What is the storage requirement of a bit mappedimage of 800 X 1200 pixels with 24 bits perpixel?

Question

asked Oct 27, 2020 196k views

1 Answer

RonanC · Answer 1 · 2020-11-03T19:13:09+0000

Answer:

The total storage requirement is 2.74 Megabytes ( 2.74 Mb )

Step-by-step explanation:

The total number of pixels are

$800*1200 = 960000\, pixels$

As there are 24 bits per pixel,

the total number of bits are calculated as

$800 * 1200 * 24 \, bits = 23,040,000 \, bits$

As,
$1\, Byte = 8 \,bits , \, 1\, Kilobyte\, = 1,024\, Bytes, \, 1 \, Megabyte \, = 1,048,576 Bytes$

The total storage requirement is 2.74658203125 Megabytes.