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JulsSmile [24]
3 years ago
15

Find the GCF of 30 * 3 and 12 * 4

Mathematics
2 answers:
11Alexandr11 [23.1K]3 years ago
8 0

Answer:

3


Step-by-step explanation:


wlad13 [49]3 years ago
5 0
30:15x2                 12:1x12
                                    2x6                         
                                    3x4           GCF:3
     3x10
     1x30
     5x6
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n=8 --> C8H18. So it is 18 atoms

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If a person uses $700 in credit with an interest of 14%,what is that person going to owe after three billing cycles with simple
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In 16% of all homes with a stay-at-home parent, the father is the stay-at-home parent (Pew Research, June 5, 2014). An independe
enot [183]

Answer:

a)

n = (\frac{z\sqrt{0.16*0.84}}{0.03})^2, in which z is related to the confidence level.

b) A sample size of 991 is needed.

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}}

In 16% of all homes with a stay-at-home parent, the father is the stay-at-home parent

This means that \pi = 0.16

a. What sample size is needed if the research firm's goal is to estimate the current proportion of homes with a stay-at-home parent in which the father is the stay-at-home parent with a margin of error of 0.03 (round up to the next whole number).

This is n for which M = 0.03. So

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

0.03 = z\sqrt{\frac{0.16*0.84}{n}}

0.03\sqrt{n} = z\sqrt{0.16*0.84}

\sqrt{n} = \frac{z\sqrt{0.16*0.84}}{0.03}

(\sqrt{n})^2 = (\frac{z\sqrt{0.16*0.84}}{0.03})^2

n = (\frac{z\sqrt{0.16*0.84}}{0.03})^2, in which z is related to the confidence level.

Question b:

99% confidence level,

So \alpha = 0.01, z is the value of Z that has a pvalue of 1 - \frac{0.01}{2} = 0.995, so Z = 2.575.

n = (\frac{z\sqrt{0.16*0.84}}{0.03})^2

n = (\frac{2.575\sqrt{0.16*0.84}}{0.03})^2

n = 990.2

Rounding up

A sample size of 991 is needed.

8 0
3 years ago
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