Mathematics, 30.10.2020 18:00 gtsaeg409

A company has 2 machines that produce widgets. An older machine produces 23% defective widgets, while the new machine produces only 8% defective widgets. Besides, the new machine produces 3 times as many widgets as the older machine does. Given that a randomly chosen widget was tested and was found to be defective, what is the probability that it was produced by the new machine

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Answer from: andershy1405

The value is P(Z' | N )= 0.92

Step-by-step explanation:

From the question we are told that

The proportion of defective widgets produced by older widget is $p = 23\% = 0.23$

The proportion of defective widgets produced by new widget is

q = 0.08

Generally since we are considering the new machine in this question it then mean that the probability that widget was defective given that it was produce by the new machine is

P(Z | N ) = 0.08

Here Z is the event that the widget was defective

N is the even that the widget was produced by the new machine

Also the probability that the widget was defective given that it was not produced by the new machine is

P(Z | N') = 0.23

Generally from the question the probability that the widget was produced by the new machine is mathematically represented as

$P( N) = \frac{3}{4}$

=> P( N) = 0.75

Generally the probability that the widget is defective and that it was produced by the new is mathematically represented as

$P( Z\ n \ N) = P(Z | N ) * P(N)$

=> $P( Z\ n \ N) = 0.08 * 0.75$

=> $P( Z\ n \ N) = 0.06$

Generally the probability that the widget is defective and that it was not produced by the new machine is

$P(Z\ n \ N') = P( Z | N' ) * [1 -P(N)]$

=> $P(Z\ n \ N') = 0.23 * [1 -0.75]$

=> $P(Z\ n \ N') =0.0575$

Generally the probability that the widget is defective is

$P(Z) = P(Z \ n \ N) + P(Z \ n \ N')$

=> P(Z) = 0.06 + 0.0575

=> P(Z) = 0.1175

Generally the probability that the widget is not defective and it produced by the new machine is mathematically represented as

$P(Z' \ n \ N ) = P(N ) - P(Z \ n \ N)$

=> $P(Z' \ n \ N ) =0.75 - 0.06$

=> $P(Z' \ n \ N ) =0.69$

Generally the probability that the widget was produced by the new machine,given that a randomly chosen widget was tested and was found to be defective is mathematically represented as

$P(Z' | N ) =\frac{P(Z' \ n \ N)}{P(N)}$

=> $P(Z' | N )= \frac{0.69}{0.75}$

=> P(Z' | N )= 0.92

Answer from: Quest

idk bring the function back im not smart

step-by-step explanation:

Answer from: Quest

answer: domain and range mean all possible values of x and y that could be substituted to the inequality. a solution means all possible values that make the inequality statement true. so while the range is still all real numbers, the solution is the shaded area, for which y will always be greater than the line of the function.