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swat32
3 years ago
5

Determine the solution set of (2x - 5)2 = 11.

Mathematics
2 answers:
zhenek [66]3 years ago
8 0
I think what you meant was

(2x - 5)² = 11 -- (1)

Square root both sides of (1), i.e.

√(2x - 5)² = ± √11 -- (2)

From (2), we have

2x - 5 = ± √11 -- (3)

By adding 5 to both sides in (3), we have

2x = 5 ± √11 -- (4)

Divide both sides of (4) by 2, and we obtain

x = (5 ± √11)/2 -- (5)

From (5), the solution set of (1) is

x = (5 + √11)/2, (5 - √11)/2 ...Ans.
Fynjy0 [20]3 years ago
3 0

Taking square root on both sides, that becomes 2x - 5 = (+/-) √11

Then:

2x = (+/-)√11 + 5

x = [5 +/- √11 ] / 2

x = 5/2 +/- (√11)/2

That is, x = 5/2 +(√11)/2 and x = 5/2 - (√11)/2

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Answer:

a) 99% of the sample means will fall between 0.933 and 0.941.

b) By the Central Limit Theorem, approximately normal, with mean 0.937 and standard deviation 0.0015.

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 zscore of a measure X is given by:

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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 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.

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Samples of 34 means that n = 34

We have that \mu = 0.937, \sigma = 0.009

By the Central Limit Theorem, s = \frac{0.009}{\sqrt{34}} = 0.0015

Within what interval will 99% of the sample means fail?

Between the (100-99)/2 = 0.5th percentile and the (100+99)/2 = 99.5th percentile.

0.5th percentile:

X when Z has a pvalue of 0.005. So X when Z = -2.575.

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

By the Central Limit Theorem

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

-2.575 = \frac{X - 0.937}{0.0015}

X - 0.937 = -2.575*0.0015

X = 0.933

99.5th percentile:

X when Z has a pvalue of 0.995. So X when Z = 2.575.

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

2.575 = \frac{X - 0.937}{0.0015}

X - 0.937 = 2.575*0.0015

X = 0.941

99% of the sample means will fall between 0.933 and 0.941.

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3 years ago
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