Complete question is:
Let C(x) be the statement “x has a cat,” let D(x) be the statement “x has a dog”, and let F(x) be the statement “x has a ferret”. Express each of these statements in terms of C(x), D(x), F(x), quantifiers, and logical connectives. Let the domain consist of all students in your class.
a) A student in your class has a cat, a dog, and a ferret.
b) All students in your class have a cat, a dog, or a ferret.
c) Some student in your class has a cat and a ferret, but not a dog.
d) No student in your class has a cat, a dog, and a ferret.
e) For each of the three animals, cats, dogs, and ferrets, there is a student in your class who has this animal as a pet.
Answer:
a) ∃x(C(x) ∧ D(x) ∧ F(x))
b) ∀x(C(x) ∨ D(x) ∨ F(x))
c) ∃x(C(x) ∧ D(x) ¬ F(x))
d) ¬∃x(C(x) ∧ D(x) ∧ F(x))
e) (∃xC(x)) ∧ (∃xD(x)) ∧ (∃xF(x))
Step-by-step explanation:
We are given;
C(x) = “x has a cat”
D(x) = “x has a dog”
F(x) = “x has a ferret”
We will make use of the following interpretation symbols;
Negation ¬ A : Not A
Disjunction A ∨ B : A or B
Conjunction A ∧ B : A and B
Existential quantification ∃ x A(x) : There exists an element x in the domain such that A(x)
Universal Quantification ∀x A(x) : A(x) for all values of x in the domain
So, using the above interpretations, we can answer the questions as;
a) ∃x(C(x) ∧ D(x) ∧ F(x))
b) ∀x(C(x) ∨ D(x) ∨ F(x))
c) ∃x(C(x) ∧ D(x) ¬ F(x))
d) ¬∃x(C(x) ∧ D(x) ∧ F(x))
e) (∃xC(x)) ∧ (∃xD(x)) ∧ (∃xF(x))
. Or we could lose the negative on the 3 in the denominator like this: 
and 
. There you go! Any one of those last three would work. It just depends upon how your teacher wants you to express your answer.