Let's denote the following:
- $ A $ as the set of students who are fans of the Buffalo Bills.
- $ B $ as the set of students who are fans of the New York Jets.
- $ |A| $ as the number of students who are fans of the Buffalo Bills.
- $ |B| $ as the number of students who are fans of the New York Jets.
- $ |A \cup B| $ as the number of students who are fans of either the Buffalo Bills or the New York Jets or both.
- $ |A \cap B| $ as the number of students who are fans of both the Buffalo Bills and the New York Jets.

From the problem, we know:
- $ |A| = 43 $
- $ |B| = 27 $
- 48 students do not like either team, so the number of students who like at least one of the teams is $ 100 - 48 = 52 $.

Using the principle of inclusion and exclusion for sets, we have:
$$ |A \cup B| = |A| + |B| - |A \cap B| $$

We know $ |A \cup B| = 52 $, so we can substitute the known values into the equation:
$$ 52 = 43 + 27 - |A \cap B| $$

Solving for $ |A \cap B| $:
$$ 52 = 70 - |A \cap B| $$
$$ |A \cap B| = 70 - 52 $$
$$ |A \cap B| = 18 $$

Therefore, the number of students who are fans of both the Buffalo Bills and the New York Jets is 18.

Question

Let's denote the following:
- $ A $ as the set of students who are fans of the Buffalo Bills.
- $ B $ as the set of students who are fans of the New York Jets.
- $ |A| $ as the number of students who are fans of the Buffalo Bills.
- $ |B| $ as the number of students who are fans of the New York Jets.
- $ |A \cup B| $ as the number of students who are fans of either the Buffalo Bills or the New York Jets or both.
- $ |A \cap B| $ as the number of students who are fans of both the Buffalo Bills and the New York Jets.

From the problem, we know:
- $ |A| = 43 $
- $ |B| = 27 $
- 48 students do not like either team, so the number of students who like at least one of the teams is $ 100 - 48 = 52 $.

Using the principle of inclusion and exclusion for sets, we have:
$$ |A \cup B| = |A| + |B| - |A \cap B| $$

We know $ |A \cup B| = 52 $, so we can substitute the known values into the equation:
$$ 52 = 43 + 27 - |A \cap B| $$

Solving for $ |A \cap B| $:
$$ 52 = 70 - |A \cap B| $$
$$ |A \cap B| = 70 - 52 $$
$$ |A \cap B| = 18 $$

Therefore, the number of students who are fans of both the Buffalo Bills and the New York Jets is 18.

Knowee AI · Accepted Answer

Let's denote the following:
- $ A $ as the set of students who are fans of the Buffalo Bills.
- $ B $ as the set of students who are fans of the New York Jets.
- $ |A| $ as the number of students who are fans of the Buffalo Bills.
- $ |B| $ as the number of students who are fans of the New York Jets.
- $ |A \cup B| $ as the number of students who are fans of either the Buffalo Bills or the New York Jets or both.
- $ |A \cap B| $ as the number of students who are fans of both the Buffalo Bills and the New York Jets.

From the problem, we know:
- $ |A| = 43 $
- $ |B| = 27 $
- 48 students do not like either team, so the number of students who like at least one of the teams is $ 100 - 48 = 52 $.

Using the principle of inclusion and exclusion for sets, we have:
$$ |A \cup B| = |A| + |B| - |A \cap B| $$

We know $ |A \cup B| = 52 $, so we can substitute the known values into the equation:
$$ 52 = 43 + 27 - |A \cap B| $$

Solving for $ |A \cap B| $:
$$ 52 = 70 - |A \cap B| $$
$$ |A \cap B| = 70 - 52 $$
$$ |A \cap B| = 18 $$

Therefore, the number of students who are fans of both the Buffalo Bills and the New York Jets is 18.

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