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inn [45]
3 years ago
6

If x/y + y/x = -1 , find the value of x^3 - y^3

Mathematics
1 answer:
marin [14]3 years ago
7 0

Answer:

  0

Step-by-step explanation:

Multiplying the first equation by xy, we have ...

  x^2 +y^2 = -xy

Factoring the expression of interest, we have ...

  x^3 -y^3 = (x -y)(x^2 +xy +y^2)

Substituting for xy using the first expression we found, this is ...

  x^3 -y^3 = (x -y)(x^2 -(x^2 +y^2) +y^2) = (x -y)(0) = 0

The value of x^3 -y^3 is 0.

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According to a 2009 Reader's Digest article, people throw away about 14% of what they buy at the grocery store. Assume this is t
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Answer:

18.67% probability that the sample proportion does not exceed 0.1

Step-by-step explanation:

Problems of normally distributed samples are solved using the z-score formula.

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.

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In this problem, we have that:

\mu = 0.14, \sigma = \sqrt{\frac{0.14*0.86}{59}} = 0.045

What is the probability that the sample proportion does not exceed 0.1

This is the pvalue of Z when X = 0.1. So

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

Z = \frac{0.1 - 0.14}{0.045}

Z = -0.89

Z = -0.89 has a pvalue of 0.1867

18.67% probability that the sample proportion does not exceed 0.1

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