Respuesta :
Treating the data as a Venn set, it is found that:
- 26 students are good in mathematics only.
- 28 students are not good in any of the three courses.
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I am going to say that:
- A is the number of students good in Math.
- B is the number of students good in English.
- C is the number of students good in Psychology.
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9 are good in all of the three courses.
This means that: [tex]A \cap B \cap C = 9[/tex]
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13 are good in both mathematics and psychology
This means that:
[tex](A \cap C) + (A \cap B \cap C) = 13[/tex]
[tex](A \cap C) + 9 = 13[/tex]
[tex](A \cap C) = 4[/tex]
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15 are good in both mathematics and in English
This means that:
[tex](A \cap B) + (A \cap B \cap C) = 15[/tex]
[tex](A \cap B) + 9 = 15[/tex]
[tex](A \cap B) = 6[/tex]
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16 are good in both English and psychology
This means that:
[tex](B \cap C) + (A \cap B \cap C) = 16[/tex]
[tex](B \cap C) + 9 = 16[/tex]
[tex](B \cap C) = 7[/tex]
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20 are good in psychology
This means that:
[tex]c + (A \cap C) + (B \cap C) + (A \cap B \cap C) = 20[/tex]
[tex]c + 4 + 7 + 9 = 20[/tex]
[tex]c = 0[/tex]
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45 are good in mathematics
This means that:
[tex]a + (A \cap B) + (A \cap C) + (A \cap B \cap C) = 45[/tex]
[tex]a + 6 + 4 + 9 = 45[/tex]
[tex]a = 26[/tex]
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Question a:
[tex]a = 26[/tex], which means that 26 students are good in mathematics only.
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Question b:
At least one is:
[tex]a + (A \cap B) + (A \cap C) + (B \cap C) + (A \cap B \cap C) = 26 + 6 + 4 + 7 + 9 = 52[/tex]
Thus, 80 - 52 = 28
28 students are not good in any of the three courses.
A similar problem is given at: https://brainly.com/question/22003843
