Question

if one turn consists of spinning each spinner once, which table shows all the possible outcomes for one turn? spinners: top spinner divided into four sections labeled 2, 4, 6, 8; bottom spinner divided into three sections labeled a, b, c. two tables (first spinner vs second spinner) with columns/rows including 2a, 2b, 4a, 4b, 6a, 6b, 8a, 8b (and variations) are shown, along with radio button options.

Explanation:

To solve this, we analyze the two spinners:

Step 1: Identify Outcomes of Each Spinner

• First Spinner (top): Numbers \( 2, 4, 6, 8 \) (4 outcomes).
• Second Spinner (bottom): Letters \( A, B, C \) (3 outcomes).

Step 2: Determine Possible Combinations

Each outcome of the first spinner pairs with each outcome of the second spinner. For example:

• If the first spinner lands on \( 2 \), the second can be \( A, B, \) or \( C \) → \( 2A, 2B, 2C \)? Wait, no—wait, the tables given have \( 2A, 2B \) (but the second spinner has \( C \) too). Wait, no—wait, the second spinner is divided into three: \( A, B, C \). So the correct pairs should include \( C \) for each first spinner number.

Wait, looking at the tables:

• The first table (left) has \( 2A, 2B, 4A, 4B, 6A, 6B, 8A, 8B \) (missing \( C \) pairs).
• The second table (right) has \( 2A, 2B, 4A, 4B, 6A, 6B, 8A, 8B \) (also missing \( C \))? Wait, no—wait, maybe the second spinner is actually \( A, B \) (but the diagram shows three sections: \( A, B, C \)). Wait, maybe the diagram is misread. Wait, the problem’s tables: let’s re-express.

Wait, the question is which table shows all possible outcomes. Let’s list all possible pairs:

First spinner: \( 2, 4, 6, 8 \) (4 numbers).
Second spinner: \( A, B, C \) (3 letters). Wait, no—wait, the second spinner in the diagram has three sections: \( A, B, C \). So total outcomes: \( 4 \times 3 = 12 \)? But the tables have 8 cells. Wait, maybe the second spinner is \( A, B \) (two sections)? Wait, the diagram shows a spinner with three sections: \( A, B, C \) (each a third? Or maybe \( B \) is larger, but the labels are \( A, B, C \)).

Wait, the tables given:

• First table (left): Rows (First Spinner) \( 2, 4, 6, 8 \); Columns (Second Spinner) \( A, B \). So cells: \( 2A, 2B, 4A, 4B, 6A, 6B, 8A, 8B \) (8 outcomes).
• Second table (right): Same structure? Wait, no—wait, the user’s image shows two tables. Wait, maybe the second spinner is actually \( A, B \) (two sections), not three. Let’s check the diagram again: the bottom spinner has three sections labeled \( A, B, C \), but maybe it’s a typo, and the tables use \( A, B \). Wait, no—wait, the problem says “which table shows all possible outcomes”. Let’s re-express:

If the first spinner has 4 outcomes (\( 2, 4, 6, 8 \)) and the second has 2 outcomes (\( A, B \)), then total outcomes are \( 4 \times 2 = 8 \), which matches the tables (8 cells). So the second spinner must have \( A, B \) (two sections, maybe the \( C \) is a mistake, or the tables are correct).

Now, the first table (left) has:

• Row \( 2 \): \( 2A, 2B \)
• Row \( 4 \): \( 4A, 4B \)
• Row \( 6 \): \( 6A, 6B \)
• Row \( 8 \): \( 8A, 8B \)

The second table (right) has:

• Row \( 2 \): \( 2A, 2B \)
• Row \( 4 \): \( 4A, 4B \)
• Row \( 6 \): \( 6A, 6B \)
• Row \( 8 \): \( 8A, 8B \)

Wait, no—wait, the first table (left) has \( 2A, 2B, 4A, 4B, 6A, 6B, 8A, 8B \) (correct, as \( 4 \times 2 = 8 \)). The second table (right) looks the same? Wait, maybe the first table (left) is the correct one, but wait—no, let’s check the labels. Wait, the first spinner is labeled “First Spinner” with rows \( 2, 4, 6, 8 \), and the second spinner is “Second Spinner” with columns \( A, B \). So the correct table is the one where each first spinner number (2,4,6,8) pairs with each second spinner letter (A,B).

Assuming the second spinner has \( A, B \) (two sections), the table with \( 2A, 2B, 4A, 4B, 6A, 6B, 8A, 8B \) is correct. From the image, the left table (first table) has these pairs.

Answer:

The left table (with rows 2,4,6,8 and columns 2A,2B; 4A,4B; 6A,6B; 8A,8B) shows all possible outcomes.

(Note: If the second spinner actually has \( A, B, C \), there would be \( 4 \times 3 = 12 \) outcomes, but the tables have 8, so the second spinner likely has \( A, B \) (two sections), making the left table correct.)