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Advocard [28]
3 years ago
14

Please solve this question?​

Mathematics
1 answer:
Alchen [17]3 years ago
3 0

Given,

\lim_{x}^{a}\dfrac{x^{2/3}-a^{2/3}}{x-a}

We can use L'Hopital's Rule to get,

\lim_{x}^{a}\dfrac{2}{3-\sqrt[3]{x}}

Now plug in a,

\boxed{\dfrac{2}{3-\sqrt[3]{a}}}

Hope this helps.

r3t40

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A normal distribution has a mean of 137 and a standard deviation of 7. Find the z-score for a data value of 121. Incorrect Round
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Answer:

The z-score for a data value of 121 is -2.29.

Step-by-step explanation:

In a set with mean \mu and standard deviation \sigma, the zscore of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

\mu = 137, \sigma = 7

Find the z-score for a data value of 121.

This is Z when X = 121. So

Z = \frac{X - \mu}{\sigma}

Z = \frac{121 - 137}{7}

Z = -2.29

The z-score for a data value of 121 is -2.29.

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Answer:

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Step-by-step explanation:

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Step-by-step explanation:

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