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lozanna [386]
3 years ago
8

How do I Simplify 4a – 7b + 2ab – a + b (I need step by step cause I'm trying to learn this)

Mathematics
1 answer:
Alex3 years ago
4 0

Step-by-step explanation:

Ok. First of all, we need to follow the order of operations: Parentheses, Exponents, Multiplication, Division, Addition, Subtraction. We don't have any parentheses, exponents, or division to resolve, so you can skip those. So now you have: Multiplication, Addition, and Subtraction. We have multiplication in each term, but each term is fully simplified, so we have Addition and Subtraction left.

With that out of the way, let's go ahead and rearrange this equation so it is easier to solve. (<em>Note: we can only rearrange terms that are positive because subtraction is not commutative. But we can turn negative terms into "positive" terms by the method shown below.</em>)

4a - 7b + 2ab - a + b

4a + (-7b) + 2ab + (-a) + b (<em>Now the terms are all positive, so we can rearrange them, but they still have the same value.</em>)

4a + (-a) + 2ab + (-7b) + b

3a + 2ab + 7b

And there we go. Our answer is fully simplified. If you can understand this, you'll be able to simplify without isolating and rearranging the terms each time.

Hopefully this was helpful and not confusing.

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Lady_Fox [76]

Answer:

0.0838 = 8.38% probability of obtaining a sample proportion less than 0.5.

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}

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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}}

Proportion of 0.6

This means that p = 0.6

Sample of 46

This means that n = 46

Mean and standard deviation:

\mu = p = 0.6

s = \sqrt{\frac{p(1-p)}{n}} = \sqrt{\frac{0.6*0.4}{46}} = 0.0722

Probability of obtaining a sample proportion less than 0.5.

p-value of Z when X = 0.5. So

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

By the Central Limit Theorem

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

Z = \frac{0.5 - 0.6}{0.0722}

Z = -1.38

Z = -1.38 has a p-value of 0.0838

0.0838 = 8.38% probability of obtaining a sample proportion less than 0.5.

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Answer:

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Step-by-step explanation:

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