Question

The table shows the results from a survey of 335 randomly selected households with pets. this survey was conducted by a new pet store
that is opening nearby.
have children
do not have children.
total
1 pet
38
53
91
2 pets
85
41
126
3 or more pets
46
72
118
total
169
166
335
the pet store uses the data to make decisions about inventory. complete the given statement.
a customer is more likely to have 1 pet and no children than they are to have

Answer 1

A customer is more likely to have 1 pet and no children than they are to have 2 pets and no children.

How to find probability?

We are given a table with 3 columns titled "have children", "do not have children" and "total".

Under the row titled "1 pet", the respective probabilities are;

P(having 1 pet and children) = 38/91 =  0.4176

P(having 1 pet and no children) = 53/91 = 0.5824

Under the row titled "2 pets", the respective probabilities are;

P(having 2 pets and children) = 85/126 = 0.6746

P(having 2 pets and no children) = 41/126 = 0.3254

Under the row titled "3 or more pets", the respective probabilities are;

P(having 3 or more pets and children) = 46/118 = 0.3898

P(having 3 or more pets and no children) = 72/118 = 0.6102

The only probability that is lower than P(having 1 pet and no children) in the same way is P(having 2 pets and no children).

Read more about Probability at; https://brainly.com/question/24756209

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Answer 2

Answer:

The answer is "3 pets and children"

Step-by-step explanation:

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