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enot [183]
3 years ago
10

Which statements are true about the ordered pair (10, 5) and the system of equations?

Mathematics
1 answer:
Alexeev081 [22]3 years ago
3 0

Answer:

C

Step-by-step explanation:

We have the system of equations:

\left\{\begin{array}{ll}2x-5y=-5 \\ x+2y=11\end{array}

And an ordered pair (10, 5).

In order for an ordered pair to satisfy any system of equations, the ordered pair must satisfy both equations.

So, we can eliminate choices A and B. Satisfying only one of the equations does not satisfy the system of equations.

Let’s test the ordered pair. Substituting the values into the first equation, we acquire:

2(10)-5(5)\stackrel{?}{=}-5

Evaluate:

20-25\stackrel{?}{=}-5

Evaluate:

-5\stackrel{\checkmark}{=} -5

So, our ordered pair satisfies the first equation.

Now, we must test it for the second equation. Substituting gives:

(10)+2(5)\stackrel{?}{=} 11

Evaluate:

20\neq 11

So, the ordered pair does not satisfy the second equation.

Since it does not satisfy both of the equations, the ordered pair is not a solution to the system because it makes at least one of the equations false.

Therefore, our answer is C.

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Step-by-step explanation:

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A simple random sample of 110 analog circuits is obtained at random from an ongoing production process in which 20% of all circu
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Answer:

64.56% probability that between 17 and 25 circuits in the sample are defective.

Step-by-step explanation:

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 = 110, p = 0.2

So

\mu = E(X) = np = 110*0.2 = 22

\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{110*0.2*0.8} = 4.1952

Probability that between 17 and 25 circuits in the sample are defective.

This is the pvalue of Z when X = 25 subtrated by the pvalue of Z when X = 17. So

X = 25

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

Z = \frac{25 - 22}{4.1952}

Z = 0.715

Z = 0.715 has a pvalue of 0.7626.

X = 17

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

Z = \frac{17 - 22}{4.1952}

Z = -1.19

Z = -1.19 has a pvalue of 0.1170.

0.7626 - 0.1170 = 0.6456

64.56% probability that between 17 and 25 circuits in the sample are defective.

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