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3 years ago

Teddy is rolling a die and writing the number that shows. The collection of the data is with replacement. (True or False)

Mathematics

1 answer:

maks197457 [2]3 years ago

5 0

Answer:

True

Step-by-step explanation:

You can only collect data if you keep rolling the dice each time

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guajiro [1.7K]

se necesitan 8 envases

denada

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3 years ago

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IrinaK [193]

Answer:

$\frac{-abx^{6}+1 }{b}$

Step-by-step explanation:

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3 years ago

A class has a total of 135 absences over 18 weeks. what is the mean number of absences per week?

Butoxors [25]

135 divided by 18 is 7.5 so I would guess 7 each week and maybe possibly 8 on some weeks

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4 years ago

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

Svet_ta [14]

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:

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

For those with defects operating at a 6 sigma level, the probability is:

$\\ 1 - p = 1 - 0.999999998027 = 0.000000001973$

Similarly, for finding no defects in a 5 sigma level, we have:

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

The probability of defects is:

$\\ 1 - p = 1 - 0.999999426697 = 0.000000573303$

Well, the defects present in a six sigma level and a five sigma level are, respectively:

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

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

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

$\\ {6\sigma} \approx \frac{2}{10^9}$ [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 2 defects per billion opportunities, whereas a five sigma process has 600 defects per billion opportunities.

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3 years ago

How many seconds are in two weeks?

podryga [215]

Answer:

This conversion of 2 weeks to seconds has been calculated by multiplying 2 weeks by 604,800 and the result is 1,209,600 seconds.

Step-by-step explanation:

5 0

3 years ago

Read 2 more answers