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3 years ago

50 points!!! please have serious answers

Mathematics

1 answer:

Darina [25.2K]3 years ago

4 0

Median: Put the list in order from least to greatest.

$20, $20, $23.50, $25, $26.50

$20, $23.50, $25

$23.50 is the median. I took one from each side until I had one left.

Mean: Get the sum of them all and then divide by the number of earnings ( in this case )

$25 + $26.50 = $51.50 $51.50 + $20 = $71.50 $71.50 + $23.50 = $95

$95 + $20 = $115 $115/5 = $23

$23 is the mean.

Mode: Is the one that finds the frequency or in lamen terms it is the number which appears most often.

There's no steps for finding the expect counting how many times that number appeared.

$20 is the mode.

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120 * 0.20 = 24

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3 years ago

Read 2 more answers

What is the product of (2x+3) and (4x^2-6x+9)?

gogolik [260]

Answer:

=8x^3+27

Step-by-step explanation:

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3 years ago

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2 years ago

Unoccupied seats on flights cause airlines to lose revenue. Suppose a large airline wants to estimate its average number of unoc

Bingel [31]

Answer:

We need a sample size of at least 719

Step-by-step explanation:

We have that to find our $\alpha$ level, that is the subtraction of 1 by the confidence interval divided by 2. So:

$\alpha = \frac{1-0.95}{2} = 0.025$

Now, we have to find z in the Ztable as such z has a pvalue of $1-\alpha$ .

So it is z with a pvalue of $1-0.025 = 0.975$ , so $z = 1.96$

Now, find the margin of error M as such

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

In which $\sigma$ is the standard deviation of the population and n is the size of the sample.

How large a sample size is required to vary population mean within 0.30 seat of the sample mean with 95% confidence interval?

This is at least n, in which n is found when $M = 0.3, \sigma = 4.103$ . So

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

$0.3 = 1.96*\frac{4.103}{\sqrt{n}}$

$0.3\sqrt{n} = 1.96*4.103$

$\sqrt{n} = \frac{1.96*4.103}{0.3}$

$(\sqrt{n})^{2} = (\frac{1.96*4.103}{0.3})^{2}$

$n = 718.57$

Rouding up

We need a sample size of at least 719

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3 years ago