please do not copy the wrong solution from the internet like the person before did

In the IEEE 754 representation, 32 bit floating point has 1 bit to represent the sign, 8 bits to represent the exponent of 2 and 23 bits to represent the mantissa. This means the mantissa can be thought of as the numerator of a fraction where the denominator is 223 8,388,608. Don't forget that the mantissa has a 24th bit at the beginning which is 1 unless the number being represented is 0. The eight bits of the exponent will be represent the number 128- exponent, which means it can represent any number between 128 and -127 Example: is represented as 0 10000010 01010101010101010101011 or2-2x1184811 and1_11184.811 larger as well. 8,388,608 has an exponent of 10-8. When the exponent is larger, the error should get For the following numbers, find the numerator stored in the mantissa as a base 10 value, including the hidden 1 at the beginning of the binary representation for the given values of the exponent and the base 10 exponent of the error. (8 points) Number exponent mantissa base 10 exponent of error -1 V50 1,000,000,000 14 1,000,000.3 19