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Nitella [24]
3 years ago
11

9.054 x 10^-6 in standard form

Mathematics
1 answer:
ki77a [65]3 years ago
8 0
0.000009054
it’s five zeros after that decimal.
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Factorise = <br> 5m + 20
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So 20=2 itmes m2 times 5

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5m=20
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divid eboth sides y 5
m=2 times 2
m=4
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A. 25%

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Suppose that your business is operating at the 4.5-Sigma quality level. If projects have an average improvement rate of 50% annu
Luda [366]

Answer:

  11.75 years

Step-by-step explanation:

If we ignore the fact that "6-sigma" quality means the error rate corresponds to about -4.5σ (3.4 ppm) and simply go with ...

  P(z ≤ -6) ≈ 9.86588×10^-10

and

  P(z ≤ -4.5) ≈ 3.39767×10^-6

the ratio of these error rates is about 0.000290372. We're multiplying the error rate by 0.5 each year, so we want to find the power of 0.50 that gives this value:

  0.50^t = 0.000290372

  t·log(0.50) = log(0.00290372) . . . . take logarithms

  t = log(0.000290372)/log(0.50) ≈ -3.537045/-0.301030

  t ≈ 11.75

It will take about 11.75 years to achieve Six Sigma quality (0.99 ppb error rate).

_____

<em>Comment on Six Sigma</em>

A 3.4 ppm error rate is customarily associated with "Six Sigma" quality. It assumes that the process may have an offset from the mean of up to 1.5 sigma, so the "six sigma" error rate is P(z ≤ (1.5 -6)) = P(z ≤ -4.5) ≈ 3.4·10^-6.

Using that same criteria for the "4.5-Sigma" quality level, we find that error rate to be P(z ≤ (1.5 -4.5)) = P(z ≤ -3) ≈ 1.35·10^-3.

Then the improvement ratio needs to be only 0.00251699, and it will take only about ...

  t ≈ log(0.00251699)/log(0.5) ≈ 8.6 . . . . years

5 0
3 years ago
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