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Pepsi [2]
3 years ago
12

John runs a computer software store. Yesterday he counted 123 people who walked by the store, 56 of whom came into the store. Of

the 56, only 23 bought something in the store. (Round your answers to two decimal places.) (a) Estimate the probability that a person who walks by the store will enter the store. (b) Estimate the probability that a person who walks into the store will buy something. (c) Estimate the probability that a person who walks by the store will come in and buy something. (d) Estimate the probability that a person who comes into the store will buy nothing.
Mathematics
1 answer:
vaieri [72.5K]3 years ago
8 0

Answer:

a) There is a 45.53% probability that a person who walks by the store will enter the store.

b) There is a 41.07% probability that a person who walks into the store will buy something.

c) There is a 18.70% probability that a person who walks by the store will come in and buy something.

d) There is a 58.93% probability that a person who comes into the store will buy nothing.

Step-by-step explanation:

This a probability problem.

The probability formula is given by:

P = \frac{D}{T}

In which P is the probability, D is the number of desired outcomes and T is the number of total outcomes.

The problem states that:

123 people walked by the store.

56 people came into the store.

23 bought something in the store.

(a) Estimate the probability that a person who walks by the store will enter the store.

123 people walked by the store and 56 entered the store, so T = 123, D = 56.

So

P = \frac{D}{T} = \frac{56}{123} = 0.4553

There is a 45.53% probability that a person who walks by the store will enter the store.

(b) Estimate the probability that a person who walks into the store will buy something.

56 people came into the store and 23 bought something, so T = 56, D = 23.

So

P = \frac{D}{T} = \frac{23}{56} = 0.4107

There is a 41.07% probability that a person who walks into the store will buy something.

(c) Estimate the probability that a person who walks by the store will come in and buy something.

123 people walked by the store and 23 came in and bought something, so T = 123, D = 23.

So

P = \frac{D}{T} = \frac{23}{123} = 0.1870

There is a 18.70% probability that a person who walks by the store will come in and buy something.

(d) Estimate the probability that a person who comes into the store will buy nothing.

Of the 56 people whom came into the store, 23 bought something. This means that 56-23 = 33 of them did not buy anything. So:

D = 33, T = 56

P = \frac{D}{T} = \frac{33}{56} = 0.5893

There is a 58.93% probability that a person who comes into the store will buy nothing.

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In an <u><em>Arithmetic Sequence</em></u> the difference between one term and the next is a constant, and this constant is called the common difference (d)

In this problem we have the ordered pairs

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The first term is a_1=-10                                    

The common difference is equal to d=3          

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