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Elina [12.6K]
3 years ago
13

I want to hurt myself...and im stressed what do i do?

Mathematics
2 answers:
kirill [66]3 years ago
8 0

Answer:

Calm down take a deep breath

No matter how hard things are never hurt yourself

Think about all the blessings you have

Step-by-step explanation:

I'm praying for you! Have a great day

Montano1993 [528]3 years ago
3 0
No no dont take any bad decisions god know what to do there is a bright future in front of u
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The length of a rectangle is 4 in longer than its width. if the perimeter of the rectangle is 28 in , find its area.
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The area is 45 inches squared

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3 years ago
15 × u= 135<br> ---------------
NARA [144]

Answer:

15u = 135

Divide both sides by 15

u = 135/15 = 9

U is equal to 9

Hope this helps!

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2 years ago
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Which is a true statement about conjunctions?
m_a_m_a [10]
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Explain how you can use base 10 blocks to find one. 54+2.37
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6 0
3 years ago
Please help! I don't understand how to solve this problem
Ilia_Sergeevich [38]

Using the z-distribution, a sample of 142,282 should be taken, which is not practical as it is too large of a sample.

<h3>What is a z-distribution confidence interval?</h3>

The confidence interval is:

\overline{x} \pm z\frac{\sigma}{\sqrt{n}}

The margin of error is:

M = z\frac{\sigma}{\sqrt{n}}

In which:

  • \overline{x} is the sample mean.
  • z is the critical value.
  • n is the sample size.
  • \sigma is the standard deviation for the population.

Assuming an uniform distribution, the standard deviation is given by:

S = \sqrt{\frac{(4 - 0)^2}{12}} = 1.1547

In this problem, we have a 95% confidence level, hence\alpha = 0.95, z is the value of Z that has a p-value of \frac{1+0.95}{2} = 0.975, so the critical value is z = 1.96.

The sample size is found solving for n when the margin of error is of M = 0.006, hence:

M = z\frac{\sigma}{\sqrt{n}}

0.006 = 1.96\frac{1.1547}{\sqrt{n}}

0.006\sqrt{n} = 1.96 \times 1.1547

\sqrt{n} = \frac{1.96 \times 1.1547}{0.006}

(\sqrt{n})^2 = \left(\frac{1.96 \times 1.1547}{0.006}\right)^2

n =  142,282.

A sample of 142,282 should be taken, which is not practical as it is too large of a sample.

More can be learned about the z-distribution at brainly.com/question/25890103

#SPJ1

8 0
1 year ago
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