The given 9-bit number is 010001111.

The number format is fixed-point with 5-bit integer and 4-bit fraction in two's complement.

The 5-bit integer part is 01000, which is 8 in decimal.

The 4-bit fraction part is 1111. To convert this to decimal, we consider each bit as a negative power of 2. So, 1111 in binary is 1*(1/2) + 1*(1/4) + 1*(1/8) + 1*(1/16) = 0.5 + 0.25 + 0.125 + 0.0625 = 0.9375 in decimal.

So, the given 9-bit number 010001111 in fixed-point with 5-bit integer and 4-bit fraction in two's complement is 8.9375 in decimal.

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The given 9-bit number is 010001111.

The number format is fixed-point with 5-bit integer and 4-bit fraction in two's complement.

The 5-bit integer part is 01000, which is 8 in decimal.

The 4-bit fraction part is 1111. To convert this to decimal, we consider each bit as a negative power of 2. So, 1111 in binary is 1*(1/2) + 1*(1/4) + 1*(1/8) + 1*(1/16) = 0.5 + 0.25 + 0.125 + 0.0625 = 0.9375 in decimal.

So, the given 9-bit number 010001111 in fixed-point with 5-bit integer and 4-bit fraction in two's complement is 8.9375 in decimal.

Knowee AI · Accepted Answer

The given 9-bit number is 010001111.

The number format is fixed-point with 5-bit integer and 4-bit fraction in two's complement.

The 5-bit integer part is 01000, which is 8 in decimal.

The 4-bit fraction part is 1111. To convert this to decimal, we consider each bit as a negative power of 2. So, 1111 in binary is 1*(1/2) + 1*(1/4) + 1*(1/8) + 1*(1/16) = 0.5 + 0.25 + 0.125 + 0.0625 = 0.9375 in decimal.

So, the given 9-bit number 010001111 in fixed-point with 5-bit integer and 4-bit fraction in two's complement is 8.9375 in decimal.

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)

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