is a repeating

decimal,which numbers 1 to 10 could be the denominator, PLEASE HELP!!!!!
Mathematics
2 answers:
pochemuha3 years ago
6 0
You could easily do that yourself, with a pencil, and about the same amount of time it took you to post the question here.

If you go through and try them . . . 1/1,  1/2,  1/3,  1/4, 1/5 . . . etc., you'll find
that the thirds, sixths, sevenths, and ninths produce repeating decimals.
The oneths, tooths, fourths, fifths, eighths, and tenths don't.
alexgriva [62]3 years ago
4 0

The <em><u>correct answer</u></em> is:

3, 6, 7, 9

Explanation:

If your denominator was 1, you would have whole numbers, not repeating decimals.

If your denominator was 2, you would have halves.  These do not repeat.

If your denominator was 4, you would have 0.25 or 0.75.

If your denominator was 5, you would have 0.2, 0.4, 0.6, or 0.8.

If your denominator was 8, you would have 0.125, 0.375, 0.625, or 0.875.

If your denominator was 10, you would have 0.1, 0.3, 0.7, or 0.9.

However, if your denominator is 3, you would have repeating 3's or 6's.  If your denominator was 6, you would have a 1 with a repeating 6 or an 8 with repeating 3's.  If your denominator was 7, you would have repeating 0.142857, repeating 0.285714, repeating 0.428571, repeating 0.571428, repeating 0.714285, or repeating 0.857142.  If your denominator was 9, you would have repeating 1's, 2's, 4's, 5's, 7's, or 8's.

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snow_lady [41]
So first you would use the PEMDAS method. You would begin by subtacting 12 and 5 which is 7 then you would 7 multiply by 12 which is 84. Lastly you would divide it by 6 which would leave you with your final answer 14
7 0
3 years ago
Periodically a bottling company gets complaints that their bottles are not holding enough liquid. To test this claim the bottlin
kompoz [17]

Answer:

The null hypothesis was not rejected at 1% and 5% level of significance, but was rejected at 10% level of significance.

Step-by-step explanation:

The complaints made by the customers of a bottling company is that their bottles are not holding enough liquid.

The company wants to test the claim.

Let the mean amount of liquid that the bottles are said to hold be, <em>μ₀</em>.

The hypothesis for this test can be defined as follows:

<em>H₀</em>: The mean amount of liquid that the bottles can hold is <em>μ₀,</em> i.e. <em>μ</em> = <em>μ₀</em>.

<em>Hₐ</em>: The mean amount of liquid that the bottles can hold is less than  <em>μ₀,</em> i.e. <em>μ</em> < <em>μ₀</em>.

The <em>p</em>-value of the test is, <em>p</em> = 0.054.

Decision rule:

The null hypothesis will be rejected if the <em>p</em>-value of the test is less than the significance level. And vice-versa.

  • Assume that the significance level of the test is, <em>α</em> = 0.01.

        The <em>p</em>-value = 0.054 > <em>α</em> = 0.01.

        The null hypothesis was failed to be rejected at 1% level of

        significance. Concluding that the mean amount of liquid that the

         bottles can hold is <em>μ₀</em>.

  • Assume that the significance level of the test is, <em>α</em> = 0.05.

        The <em>p</em>-value = 0.054 > <em>α</em> = 0.05.

        The null hypothesis was failed to be rejected at 5% level of

        significance. Concluding that the mean amount of liquid that the

         bottles can hold is <em>μ₀</em>.

  • Assume that the significance level of the test is, <em>α</em> = 0.10.

        The <em>p</em>-value = 0.054 < <em>α</em> = 0.10.

        The null hypothesis will be rejected at 10% level of

        significance. Concluding that the mean amount of liquid that the

         bottles can hold is less than <em>μ₀</em>.

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Hope this helps!
3 0
3 years ago
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