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2 years ago

How is five million, eight hundred ninety-six thousand, four hundred thirteen written as a base ten numeral?

Mathematics

1 answer:

8090 [49]2 years ago

4 0

Don’t you just write it normally? I believe the answer is 5,869,413.

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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

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.

8 0

2 years ago

Help I don't get it can u explain it to me

kap26 [50]

You are given the masses of four planets.

Add them together, and then divide your sum by 4.

That will be the "average planet mass" for these 4 planets.

3 0

2 years ago

Wright the expression 6a - 2 (a-1) in simplest form

vovikov84 [41]

6a-2(a-1)
6a-2a-2
4a-2
I think it is 4a-2

6 0

3 years ago

Please help me answer this question.

Alex787 [66]

1. Change x = 4t into x = 3t

$4t\cdot\frac{3}{4}=3t$ so

$y=(\frac{3}{4}t)^2-1=\frac{9}{16}t^2-1$

Answer A and D are wrong.

2. Change x = 4t into x = 2t

$4t\cdot\frac{1}{2}=2t$ and

$y=(\frac{1}{2}t)^2-1=\frac{t^2}{4}-1$

Answer b is correct.

3. Change x = 4t into x = 8t

$4t\cdot2=8t$ and

$y=(2t)^2-1=4t^2-1$

Answer C. is correct, E. is wrong.

7 0

2 years ago

BRAINLIEST!!!

sineoko [7]

Answer:

c) 60-40i-75i^2

Step-by-step explanation:

first you multiply out the brackets then collect like terms to get the answer

3 0

2 years ago

Read 2 more answers