Final answer:
In the given multiplication problem, 5,819 x 8, Yin needs to replace 'y' with the digit '5', which is the tens digit in the partial product when multiplying the hundreds place (8 x 8 + 1 carry-over).
Step-by-step explanation:
The student's question refers to the multiplication of a 4-digit number by a single-digit number using the standard algorithm. Specifically, Yin is multiplying 5,819 by 8. The work shown has a place where 'y' must be filled in, which implies that this is the partial product or part of the overall multiplication process.
In the standard algorithm, each digit of the 4-digit number is multiplied by the single-digit number, starting from the ones place and moving left to the tens, hundreds, and thousands places, carrying over any extra value as needed.
The algorithm would proceed as follows:
5,819
× 8
--------
46,552
The operation 8 x 9 yields 72, so we write down 2 and carry over 7. Next, 8 x 1 plus the carried over 7 equals 15, so we write down 5 and carry over 1. Following this, 8 x 8 plus the carried over 1 equals 65, so we write 5 and carry over 6. Lastly, 8 x 5 plus the carried over 6 equals 46. Writing down all these products in order gives us 46,552. Therefore, the 'y' that Yin has to replace is the digit '5', coming from the multiplication of 8 by 8 with a 1 carried over from the previous step.