Question

use the accompanying venn diagram, which shows the cardinality of each region, to answer the question below. how many elements belong to set a or set c? there are \boxed{} elements that belong to set a or set c. (simplify your answer.)

Explanation:

Step1: Identify regions in A or C

To find elements in \( A \) or \( C \), we sum regions unique to \( A \), unique to \( C \), and intersections of \( A \) & \( B \), \( A \) & \( C \), \( B \) & \( C \), and all three. Regions: \( A \) only (4), \( A \cap B \) only (5), \( A \cap C \) only (3), all three (\( A \cap B \cap C \), 2), \( C \) only (9), \( B \cap C \) only (8). Wait, no—correct regions for \( A \) or \( C \): \( A \) only (4), \( A \cap B \) (5), \( A \cap C \) (3), \( A \cap B \cap C \) (2), \( C \) only (9), \( B \cap C \) (8)? Wait, no: "or" means union, so \( n(A \cup C) = n(A) + n(C) - n(A \cap C) \). First, find \( n(A) \): \( A \) only (4) + \( A \cap B \) only (5) + \( A \cap C \) only (3) + all three (2) = \( 4 + 5 + 3 + 2 = 14 \). \( n(C) \): \( C \) only (9) + \( A \cap C \) only (3) + \( B \cap C \) only (8) + all three (2) = \( 9 + 3 + 8 + 2 = 22 \). \( n(A \cap C) \): \( A \cap C \) only (3) + all three (2) = \( 3 + 2 = 5 \). Then \( n(A \cup C) = 14 + 22 - 5 = 31 \)? Wait, no, better to sum all regions in \( A \) or \( C \): regions in \( A \) (4,5,3,2) and regions in \( C \) (9,8) but not double-counting. Wait, the Venn diagram: \( A \) has 4 (only A), 5 (A∩B only), 3 (A∩C only), 2 (all three). \( C \) has 9 (only C), 3 (A∩C only), 8 (B∩C only), 2 (all three). So \( A \cup C \) includes: 4 (A only), 5 (A∩B only), 3 (A∩C only), 2 (all three), 9 (C only), 8 (B∩C only). Wait, no—"A or C" includes any region in A, in C, or both. So list all regions in A or C:

• A only: 4
• A∩B only: 5 (in A)
• A∩C only: 3 (in A and C)
• All three: 2 (in A and C)
• C only: 9 (in C)
• B∩C only: 8 (in C)

Now sum these: \( 4 + 5 + 3 + 2 + 9 + 8 \). Wait, but wait: is B∩C only (8) in C? Yes, because it's part of C. Wait, but let's check again. The formula for union: \( n(A \cup C) = n(A) + n(C) - n(A \cap C) \). Let's compute \( n(A) \): regions in A: 4 (only A) + 5 (A∩B only) + 3 (A∩C only) + 2 (all three) = \( 4 + 5 + 3 + 2 = 14 \). \( n(C) \): 9 (only C) + 3 (A∩C only) + 8 (B∩C only) + 2 (all three) = \( 9 + 3 + 8 + 2 = 22 \). \( n(A \cap C) \): 3 (A∩C only) + 2 (all three) = \( 3 + 2 = 5 \). Then \( n(A \cup C) = 14 + 22 - 5 = 31 \). Alternatively, summing the regions: 4 (A only) + 5 (A∩B) + 3 (A∩C) + 2 (all three) + 9 (C only) + 8 (B∩C) = \( 4 + 5 = 9 \), \( 9 + 3 = 12 \), \( 12 + 2 = 14 \), \( 14 + 9 = 23 \), \( 23 + 8 = 31 \). Yes, that matches.

Step2: Sum the regions

Sum all regions belonging to A or C: \( 4 + 5 + 3 + 2 + 9 + 8 = 31 \).

Answer:

31