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3 years ago

4x+3y= -1 5x+4y= -1 solving by emilination or subtraction please help :)

Mathematics

1 answer:

bogdanovich [222]3 years ago

6 0

You would solve using elimination.
y=1
x=-1

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Factories A, B and C produce computers. Factory A produces 4 times as manycomputers as factory C, and factory B produces 7 times

Elina [12.6K]

Answer:

The probability is $P(A') = 0.485$

Step-by-step explanation:

Let assume that the number of computer produced by factory C is k = 1

So From the question we are told that

The number produced by factory A is 4k = 4

The number produced by factory B is 7k = 7

The probability of defective computers from A is $P(A) = 0.04$

The probability of defective computers from B is $P(B) = 0.02$

The probability of defective computers from C is $P(C) = 0.03$

Now the probability of factory A producing a defective computer out of the 4 computers produced is

$P(a) = 4 * P(A)$

substituting values

$P(a) = 4 * 0.04$

$P(a) = 0.16$

The probability of factory B producing a defective computer out of the 7 computers produced is

$P(b) = 7 * P(B)$

substituting values

$P(b) = 7 * 0.02$

$P(b) = 0.14$

The probability of factory C producing a defective computer out of the 1 computer produced is

$P(c) = 1 * P(C)$

substituting values

$P(c) = 1 * 0.03$

$P(b) = 0.03$

So the probability that the a computer produced from the three factory will be defective is

$P(t) = P(a) + P(b) + P(c)$

substituting values

$P(t) = 0.16 + 0.14 + 0.03$

$P(t) = 0.33$

Now the probability that the defective computer is produced from factory A is

$P(A') = \frac{P(a)}{P(t)}$

$P(A') = \frac{ 0.16}{0.33}$

$P(A') = 0.485$

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