The Gateway to Computer Science Excellence
First time here? Checkout the FAQ!
x
+1 vote
20 views

Help me out with this, any sort of help appreciated.

 

What data rate is needed to transmit an uncompressed 4" x 6" photograph every second with a resolution of 1200 dots per inch and 24 bits per dot (pixel)?

OPTIONS:

691,200 kb/s 

28.8 kb/s

8.29 Mb/s

829 Mb/s

asked in Computer Networks by (225 points) | 20 views

1 Answer

+1 vote
Total number of $dots/inch  = 1200$

In $4" \rightarrow 4 * 1200$ $dots$

In $6" \rightarrow 6 * 1200$ $dots$

Total number of dots in $ 4"$ x $ 6"$ $= 4*1200 *6*1200$ $dots$

Total number of bits  in $ 4"$ x $6"$ $= 4*1200 *6*1200 * 24$ $bits$              $\because$ $[ 1 dot = 24 bits]$

data rate = total number of bits $/$ time taken

              $=$  $4*1200 *6*1200 * 24$ bits $/ $$1$ sec                $\because$ [transmit an uncompressed 4"x 6" photograph every second]

             $ = 829440000$ bits$/$sec

             $ = 829.44$ Mb$/$s

closest option is $829$ Mb$/$s
answered by Loyal (6.3k points)
0
Thanks.


Quick search syntax
tags tag:apple
author user:martin
title title:apple
content content:apple
exclude -tag:apple
force match +apple
views views:100
score score:10
answers answers:2
is accepted isaccepted:true
is closed isclosed:true

34,851 questions
41,835 answers
119,102 comments
41,454 users