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rewona [7]
3 years ago
10

Suppose you are the security manager of a company and one of your goals is to design security mechanisms based on three security

goals (1) prevent the attack (2) detect the attack or (3) recover from the attack. Depending on the situation or application, you have to adopt one of these security goals. For each of the following statements, give an example of an application or situation in which the statement is true.
(i). Prevention is more important than detection and recovery
(ii). Detection is more important than prevention and recovery
(iii). Recovery is more important than prevention and detection
Computers and Technology
1 answer:
Tcecarenko [31]3 years ago
8 0

Answer:

(i) Prevention is more important than detection and recovery.

(ii) Detection is more important than prevention and recovery.

(iii) Recovery is more important than prevention and detection.

Explanation:

(i) Prevention is more important than detection and recovery.

Prevention of attack can be through various applications for example a walk through gates are placed in order to prevent any attacker from entering the premises and causing harm.

(ii) Detection is more important than prevention and recovery.

Detection of an attack can be done through for example a security alarm can detect an attack and inform others.

(iii) Recovery is more important than prevention and detection.

Recovery of an attack can be done by for example an insurance which will recover a portion of loss occurred during the attack.

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computer has a 32-bit instruction word broken into fields as follows: opcode, six bits; two register file address fields, five b
lina2011 [118]

Answer:

a.  2^6, or 64 opcodes.

b.  2^5, or 32 registers.

c. 2^16, or 0 to 65536.

d.  -32768 to 32768.

Explanation:

a. Following that the opcode is 6 bits, it is generally known that the maximum number of opcodes should be 2^6, or 64 opcodes.

b. Now, since the size of the register field is 5 bits, we know that 2^5 registers can be accessed, or 32 registers.

c. Unsigned immediate operand applies to the plus/minus sign of the number. Since unsigned numbers are always positive, the range is from 0 to 2^16, or 0 to 65536.

d. Considering that the signed operands can be negative, they need a 16'th bit for the sign and 15 bits for the number. This means there are 2 * (2^15) numbers, or 2^16. However, the numbers range from -32768 to 32768.

6 0
4 years ago
_________ are the special effects that you see when one slide changes to another in slide show view​
lisabon 2012 [21]

Answer:

Transition effects

Explanation:

Transition effects are the special effects you see when one slide changes to another in slide show view.

8 0
3 years ago
When describing memory, ____________ is the first component required in the process necessary for retention?
8_murik_8 [283]
Depends on how deep you're willing to go to really,
You need one of a few arrangements of flip flop circuits to keep 1-bit state.
Going deeper, you need either NAND, or NOR gates(or a bunch of other ones) and connectors.
Even deeper, you'll require diodes or transistors to build the logic gates.
Beyond that is particle physics.
8 0
3 years ago
Write code for a function with the following prototype: /* Addition that saturates to TMin or TMax */ int saturating_add(int x,
Kazeer [188]

Answer:

See explaination

Explanation:

program code.

/* PRE PROCESSOR DIRECTIVES */

#include<stdio.h>

/* PRE-DEFINED VALUES FOR TMAX AND TMIN */

#define TMax 2147483647

#define TMin (-TMax -1)

/* saturating_add(int,int) METHOD IS CALLED HERE */

int saturating_add(int firstNumber, int secondNumber)

{

/*

FOR BETTER UNDERSTANDING, LETS TAKE TEST CASE,

WHERE firstNumber = 5 AND secondNumber = 10

*/

int w = sizeof(firstNumber) << 3;

/*

sizeof(firstNumber) VALUE IS 4, SO USING BINARY LEFT SHIFT OPERATOR TO THREE PLACES,

WE HAVE NOW VALUE 32, ASSIGNED TO w

*/

/* ADDITION IS CALCULATED => 15 */

int addition = firstNumber + secondNumber;

/*

MASK INTEGER VARIABLE IS TAKEN

mask BIT IS LEFT SHIFTED TO 31 PLACES => 2^31 IS THE NEW VALUE

*/

int mask = 1 << (w - 1);

/* FIRST NUMBER MOST SIGNIFICANT BIT IS CALCULATED BY USING AND OPERATOR */

int msbFirstNumber = firstNumber & mask;

/* SECOND NUMBER MOST SIGNIFICANT BIT IS CALCULATED BY USING AND OPERATOR */

int msbSecondNumber = secondNumber & mask;

/* MOST SIGNIFICANT BIT OF ADDITION IS CALCULATED BY USING AND OPERATOR */

int msbAddition = addition & mask;

/* POSITIVE OVERFLOW IS DETERMINED */

int positiveOverflow = ~msbFirstNumber & ~msbSecondNumber & msbAddition;

/* NEGATIVE OVERFLOW IS DETERMINED */

int negativeOverflow = msbFirstNumber & msbSecondNumber & !msbAddition;

/* THE CORRESPONDING VALUE IS RETURNED AS PER THE SATURATING ADDITION RULES */

(positiveOverflow) && (addition = TMax);

(negativeOverflow) && (addition = TMin);

return addition;

}

/* MAIN FUNCTION STARTS HERE */

int main(){

/* TEST CASE */

int sum = saturating_add(5, 10);

/* DISPLAY THE RESULT OF TEST CASE */

printf("The Sum Is : %d\n\n",sum);

}

7 0
4 years ago
I got a kahoot! Totally random pls join!<br><br> Code 08408519!
OlgaM077 [116]

Answer:

say less

Explanation:

5 0
3 years ago
Read 2 more answers
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