A 1200KB file is transferred over a connection in 20 seconds. Determine the file size (in KB) that can be transferred

A 1200KB file is transferred over a connection in 20 seconds. Determine the file size (in KB) that can be transferred through this connection in 30 seconds.

A 1200 KB file is skipped in 20 seconds. We need to find out how much information will pass in 30 seconds. To do this, let’s make a proportion based on the bandwidth of the channel s / kB:
1200/20 = x / 30
x = 30 * 1200/20 = 1800kBytes.
Answer: 1800kB will pass in 30 seconds



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