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Leokris [45]
3 years ago
15

Multiply radicals. Help with #52

Mathematics
1 answer:
timofeeve [1]3 years ago
5 0

Answer:

\large\boxed{\sqrt{xy^3}\cdot\sqrt[3]{x^2y}=\sqrt[6]{x^7y^{11}}}

Step-by-step explanation:

\text{Use}\ a^\frac{1}{n}=\sqrt[n]{a}\\\\\sqrt{xy^3}\cdot\sqrt[3]{x^2y}=(xy^3)^\frac{1}{2}(x^2y)^\frac{1}{3}\\\\\text{use}\ (ab)^n=a^nb^n\ \text{and}\ (a^n)^m=a^{nm}\\\\=x^\frac{1}{2}y^{(3)\left(\frac{1}{2}\right)}x^{(2)\left(\frac{1}{3}\right)}y^\frac{1}{3}\\\\\text{use}\ a^na^m=a^{n+m}\\\\=x^{\frac{1}{2}+\frac{2}{3}}y^{\frac{3}{2}+\frac{1}{3}}\\\\\text{the common denominator is 6}

\dfrac{1}{2}=\dfrac{1\cdot3}{2\cdot3}=\dfrac{3}{6}\\\\\dfrac{2}{3}=\dfrac{2\cdot2}{3\cdot2}=\dfrac{4}{6}\\\\\dfrac{3}{2}=\dfrac{3\cdot3}{2\cdot3}=\dfrac{9}{6}\\\\\dfrac{1}{3}=\dfrac{1\cdot2}{3\cdot2}=\dfrac{2}{6}\\\\x^{\frac{1}{2}+\frac{2}{3}}y^{\frac{3}{2}+\frac{1}{2}}=x^{\frac{3}{6}+\frac{4}{6}}y^{\frac{9}{6}+\frac{2}{6}}=x^{\frac{7}{6}}y^{\frac{11}{6}}=\sqrt[6]{x^7y^{11}}

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Answer:

c. Smoking is associated with lower birth weights. When smokers are compared to nonsmokers, we are 95% confident that the mean weight of babies born to nonsmokers will be between 76.5 grams to 486.9 grams more than the mean birth weight of babies born to smokers.

Step-by-step explanation:

The difference in mean birth weights (nonsmokers minus smokers) is 281.7 grams with a margin of error of 205.2 grams with 95% confidence.

Then, we know that the 95% confidence interval is (76.5, 486.9)

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<em>a. We are 95% confident that smoking causes lower birth weights by an average of between 76.5 grams to 486.9 grams.</em>

False, it interprets the confidence interval as a probability for individual cases. The confidence interval is a range for the population mean.

<em>b. There is a 95% chance that if a woman smokes during pregnancy her baby will weigh between 76.5 grams to 486.9 grams less than if she did not smoke.</em>

False, it interprets the confidence interval as a probability for individual cases. The confidence interval is a range for the population mean.

<em>c. Smoking is associated with lower birth weights. When smokers are compared to nonsmokers, we are 95% confident that the mean weight of babies born to nonsmokers will be between 76.5 grams to 486.9 grams more than the mean birth weight of babies born to smokers.</em>

True.

<em>d. With such a large margin of error, this study does not suggest that there is a difference in mean birth weights when we compare smokers to nonsmokers.</em>

There is enough evidence, as the lower bound of the confidence is positive. This means that there is only a probability of 0.05/2=0.025 that the true mean weight difference is smaller than 76.5 grams.

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brainly.com/question/7323175

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