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NARA [144]
3 years ago
9

6+8x/2=5x what does x =

Mathematics
2 answers:
Zielflug [23.3K]3 years ago
6 0

Answer:

The answer is x=6

MaRussiya [10]3 years ago
5 0

Answer:

x = 6

Step-by-step explanation:

1st Divide the numbers: 8/2 = 4

     6+4x=5x

2nd Subtract 6 from both sides: 6+4x-6=5x-6

3rd Simplify: 4x=5x-6

4th Subtract 5x from both sides: 4x-5x=5x-6-5x

5th Simplify: -x = -6

6th Divide both sides by -1: -x ÷ -1 = -6 ÷ -1

7th Simplify: x = 6

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i^ even power

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i^4 = =1

So raised to an even power, it will be either 1 or negative 1

raised to an odd power it will be either i or -i

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3 years ago
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}

By the Central Limit Theorem

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

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

Z = 2

Z = 2 has a pvalue of 0.9772

1 - 0.9772 = 0.0228

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

4 0
3 years ago
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