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Vlada [557]
2 years ago
11

Need help with finding the value of x:-

x" alt="7x-4=3+8x" align="absmiddle" class="latex-formula">
Mathematics
2 answers:
kiruha [24]2 years ago
5 0

\underline{\boxed{\red{\:Solution}}}

==> 7x - 4 = 3 + 8x

==> 8x - 7x = -4 - 3

==> x = -7 [ Ans ]

Phoenix [80]2 years ago
3 0

The value of x in the equation is -7

<h3>What is an algebraic equation?</h3>

An algebraic equation is an expression that consists of numbers, variables with their arithmetic operations.

From the given algebraic equation:

7x - 4 = 3 + 8x

By collecting the like terms, we have:

7x - 8x = 3 + 4

-x = 7

Multiply both sides by (-) sign,

x = - 7

Learn more about algebraic equations here:

brainly.com/question/4344214

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According to question

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The average Act score follows a normal distribution, with a mean of 21.1 and a standard deviation of 5.1. What is the probabilit
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Answer:

0.43% probability that the mean IQ score of 50 randomly selected people will be more than 23

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal probability distribution

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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 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.

In this question:

\mu = 21.1, \sigma = 5.1, n = 50, s = \frac{5.1}{\sqrt{50}} = 0.7212

What is the probability that the mean IQ score of 50 randomly selected people will be more than 23

This is 1 subtracted by the pvalue of Z when X = 23. So

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

By the Central Limit Theorem

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

Z = \frac{23 - 21.1}{0.7212}

Z = 2.63

Z = 2.63 has a pvalue of 0.9957

1 - 0.9957 = 0.0043

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