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Yuliya22 [10]
3 years ago
11

Is it possible to have more than one (x,y)(x,y) pair that is a solution to both equations? Explain or show your reasoning

Mathematics
1 answer:
lesya692 [45]3 years ago
5 0

Answer with step-by-step explanation:

Yes, it is possible to have more that one (x,y) pair because of the distributive property.

6x + 4y = 34. If this is the case, we have to find 2 numbers that sum up to 34.

  • x = 3
  • y = 4

5x - 27 = 15. If we are going to calculate this equation, you need to reverse the equation. 15 + 27 = 42 = 5x. If 42 = 5x, we have to divide 42 by 5, which is 8.4

Hope this helped! (brainliest please)

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Binomial probability distribution

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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 distributions 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)}.

11% of all steel shafts produced by a certain process are nonconforming but can be reworked (rather than having to be scrapped).

This means that p = 0.11

Random sample of 200 shafts

This means that n = 200

Mean and Standard deviation:

\mu = E(x) = np = 200*0.11 = 22

\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{200*0.11*0.89} = 4.42

(a) What is the (approximate) probability that X is at most 30

Using continuity correction, this is P(X \leq 30 + 0.5) = P(X \leq 30.5), which is the pvalue of Z when X = 30.5. So

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

Z = \frac{30.5 - 22}{4.42}

Z = 1.92

Z = 1.92 has a pvalue of 0.9726.

0.9726 = 97.26% approximate probability that X is at most 30

8 0
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