Answer:
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Answer:
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ok have a great day what is your name ?
Answer:
1.) 25 ; 15 ; 15
2.) 50 ; 15 ; 50
Explanation:
In the first function written :
The variable val was initially decaled or assigned a value of 25 and that was what was printed first.
However, after the example function was written, the val variable was finally assiagned a value of 15 within the function. However, it was also declared that the global variable takes uonthe val value. Hence, the val variable initially assigned a value, of 25 changes to 15 globally.
For the second code :
From the top:
Val was assigned a value of 50 ;
Hence,
print(val) gives an output of 50
Within the function definition which prints the value of val that is assigned a value of 25 within the function.
Since tbe global variable isnt reset.
Printing Val again outputs 50;since ito is outside the function.
I believe the answer might be addresses
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Answer:
2^11
Explanation:
Physical Memory Size = 32 KB = 32 x 2^10 B
Virtual Address space = 216 B
Page size is always equal to frame size.
Page size = 16 B. Therefore, Frame size = 16 B
If there is a restriction, the number of bits is calculated like this:
number of page entries = 2^[log2(physical memory size) - log2(n bit machine)]
where
physical memory size = 32KB which is the restriction
n bit machine = frame size = 16
Hence, we have page entries = 2^[log2(32*2^10) - log2(16)] = 2ˆ[15 - 4 ] = 2ˆ11
