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Rainbow [258]
3 years ago
11

Can someone please help me and show me how they got the answer

Mathematics
2 answers:
Svetlanka [38]3 years ago
5 0

What is the question it is not showing up

slega [8]3 years ago
4 0

The answer for what?

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The vertices of AMNO are M (1,3), N (4,9), and O (7,3). The vertices of APQR are P (3,0), Q (4,2), and R (5,0) Which conclusion
Pavel [41]

Answer:

the correct answer should be C

Step-by-step explanation:

hope this helps you

3 0
3 years ago
12,00 equals how many hundreds?
shtirl [24]

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12

Step-by-step explanation:

8 0
3 years ago
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What is the y value of the solution to the system of equations shown below y = 22 - 6 y = 50 – 21​
zhannawk [14.2K]

Answer:

22-6y= 29

-6y=29-22

-6y=7

y=-7/6

3 0
3 years ago
Given that a 90% confidence interval for the mean height of all adult males in Idaho measured in inches was [62.532, 76.478]. Us
grigory [225]

Answer:

The confidence interval on this case is given by:

\bar X \pm z_{\alpha/2}\frac{\sigma}{\sqrt{n}} (1)

For this case the confidence interval is given by (62.532, 76.478)[/tex]

And we can calculate the mean with this:

\bar X = \frac{62.532+76.478}{2}= 69.505

So then the mean for this case is 69.505

Step-by-step explanation:

Previous concepts

The margin of error is the range of values below and above the sample statistic in a confidence interval.  

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".  

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

\bar X represent the sample mean  

\mu population mean (variable of interest)  

\sigma represent the population standard deviation  

n represent the sample size  

Assuming the X follows a normal distribution  

X \sim N(\mu, \sigma)

The confidence interval on this case is given by:

\bar X \pm z_{\alpha/2}\frac{\sigma}{\sqrt{n}} (1)

For this case the confidence interval is given by (62.532, 76.478)[/tex]

And we can calculate the mean with this:

\bar X = \frac{62.532+76.478}{2}= 69.505

So then the mean for this case is 69.505

5 0
3 years ago
What percent is 14 over 81
postnew [5]

Answer:

17,28 %

Step-by-step explanation:

81 ..............  100%

14 ....................x%

x = 14×100/81 = 1400/81 ≈ 17,28 %

7 0
3 years ago
Read 2 more answers
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