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The problem involves a spinner divided into three sections labeled 2, 4, and 6. Ramon will spin the spinner twice, and we need to find the most likely sum of the results from the two spins.

### Step-by-Step Solution:

1. **Possible Outcomes:**
   The spinner has three possible outcomes for each spin: 2, 4, and 6. When Ramon spins twice, the outcomes are combinations of these numbers.

2. **List of Possible Sums:**
   We calculate all possible sums of two spins:
   - \( 2 + 2 = 4 \)
   - \( 2 + 4 = 6 \)
   - \( 2 + 6 = 8 \)
   - \( 4 + 2 = 6 \)
   - \( 4 + 4 = 8 \)
   - \( 4 + 6 = 10 \)
   - \( 6 + 2 = 8 \)
   - \( 6 + 4 = 10 \)
   - \( 6 + 6 = 12 \)

3. **Frequencies of the Sums:**
   - Sum = 4 appears **1 time**
   - Sum = 6 appears **2 times**
   - Sum = 8 appears **3 times**
   - Sum = 10 appears **2 times**
   - Sum = 12 appears **1 time**

4. **Conclusion:**
   The sum 8 appears most frequently (3 times), making it the most likely sum of the two spins.

### Answer:
The most likely sum of the two spins is **8**.

Would you like more details or have any questions? 

Here are some related questions you might want to explore:
1. What is the probability of getting a sum of 8?
2. How would the answer change if the spinner had four sections?
3. What if Ramon spun the spinner three times instead of two?
4. How would you calculate the expected value of the sum of two spins?
5. Can this problem be solved using a probability distribution?

**Tip:** When dealing with probability problems like this, listing out all possible outcomes is a great first step to ensure nothing is overlooked.

Ramon is going to spin the arrow in the spinner and add the results. What is the most likely sum of the two spins?

Determine the most likely sum when spinning a spinner with sections labeled 2, 4, and 6 twice and adding the results. This problem involves probability and combinatorics, suitable for students in Grades 5-7.

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