A veterinarian’s assistant made a table that shows the animals seen in the office in one week. What is the probability that the pet seen was sick? Enter your answer as a percent rounded to the nearest hundredth.

The probability that the pet seen was sick is 53.57%.

Step-by-step explanation:

The probability of an event E is the ration of the number of favorable outcomes to the total number of outcomes.

Here,

n (E) = number of favorable outcomes

N = total number of outcomes

Denote the events as follows:

C = the pet is a cat

D = the pet is a dog

O = the pet is some other animal

S = the pet is sick.

The data provided is summarized as follows:

n (C) = 28

n (C ∩ S) = 3

n (D) = 42

n (B ∩ S) = 4

n (O) = 24

n (O ∩ S) = 8

Compute the probability that a cat was sick as follows:

Compute the probability that a dog was sick as follows:

Compute the probability that another animal was sick as follows:

Compute the probability that the pet seen was sick as follows:

P (S) = P (C ∩ S) + P (D ∩ S) + P (O ∩ S)

Thus, the probability that the pet seen was sick is 53.57%.

i think for the first question; an oligopoly

for the third i think its; d. people accept it in exchange for goods and services.

im not sure about the 2nd maybe b or d