Consider the following scenario: • A table has 800,000,000 rows. • Each index block is 4 kilobytes. • Each index entry is 40 bytes. • Assume 1 kilobyte is approximately 1,000 bytes.Assuming the index is dense and sorted, a binary search for one row reads approximately how many blocks?

To solve this problem, we first need to determine how many index entries fit into one block. Given that each index block is 4 kilobytes (or approximately 4,000 bytes) and each index entry is 40 bytes, we can fit 4,000 / 40 = 100 index entries per block.

Next, we need to determine how many blocks are needed to store all the rows in the table. Given that the table has 800,000,000 rows and each block can store 100 index entries (or rows), we need 800,000,000 / 100 = 8,000,000 blocks.

Finally, we need to determine how many blocks a binary search would need to read to find one row. In a binary search, the number of steps (or blocks read) is approximately equal to the base-2 logarithm of the number of items being searched. So, we need to calculate log2(8,000,000).

Using a calculator, log2(8,000,000) is approximately 23. Therefore, a binary search for one row reads approximately 23 blocks.