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Nady [450]
2 years ago
8

Ciscon Telecom is a mobile operator in the European Union. The company provides personalized services to its customers, and its

databases contain valuable information about the customers. The loss of customer information that is used to decide services would be extremely harmful to the organization. Which of the following strategies adopted by Ciscon is an example of risk transference
A. The company insures any possible data loss for a large sum.
B. It installs a corporate firewall to protect unauthorized access to information.
C. The company forms a special team of top executives to monitor and correct the information policies.
D. The company decides to absorb any damages that might occur.
E. It enforces a strict employee data policy and prohibits employees from unauthorized access.
Computers and Technology
2 answers:
Dmitry [639]2 years ago
8 0

Answer:

A. The company insures any possible data loss for a large sum.

Explanation:

Risk transference is the transferring of risk to a third party. It shift the burden that arise from lose of data from one party to another party through legislation or contract agreement. It is a common risk management techniques.

In this question, Ciscon telecom would be transferring the risk of losing data to a third party.

LuckyWell [14K]2 years ago
7 0

Answer:

A. The company insures any possible data loss for a large sum.

Explanation:

Risk Transference  is a common risk management technique which involves shifting of the burden of loss for a risk to another party through legislation, contract, insurance or other means.

In the case loss of customer information that is used to decide services which would be extremely harmful to Ciscon,  Ciscon will need to insure any possible data loss for a large sum as an adopted strategy of risk transference.

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Consider the following code segment.
m_a_m_a [10]

Answer:

10 20                                                                                                                          

30 40                                                                                                                          

50 60

Explanation:

Given

The above code segment

Required

What is printed, if anything

To do this, we analyze the code line by line.

Line 1: The first line creates a 4 by 3 array named mat

Line 2: for (int[] row : mat) {  -> This creates row[] array which represents each row of array mat

Line 3: for (int j = 0; j < row.length; j += 2){ -> This iterates through the even indexed elements of the row array i.e. 0 and 2

Line 4: System.out.print(row[i] + " "); -> This prints the even indexed elements of the row array.

The even indexed elements are: 10, 20, 30, 40, 50 and 60

Line 5: System.out.println(); --> This prints a new line

8 0
2 years ago
Recall that with the CSMA/CD protocol, the adapter waits K. 512 bit times after a collision, where K is drawn randomly. a. For f
julia-pushkina [17]

Complete Question:

Recall that with the CSMA/CD protocol, the adapter waits K. 512 bit times after a collision, where K is drawn randomly. a. For first collision, if K=100, how long does the adapter wait until sensing the channel again for a 1 Mbps broadcast channel? For a 10 Mbps broadcast channel?

Answer:

a) 51.2 msec.  b) 5.12 msec

Explanation:

If K=100, the time that the adapter must wait until sensing a channel after detecting a first collision, is given by the following expression:

  • Tw = K*512* bit time

The bit time, is just the inverse of the channel bandwidh, expressed in bits per second, so for the two instances posed by the question, we have:

a) BW  = 1 Mbps = 10⁶ bps

⇒ Tw = 100*512*(1/10⁶) bps = 51.2*10⁻³ sec. = 51.2 msec

b) BW = 10 Mbps = 10⁷ bps

⇒ Tw = 100*512*(1/10⁷) bps = 5.12*10⁻³ sec. = 5.12 msec

5 0
2 years ago
What is the data transmission time, assuming no errors or overhead, to send a 2 GByte file over a link with a 2 Gbps line rate?
schepotkina [342]

Answer:

The data transmission time will be 8 seconds (8.0s)

Explanation:

Using the formular:

T =A/S

T=Time = ?

Amount of Data = A = 2GByte

Speed = S = 2Gbps

The units of A and S are different

A is expressed in Bytes(B)

S is expressed in bits(b)

But :

1 Byte(B) = 8 bits(b)

T = A/S = 2GB/2Gb

since 1 Byte = 8 bits

T = 2 × G ×( B =8bits) / 2 × G × b

T = 2 × G × 8b / 2 × G × b

T = 2 × G × 8 × b / 2 × G × b

The common terms(2,G,b) cancel out

T = 8 seconds

4 0
2 years ago
Suppose you are given three data items that indicate a starting index in a list, an array of characters, and an array of link in
Artemon [7]

Answer:

see explaination

Explanation:

INCLUDE Irvine32.inc

;declare the variables

SPACE = 32

startIndex = 1

.data

statement BYTE "The correct sequence of array: ",0

; declare the chars('H','A','C','E','B','D','F','G')

chars BYTE 48h,41h,43h,45h,42h,44h,46h,47h

sizeOfArray = ($ - chars)

links DWORD 0,4,5,6,2,3,7,0

outputArray BYTE sizeOfArray DUP(?)

.code

main PROC

; assign starting index

mov eax,startIndex

; assign offset of array chars

mov ebx,OFFSET chars

; assign the size of the array

mov ecx, sizeOfArray

; assign the offset of array links

mov edi,OFFSET links

; assign the offset of new array

mov esi,OFFSET outputArray

; call the traverseLink procedure

call traverseLink

; display the characters of new array

mov edx,OFFSET statement

call writeString

; assign the offset of new array

mov esi,OFFSET outputArray

; assign the size of the array

mov ecx, sizeOfArray

; call the procedure printArray to print the ordered array

call printArray

exit

main ENDP

traverseLink PROC

forLoop:

; get the character at index eax

mov dl,[ebx+eax]

; assign it to new array

mov [esi] , dl

; get the value at the next index

mov eax,[edi+eax*4]

; goto next location of new array

inc esi

; repeat until ecx=0

loop forLoop

ret

traverseLink ENDP

printArray PROC

forLoop:

; get the character at index eax

mov al,[esi]

; print the character

call WriteChar

; increament esi to get next character

inc esi

; print the space character

mov al,SPACE

call WriteChar

loop forLoop

call Crlf

ret

printArray ENDP

END main

7 0
2 years ago
. Two blue armies are each poised on opposite hills preparing to attack a single red army in the valley. The red army can defeat
jeka57 [31]

Answer:No

Explanation:

Put no because there is nothing that she can do to stop it.

8 0
2 years ago
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