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Diano4ka-milaya [45]
3 years ago
8

4 = -7s - 10 how do you get -2? please show your work

Mathematics
1 answer:
andreyandreev [35.5K]3 years ago
4 0
First u add ten to both sides to cancel out. Now you have 14=-7s. Since you are trying to do the opposite operation, you divide by -7 on each Sides. Now you have -2=s. Hope this helped
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Hello  Hdolly81

<u><em>The LCM of 156 and 321 is 16692</em></u>

Backup:

156: 2,2,3,13,

321: 3,107

2 * 2 *  3 * 13 * 107 = 16692

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A number cube is rolled and a coin is flipped.
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(5x - y) + (2x + 6y)<br> What’s the answer
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According to a 2009 Reader's Digest article, people throw away approximately 11% of what they buy at the grocery store. Assume t
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Answer:

0.14% probability that the sample proportion exceeds 0.2

Step-by-step explanation:

I am going to use the binomial approximation to the normal to solve this question.

Binomial probability distribution

Probability of exactly x sucesses on n repeated trials, with p probability.

Can be approximated to a normal distribution, using the expected value and the standard deviation.

The expected value of the binomial distribution is:

E(X) = np

The standard deviation of the binomial distribution is:

\sqrt{V(X)} = \sqrt{np(1-p)}

Normal probability distribution

Problems of normally distributed samples can be 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.

When we are approximating a binomial distribution to a normal one, we have that \mu = E(X), \sigma = \sqrt{V(X)}.

In this problem, we have that:

n = 108, p = 0.11

So

\mu = E(X) = np = 108*0.11 = 11.88

\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{108*0.11*0.89} = 3.2516

What is the probability that the sample proportion exceeds 0.2

This is 1 subtracted by the pvalue of Z when X = 0.2*108 = 21.6. So

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

Z = \frac{21.6 - 11.8}{3.2516}

Z = 2.99

Z = 2.99 has a pvalue of 0.9986

1 - 0.9986 = 0.0014

0.14% probability that the sample proportion exceeds 0.2

8 0
4 years ago
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