Sympilfy 31/-8...........
2 answers:
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Answer:
Step-by-step explanation:
given that the Chocolate House specializes in hand-dipped chocolates for special occasions. Three employees do all of the product packaging
Clerk I II III total
Pack 0.33 0.23 0.44 1
Defective 0.02 0.025 0.015
Pack&def 0.0066 0.00575 0.0066 0.01895
a) probability that a randomly selected box of chocolates was packed by Clerk 2 and does not contain any defective chocolate
= P(II clerk) -P(II clerk and defective) =
b) the probability that a randomly selected box contains defective chocolate=P(I and def)+P(ii and def)+P(iiiand def)
=0.01895
c) Suppose a randomly selected box of chocolates is defective. The probability that it was packaged by Clerk 3
=P(clerk 3 and def)/P(defective)
=
Answer:
Step-by-step explanation:
To find the 5th term in the expansion, we first will need to apply the binomial theorem. I have attached an image of the binomial theorem formula due to not being able to type it.
After applying the binomial theorem and simplifying, you should get:
Our 5th term here is: which is equal to
~Hope this helps! Sorry if my answer is confusing at all, it's pretty difficult to explain.~
