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Ksenya-84 [330]
4 years ago
7

Vshsusisisisusjsjsuausus

Mathematics
1 answer:
spin [16.1K]4 years ago
6 0
In order for us to help you, you need to ask a question that deals with going under math section or a different subject.
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The measures of 13 angles of a 14-gon add up to 2014. Find the fourteenth angle measure
lisabon 2012 [21]

Answer:

154.92

Step-by-step explanation:

2014/13=154.92

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3 years ago
Subtract the following complex numbers: (4+4i)-(13+17i) ??
Lady bird [3.3K]

Before combining the real parts and the imaginary parts, line them up vertically as follows:

 4   +   4i

-13   -  17i

--------------            Now add up each column.

 -9 - 13i   (Answer A)

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PLEASE HELP ASAP!!
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my answer is C  $69.87

Step-by-step explanation:

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Mrs. jack bought 150 T-shirt for $1920 from a factory calculate the cost of one T-shirt
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Answer: $12.80

Step-by-step explanation:

$1980/ 150 t shirts =$12.80 per shirt

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3 years ago
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An amusement park would like to determine if they want to add in a new roller coaster. In order to decide whether or not they sh
Rzqust [24]

Using the z-distribution, it is found that the 90% confidence interval is given by: (0.6350, 0.6984).

<h3>What is a confidence interval of proportions?</h3>

A confidence interval of proportions is given by:

\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}

In which:

  • \pi is the sample proportion.
  • z is the critical value.
  • n is the sample size.

In this problem, we have a 90% confidence level, hence\alpha = 0.9, z is the value of Z that has a p-value of \frac{1+0.9}{2} = 0.95, so the critical value is z = 1.645.

The sample size and the estimate are given by:

n = 600, \pi = \frac{400}{600} = 0.6667

Hence, the bounds of the interval are given by:

\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.6667 - 1.645\sqrt{\frac{0.6667(0.3333)}{600}} = 0.6350

\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.6667 + 1.645\sqrt{\frac{0.6667(0.3333)}{600}} = 0.6984

The 90% confidence interval is given by: (0.6350, 0.6984).

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

#SPJ1

5 0
2 years ago
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