Given an 8-bit grayscale bitmap that is 256 x 512, how many bytes will be requir

Question

Given an 8-bit grayscale bitmap that is 256 x 512, how many bytes will be required to hold the image data only? If it is an 8-bit color bitmap? If it is a 24-bit bitmap?

Given a mono wave file that is 8 bits per sample and is sampled at 16000 samples/second, how many bytes are required to hold the audio data if the message is 2 minutes long? Given a 16-bit stereo wave file?

A 16-bit grayscale file has _____ levels of luminance.

If an image has 4 bits for each RGB channel, how many different color combinations are possible?

Explanation / Answer

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1. 8-bit grayscale bitmap image and image size - 256x512

Bytes per pixel = 1 byte

256x512 bytes = 131072 bytes.

If it is 24-bit bitmap image = 3x131072 = 393216 bytes.

The monowave sample rate = 8bits/sample

Sampling rate = 16000 samples/second

Bits per second comes out to be = 8*16000 = 128000 = 128kbps.

Bytes = 128000*2*60/8 = 1920000 = 1920KB = 1.875 MB

Now, if it is a stereo wave

Bits per second comes out to be = 16*16000 = 256000 = 256kbps

Bytes = number of channels * bits persond * times(in sec) / 8 = 2*(256000*2*60/8) = 2*3840000 = 7680KB = 7.5MB

A 16-bit grayscal file has 65536 levels of luminance

If an image has 4 bits for RGB channel then it will hav 24 total number of combination for each of the channel. That is it will have 16 different combination for R, G, B. Total number of combination 16x16x16 = 4096 combinations.

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