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Alchen [17]
3 years ago
6

A process on a machine has a standard deviation of 0.04 inches. If a product with specifications of +/- 0.07 inches has to be pr

ocessed on that machine, is the machine capable of performing that job? Group of answer choices
Mathematics
1 answer:
polet [3.4K]3 years ago
5 0

Answer:

c.) No, because Cp < 1

Step-by-step explanation:

Let us assume that there is centered in the process

and, the width is 2 times the specification

So, the width is

= 2 \times 0.07

= 0.14

Now capability index = Cp is

= \frac{specification\ width}{6 \times standard\ deviation}\\\\= \frac{0.14}{6\times 0.04}

= 0.58333

And the capability index should be minimum 1 for the process

And as we can see that Cp is 0.58333 which is less than 1 so the machine is not able to perform the job

Hene, the correct option is c.

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Given rs || qp, what is the value of x?
myrzilka [38]

In the given figure,

Angle 108° + 12x = 180°

because they for pair of co- interior angles which are supplementary.

so,

  • 12x + 108\degree = 180\degree

  • 12x = 180\degree - 108\degree

  • 12x = 72\degree

  • x =  \dfrac{72\degree}{12}

  • x = 6\degree

hence , value of x = 6°

5 0
2 years ago
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Use a protractor to draw a pentagon so that the bottom side is horizontal and the left and right sides are perpendicular to the
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3 years ago
Solve the given initial-value problem. x' = 1 2 0 1 − 1 2 x, x(0) = 2 7
Ilia_Sergeevich [38]
I'll go out on a limb and guess the system is

\mathbf x'=\begin{bmatrix}\frac12&0\\1&-\frac12\end{bmatrix}\mathbf x

with initial condition \mathbf x(0)=\begin{bmatrix}2&7\end{bmatrix}^\top. The coefficient matrix has eigenvalues \lambda such that

\begin{vmatrix}\frac12-\lambda&0\\1&-\frac12-\lambda\end{vmatrix}=\lambda^2-\dfrac14=0\implies\lambda=\pm\dfrac12

The corresponding eigenvectors \eta are such that

\lambda=\dfrac12\implies\begin{bmatrix}\frac12-\frac12&0\\1&-\frac12-\frac12\end{bmatrix}\eta=\begin{bmatrix}0&0\\1&-1\end{bmatrix}\eta=\begin{bmatrix}0\\0\end{bmatrix}
\implies\eta=\begin{bmatrix}1\\1\end{bmatrix}

\lambda=-\dfrac12\implies\begin{bmatrix}\frac12+\frac12&0\\1&-\frac12+\frac12\end{bmatrix}\eta=\begin{bmatrix}1&0\\1&0\end{bmatrix}\eta=\begin{bmatrix}0\\0\end{bmatrix}
\implies\eta=\begin{bmatrix}0\\1\end{bmatrix}

So the characteristic solution to the ODE system is

\mathbf x(t)=C_1\begin{bmatrix}1\\1\end{bmatrix}e^{t/2}+C_2\begin{bmatrix}0\\1\end{bmatrix}e^{-t/2}

When t=0, we have

\begin{bmatrix}2\\7\end{bmatrix}=C_1\begin{bmatrix}1\\1\end{bmatrix}+C_2\begin{bmatrix}0\\1\end{bmatrix}=\begin{bmatrix}C_1\\C_1+C_2\end{bmatrix}

from which it follows that C_1=2 and C_2=5, making the particular solution to the IVP

\mathbf x(t)=2\begin{bmatrix}1\\1\end{bmatrix}e^{t/2}+5\begin{bmatrix}0\\1\end{bmatrix}e^{-t/2}

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5 0
4 years ago
In ΔQRS, the measure of ∠S=90°, QR = 82 feet, and RS = 60 feet. Find the measure of ∠R to the nearest tenth of a degree.
dsp73

Answer:

43.0°

Step-by-step explanation:

cos∅=adj/hyp

cos∅=60/82

∅=cos-1 60/82

∅=cos-1  0.731707317073

∅=42.97028393174°

Round: 43.0°

8 0
3 years ago
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Answer please.....<br> Thanks!!!!
Tems11 [23]
A. x= 11/5
b. x= 6
c. x= -12/2
d. x= 6/5
3 0
3 years ago
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