(r3 +6r2 −21r−18)÷(r−3) solve using synthetic division
1 answer:
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Answer:
0.5996 is the probability that none contain high levels of contamination.
Step-by-step explanation:
We are given the following information:
We treat lab specimen containing high levels of contamination as a success.
P( lab specimen containing high levels of contamination) = 0.12
Then the number of lab specimens follows a binomial distribution, where
where n is the total number of observations, x is the number of success, p is the probability of success.
Now, we are given n = 4
We have to find the probability that none of the lab specimen consist of high level of contamination.
We have to evaluate:
0.5996 is the probability that none contain high levels of contamination.