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-Dominant- [34]
2 years ago
5

Which expression is equivalent to 4√24x∧6y÷ 128x∧4y∧5

Mathematics
1 answer:
LenKa [72]2 years ago
5 0

Answer:

Option D.  \sqrt[4]{\frac{3x^{2}}{2y}}

Step-by-step explanation:

\sqrt[4]{\frac{24x^{6}y}{128x^{4}y^{5}}}

\sqrt[4]{(\frac{24}{128})\times (\frac{x^{6}}{x^{4}})\times (\frac{y}{y^{5}})}

= \sqrt[4]{(\frac{3}{16})\times {(x)^{6-4}}\times{(y)^{1-5}}}

= \sqrt[4]{(\frac{3}{16})\times x^{2}y^{-4}}

= \sqrt[4]{\frac{3}{(2)^{4}}\times x\times y^{-4}}

= \sqrt[4]{(3\times x^{2)\times (\frac{y^{-1}}{2})^{4}}}

= \frac{y^{-1}}{2}\sqrt[4]{3x^{2}}

= \sqrt[4]{\frac{3x^{2}}{2y}}

Option D. \sqrt[4]{\frac{3x^{2}}{2y}}  is the correct answer.

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Answer:

Equal df

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Thus we find both have equal degrees of freedom.

3 0
2 years ago
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2 years ago
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4 0
1 year ago
A random sample of 28 statistics tutorials was selected from the past 5 years and the percentage of students absent from each on
SVEN [57.7K]

Answer:

Step-by-step explanation:

Hello!

X: number of absences per tutorial per student over the past 5 years(percentage)

X≈N(μ;σ²)

You have to construct a 90% to estimate the population mean of the percentage of absences per tutorial of the students over the past 5 years.

The formula for the CI is:

X[bar] ± Z_{1-\alpha /2} * \frac{S}{\sqrt{n} }

⇒ The population standard deviation is unknown and since the distribution is approximate, I'll use the estimation of the standard deviation in place of the population parameter.

Number of Absences 13.9 16.4 12.3 13.2 8.4 4.4 10.3 8.8 4.8 10.9 15.9 9.7 4.5 11.5 5.7 10.8 9.7 8.2 10.3 12.2 10.6 16.2 15.2 1.7 11.7 11.9 10.0 12.4

X[bar]= 10.41

S= 3.71

Z_{1-\alpha /2}= Z_{0.95}= 1.645

[10.41±1.645*(\frac{3.71}{\sqrt{28} } )]

[9.26; 11.56]

Using a confidence level of 90% you'd expect that the interval [9.26; 11.56]% contains the value of the population mean of the percentage of absences per tutorial of the students over the past 5 years.

I hope this helps!

7 0
2 years ago
You have found a home that you are interested in purchasing. Instead of a conventional loan, you agree to pay the premium for th
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2 years ago
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