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makvit [3.9K]
1 year ago
13

Write equivalent fractions for 1/4,1/2,and 7/9 using the least common denominator.

Mathematics
1 answer:
torisob [31]1 year ago
3 0

In this problem, we want to write equivalent fractions that share a common denominator.

To find the common denominator, we need to find the lowest common multiple of the denominator for each fraction. So, we need to find the LCM (lowest common multiple) for 4, 2, and 9.

One method is to list multiples of each number until you find one that works. My advice is to start with the largest number, as the lists of smaller numbers can get pretty long.

Let's list the first five multiples of 9:

9,18,27,36,45

We see that 9 is not a multiple of 2 or 4.

We see that 18 is a multiple of 2 (2 times 9), but not 4.

As we move through the list, we can check the multiples until it works. In this case, 36 is our common multiple:

\begin{gathered} 2\cdot18=36 \\ 4\cdot9=36 \end{gathered}

Now we want to rewrite each of our fractions using 36 as our denominator.

\begin{gathered} \frac{1}{4}\rightarrow\frac{?}{36} \\  \\ \frac{1\cdot9}{4\cdot9}=\frac{9}{36} \end{gathered}

Next, we'll do 1/2:

\begin{gathered} \frac{1}{2}\rightarrow\frac{?}{36} \\  \\ \frac{1\cdot18}{2\cdot18}=\frac{18}{36} \end{gathered}

Finally, we'll find the last fraction for 7/9:

\begin{gathered} \frac{7}{9}\rightarrow\frac{?}{36} \\  \\ \frac{7\cdot4}{9\cdot4}=\frac{28}{36} \end{gathered}

So now we have all our equivalent fractions for 1/4, 1/2, and 7/9.

\frac{1}{4}=\frac{9}{36}\frac{1}{2}=\frac{18}{36}\frac{7}{9}=\frac{28}{36}

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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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Step-by-step explanation:

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