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aksik [14]
3 years ago
13

Suppose that a process is currently operating at a 3.5-sigma quality level, and it is planned to use improvement projects to mov

e this process to a 6-sigma level? What project improvement rate would be necessary to achieve that new performance in 2 years?
Mathematics
1 answer:
jenyasd209 [6]3 years ago
7 0

Answer:

11,373 PPM

Step-by-step explanation:

The following  formula is used to compute the Proportion of error-free units given the sigma level in Excel.

<em>  The proportion of error-free units = NORMSDIST(Sigma Level - 1.5) </em>

Suppose that a process is currently operating at a 3.5-sigma quality level.

The proportion of error-free units for this process is

                           NORMSDIST(3.5 - 1.5) = 0.977250

Therefore, the rejection rate is equal to

                  1 - 0.977250 = 0.022750

or 22750 PPM.

Now, consider a 6-sigma level process.

The proportion of error-free units equals

                     NORMSDIST(6.0 - 1.5) = 0.9999966

and the rejection rate is

                             1 - 0.9999966 = 3.4 PPM

Hence, reduction is PPM required per year , that is

                              \dfrac{22750 - 3.4}{2} = 11,373 PPM

reduction per year

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What is the difference between 12/8 and 3/4?
Katen [24]

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3/4

Step-by-step explanation:

Difference means subtraction

12/8 - 3/4

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3 years ago
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Given the center of the circle (-3,4) and a point on the circle (-6,2), (10,4) is on the circle
Anastasy [175]

Answer:

Part 1) False

Part 2) False

Step-by-step explanation:

we know that

The equation of the circle in standard form is equal to

(x-h)^{2} +(y-k)^{2}=r^{2}

where

(h,k) is the center and r is the radius

In this problem the distance between the center and a point on the circle is equal to the radius

The formula to calculate the distance between two points is equal to

d=\sqrt{(y2-y1)^{2}+(x2-x1)^{2}}

Part 1) given the center of the circle (-3,4) and a point on the circle (-6,2), (10,4) is on the circle.

true or false

substitute the center of the circle in the equation in standard form

(x+3)^{2} +(y-4)^{2}=r^{2}

Find the distance (radius) between the center (-3,4) and (-6,2)

substitute in the formula of distance

r=\sqrt{(2-4)^{2}+(-6+3)^{2}}

r=\sqrt{(-2)^{2}+(-3)^{2}}

r=\sqrt{13}\ units

The equation of the circle is equal to

(x+3)^{2} +(y-4)^{2}=(\sqrt{13}){2}

(x+3)^{2} +(y-4)^{2}=13

Verify if the point (10,4) is on the circle

we know that

If a ordered pair is on the circle, then the ordered pair must satisfy the equation of the circle

For x=10,y=4

substitute

(10+3)^{2} +(4-4)^{2}=13

(13)^{2} +(0)^{2}=13

169=13 -----> is not true

therefore

The point is not on the circle

The statement is false

Part 2) given the center of the circle (1,3) and a point on the circle (2,6), (11,5) is on the circle.

true or false

substitute the center of the circle in the equation in standard form

(x-1)^{2} +(y-3)^{2}=r^{2}

Find the distance (radius) between the center (1,3) and (2,6)

substitute in the formula of distance

r=\sqrt{(6-3)^{2}+(2-1)^{2}}

r=\sqrt{(3)^{2}+(1)^{2}}

r=\sqrt{10}\ units

The equation of the circle is equal to

(x-1)^{2} +(y-3)^{2}=(\sqrt{10}){2}

(x-1)^{2} +(y-3)^{2}=10

Verify if the point (11,5) is on the circle

we know that

If a ordered pair is on the circle, then the ordered pair must satisfy the equation of the circle

For x=11,y=5

substitute

(11-1)^{2} +(5-3)^{2}=10

(10)^{2} +(2)^{2}=10

104=10 -----> is not true

therefore

The point is not on the circle

The statement is false

7 0
3 years ago
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