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Taya2010 [7]
2 years ago
14

GCF for 44 77 210 126

Mathematics
1 answer:
alexgriva [62]2 years ago
5 0

The GCF for 44, 77,210, and 126 is 1 . I know this because other than one they have none of the same factors that go into each of the numbers.

Hope this helps!! Please rate em the brainliest answer!!!

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Answer: $1.35

Step-by-step explanation:

1.29 * 5% = 1.29 * 0.05 = 0.0645

0.0645 rounds down to 0.06

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The statistical difference between a process operating at a 5 sigma level and a process operating at a 6 sigma level is markedly
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Answer:

True

Step-by-step explanation:

A six sigma level has a lower and upper specification limits between \\ (\mu - 6\sigma) and \\ (\mu + 6\sigma). It means that the probability of finding no defects in a process is, considering 12 significant figures, for values symmetrically covered for standard deviations from the mean of a normal distribution:

\\ p = F(\mu + 6\sigma) - F(\mu - 6\sigma) = 0.999999998027

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Similarly, for finding <em>no defects</em> in a 5 sigma level, we have:

\\ p = F(\mu + 5\sigma) - F(\mu - 5\sigma) = 0.999999426697.

The probability of defects is:

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\\ {6\sigma} = 0.000000001973 = 1.973 * 10^{-9} \approx \frac{2}{10^9} \approx \frac{2}{1000000000}

\\ {5\sigma} = 0.000000573303 = 5.73303 * 10^{-7} \approx \frac{6}{10^7} \approx \frac{6}{10000000}  

Then, comparing both fractions, we can confirm that a <em>6 sigma level is markedly different when it comes to the number of defects present:</em>

\\ {6\sigma} \approx \frac{2}{10^9} [1]

\\ {5\sigma} \approx \frac{6}{10^7} = \frac{6}{10^7}*\frac{10^2}{10^2}=\frac{600}{10^9} [2]

Comparing [1] and [2], a six sigma process has <em>2 defects per billion</em> opportunities, whereas a five sigma process has <em>600 defects per billion</em> opportunities.

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