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Two cameras are streaming 6 Mbit/s each. Approximately how much storage is required to record one hour of video from both streams?
Correct Answer: B
To calculate the storage required to record one hour of video from two cameras streaming at 6 Mbit/s each, we use the following steps: * Total bitrate for both cameras: 6 Mbit/s×2=12 Mbit/s6 \text{ Mbit/s} \times 2 = 12 \text{ Mbit/s}6 Mbit/s×2=12 Mbit/s * Convert Mbit/s to Mbyte/s: 12 Mbit/s÷8=1.5 Mbyte/s12 \text{ Mbit/s} \div 8 = 1.5 \text{ Mbyte/s}12 Mbit/s÷8=1.5 Mbyte/s * Calculate the total data for one hour: 1.5 Mbyte/s×3600 seconds=5400 Mbytes1.5 \text{ Mbyte/s} \times 3600 \text{ seconds} = 5400 \text{ Mbytes}1.5 Mbyte/s×3600 seconds=5400 Mbytes * Convert Mbytes to Gbytes: 5400 Mbytes÷1024=5.27 Gbytes5400 \text{ Mbytes} \div 1024 = 5.27 \text{ Gbytes}5400 Mbytes÷1024=5.27 Gbytes Approximately, the storage required is 5.4 GB. Reference: Axis Communications. "Calculating storage requirements." Axis Storage Calculation.