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andre [41]
2 years ago
11

Three primary types of data that a forensic investigator must collect, in this order: 1.Volatile data 2.Temporary data 3.Persist

ent data
Computers and Technology
1 answer:
posledela2 years ago
7 0

Answer:

1. volatile data

2. temporary data

3. persistent data

seems like you already got it correct

Explanation:

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The two types of one-time-password tokens are ______ and ______.
Nutka1998 [239]

Answer:

event based and time based

Explanation:

OTP tokens come in two types: event-based (HOTP) and time-based (TOTP). Event-based OTP tokens generate new codes at the press of the button and the code is valid until it is used by the application. Time-based OTP tokens generate codes that are valid only for a certain amount of time (eg, 30 or 60 seconds), after which a new code must be generated

5 0
4 years ago
Read 2 more answers
Drag the words into the correct boxes
tresset_1 [31]

Explanation:

if you have any doubts or queries regarding the answer please feel free to ask

4 0
2 years ago
What is an example of the most important role of a systems analyst in any corporation?
dsp73

Answer: The system analyst is one of the most important members in any organisation. These system analyst has to analyse different data of the organisation which would help to bring out the different any new business policy changes or any kind of improvement.

Explanation:

An example to know this better would be the system analyst of a telecom company. Here the role of the system analyst would be bring out the design and implementation of new telecom information system and also should be aware of previous data of the organization. The system analyst would also be responsible to bring out the new business policies based on latest telecom standards and ensure the systems conforms to the latest standards.

5 0
3 years ago
Everybody at a company is assigned a unique 9 digit ID. How many unique IDs exist?
timama [110]

Answer:

3,265,920 unique ID exist.

Explanation:

The nine digits are from 0 to 9, there are ten bits from 0 -9,

0, 1, 2, 3, 4, 5, 6, 7, 8, 9

The first is select from the highest bit (9), and the second is selected at random from 0 - 9, the third bit to the last must be unique and different from each other;

number of unique IDs = 9 × 9 × 8 × 7 × 6 × 5 × 4 × 3 × 2

Multiplying the nine bits of unique IDs = 3,265,920.

3 0
3 years ago
g Write a program that prompts the user for an integer n between 1 and 100. If the number is outside the range, it prints an err
grin007 [14]

Answer:

The cpp program is given below.

#include<iostream>

#include<iomanip>

using namespace std;

int main() {

   

   // variables declared

   int n;

   int sum=0;

   float avg;

   

   do

   {

       // user input taken for number    

       cout<< "Enter a number between 1 and 100 (inclusive): ";

       cin>>n;

       

       if(n<1 || n>100)

           cout<<" Number is out of range. Enter valid number."<<endl;

       

   }while(n<1 || n>100);

   

   cout<<" "<<endl;

   

   // printing even numbers between num and 50  

   for(int num=1; num<=n; num++)

   {

       sum = sum + num;

   }

   

   avg = sum/n;

   

   // displaying sum and average

   cout<<"Sum of numbers between 1 and "<<n<<" is "<<sum<<endl;

   cout<<"Average of numbers between 1 and "<<n<<" is ";

   printf("%.2f", avg);

   

       return 0;

}

OUTPUT

Enter a number between 1 and 100 (inclusive): 123

Number is out of range. Enter valid number.

Enter a number between 1 and 100 (inclusive): 56

 

Sum of numbers between 1 and 56 is 1596

Average of numbers between 1 and 56 is 28.00

Explanation:

The program is explained below.

1. Two integer variables are declared to hold the number, n, and to hold the sum of numbers from 1 to n, sum. The variable sum is initialized to 0.

2. One float variable, avg, is declared to hold average of numbers from 1 to n.

3. User input is taken for n inside do-while loop. The loop executes till user enters value between 1 and 100. Otherwise, error message is printed.

4. The for loop executes over variable num, which runs from 1 to user-entered value of n.

5. Inside for loop, all the values of num are added to sum.

sum = sum + num;

6. Outside for loop, average is computed and stored in avg.

avg = sum/n;

7. The average is printed with two numbers after decimal using the following code.

printf("%.2f", avg);

8. The program ends with return statement.

9. All the code is written inside main() and no classes are involved.

3 0
3 years ago
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