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Softa [21]
3 years ago
7

An office supply company manufactures paper clips, and even tolerates a small proportion of those paper clips being ‘defective’

(or incorrectly shaped and/or twisted) in its outgoing product. (The company reasons that paper clips are so cheap, users will simply discard the occasional defective paper clip they might find in a box.) The average proportion of ‘defective’ paper clips is known to be 2% when the paper clip manufacturing process is ‘in control’. To monitor this issue, what should be the value of the upper control limit of a p-chart if the company plans to include 25 paper clips in each of its samples and use z-value of 3.0 to construct the chart? g
Mathematics
1 answer:
vovikov84 [41]3 years ago
4 0

Answer:

0.104 (10.4%)

Step-by-step explanation:

UCL = \bar{p}+z(\sigma)

\sigma = p(1-)) Vn = 1.02(1-202)) 5 = 0.028

\thereforeUCL = .02+(3x0.028) = 0.104

\thereforeUCL= 10.4%

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Which of the following is the result of the equation below after completing the square and factoring?
Tems11 [23]

Answer:

D)  (x +  \frac{5}{2})^{2}   =  \frac{9}{4}

Step-by-step explanation:

<u><em>Step(i):-</em></u>

Given equation

             <em>   x² + 5 x + 8 = 4</em>

    ⇒        x^{2} + 2 X \frac{5}{2} x + (\frac{5}{2} )^{2} -  (\frac{5}{2} )^{2}+ 8 = 4

<u><em>Step(ii):</em></u>-

    By using (a + b)² = a² + 2 a b + b²

  ⇒        (x +  \frac{5}{2})^{2}   -  (\frac{5}{2} )^{2}+ 8 = 4

  ⇒        (x +  \frac{5}{2})^{2}   = 4 + (\frac{5}{2} )^{2} -8

⇒           (x +  \frac{5}{2})^{2}   =  (\frac{25}{4} ) -4

⇒           (x +  \frac{5}{2})^{2}   =  (\frac{25-16}{4} )

⇒           (x +  \frac{5}{2})^{2}   =  \frac{9}{4}

<u><em>Final answer:-</em></u>

(x +  \frac{5}{2})^{2}   =  \frac{9}{4}

 

7 0
3 years ago
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