A packet of 10 Kb is to be downloaded from a web server.
Find the time needed to download this packet using:
a) A dial up telephone connection at 28 Kbps
b) A cable modem at 28 Mbps.
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Answer Posted / john
I agree this is the correct one.
10kb=10*1024bytes=10240 bytes
28Kilobitsps=28/8KiloBytesps=3.5KBps=3.5*1024Bytesps=3584Bps
time=10240/3584=2.85 s
do same with 28.8Mbps=28.8*1024*1024.
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