What is the maximum decimal value of a binary octet represented as 1111 1111?

What is the maximum decimal value of a binary octet represented as 1111 1111?

• A. 128
• B. 255
• C. 512
• D. 1024

Answer: (B)

A binary octet consists of 8 bits, and it can represent values from 0 to 255 in decimal. In the representation of 1111 1111, each digit (bit) corresponds to a specific power of 2, specifically: - The first bit (from the right) is \(2^0\) - The second bit is \(2^1\) - The third bit is \(2^2\) - The fourth bit is \(2^3\) - The fifth bit is \(2^4\) - The sixth bit is \(2^5\) - The seventh bit is \(2^6\) - The eighth bit is \(2^7\) In the binary representation 1111 1111, all bits are set to 1. Therefore, the calculation to convert it to decimal involves summing the values of each bit where a 1 is present: \[ 2^7 + 2^6 + 2^5 + 2^4 + 2^3 + 2^2 + 2^1 + 2^0 = 128 + 64 + 32 + 16 + 8 + 4 + 2 +

A binary octet consists of 8 bits, and it can represent values from 0 to 255 in decimal. In the representation of 1111 1111, each digit (bit) corresponds to a specific power of 2, specifically:

• The first bit (from the right) is (2^0)

• The second bit is (2^1)

• The third bit is (2^2)

• The fourth bit is (2^3)

• The fifth bit is (2^4)

• The sixth bit is (2^5)

• The seventh bit is (2^6)

• The eighth bit is (2^7)

In the binary representation 1111 1111, all bits are set to 1. Therefore, the calculation to convert it to decimal involves summing the values of each bit where a 1 is present:

[

2^7 + 2^6 + 2^5 + 2^4 + 2^3 + 2^2 + 2^1 + 2^0 = 128 + 64 + 32 + 16 + 8 + 4 + 2 +