Answer: 1,500,000 bytes.
Explanation:
If we assume that we have an image of 4000 pixels (picture elements) wide, by 3000 pixels height, a complete uncompressed image will be represented by 4000*3000= 12,000,000 pixels.
Now, if we are talking of a binary image, this means that each pixel will have, as a maximum, two possible values, so we will need only one bit per pixel.
This means that we will need to store 12,000,000 bits.
As we know, 1 byte=8 bits.
So, we will need 12,000, 000/8 bytes ⇒ 1,500,000 bytes in order to store an uncompressed binary image of size 4000 x 3000 pixels.
I think you should take a look into this link: https://www.wired.com/2009/12/ye-cybercrimes/
Here are the 10 most <span>destructive cybercrimes that you can take as areal example of cybercrime.</span>
Answer:
SONET was the only one on the list created near the 1980s (1985)
Good luck!
<em>~Awwsome</em>
The read-write type of memory helps in reading as well as writing data. This computer memory is used by users to continually update the data (to access (read from) or alter (write to) ) that is held on hardware storage devices. Internal or external hard disk drives, rewritable CDs or small flash drives can be all physical setups of read-write memory.
Answer:
1.
DIM myArray(10) as INTEGER
LET A = 0
FOR I = 1 TO 10 STEP 2
INPUT “INPUT NUMBER”; myArray(i)
LET A = A + myArray(i)
NEXT
PRINT A
END
2.
REM PROGRAM FOR CALCULATING THE SIMPLE INTEREST
CLS
INPUT “INPUT THE PRINCIPAL”; P
INPUT “INPUT THE TIME”; T
INPUT “INPUT THE RATE”;R
SI = P* T * R / 100
PRINT “SIMPLE INTEREST =”; SI
END
Explanation:
Please find the respective programs in the answer section.
