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Korvikt [17]
3 years ago
14

Which numbers are factors of 42

Mathematics
2 answers:
ValentinkaMS [17]3 years ago
8 0

Answer:

Factors of 42: 1, 2, 3, 6, 7, 14, 21, 42.

hope this helps good luck

skad [1K]3 years ago
3 0

Answer:

{2, 3, 7}

Step-by-step explanation:

Actually there are several pairs of integers whose product is 42:

1*42, 2*21, 3*14, 6*7.

Some of these factors are themselves factorable.

For example:  42 = 2*21 = 2*3*7, plus the obvious factor 1, are the unique factors of 42.

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A researcher wishes to estimate the percentage of adults who support abolishing the penny. What size sample should be obtained i
Scorpion4ik [409]

Answer:

a) We need a sample size of at least 3109.

b) We need a sample size of at least 4145.

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}}

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.

(a) he uses a previous estimate of 25​%?

we need a sample of size at least n.

n is found when M = 0.02, \pi = 0.25. So

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

0.02 = 2.575\sqrt{\frac{0.25*0.75}{n}}

0.02\sqrt{n} = 2.575\sqrt{0.25*0.75}

\sqrt{n} = \frac{2.575\sqrt{0.25*0.75}}{0.02}

(\sqrt{n})^{2} = (\frac{2.575\sqrt{0.25*0.75}}{0.02})^{2}

n = 3108.1

We need a sample size of at least 3109.

(b) he does not use any prior​ estimates?

When we do not use any prior estimate, we use \pi = 0.5

So

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

0.02 = 2.575\sqrt{\frac{0.5*0.5}{n}}

0.02\sqrt{n} = 2.575\sqrt{0.5*0.5}

\sqrt{n} = \frac{2.575\sqrt{0.5*0.5}}{0.02}

(\sqrt{n})^{2} = (\frac{2.575\sqrt{0.5*0.5}}{0.02})^{2}

n = 4144.1

Rounding up

We need a sample size of at least 4145.

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3 years ago
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