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Advocard [28]
2 years ago
8

When -3v = -18 is solved, the result is:

Mathematics
2 answers:
mel-nik [20]2 years ago
7 0

The result of this mathematical sentence is <u>v=6</u>

<u />

<u />

<h3>First degree equation</h3>

First degree equation is a mathematical sentence - having at least one unknown. Its representation is given by <u>ax + b = 0</u>. To calculate an equation, as it is in the expression in the statement, simply: isolate x and divide the second to the first member.

*Note: Equal signs, the result will be equated as positive.

-3v = -18

v = -18/-3

<u>v = 6</u>

Therefore, the correct value of this equation is <u>v = 6</u>.

rjkz [21]2 years ago
4 0

Answer:

When -3v = -18 is solved, the result is:

21v

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Step-by-step explanation:

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Ei is then the event that "the piece was printed by Di" and "the piece is defective".

You need to determine the probability of randomly selecting a defective part printed by each one of these printers, i.e. you need to find the probability of the part being printed by the i printer given that is defective, symbolically: P(DiIE)

Where "E" represents the event "the piece is defective" and its probability represents the total error rate of the production line:

P(E)= P(E1)+P(E2)+P(E3)+P(E4)= 0.01+0.02+0.01+0.02= 0.06

This is a conditional probability and you can calculate it as:

P(A/B)= \frac{P(AnB)}{P(B)}

To reach the asked probability, first, you need to calculate the probability of the intersection between the two events, that is, the probability of the piece being printed by the Di printer and being defective Ei.

P(D1∩E)= P(E1)= 0.01

P(D2∩E)= P(E2)= 0.02

P(D3∩E)= P(E3)= 0.01

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P(D1/E)= \frac{P(E1)}{P(E)} = \frac{0.01}{0.06}= 0.17

P(D2/E)= \frac{P(E2)}{P(E)} = \frac{0.02}{0.06}= 0.33

P(D3/E)= \frac{P(E3)}{P(E)} = \frac{0.01}{0.06}= 0.17

P(D4/E)= \frac{P(E4)}{P(E)} = \frac{0.02}{0.06}= 0.33

P(D2)= 0.25 and P(D2/E)= 0.33 ⇒ The prior probability of D2 is smaller than the posterior probability.

The fact that P(D2) ≠ P(D2/E) means that both events are nor independent and the occurrence of the piece bein defective modifies the probability of it being printed by the second printer (D2)

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