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3 years ago

11

6u-2u+4 simplified please

2 answers:

lubasha [3.4K]3 years ago

8 0

Answer:

4u+4

Step-by-step explanation:

olga_2 [115]3 years ago

6 0

Answer:

4u+4

Step-by-step explanation:

6-2+4

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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.
$P(X = 3) = \binom{3}{3}(p)^{3}{q}^{3 - 3}$
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