MIPS

please explain thoroughly how to do this, as i'm very confused

6 (Chapter 3.5) IEEE 754-2008 contains a half precision that is only 16 bits wide. The leftmost bit is still the sign bit, the exponent is 5 bits wide and has a bias of 15, and the mantissa is 10 bits long. A hidden 1 is assumed. Write down the bit pattern to represent -1.5625 * 10-Passuming a version of this format. Calculate the sum of 2.6125*10 and 4.150390625 * 10-by hand, assuming both numbers are stored in the 16-bit half precision described above. Assume 1 guard, 1 round bit, and 1 sticky bit, and round to the nearest even. Show all the steps. (10 points)