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Maurinko [17]
3 years ago
11

Simplify leaving your answer with positive exponents.

Mathematics
1 answer:
vaieri [72.5K]3 years ago
4 0
\bf \left( \cfrac{5m^{\frac{4}{3}}}{n^{\frac{2}{3}}} \right)^{-\frac{2}{3}}\left(\cfrac{m^4}{8n^5}  \right)^{-\frac{4}{3}}\implies 
\left( \cfrac{n^{\frac{2}{3}}}{5m^{\frac{4}{3}}} \right)^{\frac{2}{3}}\left(\cfrac{8n^5}{m^4}  \right)^{\frac{4}{3}}\impliedby  
\begin{array}{llll}
\textit{and now we}\\
\textit{distribute}\\
\textit{the exponent}
\end{array}

\bf \left( \cfrac{n^{\frac{2}{3}\cdot \frac{2}{3}}}{5^{\frac{2}{3}}m^{\frac{4}{3}\cdot \frac{2}{3}}} \right)\left( \cfrac{8^{\frac{4}{3}}n^{5\cdot \frac{4}{3}}}{m^{4\cdot \frac{4}{3}}} \right)\implies \cfrac{n^{\frac{4}{9}}}{5^{\frac{2}{3}}m^{\frac{8}{9}}}\cdot \cfrac{8^{\frac{4}{3}}n^{\frac{20}{3}}}{m^{\frac{16}{3}}}\implies \cfrac{8^{\frac{4}{3}}n^{\frac{4}{9}+\frac{20}{3}}}{5^{\frac{2}{3}}m^{\frac{8}{9}+\frac{16}{3}}}

\bf \cfrac{8^{\frac{4}{3}}n^{\frac{4+60}{9}}}{5^{\frac{2}{3}}m^{\frac{8+48}{9}}}\implies \cfrac{8^{\frac{4}{3}}n^{\frac{64}{9}}}{5^{\frac{2}{3}}m^{\frac{56}{9}}}
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Answer:

Step-by-step explanation:

Hello!

You have the information of the net gain of weight of rats of three independent groups that were fed with three different formulas (A, B and C)

The objective is to determine if the population variance of the weight gain of the rats fed with formula C (feed by Javier) is the same as the population variance of the weight gain of the rats fed with formula A (feed by Linda)

Then you have two variables of interest:

X₁: Net weight gain of the rats fed with formula A (gr)

X₂: Net weight gain of the rats fed with formula C (gr)

Assuming that both variables have a normal distribution, the statistical test for this analysis is a variance ratio.

The hypotheses are:

H₀: σ₁²=σ₂²

H₁: σ₁²≠σ₂²

α: 0.10

F= \frac{S_1^2}{S_2^2} * \frac{Sigma_1^2}{Sigma_2^2}~~F_{n_1-1;n_2-1}

The sample variances are:

S₁²= 10.09gr

S₂²= 27.16

F_{H_0}= \frac{10.09}{27.16} *1= 0.37

  • <em>Remember, another way of expressing the statistical hypotheses for this test is H₀: σ₁²/σ₂²= 1 vs. H₁: σ₁²/σ₂²≠ 1, the theoretical value for the ration of the population variances is one (considering that the null hypothesis is true).</em>

The p-value of the test is 0.3605

Using the p.value approach, the decision rule is

If p-value ≤ α, reject the null hypothesis.

If p-value > α, do not reject the null hypothesis.

The p-value is greater than α, the decision is to not reject the null hypothesis.

Using a level of 10%, there is no significant evidence to reject the null hypothesis. Then you can conclude that the population variance of the net weight gain of the rats fed by Javier is statistically the same as the population net weight gain of the rats fed by Linda.

I hope this helps!

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2 times 4 equals 8, which is the part where she needs to buy 4 times as many pencils as erasers.

8 times 2 is 16 and 1 times 2 is 2 add both 16 and 2 together and you get 18 so she spends all $18

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