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vekshin1
2 years ago
6

PLEASE HELP PLEASE PLEASE

Mathematics
1 answer:
Alecsey [184]2 years ago
4 0
<h2>Coordinate Pairs</h2>

Coordinate pairs are organized like (x,y).

  • x tells us the location of the point in relation to the x-axis, the axis that is horizontal.
  • y tells us the location of the point in relation to the y-axis, the axis that is vertical.

To determine a coordinate pair, we can determine each coordinate individually, then put them together.

<h2>Solving the Question</h2>

Notice how the red point sits on the very edge of the graph.

When we look at the x-axis, we can see that it occurs at the number 0 on the x-axis. In other words, the red point occurs when x=0.

When we look at the y-axis, we can see that it lines up with the number 2. In other words, the red point occurs when y=2.

Therefore, when we put the two coordinates together like (x,y), we get (0,2).

<h2>Answer</h2>

(0,2)

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Answer:

40.13% probability that 9 randomly selected packages will have an average weight in excess of 16.025 ounces

Step-by-step explanation:

To solve this question, we have to understand the normal probability distribution and the central limit theorem.

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.

Central limit theorem:

The Central Limit Theorem estabilishes that, for a random normally distributed variable X, with mean \mu and standard deviation \sigma, the sample means with size n can be approximated to a normal distribution with mean \mu and standard deviation s = \frac{\sigma}{\sqrt{n}}

In this problem, we have that:

\mu = 16, \sigma = 0.3, n = 9, s = \frac{0.3}{\sqrt{9}} = 0.1

What is the probability that 9 randomly selected packages will have an average weight in excess of 16.025 ounces

This is 1 subtracted by the pvalue of Z when X = 16.025. So

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

By the Central Limit Theorem

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

Z = \frac{16.025 - 16}{0.1}

Z = 0.25

Z = 0.25 has a pvalue of 0.5987

1 - 0.5987 = 0.4013

40.13% probability that 9 randomly selected packages will have an average weight in excess of 16.025 ounces

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