How much time is reqd for downloading the software?
i. The Data transfer rate is 6 kbps
ii. The size of the software is 4.5 megabytes

• size of the software=4.5mb
  =4.5*1024=4608kb (since 1 mb=1024kb)
  data transfer rate=6kbps (given)
  so, time required for downloading the softwre
  =4608/6
  =768 sec
  =12 mints 48 secs
  
• 12 years agoHelpfull: Yes(54) No(2)
• 1 MB = 1024 KB
  so,4.5MB = 1024*4.5 = 4608 KB
  SO,if Data transfer rate is 6KB per second
  then 4608/6 = 768seconds
  = 12min and 8 sec
  
• 12 years agoHelpfull: Yes(14) No(16)
• 6kbps means 6 kilo bits per sec
  refer google for expansion of kbps
  'b' is lower case...so it stands for bits not byte...as per my knowledge goes
  so convert 4.5 mega bytes into kbps....4.5megabytes=4.5*1024k*8
  by dividing abovr value with 6kbps we get 6*1024=6144
• 12 years agoHelpfull: Yes(11) No(7)
• SIZ OF SFTWR=4.5*1024=4608 BECAUSE 1MB =1024KB
  THEN TIME TO TRASFER DT=4608/6=768(BECAUSE DT TRSFR RATE IS 6 ACORDING TO QUES)
  THEN 768 SEC IS CONVERT IN MINIT IS 768/60=12 MIN ANDD 48 SEC.............
• 12 years agoHelpfull: Yes(6) No(2)
• This Question is related to data sufficiency...
  
  both (i)(ii) is needed
• 10 years agoHelpfull: Yes(5) No(0)
• answer is 12min and 8 sec
• 12 years agoHelpfull: Yes(2) No(9)
• 
  1 MB = 1024 KB
  so,4.5MB = 1024*4.5 = 4608 KB
  SO,if Data transfer rate is 6KB per second
  then 4608/6 = 768seconds
  = 12min and 8 sec
• 12 years agoHelpfull: Yes(1) No(9)
• 1mb=1024kb
  so
  4.5mb=1024*4.5=4608kb
  
  because 6kb in 60 sec so time required to transfer data
  4608/6=768 sec
  now 768/60=12m and 8 sec
• 7 years agoHelpfull: Yes(0) No(0)