The reliability of a two-component product if the components are in parallel is 0.99.
In this question,
The probability of failure-free operation of a system with several parallel elements is always higher than that of the best element in the system. Reliability can be increased if the same function is done by two or more elements arranged in parallel.
A system contains two components that are arranged in parallel, they are 0.95 and 0.80.
Therefore the system reliability can be calculated as follows
⇒ 1 - ( 1 - 0.95 ) × ( 1 - 0.80 )
⇒ 1 - (0.05 × 0.20)
⇒ 1 - 0.01
⇒ 0.99
Hence we can conclude that the reliability of a two-component product if the components are in parallel is 0.99.
Learn more about reliability of components here
brainly.com/question/20314118
#SPJ4
Answer:
A
Step-by-step explanation:
The common difference is simply the constant we add/subtract each time. In this case, the common difference is 3/2 as can be seen from the recursive formula. In other words, each new
will be 3/2 larger than the previous.
The answer is A.
Im pretty sure the Correct answer is D
I Could be Wrong Tho.
Answer:
The answer is =A2*B2*C2
Step-by-step explanation:
Answer:
There are 685464 ways of selecting the 5-card hand
Step-by-step explanation:
Since the hand has 5 cards and there should be at least 1 card for each suit, then there should be 3 suits that appear once in the hand, and one suit that apperas twice.
In order to create a possible hand, first we select the suit that will appear twice. There are 4 possibilities for this. For that suit, we select the 2 cards that appear with the respective suit. Since there are 13 cards for each suit, then we have
possibilities. Then we pick one card of all remaining 3 suits. We have 13 ways to pick a card in each case.
This gives us a total of 4*78*13³ = 685464 possibilities to select the hand.