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saul85 [17]
2 years ago
7

Hisoka is creating a summary document for new employees about their options for different mobile devices. One part of his report

covers encryption. What would Hisoka NOT include in his document
Computers and Technology
1 answer:
algol132 years ago
3 0

Hisoka would not include in his document that is Apple users file-based encryption to offer a higher level of security.

<h3>Does Apple use the file based encryption?</h3>

It is said that iOS and iPad OS devices are known to often use a file encryption system known to be Data Protection.

Therefore,  Hisoka would not include in his document that is Apple users file-based encryption to offer a higher level of security.

Hence option A is correct.

Learn more about encryption from

brainly.com/question/9979590

#SPJ1

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c

Explanation:

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3 years ago
(a) How many locations of memory can you address with 12-bit memory address? (b) How many bits are required to address a 2-Mega-
I am Lyosha [343]

Answer:

Follows are the solution to this question:

Explanation:

In point a:

Let,

The address of 1-bit  memory  to add in 2 location:

\to \frac{0}{1}  =2^1  \ (\frac{m}{m}  \ location)

The address of 2-bit  memory to add in 4 location:

\to \frac{\frac{00}{01}}{\frac{10}{11}}  =2^2  \ (\frac{m}{m}  \ location)

similarly,

Complete 'n'-bit memory address' location number is = 2^n.Here, 12-bit memory address, i.e. n = 12, hence the numeral. of the addressable locations of the memory:

= 2^n \\\\ = 2^{12} \\\\ = 4096

In point b:

\to Let \  Mega= 10^6

              =10^3\times 10^3\\\\= 2^{10} \times 2^{10}

So,

\to 2 \ Mega =2 \times 2^{20}

                 = 2^1 \times 2^{20}\\\\= 2^{21}

The memory position for '2^n' could be 'n' m bits'  

It can use 2^{21} bits to address the memory location of 21.  

That is to say, the 2-mega-location memory needs '21' bits.  

Memory Length = 21 bit Address

In point c:

i^{th} element array addresses are given by:

\to address [i] = B+w \times (i-LB)

_{where}, \\\\B = \text {Base  address}\\w= \text{size of the element}\\L B = \text{lower array bound}

\to B=\$ 52\\\to w= 4 byte\\ \to L B= 0\\\to address  = 10

\to address  [10] = \$ 52 + 4 \times (10-0)\\

                       =   \$ 52   + 40 \ bytes\\

1 term is 4 bytes in 'MIPS,' that is:

= \$ 52  + 10 \ words\\\\ = \$ 512

In point d:

\to  base \ address = \$ t 5

When MIPS is 1 word which equals to 32 bit :

In Unicode, its value is = 2 byte

In ASCII code its value is = 1 byte

both sizes are  < 4 byte

Calculating address:

\to address  [5] = \$ t5 + 4 \times (5-0)\\

                     = \$ t5 + 4 \times 5\\ \\ = \$ t5 + 20 \\\\= \$ t5 + 20  \ bytes  \\\\= \$ t5 + 5 \ words  \\\\= \$ t 10  \ words  \\\\

3 0
3 years ago
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