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spin [16.1K]
3 years ago
4

Solve x2 – 8x = 3 by completing the square. Which is the solution set of the equation?

Mathematics
2 answers:
alexandr402 [8]3 years ago
6 0

Answer:

x = 4 ± \sqrt{19}

Step-by-step explanation:

Given

x² - 8x = 3

To complete the square

add ( half the coefficient of the x- term )² to both sides

x² + 2(- 4)x + 16 = 3 + 16

(x - 4)² = 19 ( take the square root of both sides )

x - 4 = ± \sqrt{19} ( add 4 to both sides )

x = 4 ± \sqrt{19}, that is

x = 4 - \sqrt{19}, 4 + \sqrt{19}

xxMikexx [17]3 years ago
6 0

Answer:

the answer is the first option

Step-by-step explanation:

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Suppose a large consignment of televisions contained 11% defectives. If a sample of size 237 is selected, what is the probabilit
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0.8612 = 86.12% probability that the sample proportion will differ from the population proportion by less than 3%

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 normal distributions can be solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the z-score 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 p-value, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

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

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

For a proportion p in a sample of size n, the sampling distribution of the sample proportion will be approximately normal with mean \mu = p and standard deviation s = \sqrt{\frac{p(1-p)}{n}}

Suppose a large consignment of televisions contained 11% defectives.

This means that p = 0.11

Sample of size 237

This means that n = 237

Mean and standard deviation:

\mu = p = 0.11

s = \sqrt{\frac{p(1-p)}{n}} = \sqrt{\frac{0.11*0.89}{237}} = 0.0203

What is the probability that the sample proportion will differ from the population proportion by less than 3%?

P-value of Z when X = 0.11 + 0.03 = 0.14 subtracted by the p-value of Z when X = 0.11 - 0.03 = 0.08. So

X = 0.14

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

By the Central Limit Theorem

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

Z = \frac{0.14 - 0.11}{0.0203}

Z = 1.48

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

Z = \frac{0.08 - 0.11}{0.0203}

Z = -1.48

Z = -1.48 has a p-value of 0.0694

0.9306 - 0.0694 = 0.8612

0.8612 = 86.12% probability that the sample proportion will differ from the population proportion by less than 3%

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