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sattari [20]
2 years ago
12

Suppose a shipment of 140 electronic components contains 3 defective components. to determine whether the shipment should be​ ac

cepted, a​ quality-control engineer randomly selects 3 of the components and tests them. if 1 or more of the components is​ defective, the shipment is rejected. what is the probability that the shipment is​ rejected?
Mathematics
2 answers:
Westkost [7]2 years ago
7 0
If its out of the defective components its 1 out of 3 if its the entire shipment its 46% chance it is rejected
 
MA_775_DIABLO [31]2 years ago
4 0
The chance that one of the defective components is \frac{3}{140} or 2.14285714% that the qc person will test a defective unit.
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Step-by-step explanation:

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Julli [10]

\huge\text{Hey there!}


\huge\textsf{Breaking the meaning down in simpler terms}
\huge\textbf{Median:}\\\large\text{is understood to be \bf the number in the center (also known as}\\\large\textbf{the middle number)}


\huge\textbf{Mean:}\\\large\text{is understood to be the \bf total.}

\huge\textsf{Formulas for each term}

\huge\text{Median:}\\\large\textsf{To find the median you have to put each of your numbers in your data}\\\large\textsf{set from LEAST (smallest) to GREATEST (biggest). }


\huge\text{Mean:}\\\mathsf{\dfrac{sum\ of\ all\ terms}{number\ of\ terms}= mean}


\huge\textbf{SOLVING FOR THE QUESTION}

\huge\textsf{Median:}\\\\\large\textbf{Original data set: }\large\text{27, 52, 64, 41, 33, 38, 42, 60, 72, 68}\\\\\large\textbf{Conversion: }\large\text{27, 33, 38, 41, 42, 52, 64, 68, 72}\\\\\large\textsf{Make sure it is even between both sides of the data plot.}\\\\\large\text{It seems to even on BOTH sides of  \boxed{\textsf{14}} so it could possibly be your}\\\large\text{median.}

\huge\textsf{Mean:}\\\\\large\textbf{Original data set: }\large\text{27, 52, 64, 41, 33, 38, 42, 60, 72, 68}\\\\\large\textsf{Your equation: }\mathsf{\dfrac{27 + 52 + 64 + 41 + 33 + 38 + 42 + 60 +72 + 68}{10}}\\\\\mathsf{\dfrac{79 + 64 + 41 + 33 + 38 + 42 + 60 + 72 + 68}{10}}\\\\\mathsf{\dfrac{143 + 41 + 33 + 38 + 42 + 60 + 72 + 68}{10}}\\\\\mathsf{\dfrac{184 + 33 + 38 + 42 + 60 + 72 + 68}{10}}\\\\\mathsf{\dfrac{217 + 38 + 42 + 60 + 72 + 68}{10}}\\\\\mathsf{\dfrac{255 + 42 + 60 + 72 + 68}{10}}

\mathsf{\dfrac{297 + 60 + 72 + 68}{10}}\\\\\mathsf{\dfrac{357 + 72 + 68}{10}}\\\\\mathsf{\dfrac{429 + 68}{10}}\\\\\mathsf{\dfrac{497}{10}}\\\\\mathsf{\approx 49.70}\\\\\large\text{From the looks of it \boxed{\rm{\dfrac{497}{10}}}\ or \boxed{\text{49.70}} could possibly be your mean.}


\huge\text{Good luck on your assignment \& enjoy your day!}

<h3>
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8 0
2 years ago
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Tema [17]

Answer:

Yes, it is.

Step-by-step explanation:

The second 8 in 8.8 is greater than the 0 in 8.01.

6 0
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Vlad1618 [11]

Answer:

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Step-by-step explanation:

The fact that the figure is a cube means all sides, faces, and corners are equal.

V=BHW

V=49cm^{2}(W)

V=49cm^{2}(49cm)\\ V=2401cm^{3}

3 0
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