114. The total transfer time for a disk I/O operation magnetic is equal to 30 msec. Knowing that: the overall seek time is equal to 18 msec, the overall rotational delay is equal to 7 msec and that the transfer rate is equal to 1.5 Gbit/sec, what is the total amount of data transferred? (Remember that 1 B = 1 byte = 8 bits and 1 MB = 10^3 KB = 10^6 BXtpat-4 e)

Database System Concepts
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ISBN:9780078022159
Author:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan
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Answer of the below question is 937.5KB but i don't know how he got it 

the formula used is i think 

data transfer time  = seek time + rotational delay + transfer time

114. The total transfer time for a disk I/O operation
magnetic is equal to 30 msec. Knowing that: the overall seek time is equal to 18
msec, the overall rotational delay is equal to 7 msec and that the transfer rate is
equal to 1.5 Gbit/sec, what is the total amount of data transferred? (Remember
that 1 B = 1 byte = 8 bits and 1 MB = 10^3 KB = 10^6 BXcutpat
tes)
Transcribed Image Text:114. The total transfer time for a disk I/O operation magnetic is equal to 30 msec. Knowing that: the overall seek time is equal to 18 msec, the overall rotational delay is equal to 7 msec and that the transfer rate is equal to 1.5 Gbit/sec, what is the total amount of data transferred? (Remember that 1 B = 1 byte = 8 bits and 1 MB = 10^3 KB = 10^6 BXcutpat tes)
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