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photoshop1234 [79]

3 years ago

13

Please answer asap it’s due in 1 min

1 answer:

sweet-ann [11.9K]3 years ago

3 0

They are corresponding angles. The angle rule of corresponding angles or the corresponding angles postulate states that the corresponding angles are equal if a transversal cuts two parallel lines. Corresponding angles are equal if the transversal line crosses at least two parallel lines:)

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Write the explicit formula for the arithmetic sequence.<br><br> 2, -2, -6, -10, -14, ...

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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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