Answer:
a. 1.6 Kbytes/min
b. 10.417 Mbytes/min
Explanation:
a. The DSL modem operates at 768 Kbits/sec.
But,
8 bits = 1 byte and 1 Kbit = 1 000 bits, so that:
= [tex]\frac{768 000}{8}[/tex]
= 96 000 bytes
Therefore, the modem operates at 96 Kbytes/sec.
The byte to be received in 1 minute can be calculated thus;
Since 60 seconds = 1 minute, then:
= [tex]\frac{96000}{60}[/tex]
= 1600
= 1.6 Kbytes/min
The modem receives 1.6 Kbytes/min
b. The USB sends 5 Gbits/sec.
But, 8 bits = 1 byte and 1 Gbit = 1000000000 bits so that:
= [tex]\frac{5000000000}{8}[/tex]
= 625000000
= 0.625 Gbytes
The USB sends 0.625 Gbyte/sec.
Since 60 seconds = 1 minute, then:
= [tex]\frac{625000000}{60}[/tex]
= 10416666.67
= 10.417 Mbytes/min
