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ratelena [41]
3 years ago
9

Find the GCF. 12, 18, and 84

Mathematics
2 answers:
kenny6666 [7]3 years ago
6 0

For this case we have that by definition the Greast Common Factor is the biggest factor that two or more numbers have in common.

So, we look for the factors of 12, 18 and 84:

12: 1,2,3,4,6,12

18: 1,2,3,6,9,18

84: 1,2,3,4,6

Thus, we can see that the GCF of 12, 18 and 84 is 6.

Answer:

GCF: 6

Dmitry [639]3 years ago
4 0

Answer: 6

Step-by-step explanation:

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2. When a large truckload of mangoes arrives at a packing plant, a random sample of 150 is selected and examined for
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a) The 90% confidence interval of the percentage of all mangoes on the truck that fail to meet the standards is: (7.55%, 12.45%).

b) The margin of error is: 2.45%.

c) The 90% confidence is the level of confidence that the true population percentage is in the interval.

d) The needed sample size is: 271.

<h3>What is a confidence interval of proportions?</h3>

A confidence interval of proportions has the bounds given by the rule presented as follows:

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

The margin of error is given by:

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

The variables are listed as follows:

  • \pi is the sample proportion, which is also the estimate of the parameter.
  • z is the critical value.
  • n is the sample size.

The confidence level is of 90%, hence the critical value z is the value of Z that has a p-value of \frac{1+0.90}{2} = 0.95, so the critical value is z = 1.645.

The sample size and the estimate are given as follows:

n = 150, \pi = \frac{15}{150} = 0.1

The margin of error is of:

M = z\sqrt{\frac{0.1(0.9)}{150}} = 0.0245 = 2.45\%

The interval is given by the estimate plus/minus the margin of error, hence:

  • The lower bound is: 10 - 2.45 = 7.55%.
  • The upper bound is: 10 + 2.45 = 12.45%.

For a margin of error of 3% = 0.03, the needed sample size is obtained as follows:

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

0.03 = 1.645\sqrt{\frac{0.1(0.9)}{n}}

0.03\sqrt{n} = 1.645\sqrt{0.1(0.9)}

\sqrt{n} = \frac{1.645\sqrt{0.1(0.9)}}{0.03}

(\sqrt{n}})^2 = \left(\frac{1.645\sqrt{0.1(0.9)}}{0.03}\right)^2

n = 271 (rounded up).

More can be learned about the z-distribution at brainly.com/question/25890103

#SPJ1

3 0
1 year ago
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