Consider a multidimensional array A stored in row-major order with 22 rows and 6 columns whose subscripts start at 0. If each element occupies eight bytes, then how many bytes from the start address of A will element A[3, 5] be stored? a. 284 b. 136 c. 184d. 568

Option c is the correct answer for the above question.

Explanation:

If the array will store in row-major order, then it will be a store like the first row, then second row and then third row.If the user wants to conclude the address of the A[3,5], then the address of the A[3,5] will be 184.It is because the 3 states the row number and the 5 states the column number.The above question states that every value take 8 bit. and there are 4 rows from 0 to 3 and 6 column (0 to 5) for every row (except the fourth rows because it takes 5 columns from 0 to 4) before the A[3,5].So there are 23 value before A[3,5] which are as-- [00,01,02,03,04,05][10,11,12,13,14,15][20,21,22,23,24,25][30,31,32,33,34].So when every value takes 8 bit then 23 value takes 184 bits(23*8).So the address of A[3,4] is 184.Hence option c is the correct and the other is not because of 184 stated from the c option only.

Answercrankcase;

It took 1.75 sec to get to the batter