50 students in a certain class were enrolled in three subjects as follows: 30 enrolled in Math 25 enrolled in English 25 enrolled in P.E. 14 enrolled in P.E. and Math 12 enrolled in P.E. and English 11 enrolled in Math and English 5 enrolled in the three subjects Questions: How many students were enrolled in exactly one subject? _____________ How many students were enrolled in exactly two subjects? ____________ How many students were enrolled in Math only? ____________ How many students were enrolled in the three subjects? _____________ How many students did not enrolled in any of the three subjects?

Answer:We are given:Total students Students enrolled in Math Students enrolled in English Students enrolled in P.E. Students enrolled in both P.E. and Math Students enrolled in both P.E. and English Students enrolled in both Math and English Students enrolled in all three subjects 1. Number of students enrolled in exactly one subjectFirst, we use the principle of inclusion-exclusion to find the total number of students enrolled in at least one subject:n(M \cup E \cup P) = n(M) + n(E) + n(P) - n(M \cap E) - n(M \cap P) - n(E \cap P) + n(M \cap E \cap P)Substitute the given values:n(M \cup E \cup P) = 30 + 25 + 25 - 11 - 14 - 12 + 5n(M \cup E \cup P) = 80 - 37 + 5 = 48This means 48 students are enrolled in at least one subject. Thus, the number of students not enrolled in any of the subjects is:n(\text{Not enrolled in any}) = n - n(M \cup E \cup P) = 50 - 48 = 22. Number of students enrolled in exactly one subjectTo find this, we need to find:Students enrolled only in MathStudents enrolled only in EnglishStudents enrolled only in P.E.Only in Math:n(\text{Math only}) = n(M) - n(M \cap E) - n(M \cap P) + n(M \cap E \cap P)Substitute the given values:n(\text{Math only}) = 30 - 11 - 14 + 5 = 10Only in English:n(\text{English only}) = n(E) - n(M \cap E) - n(E \cap P) + n(M \cap E \cap P)Substitute the given values:n(\text{English only}) = 25 - 11 - 12 + 5 = 7Only in P.E.:n(\text{P.E. only}) = n(P) - n(P \cap M) - n(P \cap E) + n(M \cap E \cap P)Substitute the given values:n(\text{P.E. only}) = 25 - 14 - 12 + 5 = 4So, the number of students enrolled in exactly one subject is:10 + 7 + 4 = 213. Number of students enrolled in exactly two subjectsThis can be calculated as follows:Math and English only:n(M \cap E \text{ only}) = n(M \cap E) - n(M \cap E \cap P) = 11 - 5 = 6Math and P.E. only:n(M \cap P \text{ only}) = n(M \cap P) - n(M \cap E \cap P) = 14 - 5 = 9English and P.E. only:n(E \cap P \text{ only}) = n(E \cap P) - n(M \cap E \cap P) = 12 - 5 = 7So, the number of students enrolled in exactly two subjects is:6 + 9 + 7 = 224. Number of students enrolled in all three subjectsFrom the given information:n(M \cap E \cap P) = 5Summary:1. Number of students enrolled in exactly one subject: 212. Number of students enrolled in exactly two subjects: 223. Number of students enrolled in Math only: 104. Number of students enrolled in all three subjects: 55. Number of students not enrolled in any of the three subjects: 2I only use chat gpt its too hard lol

Answered by phylleraquel | 2024-09-10