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mafiozo [28]
4 years ago
6

Can u pls help me with this thanks

Mathematics
1 answer:
aleksandrvk [35]4 years ago
7 0

<em>Hey</em><em>!</em><em>!</em><em>!</em>

<em><</em><em>a</em><em> </em><em>and</em><em> </em><em><</em><em>b</em><em> </em><em> </em><em>are</em><em> </em><em><u> </u></em><em><u>co</u></em><em><u>mpl</u></em><em><u>e</u></em><em><u>m</u></em><em><u>e</u></em><em><u>n</u></em><em><u>t</u></em><em><u>a</u></em><em><u>r</u></em><em><u>y</u></em><em><u> </u></em><em><u>angles</u></em><em><u>.</u></em>

<em><u>Those</u></em><em><u> </u></em><em><u>angles</u></em><em><u> </u></em><em><u>which</u></em><em><u> </u></em><em><u>are</u></em><em><u> </u></em><em><u>exactly</u></em><em><u> </u></em><em><u>9</u></em><em><u>0</u></em><em><u> </u></em><em><u>degree</u></em><em><u> </u></em><em><u>are</u></em><em><u> </u></em><em><u>called</u></em><em><u> </u></em><em><u>complementary</u></em><em><u> </u></em><em><u>angles</u></em><em><u>.</u></em>

<em><u>hope</u></em><em><u> </u></em><em><u>it</u></em><em><u> </u></em><em><u>helps</u></em><em><u>.</u></em><em><u>.</u></em>

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four component system Assume A, B, C, and D function independently. If the probabilities that A, B, C, and D fail are 0.1, 0.2,
ArbitrLikvidat [17]

Answer:

then the probability of failure goes between 0.00003 (0.003%) and 0.5212 (52.12%) depending on the system configuration

Step-by-step explanation:

the solution depends on the system configuration, that is , if some component ( lets say A) is run in parallel from other , or is in series

if a component is run in parallel then the system fails only if all the components in parallel fails

but if the system is connected in series , the system will fail only if one of the components the serie fails.

Therefore denoting the events A= fails A , B= fails B , C= fails C , D= fails D , we have:

- lower bound of probability of failure = all components are in parallel

probability of failure P(A∩B∩C∩D)=P(A)*P(B)*P(C)*P(D)= 0.1 * 0.2 * 0.05 * 0.3 = 0.00003 (0.003%)

- upper bound of probability of failure = all components are in parallel

probability of failure P(A∪B∪C∪D)= P(A) + P(B) + P(C) +P(D) - P(A ∩ B) - P(A ∩ C) - P(A ∩ D)- P(B ∩ C) - P(B ∩ D) - P(C ∩ D) + P(A ∩ B ∩ C) + P(A ∩ B ∩ D) + P(A ∩ C ∩ D) + P(B ∩ C ∩ D) - P(A ∩ B ∩ C ∩ D) = (P(A) + P(B) + P(C) +P(D)) - ( P(A)*P(B) + P(A)*P(C) + P(A)*P(D) + P(B)*P(C) + P(B)*P(D) + P(C)*P(D) ) + P(A)*P(B)*P(C)  + P(A)*P(B)*P(D)+  P(A)*P(C)*P(D)+  P(B)*P(C)*P(D) -  P(A)*P(B)*P(C)*P(D)

replacing values

P(A∪B∪C∪D)= 0.5212 (52.12%)

then the probability of failure goes between 0.00003 (0.003%) and 0.5212 (52.12%) depending on the system configuration

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state the y-intercept and the slope of the following linear equations.

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