In the IEEE 754 representation, 32 bit floating point has 1 bit to represent the sign, 8 bits to can be thought of as the numerator of a fraction where the denominator is 223 = 8,388,608. The eight bits of the exponent will be represent the exponent-127, which means it can represent any number between 128 and -127. Don't forget that the mantissa has a 24th bit at the beginning which is 1 unless the exponent = 0, which represents 2-127, the smallest exponent possible, then the hidden 24th bit is a 0. 2-2 y 11,184,811 Example: is represented as 001111101 01010101010101010101011 or 2-2x de 8,388,608 and ? - 11.184.811 has an exponent of 10-8. When the exponent is larger, the error should get larger as well. For the following numbers, find the numerator stored in the mantissa as a base 10 value, the exponent and the base 10 exponent of the error. (12 points) Number base 2 exponent mantissa numerator base 10 exponent of error 183 2,598,960 300,000.3 In the IEEE 754 representation, 32 bit floating point has 1 bit to represent the sign, 8 bits to can be thought of as the numerator of a fraction where the denominator is 223 = 8,388,608. The eight bits of the exponent will be represent the exponent-127, which means it can represent any number between 128 and -127. Don't forget that the mantissa has a 24th bit at the beginning which is 1 unless the exponent = 0, which represents 2-127, the smallest exponent possible, then the hidden 24th bit is a 0. 2-2 y 11,184,811 Example: is represented as 001111101 01010101010101010101011 or 2-2x de 8,388,608 and ? - 11.184.811 has an exponent of 10-8. When the exponent is larger, the error should get larger as well. For the following numbers, find the numerator stored in the mantissa as a base 10 value, the exponent and the base 10 exponent of the error. (12 points) Number base 2 exponent mantissa numerator base 10 exponent of error 183 2,598,960 300,000.3