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November 26, 2014

November 26, 2014

Posted by **freddo** on Thursday, January 13, 2011 at 3:07pm.

Can someone check this for me please..

18 MB of data are to be transmitted over a a megabyte per second connection. How long should the complete transaction take, to the nearest 0.1s.

Size of file / data rate so,

18MB x 10^6 x 8 bits = 144,000,00

4 megabits per second = 4 x 10^6 = 4,000,000 bits per second

so the download time will be

144,000,000 / 4,000,000 = 36 seconds

my answer is 36 seconds is that right

next ,....

a 50 MB file compressed using a ratio of 10 what is the compressed file size

50MB / 10 = 5 MB

Is this correct ???

Thanks you

- IT computer memory -
**PC**, Thursday, January 13, 2011 at 4:30pmYou have to double check the question:

"18 MB of data are to be transmitted over a a megabyte per second connection"

Is it "a" mega*byte*connection" or is it "4 mega*bit*connection"?

- IT computer memory -
**freddo**, Thursday, January 13, 2011 at 5:15pmHi

Sorry wasnt too clear

This should read 18MB of data are to be transmitted over a 4 megabits per second connection ... how long would it take to complete the transaction

- IT computer memory -
**freddo**, Thursday, January 13, 2011 at 5:15pmto the nearest 0.1 s

- IT computer memory -
**Ga**, Tuesday, February 1, 2011 at 4:58pmFor the second part of your question...

Compressed file size = uncompressed / compression ratio

so

Compressed file size = 50MB / 10

Compressed file size = 5MB

I think

- IT computer memory -
**Andy**, Sunday, February 13, 2011 at 10:13amNo, incorrect.

Be carefull with this, your:

18MB x 10^6 x 8 bits = 144,000,00

Should in fact be:

18mb x 2^20 x 8 bits = 150,994,994

Your second part is correct, so its

4 megabits per second = 4 x 10^6 = 4,000,000

Answer:

150994994 / 4000000 = 37.75 seconds.

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