\( + \left. \begin{array} { l l l l | l | l } { 5 } & { \square } & { 4 } & { } & { 8 } \\ { + } & { 7 } & { 1 } & { 5 } & { } \\ \hline \end{array} \right. \)

To solve the addition problem represented in the table, we need to fill in the square (denoted as \( \square \)) and ensure that the addition is correct. The problem can be interpreted as follows: ``` 5 ☐ 4 + 7 1 5 ----------- 8 ``` Let's break it down step by step: 1. **Align the numbers for addition**: ``` 5 ☐ 4 + 7 1 5 --------- 8 ``` 2. **Start from the rightmost column**: - The rightmost column has \( 4 + 5 = 9 \). Since we only have a single digit in the result (which is 8), we carry over 1 to the next column. 3. **Move to the next column**: - The second column now has \( \square + 1 + 1 = \square + 2 \) (the 1 is from the carry). - We need this to equal 8 (the result in the second column). - Therefore, we set up the equation: \[ \square + 2 = 8 \] - Solving for \( \square \): \[ \square = 8 - 2 = 6 \] 4. **Check the leftmost column**: - The leftmost column has \( 5 + 7 = 12 \). Since we have a carry of 1 from the previous column, we have \( 12 + 1 = 13 \). - The leftmost digit of the result is 1 (from 13), which is consistent. Thus, the value of \( \square \) is \( 6 \). The completed addition looks like this: ``` 5 6 4 + 7 1 5 ------- 1 3 9 ``` So, the final answer is: \[ \boxed{6} \]