Respuesta :
Answer:
4 students
Step-by-step explanation:
The total number of students is given by:
[tex]|A|+|B|+|C|-|A\cap B|-|A\cap C|-|B\cap C| + |A\cap B \cap C| = N[/tex]
Where N is the total number of students, A is the number of students that have taken Math 2A, B is the number of students that have taken Math 2B, and C is the number of students that have taken Math 2C:
[tex]|A|+|B|+|C|-|A\cap B|-|A\cap C|-|B\cap C| + |A\cap B \cap C| = N\\51+80+70-15-20-13 + |A\cap B \cap C| = 157\\ |A\cap B \cap C| =4[/tex]
4 students in Group A have taken all three classes.
*The number of students that have taken two classes has be subtracted so those students do not get counted twice
The number of students in group A that are taking all the three classes of Math 2A, 2B, and 2C is 4.
What is an equation?
An equation is used to compare two expressions, with the help of an equal sign '='.
We know that the total number of students(N=157), can be written as,
[tex]N = |A| + |B| + |C| - |A\cap B|-|B\cap C|-|A\cap C|+|A\cap B \cap C|[/tex]
As it was given to us,
- 51 students in Group A have taken Math 2A(|A|),
- 80 have taken Math 2B(|B|), and
- 70 have taken Math 2C(|C|).
- 15 have taken both Math 2A and 2B(|A∩B|),
- 20 have taken both Math 2A and 2C(|A∩C|), and
- 13 have taken both Math 2B and 2C(|B∩C|).
Substituting the values in the above equation of the total number of students,
[tex]157 = 51+80+70-15-13-20+|A\cap B \cap C|\\\\157 = 153+|A\cap B \cap C|\\\\|A\cap B \cap C| = 157-153\\\\|A\cap B \cap C| = 4[/tex]
Hence, the number of students in group A that are taking all the three classes of Math 2A, 2B, and 2C is 4.
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