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2 years ago

The statistical definition of Six Sigma allows for 3.4 defects per million. This is achieved by what Cpk index value

Mathematics

1 answer:

natali 33 [55]2 years ago

4 0

Answer:

Step-by-step explanation:

In general, the higher the Cpk, the better. A Cpk value less than 1.0 is considered poor and the process is not capable. A value between 1.0 and 1.33 is considered barely capable, and a value greater than 1.33 is considered capable. But, you should aim for a Cpk value of 2.00 or higher where possible.

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What is the answer? 8 x 4 + 9^2

Anuta_ua [19.1K]

8 x 4 is equal to 32 and 9^2 is 81. You add them together and get 113.

6 0

2 years ago

A cylinder has a volume of 360 in^3. If the height and radius are multiplied by 2.5, what will the resulting volume be?

4vir4ik [10]

$\bf \textit{volume of a cylinder}\\\\
V=\pi r^2 h\quad 
\begin{cases}
r=radius\\
h=height\\
-----\\
r=2.5r\\
h=2.5h
\end{cases}\implies V=\pi (2.5r)^2(2.5h)
\\\\\\
V=\pi \cdot 2.5^2\cdot r^2\cdot 2.5h\implies V=(2.5^2\cdot 2.5)\pi r^2 h
\\\\\\
V=15.625\pi r^2h\impliedby \textit{15.625 of the original}
\\\\\\
\textit{if the original was 360}\qquad 15.265\cdot 360$

5 0

3 years ago

When you subtract one equation from the other what is the value of x

sweet [91]

Answer:

uh depends on the equations

6 0

3 years ago

(a) A lamp has two bulbs, each of a type with average lifetime 1400 hours. Assuming that we can model the probability of failure

Temka [501]

Answer:

For first lamp ; The resultant probability is 0.703

For both lamps; The resultant probability is 0.3614

Step-by-step explanation:

Let X be the lifetime hours of two bulbs

X∼exp(1/1400)

f(x)=1/1400e−1/1400x

P(X<x)=1−e−1/1400x

X∼exp⁡(1/1400)

f(x)=1/1400 e−1/1400x

P(X<x)=1−e−1/1400x

The probability that both of the lamp bulbs fail within 1700 hours is calculated below,

P(X≤1700)=1−e−1/1400×1700

=1−e−1.21=0.703

The resultant probability is 0.703

Let Y be a lifetime of another lamp two bulbs

Then the Z = X + Y will follow gamma distribution that is,

X+Y=Z∼gamma(2,1/1400)

2λZ∼

X+Y=Z∼gamma(2,1/1400)

2λZ∼χ2α2

The probability that both of the lamp bulbs fail within a total of 1700 hours is calculated below,

P(Z≤1700)=P(1/700Z≤1.67)=

P(χ24≤1.67)=0.3614

The resultant probability is 0.3614

8 0

2 years ago

7-7(5+x)-9x in simplified

Luden [163]

Answer:

-28 - 16x

Step-by-step explanation:

7-7(5+x)-9x

Distribute

7 - 35 -7x -9x

Combine like terms

-28 - 16x

5 0

2 years ago