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garik1379 [7]
3 years ago
13

The identification of critical information is a key part of the OPSEC process because: a) It is required by the JTF commander to

formulate operational COAs in the planning process. b) It is the fifth and final step of the OPSEC process. c) It focuses the remainder of the OPSEC process on protecting vital information rather than attempting to protect all unclassified information. d) None of the answers are correct. e) It focuses the OPSEC process on protecting all unclassified information.
Computers and Technology
1 answer:
Mazyrski [523]3 years ago
8 0

Answer:

The answer is "Option C".

Explanation:

It is the complete information regarding oriented motives, abilities, and strategies, in which opponents need to prepare or operate efficiently to ensure defeat or unreasonable implications toward favorable task achievement. In this information works on maintaining vital information for all the remaining part for the OPSEC form instead of protecting all classified data, and wrong choices can be described as follows:

  • In option a, It can't include, JTF commands that's why it is wrong.
  • In option b, The final stage is developed and apply, that's why it is incorrect.
  • In option e, It is wrong because it can't protect all unclassified data.
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Answer:

Check the explanation

Explanation:

Here is the program with function definition and two sample calls.

Code:

#include <iostream>

using namespace std;

//checkMe FUNCTION which takes values a, b and c

void checkMe(char &a, int &b, int &c)

{

//if sum of b and c is negative and a is 'n', b and c are set to 0, otherwise a is set to 'p'

if((b+c)<0 && a=='n')

{

b = 0;

c = 0;

}

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a = 'p';

}

}

int main()

{

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char a = 'n';

int b = 5;

int c = 6;

 

checkMe(a, b, c);

 

cout<<a<<" "<<b<<" "<<c<<endl;

 

//second test case when if part is executed

a = 'n';

b = -4;

c = -5;

 

checkMe(a, b, c);

 

cout<<a<<" "<<b<<" "<<c<<endl;

return 0;

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Kindly check the Output below:

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