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Answer
\(1011101_2\) in decimal is \(93\).
Solution
We are given the binary number
\[
1011101_2.
\]
We convert it to decimal by expanding it as a sum of powers of 2. Write the number with its digits and corresponding exponents:
\[
\begin{array}{ccccccc}
1 & 0 & 1 & 1 & 1 & 0 & 1 \\
2^6 & 2^5 & 2^4 & 2^3 & 2^2 & 2^1 & 2^0 \\
\end{array}
\]
Multiply each binary digit by its corresponding power of 2:
\[
\begin{aligned}
1 \times 2^6 &= 1 \times 64 = 64, \\
0 \times 2^5 &= 0 \times 32 = 0, \\
1 \times 2^4 &= 1 \times 16 = 16, \\
1 \times 2^3 &= 1 \times 8 = 8, \\
1 \times 2^2 &= 1 \times 4 = 4, \\
0 \times 2^1 &= 0 \times 2 = 0, \\
1 \times 2^0 &= 1 \times 1 = 1.
\end{aligned}
\]
Now, sum these values:
\[
64 + 0 + 16 + 8 + 4 + 0 + 1 = 93.
\]
Thus, the binary number \(1011101_2\) in decimal is
\[
93_{10}.
\]
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Simplify this solution Beyond the Answer
To convert the binary number \( 1011101_{2} \) to decimal, each digit is multiplied by 2 raised to the power of its position from the right (starting at 0). So we have: \[ 1 \times 2^6 + 0 \times 2^5 + 1 \times 2^4 + 1 \times 2^3 + 1 \times 2^2 + 0 \times 2^1 + 1 \times 2^0 \] Calculating each term gives us: \[ 1 \times 64 + 0 \times 32 + 1 \times 16 + 1 \times 8 + 1 \times 4 + 0 \times 2 + 1 \times 1 = 64 + 0 + 16 + 8 + 4 + 0 + 1 = 93 \] Thus, \( 1011101_{2} \) equals \( 93_{10} \) in decimal.
