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Oxana [17]
3 years ago
11

How to create a pattern with the rule n 3?

Mathematics
1 answer:
SCORPION-xisa [38]3 years ago
5 0
If you copied the "n 3" part, it's very likely that your job was to create a pattern with either the rule n^3 or n*3.

In the case of the former, we can start with the initial number of 1 and increase by 1.
In that way, using the rule n^3 would create this pattern of numbers: 1, 8, 27, 64, and so on. Or stated in another way 1*1*1, 2*2*2, 3*3*3, 4*4*4 ...

In the case of the latter, we can start with the initial number of 1 and increase it by 1. 
In this way, using the rule of n*3 would create this pattern of numbers: 3, 6, 9, 12, 15, 18 and o son. Or stated in another way 1*3, 2*3, 3*3, 4*3, 5*3 ...
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The level of nitrogen oxides (NOX) in a exhaust of cars of a particular model varies normally with mean 0.25 grams per miles and
antoniya [11.8K]

Answer:

a) 15.87% probability that a single car of this model fails to meet the NOX requirement.

b) 2.28% probability that the average NOX level of these cars are above 0.3 g/mi limit

Step-by-step explanation:

We use the normal probability distribution and the central limit theorem to solve this question.

Normal probability distribution:

Problems of normally distributed samples are 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, a large sample size 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 = 0.25, \sigma = 0.05

a. What is the probability that a single car of this model fails to meet the NOX requirement?

Emissions higher than 0.3, which is 1 subtracted by the pvalue of Z when X = 0.3. So

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

Z = \frac{0.3 - 0.25}{0.05}

Z = 1

Z = 1 has a pvalue of 0.8417.

1 - 0.8413 = 0.1587.

15.87% probability that a single car of this model fails to meet the NOX requirement.

b. A company has 4 cars of this model in its fleet. What is the probability that the average NOX level of these cars are above 0.3 g/mi limit?

Now we have n = 4, s = \frac{0.05}{\sqrt{4}} = 0.025

The probability is 1 subtracted by the pvalue of Z when X = 0.3. So

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

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

Z = \frac{0.3 - 0.25}{0.025}

Z = 2

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1 - 0.9772 = 0.0228

2.28% probability that the average NOX level of these cars are above 0.3 g/mi limit

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