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irga5000 [103]
3 years ago
10

ASAP, PLEASE!! Is the system of equations consistent and independent, consistent and dependent, or inconsistent? y = -3x + 1...

2y = -6x +2.
A. Consistent and independent

B. Consistent and dependent

C. Inconsistent
Mathematics
1 answer:
Aliun [14]3 years ago
7 0
I think this system is consistent and dependent
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Which gives the best estimate and the exact product of (5.9)(8.28)? A: estimate: 40; exact product: 48.852
Evgesh-ka [11]
I think B gives the best estimate and exact product because

1). 5.9 x 8.28 is very similar to 6 x 8 which equals 48.

and 

2. 5.9 x 8.28 = 48.852.

Hopefully that made sense. :)
5 0
4 years ago
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5/17 times 3/8 as a fraction
adelina 88 [10]
It would be: 5/17 * 3/8 = 5*3 / 17*8 = 15/136

In short, Your Answer would be 15/136

Hope this helps!
8 0
4 years ago
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Use the graph to fill in the blank with the correct number.
WITCHER [35]

Answer:

f(0) = 1

Step-by-step explanation:

f(x) indicates the "output" value present at the "input" (Output referring to the y value and input referring to the x value). So f(0) would mean the "y" value present at x = 0; however, there are two circles at x = 0. One of the dots is open(hollow) meaning that the "x" and "y" values present there are excluded. This leaves the closed(filled) dot which means that value is included and is the answer.

6 0
4 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
3 years ago
What expression is equivalent to 18c^8 d^9/9c^3 d^6
Annette [7]
First, since all of the terms are multiplied, you can cancel them out to simplify them:

18/9 = 2

When a term has exponents on the top and bottom of a fraction, you subtract the bottom from the top:

c^8/c^3 = c^5

d^9/d^6 = d^3

So now, we can get rid of the fraction bar and put all of these terms into one expression:

2(c^5)(d^3)
7 0
4 years ago
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