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8_murik_8 [283]
2 years ago
12

I need the value of x. 10x + 4 - 2x = 6x + 9 + 7

Mathematics
2 answers:
erastova [34]2 years ago
3 0

Answer:

x = 6

Step-by-step explanation:

nydimaria [60]2 years ago
3 0

Answer:

x=6

Step-by-step explanation:

Step 1: Simplify both sides of the equation.

10x+4−2x=6x+9+7

10x+4+−2x=6x+9+7

(10x+−2x)+(4)=(6x)+(9+7)(Combine Like Terms)

8x+4=6x+16

8x+4=6x+16

Step 2: Subtract 6x from both sides.

8x+4−6x=6x+16−6x

2x+4=16

Step 3: Subtract 4 from both sides.

2x+4−4=16−4

2x=12

Step 4: Divide both sides by 2.

2x=12

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The mean cost of a five pound bag of shrimp is 50 dollars with a variance of 64. If a sample of 43 bags of shrimp is randomly se
anyanavicka [17]

Answer:

0.4122 = 41.22% probability that the sample mean would differ from the true mean by greater than 1 dollar

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 are solved using the z-score formula.

In a set with mean \mu and standard deviation(which is the square root of the variance) \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 \mu and standard deviation s = \frac{\sigma}{\sqrt{n}}

In this problem, we have that:

\mu = 50, \sigma = \sqrt{64} = 8, n = 43, s = \frac{8}{\sqrt{43}} = 1.22

What is the probability that the sample mean would differ from the true mean by greater than 1 dollar?

Either it differs by 1 dollar or less, or it differs by more than one dollar. The sum of the probabilities of these events is decimal 1.

Probability it differs by 1 dollar or less:

pvalue of Z when X = 50+1 = 51 subtracted by the pvalue of Z when X = 50 - 1 = 49.

X = 51

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

By the Central Limit Theorem

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

Z = \frac{51-50}{1.22}

Z = 0.82

Z = 0.82 has a pvalue of 0.7939

X = 49

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

Z = \frac{49-50}{1.22}

Z = -0.82

Z = -0.82 has a pvalue of 0.2061

0.7939 - 0.2061 = 0.5878

Probability it differs by more than 1 dollar:

p + 0.5878 = 1

p = 0.4122

0.4122 = 41.22% probability that the sample mean would differ from the true mean by greater than 1 dollar

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