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dlinn [17]
3 years ago
15

Which expression is equivalent to (m^5n/pq^2)^4

Mathematics
2 answers:
tatuchka [14]3 years ago
8 0
\left( \cfrac{m^5n}{pq^2}\right)^4 =  \cfrac{(m^5n)^4}{(pq^2)^4}=  \cfrac{m^{20}n^4}{p^4q^8}
erica [24]3 years ago
5 0

Answer

Find the expression is equivalent to

(\frac{m^{5}n}{pq^{2}})^{4}

To prove

As the expression is given in the question as follow .

=(\frac{m^{5}n}{pq^{2}})^{4}

By using the exponent properties of the raise a power to a power

(x^{a})^{b} = x^{ab}

than the above expression becomes

=\frac{(m^{5}n)^{4}}{(pq^{2})^{4}}\\ =\frac{(m^{5})^{4}n^{4}}{p^{4}(q^{2})^{4}}

=\frac{m^{20}n^{4}}{p^{4}q^{8}}

Thus the expression is equivalent to

=(\frac{m^{20}n^{4}}{p^{4}q^{8}})





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Using the t-distribution, it is found that the test statistic for the hypotheses test is given by:

t = \frac{-1.75 - 0}{\frac{7.12}{\sqrt{35}}}

<h3>What are the hypotheses tested?</h3>

At the null hypotheses, it i tested if the mean difference remains the same, that is:

H_0: \mu  = 0

At the alternative hypotheses, it is tested if it has decreased, hence:

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<h3>What is the test statistic?</h3>

The test statistic is given by:

t = \frac{\overline{x} - \mu}{\frac{s}{\sqrt{n}}}

The parameters are:

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The parameters are given as follows:

\overline{x} = -1.75, s = 7.12, n = 35.

Hence the test statistic is given by:

t = \frac{\overline{x} - \mu}{\frac{s}{\sqrt{n}}}

t = \frac{-1.75 - 0}{\frac{7.12}{\sqrt{35}}}

More can be learned about the t-distribution at brainly.com/question/16162795

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