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stira [4]
2 years ago
6

Helppppp ASAPPPP!!!!

Computers and Technology
1 answer:
Kobotan [32]2 years ago
5 0

Answer:

I think it might be archive the less used data

Explanation:

archive doesnt mean delete but it does put away whatever you dont want to look at, so she still has all of the data with more space.

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"If someone really wants to get at the information, it is not difficult if they can gain physical access to the computer or hard
Jobisdone [24]

Answer:

It is  just true in most cases

Explanation:

This statement is just true in most cases because in some cases having a physical access to the computer or hard drive this does not guarantee access to the information contained in the hard drive or computer because the data/information in the hard drive or computer might as well be encrypted and would require a password in order to de-crypt the information.

Having a physical access is one step closer in some cases but not the only step required .

4 0
3 years ago
The average lease payment for a new vehicle is just over $450 per month for a three-year...
hichkok12 [17]

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it depends on your insurance company

Explanation:

7 0
3 years ago
Which of the following is true about simulation games? A. Simulation games involve competing in a sport against other players. B
WARRIOR [948]
<span>Simulation games recreate a real-world environment. </span>
6 0
3 years ago
Read 2 more answers
The use of space bar​
Anna11 [10]
The question is unclear or unfinished please send again the question
5 0
2 years ago
Compilers can have a profound impact on the performance of an application. Assume that for a program, compiler A results in a dy
Naddik [55]

Answer:

The answer is "1.25"

Explanation:

{CPU \ time}= {instructions \times  CPI \times  cycle\  time}

\therefore\\\CPI= \frac{CPU \ time}{instructions \times cycle \ time} \\\\cycle \ time = 1 \\\\ns = 10^{-9} s \\

Also for this context, it executes the time =  CPU time. So, the compiler A, we has

CPI_{A}= \frac{CPU \ time_{A}} {instructions_{A} \times cycle \ time}= \frac{1.1 s}{10^{9} \times 10^{-9} s}= 1.1

For compiler B, we have

CPI_{B}= \frac{CPU \ times_{B}} {instructions_{B} \times cycle \ time}

         = \frac{1.5\ s}{ 1.2 \times 10^{9} \times 10^{-9} \ s}\\\\= \frac{1.5}{ 1.2 }\\\\= 1.25

3 0
3 years ago
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