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katen-ka-za [31]
3 years ago
6

A clothing vendor estimates that 78 out of every 100 of its online customers do not live within 50 miles of one of its physical

stores. Using this​ estimate, what is the probability that a randomly selected online customer does not live within 50 miles of a physical​ store?
Mathematics
1 answer:
Korolek [52]3 years ago
4 0

Answer:

78% probability that a randomly selected online customer does not live within 50 miles of a physical​ store.

Step-by-step explanation:

A probability is the number of desired outcomes divided by the number of total outcomes.

In this problem, we have that:

Total outcomes:

100 customers

Desired outcomes:

A clothing vendor estimates that 78 out of every 100 of its online customers do not live within 50 miles of one of its physical stores. So the number of desired outcomes is 78 customers.

Using this​ estimate, what is the probability that a randomly selected online customer does not live within 50 miles of a physical​ store?

p = \frac{78}{100} = 0.78

78% probability that a randomly selected online customer does not live within 50 miles of a physical​ store.

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A. \left[\begin{array}{cc}0.93&0.07&\\0.01&0.99&\end{array}\right]

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Step-by-step explanation:

(a) The transition matrix for the information is

                C        S

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(b) the probability vector for the information is

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(c) we simply multiply the above  two matrices to find the percent of the population can be expected to be in each category after one year after

\left[\begin{array}{cc}0.85&0.15\end{array}\right] \left[\begin{array}{cc}0.93&0.07&\\0.01&0.99&\end{array}\right] = \left[\begin{array}{cc}0.792&0.208&\end{array}\right]

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