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S_A_V [24]
3 years ago
14

Which set of order pairs does NOT represent a function?

Mathematics
1 answer:
Anon25 [30]3 years ago
4 0
The set for B does not represent a function because if plotted, the graph does not pass the vertical line test in which a vertical line is moved along the graph.
If there is more than one point along the vertical line, it is not a function. 

A way to check if it passes the vertical line test without graphing it and actually doing the test is to make sure that all of the values for x are different. 
If they are all different, it is a function and if they are not all different, then it is not a function. 

Hope this helps!
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When you multiply or divide both sides of an inequality by a negative number, the inequality reverses.

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4 years ago
Add a cell phone assembly plant, 80% of the cell phone keypad pass inspection. A random sample of 181 keypads is analyzed. Find
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0% probability that less than 76 in the sample keypads pass inspection.

Step-by-step explanation:

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

Binomial probability distribution

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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 = 181, p = 0.8

So

\mu = E(X) = np = 181*0.8 = 144.8

\sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{181*0.8*0.2} = 5.38

Find the probability that less than 76 in the sample keypads pass inspection.

Using continuity correction, this is P(X < 76 - 0.5) = P(X < 75.5), which is the pvalue of Z when X = 75.5.

So

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

Z = \frac{75.5 - 144.8}{5.38}

Z = -12.88

Z = -12.88 has a pvalue of 0

0% probability that less than 76 in the sample keypads pass inspection.

6 0
4 years ago
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