From Peak-to-Peak Voltage:
$$ V_P = \frac{1}{2} \cdot V_{pp} = 0.5 \cdot V_{pp} $$
From RMS Voltage:
$$ V_P = \sqrt{2} \cdot V_{rms} \approx 1.414 \cdot V_{rms} $$
From Average Voltage:
$$ V_P = \frac{\pi}{2} \cdot V_{avg} \approx 1.571 \cdot V_{avg} $$
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