Answer:
see explanation
Step-by-step explanation:
1
4x = 3x + 5 ( subtract 3x from both sides )
x = 5
2
7m = 8 + 5m ( subtract 5m from both sides )
2m = 8 ( divide both sides by 2 )
m = 4
3
2a = 9 - a ( add a to both sides )
3a = 9 ( divide both sides by 3 )
a = 3
4
d = 3 ( multiply both sides by 4 to clear the fraction )
d = 3 × 4 = 12
5
5y + 6 = 21 ( subtract 6 from both sides )
5y = 15 ( divide both sides by 5 )
y = 3
6
2n - 3 = 6 ( add 3 to both sides )
2n = 9 ( divide both sides by 2 )
n = 4.5
Answer:
Step-by-step explanation:
For each component, there are only two possible outcomes. Either it fails, or it does not. The components are independent. We want to know how many outcomes until r failures. The expected value is given by
In which r is the number of failures we want and p is the probability of a failure.
In this problem, we have that:
r = 1 because we want the first failed unit.
The expected number of systems inspected until the first failed unit is 2.5
Answer:
True
Step-by-step explanation:
A six sigma level has a lower and upper specification limits between and
. 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:
For those with defects <em>operating at a 6 sigma level, </em>the probability is:
Similarly, for finding <em>no defects</em> in a 5 sigma level, we have:
.
The probability of defects is:
Well, the defects present in a six sigma level and a five sigma level are, respectively:
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>
[1]
[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.