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lilavasa [31]
2 years ago
13

A certain number has exactly 8 factors including 1 and itself. 2 of its factors are 21 and 35 what is the number?

Mathematics
1 answer:
vodomira [7]2 years ago
6 0
21 = 3.7
35 = 5.7

probably the factors have 3.5.7
= 105

let's try :
1 . 105
3 . 35
5 . 21
7 . 15
Yep, correct.
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A sample is random sample is valid
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When you ask a question how do you mark Brainliest?<br> If someone tells me they will get brainleist
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2 years ago
Solve the following equation for x to find the total number of people who became members of a social networking site for a certa
Aleksandr [31]

Answer:

x = 360

Step-by-step explanation:

Reorder the terms:

x = 72 + 0.8x

Solving

x = 72 + 0.8x

Solving for variable 'x'.

Move all terms containing x to the left, all other terms to the right.

Add '-0.8x' to each side of the equation.

x + -0.8x = 72 + 0.8x + -0.8x

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0.2x = 72 + 0.8x + -0.8x

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0.2x = 72

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4 0
3 years ago
noted psychic was tested for extrasensory perception (ESP). The psychic was presented with 400 cards face down and asked to dete
adelina 88 [10]

Answer:

A sample of 11352 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 given by:

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

For this problem, we have that:

400 cards, 120 cases were correct. So

p = \frac{120}{400} = 0.3

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.

How large a sample n would you need to estimate p with margin of error 0.01 and 95% confidence

We would need a sample of size n, and n is found when M = 0.01. So

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

0.01 = 2.325\sqrt{\frac{0.3*0.7}{n}}

0.01\sqrt{n} = 2.325\sqrt{0.3*0.7}

\sqrt{n} = \frac{2.325\sqrt{0.3*0.7}}{0.01}

(\sqrt{n})^2 = (\frac{2.325\sqrt{0.3*0.7}}{0.01})^2

n = 11351.8

Rounding up

A sample of 11352 is needed.

5 0
3 years ago
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