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Fiesta28 [93]
3 years ago
9

Determine the number of bytes necessary to store an uncompressed binary image of size 4000 × 3000 pixels.

Computers and Technology
1 answer:
miss Akunina [59]3 years ago
5 0

Answer:

12,000,000 bytes.

Explanation:

If we want to store an uncompressed binary image, assuming that we will use 256 different levels to represent all tones in a monochromatic image, each pixel will be stored as an 8-bit sample, i.e. one byte per pixel.

The complete image will have 4000 pixels wide by 3000 pixels height, so an entire image will need 3000*4000= 12,000,000 pixels.

As we will use 8 bits= 1 Byte per pixel, we will need to store 12,000, 000 bytes for a complete uncompressed monochromatic (white & black) image.

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Dakota's friend Stephen created a cypher using the QWERTY keyboard, but Dakota is confused. Stephen said to move one to the righ
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Answer:

people are so confusing nowadays... Mr. Kopin told me to get him a coffee. I'm not his servant!

Explanation: QWERTY Q=P A=L Z=M and the others are what is to the left of them.

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3 years ago
What are the possible consequences of plagiarism?
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Answer:

Students who plagiarize or otherwise engage in academic dishonesty face serious consequences. Sanctions may include, but are not limited to, failure on an assignment, grade reduction or course failure, suspension, and possibly dismissal.

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3 years ago
Consider sending a 2400-byte datagram into a link that has an mtu of 700 bytes. suppose the original datagram is stamped with th
Feliz [49]

Explanation:

Let, DG is the datagram so, DG= 2400.

Let, FV is the Value of Fragment and F is the Flag and FO is the Fragmentation Offset.

Let, M is the MTU so, M=700.

Let, IP is the IP header so, IP= 20.

Let, id is the identification number so, id=422

Required numbers of the fragment = [\frac{DG-IP}{M-IP} ]

Insert values in the formula = [\frac{2400-20}{700-20} ]

Then,        = [\frac{2380}{680} ] = [3.5]

The generated numbers of the fragment is 4

  • If FV = 1 then, bytes in data field of DG= 720-20 = 680 and id=422 and FO=0 and F=1.
  • If FV = 2 then, bytes in data field of DG= 720-20 = 680 and id=422 and FO=85(85*8=680 bytes) and F=1.
  • If FV = 3 then, bytes in data field of DG= 720-20 = 680 and id=422 and FO=170(170*8=1360 bytes) and F=1.
  • If FV = 4 then, bytes in data field of DG= 2380-3(680) = 340 and id=422 and FO=255(255*8=2040 bytes) and F=0.

3 0
3 years ago
The ability of a language to let a programmer develop a program on computer system that can be run on other systems is called
kakasveta [241]
It is called 'portability'.
6 0
3 years ago
When is it most appropriate to send an automatic reply? Check all that apply.
konstantin123 [22]

Answer:

2, 4, 5, 6 are correct

Explanation:

1 and 3 are wrong on edg

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3 years ago
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