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myrzilka [38]
3 years ago
14

a party rental company has chairs and tables for rent the total cost to rent 2 chairs and 3 tables is 29 dollars the total cost

to rent 6 chairs and 5 tables is 53 dollars what is the cost to rent each chairs and each tables​
Mathematics
2 answers:
melomori [17]3 years ago
5 0

Answer:

78

Step-by-step explanation:

Anit [1.1K]3 years ago
4 0

Answer:

78

Step-by-step explanation:

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If a line has a slope of 4 and contains the point (-2,5), what is its equation in slope form?
melomori [17]

Answer:

y-y1=m(x-x1)

y-5=4(x + 5)

y-5=4x+20

5. 5

y = 4x + 25

6 0
2 years ago
How do I solve this problem??
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They're the same like integers.
First you should interpret it.
For example:
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A and b are identical batteries, which is brighter
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OLEGan [10]
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4 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
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