Interested to Buy Any Domain ? << Click Here >> for more details...
A packet of 10 Kb is to be downloaded from a web server.
Find the time needed to download this packet using:
a) A dial up telephone connection at 28 Kbps
b) A cable modem at 28 Mbps.
- 5 Answers
- 9792 Views
- Ericsson, I also Faced
- E-Mail Answers
Answer Posted / john
I agree this is the correct one.
10kb=10*1024bytes=10240 bytes
28Kilobitsps=28/8KiloBytesps=3.5KBps=3.5*1024Bytesps=3584Bps
time=10240/3584=2.85 s
do same with 28.8Mbps=28.8*1024*1024.
| Is This Answer Correct ? | 2 Yes | 3 No |
Post New Answer View All Answers
What is the j std 001?
Is arp a tcp or udp?
What is the port number of telnet and dns?
What is Tcp/ip Model?
What is the advantage of packet switching?
How to identify ip address?
What are the services provided by Session layer?
What is port 8080 typically used for?
What is the differance between HTTP and TCP/IP Protocols?
What is an ip address?
Is port 80 open on my network?
What is the use of tcp/ip?
What is the core naming mechanism, domain name system (dns)?
What is the equivalent layer or layers of the tcp/ip application layer in terms of osi reference model?
What is codec ? What is the use of that?
