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hoa [83]
3 years ago
5

A trust domain is defined as Select one: a. The agreed upon, trusted third party b. A scenario where one user needs to validate

the other’s certificate c. A construct of systems, personnel, applications, protocols, technologies, and policies that work together to provide a certain level of protection d. A scenario in which the certificate’s issuer and subject fields hold the same information
Computers and Technology
2 answers:
soldier1979 [14.2K]3 years ago
7 0

Answer:

A construct of systems, personnel, applications, protocols, technologies, and policies that work together to provide a certain level of protection

Explanation:

Creation of dynamic Trust Domains can be realized through collaborative, secure sharing solutions. The local system is able to trust a domain to authenticate users. If an application or a user is authenticated by a trusted domain, all domains accept the authenticating domain. For instance, if a system trust domain X, it will also trust all domains that domain X trusts. It is a two-way trust relationship and is transitive.

RUDIKE [14]3 years ago
3 0

Answer:

A construct of systems, personnel, applications, protocols, technologies and policies that work together to provide a certain level of protection

Explanation:

A trust domain is a certificate that entities or users gain and is granted by a trusted domain this is that when you enter another software or other systems you will be recognized as a trusted character or user and gives protection between systems and softwares and provides protection for the users.

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Class.object.method is an example of using the ______<br> to access the elements of a class.
Aleonysh [2.5K]

Answer:

dot operator

I hope I helped ^-^

6 0
2 years ago
How many bits can a memory chip with the below configuration support? For full credit, show how you got the answer.
tatiyna

Answer:

Total Memory= 4 KB = 4096 bytes = 32768 bits

Explanation:

<em><u>1. Data lines are 8 From D0 to D7</u></em>

so

Total memory at single address locations is 8 bits.

<em><u>2. Address lines are 12 (A0 to A11)</u></em>

There are 12 address lines but 3 out 12 are for selction of chip or memory bank.

so only 9 pins are there to address the locations on one chip.

Total No. of address locations on single chip = 2^9 = 512 locations

as 1 location is 1 byte so total memory of single chip is 512 bytes.

<u><em>3. Total Memory Bank </em></u>

There are total 3 selection pins for memory bank.

so

Total chips = 2^3 = 8.

<em><u>4. Total Memory </u></em>

Total size of 1 chip = 512 bytes

Total size of 8 chip = 8x512 bytes = 4096 bytes = 4096/1024 kb = 4 kb

<em>So total memory of system is 4 Kb = 4096 bytes = 32768 bits</em>

5 0
3 years ago
Given an array arr, of type int, along with two int variables i and j, write some code that swaps the values of arr[i] and arr[j
Inessa05 [86]

You'll need a helper variable for this, so depending on your programming language, the solution becomes:

int helper;

helper = arr[i];

arr[i] = arr[j];

arr[j] = helper;

3 0
3 years ago
Write a procedure named Str_find that searches for the first matching occurrence of a source string inside a target string and r
kirill115 [55]

Answer: Provided in the explanation section

Explanation:

Str_find PROTO, pTarget:PTR BYTE, pSource:PTR BYTE

.data

target BYTE "01ABAAAAAABABCC45ABC9012",0

source BYTE "AAABA",0

str1 BYTE "Source string found at position ",0

str2 BYTE " in Target string (counting from zero).",0Ah,0Ah,0Dh,0

str3 BYTE "Unable to find Source string in Target string.",0Ah,0Ah,0Dh,0

stop DWORD ?

lenTarget DWORD ?

lenSource DWORD ?

position DWORD ?

.code

main PROC

  INVOKE Str_find,ADDR target, ADDR source

  mov position,eax

  jz wasfound           ; ZF=1 indicates string found

  mov edx,OFFSET str3   ; string not found

  call WriteString

  jmp   quit

wasfound:                   ; display message

  mov edx,OFFSET str1

  call WriteString

  mov eax,position       ; write position value

  call WriteDec

  mov edx,OFFSET str2

  call WriteString

quit:

  exit

main ENDP

;--------------------------------------------------------

Str_find PROC, pTarget:PTR BYTE, ;PTR to Target string

pSource:PTR BYTE ;PTR to Source string

;

; Searches for the first matching occurrence of a source

; string inside a target string.

; Receives: pointer to the source string and a pointer

;    to the target string.

; Returns: If a match is found, ZF=1 and EAX points to

; the offset of the match in the target string.

; IF ZF=0, no match was found.

;--------------------------------------------------------

  INVOKE Str_length,pTarget   ; get length of target

  mov lenTarget,eax

  INVOKE Str_length,pSource   ; get length of source

  mov lenSource,eax

  mov edi,OFFSET target       ; point to target

  mov esi,OFFSET source       ; point to source

; Compute place in target to stop search

  mov eax,edi    ; stop = (offset target)

  add eax,lenTarget    ; + (length of target)

  sub eax,lenSource    ; - (length of source)

  inc eax    ; + 1

  mov stop,eax           ; save the stopping position

; Compare source string to current target

  cld

  mov ecx,lenSource    ; length of source string

L1:

  pushad

  repe cmpsb           ; compare all bytes

  popad

  je found           ; if found, exit now

  inc edi               ; move to next target position

  cmp edi,stop           ; has EDI reached stop position?

  jae notfound           ; yes: exit

  jmp L1               ; not: continue loop

notfound:                   ; string not found

  or eax,1           ; ZF=0 indicates failure

  jmp done

found:                   ; string found

  mov eax,edi           ; compute position in target of find

  sub eax,pTarget

  cmp eax,eax    ; ZF=1 indicates success

done:

  ret

Str_find ENDP

END main

cheers i hoped this helped !!

6 0
3 years ago
Several students are making presentations for a science class. The chart describes the concept that each student will present to
Sedbober [7]

Question

Student presentation concepts from another source

Letitia - a step-by-step process of breaking down glucose into energy

Mana - repeating process of the steps of the cell division, going from interphase to division, and back to interphase.

Paul - a series of organisms that show similar characteristics and compare them to those that do not

Which best describes the SmartArt graphic layout each student should use?

Answer:

Letitia would use a process, Maria would use a cycle, and Paul would use a relationship.

Explanation:

Letita's assignment requires her to show how glucose can be broken down into energy, this requires a series of steps in s linear fashion, meaning it would be best suited to a process-style diagram.

Marta's project involves showing the steps of the cell cycle: the clue is in the name. This process is not linear, and instead repeats itself, meaning it fits to a cycle diagram (i.e. a circular diagram)

Paul is comparing and contrasting different organisms. Therefore, he should use a relationship diagram to show shared characteristics as well as features that differ.

4 0
3 years ago
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