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gogolik [260]
4 years ago
8

the easiest and cheapest time to alter the information system is in the ________ phase of the sdlc. a. requirements analysis b.

system definition c. component design d. implementation
Computers and Technology
1 answer:
nignag [31]4 years ago
7 0
I would think its the "<span>requirements analysis" phase.

Because the </span>requirements analysis<span>,</span><span> is the process of determining user expectations for a new or modified product.</span>
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What are the first and the last physical memory addressesaccessible using
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Answer

For First physical memory address,we add 00000 in segment values.

For Last physical memory address,we add 0FFFF in segment values.

<u>NOTE</u>-For addition of hexadecimal numbers ,you first have to convert it into binary then add them,after this convert back it in hexadecimal.

a)1000

For First physical memory address, we add 00000 in segment value

We add 0 at the least significant bit while calculating.

          1000<u>0</u> +00000 = 1000<u>0</u> (from note)

For Last physical memory address,We add 0 at the least significant bit while calculating.

   we add 0FFFF in segment value

           1000<u>0</u> +0FFFF =1FFFF    (from note)

b)0FFF

For First physical memory address,we add 00000 in segment value

We add 0 at the least significant bit while calculating.

     0FFF<u>0</u> +00000=0FFF0 (from note)

For Last physical memory address,We add 0 at the least significant bit while calculating.

   we add 0FFFF in segment value

           0FFF<u>0</u> +0FFFF =1FFEF (from note)

c)0001

For First physical memory address,we add 00000 in segment value

We add 0 at the least significant bit while calculating.

     0001<u>0</u> +00000=00010 (from note)

For Last physical memory address,We add 0 at the least significant bit while calculating.

   we add 0FFFF in segment value

           0001<u>0</u> +0FFFF =1000F (from note)

d) E000

For First physical memory address,we add 00000 in segment value

 We add 0 at the least significant bit while calculating.

    E000<u>0</u> +00000=E0000 (from note)

For Last physical memory address,We add 0 at the least significant bit while calculating.

   we add 0FFFF in segment value

           E000<u>0</u> +0FFFF =EFFFF (from note)

e) 1002

For First physical memory address,we add 00000 in segment value

 We add 0 at the least significant bit while calculating.

    1002<u>0</u> +00000=10020  (from note)

For Last physical memory address,We add 0 at the least significant bit while calculating.

   we add 0FFFF in segment value

           1002<u>0</u> +0FFFF =2001F (from note)

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