To determine how much power remains at the output of a cable with a loss of 6 dB, we can use the formula that relates decibels to power:
\[ \text{Loss (dB)} = 10 \times \log_{10} \left( \frac{P_{\text{input}}}{P_{\text{output}}} \right) \]
Given that the loss is 6 dB, we can rearrange the formula to find the ratio of input power to output power:
\[ 6 = 10 \times \log_{10} \left( \frac{P_{\text{input}}}{P_{\text{output}}} \right) \]
Dividing both sides by 10:
\[ 0.6 = \log_{10} \left( \frac{P_{\text{input}}}{P_{\text{output}}} \right) \]
Now, we exponentiate both sides:
\[ 10^{0.6} = \frac{P_{\text{input}}}{P_{\text{output}}} \]
Calculating \(10^{0.6}\):
\[ 10^{0.6} \approx 3.981 \]
This indicates that:
\[ \frac{P_{\text{input}}}{P_{\text{output}}} \approx 3.981 \]
To find the output power as a percentage of the input power, we can express it as:
\[ \frac{P_{\text{output}}}{P_{\text{input}}} = \frac{1}{3.981} \approx 0.251 \]
Converting this to a percentage:
\[ \frac{P_{\text{output}}}{P_{\text{input}}} \times 100 \approx 25.1% \]
Thus, the percentage of input power that appears at the output is approximately 25%.
So the correct choice from the given options is:
25%