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Wewaii [24]
3 years ago
10

Determine whether the Mean Value Theorem applies to the function f(x)=2x^1/5 on the interval [-32, 32]

Mathematics
1 answer:
Anettt [7]3 years ago
5 0

Answer

given,

f(x) = 2 x^{1/5}

interval = [-32, 32]        

differentiating the given equation

f'(x) = \dfrac{d}{dx}(2 x^{1/5})

f'(x) = \dfrac{2}{5x^{4/5}}

hence, the above solution is not defined at x = 0

and x = 0 lie in the given interval i.e. [-32, 32]

so, at x = 0 the function is not differentiable.

Hence, mean value theorem does not apply to the given function.

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When finding the margin of error for the mean of a normally distributed population from a sample, what is the critical probabili
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Critical probability is the essentially cut-off value. The critical probability when the confidence level of 58% is 0.79.

<h3>What is the critical probability?</h3>

Critical probability is the essentially cut-off value that defines the region where the test statistic is unlikely to lie.

As it is given that the confidence level is 58%. therefore, in order to calculate the critical probability, we need to calculate the margin of error within a set of data, and  it is given by the formula

\rm Critical\ Probability, (P*) = 1-\dfrac{\alpha }{2}

where the value of the α is expressed as,

\alpha= 1 -\dfrac{\rm Confidence\ interval}{100}

Now, as the confidence interval is given to us, therefore, the value of the alpha can be written as,

\alpha= 1 -\dfrac{\rm 58\%}{100} = 0.42

Further, the critical probability, assuming a confidence level of 58% is,

\rm Critical\ Probability, (P*) = 1-\dfrac{\alpha }{2}\\\\\rm Critical\ Probability, (P*) = 1-\dfrac{0.42}{2} = 0.79

Hence, the critical probability is 0.79.

Learn more about Critical Probability:

brainly.com/question/5625386

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Answer:

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Step-by-step explanation:

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A recipe for trail mix uses 7 ounces of almonds with 5 ounces of raisins. How many ounces of almonds would be in a one pound bag
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