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3 years ago

12

Please show work and answer for 35

1 answer:

Anuta_ua [19.1K]3 years ago

8 0

Answer: It should be in the comments

Step-by-step explanation: I’m sorry if you can’t rlly understand

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Let p be 0.831 denote the percentage of defective welds and q be 0.169 denote the percentage of non-defective welds.

Using the binomial distribution, we want all three to be defective.
$P(X = 3) = \binom{3}{3}(p)^{3}{q}^{3 - 3}$
$P(X = 3) = \binom{3}{3}(0.831)^{3}(1)$
$P(X = 3) = (0.831)^{3} \approx 0.573856191$

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