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3 years ago

12

105-30-45-3028+10000000000000038492=?

1 answer:

Volgvan3 years ago

8 0

Answer:

105-30-45-3028+10000000000000038492= 1.000000000000003e19

Step-by-step explanation:

I did the math

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Can u tell exolain this to me in 5 sentences

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olga_2 [115]

Answer:

a

The probability that the selected joint was judged to be defective by neither of the two inspectors is $P(A' n B' ) = 0.8855$

b

The probability that the selected joint was judged to be defective by inspector B but not by inspector A is $P(A' n B) =0.0403$

Step-by-step explanation:

From the question we are told that

The sample size is $n_s = 10000$

The number of outcome for inspector A is $n__{A}} = 742$

The number of outcome for inspector B is $n__{B}} = 745$

The number of joints judged defective by both inspector is $n(A u B) = 1145$

The the probability that the selected joint was judged to be defective by neither of the two inspectors is mathematically represented as

$P(A' n B' ) = 1 - P(A u B)$

Now

$P(A\ u \ B) = \frac{n(Au B)}{n_s }$

substituting values

$P(A\ u \ B) = \frac{1145}{ 10 000 }$

So

$P(A' n B' ) = 1 - \frac{1145}{10 000}$

$P(A' n B' ) = 0.8855$

the probability that the selected joint was judged to be defective by inspector B but not by inspector A is mathematically represented as

$P(A' n B) = P(A \ u \ B) -P(A)$

Now

$P(A) = \frac{n__{A}}{n_s}$

substituting values

$P(A) = \frac{742}{10 000}$

So

$P(A' n B) = \frac{1145}{10 000} - \frac{742}{10 000}$

$P(A' n B) =0.0403$

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