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Nata [24]
3 years ago
5

Simpson is solving the systems of equations below by substitution. Show Mr. Simpsons first step if he substitutes the first equa

tion into the second equation. DO
SOLVE!


y = 5x + 12

3y + 4x = 150


I’m struggling on this one.
Mathematics
1 answer:
In-s [12.5K]3 years ago
3 0

Answer:

x=6,y=42

Step-by-step explanation:

The equations are

y=5x+12

3y+4x=150

Substituting the first equation in the second we get

3(5x+12)+4x=150

\Rightarrow 15x+36+4x=150

\Rightarrow x=\dfrac{150-36}{19}

\Rightarrow x=6

y=5x+12=5\times 6+12\\\Rightarrow y=42

So, the required values are x=6,y=42

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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:

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For those with defects <em>operating at a 6 sigma level, </em>the probability is:

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

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The probability of defects is:

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\\ {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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