Which property of exponents must be used first to solve this expression? (xy^2)^1/3

Mathematics

1 answer:

Pavel [41]3 years ago

5 0

^^To solve you must use the fractional exponent rule.

I dont quite understand what the question is asking but thats what I have to help :)

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If u, v, and w are nonzero vectors in r 2 , is w a linear combination of u and v?

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Not necessarily. $\mathbf u$ and $\mathbf v$ may be linearly dependent, so that their span forms a subspace of $\mathbb R^2$ that does not contain every vector in $\mathbb R^2$ .

For example, we could have $\mathbf u=(0,1)$ and $\mathbf v=(0,-1)$ . Any vector $\mathbf w$ of the form $(r,0)$ , where $r\neq0$ , is impossible to obtain as a linear combination of these $\mathbf u$ and $\mathbf v$ , since

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unless $r=0$ and $c_1=c_2$ .

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Answer:

Following are the solution to the given question:

Step-by-step explanation:

Please find the complete question in the attached file.

$H_0: \sigma^{2} \leq 0.0008\\\\H_a: \sigma^{2} > 0.0008$

The testing states value is:

$\to x^2=\frac{(n-1)s^2}{\sigma^2}=32.6250$

therefor the $\rho - \ value = 0.2931$

Through out the above equation its values Doesn't rejects the H_0 value, and its sample value doesn't support the claim that although the configuration of its dependent variable has been infringed.