. 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

• 1 mb=1024kb
  4.5 mb=4608kb
  
  time reuired=768 seconds
• 10 years agoHelpfull: Yes(17) No(1)
• 4.5 MB = 1024*4.5 = 4608 KB
  
  transfer rate is 6 KB per Second
  so time = 4608/6=768 seconds
• 10 years agoHelpfull: Yes(6) No(1)
• 1kbps=1 kilo bit per second
  So ans is 4.5*1024*8/6=6144sec
  
  
• 10 years agoHelpfull: Yes(6) No(4)
• 4.5*1024/6
  768 sec
  
• 10 years agoHelpfull: Yes(2) No(0)
• 1 MB=1024KB.
  4.5 MB=4608KB.
  6KB/S.
  Than
  TIME=4608/6=786S.
• 10 years agoHelpfull: Yes(2) No(1)
• (4.5*1024)/6 =768 sec
  
• 10 years agoHelpfull: Yes(2) No(0)
• 1mb = 1024kb
  so 4.5mb= 1024*4.5=4608kb
  time neded= 4608/6=768 sec
• 10 years agoHelpfull: Yes(2) No(0)
• ans.12min,48sec.
  4.5*1024=4608/6
  768 sec
  768/60 =12min,48sec
  
• 10 years agoHelpfull: Yes(2) No(0)
• 1mb=1024kb
  so
  4.5 mb=(4.5*1024)kb
  now fur 6 kb takes 1 second
  therefore (4.5*1024) takes (4.5*1024/6)=768sec
  =12min 48 sec
• 10 years agoHelpfull: Yes(1) No(0)
• size of software=4.5MB=4.5*1024KB=4608KB
  Data Transfer Rate=6kbps
  So time required for downloading=4608/6=768s
• 10 years agoHelpfull: Yes(0) No(0)