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Bond [772]
3 years ago
6

The transaction history at an electronic goods store indicates that 21 percent of customers purchase the extended warranty when

they buy an eligible item. Suppose customers who buy eligible
items are chosen at random one at a time, until one is found who purchased the extended warranty. Let the random variable X represent the number of customers it takes to find one who
purchased the extended warranty. Assume customers' decisions on whether to purchase the extended warranty are independentWhich of the following is closest to the probability that X > 3;
that is, the probability that it takes more than 3 customers who buy an eligible item to find one who purchased the extended warranty?
Mathematics
1 answer:
sweet [91]3 years ago
6 0

Answer:

.131

Step-by-step explanation

(.79)(.79)(.21) = .131

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All you have to do is substitute the values in for y to see if they are true.

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So, the values that belong to 9 ≤ 6 - y are -3, -4, and -6.


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3 years ago
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NNADVOKAT [17]

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Step-by-step explanation:

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Answer:

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Step-by-step explanation:

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To solve this question we just need to find the zeros of the function f(x), and we do that by making f(x) = 0 and finding the values of x:

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Now we can use Bhaskara's formula:

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As the options A, B, C and D are not clear in the question text, you can just mark the option that has this result.

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