$$ \begin{aligned} 101.011_{ ext {two }} & =1 imes 2^{2}+0 imes 2^{1}+1 imes 2^{0}+0 imes 2^{-1}+1 imes 2^{-2}+1 imes 2^{3} \ & =4+1+1 / 4+1 / 8 ext { in base ten } \ & =5.375_{ ext {ten }}\end{aligned} $$

Question

$$ \begin{aligned} 101.011_{	ext {two }} & =1 	imes 2^{2}+0 	imes 2^{1}+1 	imes 2^{0}+0 	imes 2^{-1}+1 	imes 2^{-2}+1 	imes 2^{3} \ & =4+1+1 / 4+1 / 8 	ext { in base ten } \ & =5.375_{	ext {ten }}\end{aligned} $$

UpStudy AI · Accepted Answer

Let's break down the conversion of the binary number $$ 101.011_{\text{two}} $$ to its decimal (base ten) equivalent step by step.

1. **Understanding the binary number**: The binary number $$ 101.011_{\text{two}} $$ can be separated into two parts: the integer part $$ 101_{\text{two}} $$ and the fractional part $$ 011_{\text{two}} $$.

2. **Convert the integer part**:
   - The integer part $$ 101_{\text{two}} $$ can be calculated as follows:
   $$
   1 \times 2^2 + 0 \times 2^1 + 1 \times 2^0
   $$
   - Calculating each term:
     - $$ 1 \times 2^2 = 4 $$
     - $$ 0 \times 2^1 = 0 $$
     - $$ 1 \times 2^0 = 1 $$
   - Adding these values together:
   $$
   4 + 0 + 1 = 5
   $$

3. **Convert the fractional part**:
   - The fractional part $$ 011_{\text{two}} $$ can be calculated as follows:
   $$
   0 \times 2^{-1} + 1 \times 2^{-2} + 1 \times 2^{-3}
   $$
   - Calculating each term:
     - $$ 0 \times 2^{-1} = 0 $$
     - $$ 1 \times 2^{-2} = \frac{1}{4} = 0.25 $$
     - $$ 1 \times 2^{-3} = \frac{1}{8} = 0.125 $$
   - Adding these values together:
   $$
   0 + 0.25 + 0.125 = 0.375
   $$

4. **Combine both parts**:
   - Now, we combine the integer part and the fractional part:
   $$
   5 + 0.375 = 5.375
   $$

Thus, the binary number $$ 101.011_{\text{two}} $$ is equal to $$ 5.375_{\text{ten}} $$ in decimal (base ten).
$$ 101.011_{\text{two}} = 5.375_{\text{ten}} $$

\( \begin{aligned} 101.011_{\text {two }} & =1 \times 2^{2}+0 \times 2^{1}+1 \times 2^{0}+0 \times 2^{-1}+1 \times 2^{-2}+1 \times 2^{3} \\ & =4+1+1 / 4+1 / 8 \text { in base ten } \\ & =5.375_{\text {ten }}\end{aligned} \)

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