A high school student took two college entrance exams, scoring 1070 on the SAT and 25 on the ACT. Suppose that SAT scores have a

Answer:

• Yes; refer to the question's clarification and the detailed answer below for more information.

Explanation:

Sets are in bijection if there is a bijective function connecting them.

This implies that the question can be interpreted as: can sets A and B be related via a bijective function?

A bijective function must be both injective and surjective, which means it is reversible.

Injective indicates that each element in the domain corresponds uniquely to an element in the codomain.

Surjective denotes that every element in the codomain relates to exactly one element in the domain.

This ensures there are no unmatched elements.

If we consider set A as the function's domain and set B as the codomain.

The core question then becomes: can the set {17, 34, 51, 68, 85,...} be in bijection with the set {11, 22, 33, 44, 55,...}?

Notice that the two sets share no common elements, as 17 and 34 are coprime.

A bijective function can only be established between sets A and B or B and A if both sets possess the same cardinality (number of elements).

Indeed, the cardinality of sets A and B is identical: both sets are of equal size, affirming that a bijective relationship exists between them.

Response:

a. 55 cars

b. 25 cars

Detailed explanation:

Let’s denote the quantity of cars with stereo systems as N(ss), those with air conditioning as N(ac), and those with sunroofs as N(sr).

We find that:

N(ss) = 30

N(ac) = 30

N(sr) = 40

N(ss and ac and sr) = 15

N(at least two) = 30

a.

To calculate how many cars possess at least one feature (N(at least one) or N(ss or ac or sr)), we apply:

N(ss or ac or sr) = N(ss) + N(ac) + N(sr) - N(ss and ac) - N(ss and sr) - N(ac and sr) + N(ss and ac and sr)

N(ss or ac or sr) = 30 + 30 + 40 - (N(at least two) + 2*N(ss and ac and sr)) + 15

Substituting, we find N(ss or ac or sr) = 30 + 30 + 40 - (30 + 2*15) + 15 = 55

b.

For those cars that have exactly one feature, we have:

N(only one) = N(at least one) - N(at least two)

N(only one) = 55 - 30 = 25