Study the following 9 bit number 010011000 The number format is fixed-point with 3 bit integer bits, unsigned. Suppose you add this number to itself. What is the actual round-off error?

Study the following 9 bit number: 010001111The number format is fixed-point with 5 bit integer, 4-bit fraction, two's complement. Convert this to decimal. Use up to 4 decimal places (round if necessary)

OK, let's see if I can explain this. When you round off, you change the value of the number,except if you round off a zero. Following the old rules, you can round a number down in valuefour times (rounding with one, two, three, four) compared to rounding it upwards five times(five, six, seven, eight, nine). Remember that "rounding off" a zero does not change the value ofthe number being rounded off.

Consider the following IEEE 754 32-bit floating point value:01001111 01001000 00000000 00000000Enter S, its sign, as either + or -. Enter E, its exponent part's value, in base 10. Enter F, its fraction part's value, in base 10 in the form 0.XXXX where X is a digit

In the 32-bit floating point system (for float variables) that was explored using the floatbits.c program the following bit patterns are observed for the three given values, expressed as four 8-bit bytes: 2.0 = 01000000 00000000 00000000 00000000 3.0 = 01000000 01000000 00000000 00000000 4.0 = 01000000 10000000 00000000 00000000The bit pattern corresponding to the decimal number 5.0 is most likely to be:Group of answer choices01000000 10100000 00000000 0000000001000001 10100000 00000000 0000000001000000 11000000 00000000 0000000001000000 10101000 00000000 0000000001000000 10010000 00000000 00000000

Consider the following IEEE 754 32-bit floating point value:01001111 01001000 00000000 00000000Enter S, its sign, as either + or -. Enter E, its exponent part's value, in base 10.