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Oksana_A [137]
1 year ago
15

Point M is on line segment \overline{LN}

Mathematics
1 answer:
MissTica1 year ago
3 0

Considering the relation built the presence of point M on line LN, the numerical length of LN is of 9 units.

<h3>What is the relation from the presence of point M on the line LN?</h3>

Point M splits line LN into two parts, LM and MN, hence the total length is given by:

LN = LM + MN.

From the given data, we have that:

  • LN = 2x - 5.
  • LM = 3.
  • MN = x - 1.

Hence we first solve for x.

LN = LM + MN.

2x - 5 = 3 + x - 1

x = 7.

Hence the total length is:

LN = 2x - 5 = 2 x 7 - 5 = 14 - 5 = 9 units.

More can be learned about relations and lines at brainly.com/question/2306122

#SPJ1

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nignag [31]
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First one: (0, -3)
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Second one: (2,0)
3x-2y=6
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3 years ago
Use the following probability model to answer the question.
Arisa [49]

Answer:

Step-by-step explanation:

Number of Vehicles Probability

0 0.04

1 0.41

2 0.38

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Answer:

0.5034 = 50.34% probability that the sample mean would differ from the true mean by less than 1.1 months

Step-by-step explanation:

To solve this question, we need 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 variable X, with mean \mu and standard deviation \sigma, the sample means with size n of at least 30 can be approximated to a normal distribution with mean

In this problem, we have that:

\mu = 119, \sigma = 14, n = 74, s = \frac{14}{\sqrt{74}} = 1.6275

What is the probability that the sample mean would differ from the true mean by less than 1.11 months?

This is the pvalue of Z when X = 119 + 1.1 = 120.1 subtracted by the pvalue of Z when X = 119 - 1.1 = 117.9. So

X = 120.1

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

By the Central Limit Theorem

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

Z = \frac{120.1 - 119}{1.6275}

Z = 0.68

Z = 0.68 has a pvalue of 0.7517

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

Z = \frac{117.9 - 119}{1.6275}

Z = -0.68

Z = -0.68 has a pvalue of 0.2483

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0.5034 = 50.34% probability that the sample mean would differ from the true mean by less than 1.1 months

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3 years ago
How would I be able to get the step by step prosses of this and the answer of y?
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Divide both sides by -4 to isolate variable y:

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(Negative ÷ Negative = Positive)

3 = 1y

3 = y

Check your work:

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Use PEMDAS:

-7 = -12 + 5

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Correct!

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Hopefully that is correct
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