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labwork [276]
3 years ago
11

HELP HELP HELP HELP HELP HELP

Mathematics
2 answers:
Aleksandr [31]3 years ago
7 0

Answer:

B

Step-by-step explanation:

For this problem, you would only need to figure one of them out since all the numbers vary. I calculated the mean. Mean is the same as the average. First, add all the numbers up. I got 112. Then, divide 112 by the amount of numbers you added (8). 112/8 is 14, and the mean=14 is b.

mixas84 [53]3 years ago
4 0

Answer:

mean: 14

median: 13.5

range: 8

Step-by-step explanation:

Hope this helps! Sorry if I am late.

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.. √(13^2 -5^2) = √(169 -25) = √144 = 12
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What is the value of the expression below when y = 4 and y = 2? 4x + 10y
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Step-by-step explanation:

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3 years ago
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7 0
3 years ago
Read 2 more answers
Write the expresion in simplest form<br> (-11/2x+30)-2(-11/4x-5/2)
qwelly [4]

The simplified expression of (-11/2x+30)-2(-11/4x-5/2) is 35

<h3>How to simplify the expression?</h3>

The expression is given as:

(-11/2x+30)-2(-11/4x-5/2)

Expand the bracket

-11/2x + 30 + 11/2x + 5

Collect the like terms

11/2x -11/2x + 30 + 5

Evaluate the like terms

35

Hence, the simplified expression of (-11/2x+30)-2(-11/4x-5/2) is 35

Read more about expressions at:

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4 0
2 years ago
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

8 0
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