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sasho [114]
3 years ago
9

Directions: Simplify the following monomials.SHOW ALL STEPS!

Mathematics
2 answers:
Simora [160]3 years ago
7 0

19.)

\frac{ - 4x - 3 + 96 {a}^{2} b}{6b}

<h2>20.) </h2><h2>= 9 {x}^{2}  {y}^{7}  - 4 {x}^{2}  {y}^{2}</h2>

<em>hope </em><em>it</em><em> helps</em>

<em>#</em><em>c</em><em>a</em><em>r</em><em>r</em><em>y</em><em> </em><em>on</em><em> learning</em>

<em>follow</em><em> me</em><em> </em><em>and</em><em> mark</em><em> me</em><em> as</em><em> brainlist</em><em> plss</em>

Softa [21]3 years ago
5 0

HERE IS YOUR ANS AFTER A LOT OF RESEARCH ! ^-^

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Question 15
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Answer:

The correct option is (C).

Step-by-step explanation:

According to the Central Limit Theorem if we have n unknown population with mean <em>μ</em> and standard deviation <em>σ</em> and appropriately huge random samples (<em>n</em> > 30) are selected from the population with replacement, then the distribution of the sample means will be approximately normally distributed.

Then, the mean of the distribution of sample means is given by,

\mu_{\bar x}=\mu

And the standard deviation of the distribution of sample means is given by,

\sigma_{\bar x}=\frac{\sigma}{\sqrt{n}}

The information provided is:

<em>μ</em> = 3.5

<em>σ</em> = 1.7078

<em>n</em> = 400 = number of times the experiment is repeated.

As the sample size is quite large, i.e. <em>n</em> = 400 > 30 the central limit theorem can be used to approximate the sampling distribution of the sample mean.

The mean of the distribution of sample means is:

\mu_{\bar x}=\mu=3.5

The standard deviation of the distribution of sample mean is:

\sigma_{\bar x}=\frac{\sigma}{\sqrt{n}}=\frac{1.7078}{\sqrt{400}}=0.0854

The distribution of the sample mean is:

\bar X\sim N(\mu = 3.5,\ \sigma=0.0854).

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Step-by-step explanation:

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