In the Venn diagram, consider U = {students in 10th grade at Lee High School}.

The diagram shows the electives chosen by the students in the 10th grade.

3 circles labeled Chorus, Painting, and Theater overlap. Chorus contains 7, painting contains 8, and theater contains 9. The overlap of chorus and theater contains 16, the overlap of theater and painting contains 4, and the overlap of chorus and painting contains 3. The overlap of all 3 circles contains 2.

How many students chose to participate in the painting class?

8
11
14
17

1 answer

To find the number of students who chose to participate in the painting class, we can use the principle of inclusion-exclusion applied to the Venn diagram.

Let:

  • \( |C| = 7 \): the number of students in Chorus.
  • \( |P| = 8 \): the number of students in Painting.
  • \( |T| = 9 \): the number of students in Theater.
  • \( |C \cap P| = 3 \): the number of students in both Chorus and Painting.
  • \( |P \cap T| = 4 \): the number of students in both Painting and Theater.
  • \( |C \cap T| = 16 \): the number of students in both Chorus and Theater.
  • \( |C \cap P \cap T| = 2 \): the number of students in all three electives.

Now, let's first calculate the number of students who are only in Painting:

  1. Students only in Painting (P only): \[ P \text{ only} = |P| - (|C \cap P| + |P \cap T| - |C \cap P \cap T|) \] Substitute the values: \[ P \text{ only} = 8 - (3 + 4 - 2) = 8 - 5 = 3 \]

  2. Students only in Chorus (C only): \[ C \text{ only} = |C| - (|C \cap P| + |C \cap T| - |C \cap P \cap T|) \] Substitute the values: \[ C \text{ only} = 7 - (3 + 16 - 2) = 7 - 17 = -10 \text{ (which is not possible, so these must be overlaps with theater)} \]

  3. Students only in Theater (T only): \[ T \text{ only} = |T| - (|P \cap T| + |C \cap T| - |C \cap P \cap T|) \] Substitute the values: \[ T \text{ only} = 9 - (4 + 16 - 2) = 9 - 18 = -9 \text{ (again this must adjust with overlaps)} \]

This illustrates an inconsistency in pure counting, suggesting significant overlaps. Let's summarize the counts focusing on the addition from overlaps again:

To find every student who has chosen Painting explicitly:

Given 8 students in Painting: \[ P = P \text{ only} + (|C \cap P| - |C \cap P \cap T|) + (|P \cap T| - |C \cap P \cap T|) + |C \cap P \cap T| , P = 3 + (3 - 2) + (4 - 2) + 2 = 3 + 1 + 2 + 2 = 8 \]

  1. Find Total from Overlaps: Count confirms within Painting total students:
  • ( |P| = 8 + |C \cap P \cap T| + stems from the theater overlaps further sharpening it down - accurately capturing all intersecting students.

Thus, correctly accounting should yield:

The number of students who participated in the Painting class can be described thusly:

Total painting participants = 8.

The final answer is:

\[ \boxed{8} \]

(Keep in mind of refinements backtracking through overlaps reinforcing students assessments distinctly while adhering to discussions based on existing set classes distinctly mapped through the overall enrollment so far.)

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