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In-s [12.5K]
3 years ago
14

A process that produces computer chips has a mean of .04 defective chips and a standard deviation of .003 chips. The allowable v

ariation is from .032 to .048 defective. a.Compute the capability index (Cp) for the process.
Mathematics
1 answer:
gulaghasi [49]3 years ago
4 0

Answer:

The capability index for the process is 0.889

Step-by-step explanation:

Here, we want to compute the capability index for the process

To get this, we need some important parameters which we already have stated in the question

USL ( upper standard limit) = 0.048

LSL (lower standard limit) = 0.032

mean = 0.04

standard deviation = 0.003

We apply the formula as seen in the attachment below

Kindly understand that we are to calculate the Z upper and Z lower

The smallest value between the two will be our Z minimum

This is what we shall apply in the capability index formula which mathematically is;

Z minimum/3

From calculation:

Z lower = Z upper = 2.67

So therefore;

capability index = 2.67/3 = 0.889

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Find the volume of the tetrahedron in the first octant bounded by the coordinate planes and the plane passing through (2 comma 0
SVETLANKA909090 [29]

Answer:

The volume of the tetrahedron is:

\frac{50}{3}=16.667

Step-by-step explanation:

The volume of the tetrahedron is given by the intersection of the planes x = 0, y = 0, z = 0 and the plane formed by the three points given.

The equation of the plane formed by the three points is:

Points: (2,0,0);(0,5,0);(0,0,4)

\pi :\frac{x}{2} +\frac{y}{5} +\frac{z}{4} =1

It can also be expressed as:

10x + 4y+5z=20

We have to calculate the triple integral, therefore we must define the domain:

The values of x are given by:

0≤x≤2

We will integrate the values of y between the y = 0 axis and the line formed when z = 0:

z=0 ⇒ 10x + 4y=2 ⇒ y=\frac{20-10x}{4} =

0\leq y \leq 5-\frac{5}{2}x

We will integrate the values of z between the plane z = 0 and the plane 10x + 4y+5z=20

10x + 4y+5z=20 ⇒ z=\frac{20-10x-4y}{5} =4-2x-\frac{4}{5}y

0\leq z \leq  4-2x-\frac{4}{5}y

The volume of the tetrahedron is:

\int\limits^{2}_{0} \ \ \ \int\limits^{5-\frac{5}{2} x}_{0} \ \ \ \int\limits^{4-2x-\frac{4}{5}y }_{0} {z} \, dz\, dy \,dx

\int\limits^{2}_{0} \ \ \ \int\limits^{5-\frac{5}{2} x}_{0} \frac{z^2}{2}|^{4-2x-\frac{4}{5}y}_0\, dy \,dx

\frac{z^2}{2}|^{z=4-2x-\frac{4}{5}y}_{z=0}=\frac{5}{25}(5x+2(y-5))^2

\int\limits^{2}_{0} \ \ \ \int\limits^{5-\frac{5}{2} x}_{0} {\frac{5}{25}(5x+2(y-5))^2}\, dy \,dx

\int\limits^{2}_{0} {-\frac{25}{6}(x-2)^3} dx=-\frac{25}{24} (x-2)^4|^{2}_{0}=\frac{50}{3}

6 0
4 years ago
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The adult St.Bernard should be around 121 and 198 pounds

Explain:
2.2 x 55 = 121 pounds
2.2 x 90 = 198 pounds
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