1. The problem is to understand the conversion of the binary number $10111_2$ to its decimal equivalent and to discuss the representation of decimal floating point numbers in octal and hexadecimal forms. 2. First, let's convert the binary number $10111_2$ to decimal. 3. Each digit in the binary number represents a power of 2, starting from the right (least significant bit): $$ \begin{aligned} &1 \times 2^4 + 0 \times 2^3 + 1 \times 2^2 + 1 \times 2^1 + 1 \times 2^0 \\ &= 1 \times 16 + 0 \times 8 + 1 \times 4 + 1 \times 2 + 1 \times 1 \\ &= 16 + 0 + 4 + 2 + 1 = 23 \end{aligned} $$ 4. So, the binary number $10111_2$ equals $23_{10}$ in decimal. 5. Regarding the decimal floating point number $3.75_{10}$, it can be represented in octal and hexadecimal as follows: 6. Convert the integer part (3) and fractional part (0.75) separately. 7. Integer part 3 in octal is $3_8$ and in hexadecimal is $3_{16}$. 8. For the fractional part 0.75: - Multiply by 8 for octal: $0.75 \times 8 = 6.0$, so fractional octal part is $0.6_8$. - Multiply by 16 for hexadecimal: $0.75 \times 16 = 12.0$, which corresponds to $C$ in hexadecimal. 9. Therefore, $3.75_{10}$ is $3.6_8$ in octal and $3.C_{16}$ in hexadecimal. 10. This shows how decimal floating point numbers can be represented in octal and hexadecimal systems.