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PtichkaEL [24]
3 years ago
9

What is the name for a measurement of how two variables are related?

Mathematics
2 answers:
lorasvet [3.4K]3 years ago
7 0

Answer:

correlation

Step-by-step explanation:

Correlation is a statistical technique that is used to measure and describe a relationship between two variables.

astraxan [27]3 years ago
7 0
What is Correlation? Correlation is a statistical technique that is used to measure and describe a relationship between two variables. Usually the two variables are simply observed, not manipulated.
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Please help I need to get this done
hichkok12 [17]

Answer:

48

Step-by-step explanation:

So you take 7x-92+6x+12=180

And then you get x=20

Plug in 20 for x for <LOM

And you get and you get 48

Hope this helped!!

3 0
3 years ago
A candidate for a US Representative seat from Indiana hires a polling firm to gauge her percentage of support among voters in he
UkoKoshka [18]

Answer:

a) The minimum sample size is 601.

b) The minimum sample size is 2401.

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of \pi, and a confidence level of 1-\alpha, we have the following confidence interval of proportions.

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

In which

z is the zscore that has a pvalue of 1 - \frac{\alpha}{2}.

The margin of error is:

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

For this problem, we have that:

We dont know the true proportion, so we use \pi = 0.5, which is when we are are going to need the largest sample size.

95% confidence level

So \alpha = 0.05, z is the value of Z that has a pvalue of 1 - \frac{0.05}{2} = 0.975, so Z = 1.96.

a. If a 95% confidence interval with a margin of error of no more than 0.04 is desired, give a close estimate of the minimum sample size that will guarantee that the desired margin of error is achieved. (Remember to round up any result, if necessary.)

This is n for which M = 0.04. So

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

0.04 = 1.96\sqrt{\frac{0.5*0.5}{n}}

0.04\sqrt{n} = 1.96*0.5

\sqrt{n} = \frac{1.96*0.5}{0.04}

(\sqrt{n})^2 = (\frac{1.96*0.5}{0.04})^2

n = 600.25

Rounding up

The minimum sample size is 601.

b. If a 95% confidence interval with a margin of error of no more than 0.02 is desired, give a close estimate of the minimum sample size necessary to achieve the desired margin of error.

Now we want n for which M = 0.02. So

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

0.02 = 1.96\sqrt{\frac{0.5*0.5}{n}}

0.02\sqrt{n} = 1.96*0.5

\sqrt{n} = \frac{1.96*0.5}{0.02}

(\sqrt{n})^2 = (\frac{1.96*0.5}{0.02})^2

n = 2401

The minimum sample size is 2401.

4 0
3 years ago
Colby and Sarah sold newspapers this past summer. Colby earned $4.50 for each newspaper he sold. Sarah's earnings were $35 less
Furkat [3]
Y represent the total number of newspapers sold by Colby.
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Answer:

what you marked blue is right

Step-by-step explanation:

I think

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3(Y-4) - 2(y-9) + 5(y+6)=0​
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Step-by-step explanation:

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