A data structure used to bind an authenticated individual to a public key is the definition of ________. digital certificate
baseline
internet layer
data link layer

Answers

Answer 1
Answer:

Answer:

Digital Certificate is the correct answer of this question.

Explanation:

Digital certificates are always for encryption and identification for transmitting public keys.The concept of digital certificate is a data structure used for linking an authenticated person to a public key. It is used to cryptographically attach public key rights to the organization that controls it.

For example:- Verisign, Entrust, etc.

  • An appendix to an electronic document that is used for authentication purposes.
  • A digital certificate's most frequent use is to confirm that an user sending a message is who he or she appears to be, and provide the recipient with the means to encode a response.


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Access control lists (ACLs) are used to permit and deny traffic in an IP router.A. TrueB. False
Insert the missing code in the following code fragment. This fragment is intended to implement a method to set the value stored in an instance variable. Public class Employee { Private String empID; Private boolean hourly; ) . . _______ { Hourly = isHourly; } } A) public void setHourly(String isHourly) B) public void getHourly() C) public boolean getHourly() D) public boolean setHourly(boolean isHourly)
Laptop components that can be replaced relatively easily include the ____ and the ____first partscreen power supplyRAM second part hard diskkeyboard touchpad​
Egovernment involves the use of strategies and technologies to transform government by improving the delivery of services and enhancing the quality of interaction between the citizen-consumer within all branches of government. Group of answer choices
Write a program to read a list of exam scores given as integer percentages in the range 0-100. Display the total number of grades in each letter grade defined as follows:90-100 is an A, 80-89 is a B, 70-79 is a C, 60-69 is a D, 0-59 is an F. Use a negative score as a sentinel to indicate the end of the input. (The negative value is used just to end the loop, do not use it in the calculations). Then output the highest and lowest score, and the average score.For example if the input is: 72 98 87 50 70 86 85 78 73 72 72 66 63 85 -1the output would be:Total number of grades = 14Number of As =1Number of Bs = 4Number of Cs = 6Number of Ds = 2Number of Fs = 1The highest score is 98The lowest score is 50The average is 75.5This is what I have so far and it is not working correctly:public static void main(String[] args) {// scannerScanner scnr =new Scanner (System.in);//ints grades and countint x;int A = 0;int B = 0;int C = 0;int D = 0;int F = 0;int count = 1;//int min max totalint min, max;int total = 0 ;//doubledouble average;//prompt user for inputSystem.out.print("Please enter the exam scores as integer ");System.out.print("percentages in the rage 0-100. ");System.out.println("Please end the list with a negative integer.");//scnrx = scnr.nextInt();min = x;max = x;//while loopwhile (x >= 0){x = scnr.nextInt();if (x >= 0){total = total + x;count++;if (x < min)min = x;if (x > min)max = x; }while (x >= 90 && x <= 100) {x = scnr.nextInt();A++;//Grade Bif (x >= 80 && x <= 89)B++;//Grade Cif (x >= 70 && x <= 79)C++;//Grade Dif (x >= 60 && x <= 69)D++;//Grade Fif (x >= 0 && x <= 59)F++;}}// averageaverage = total/count;//results/outputSystem.out.println("Total number of grades: " + count);System.out.println("Number of A's: " + A);System.out.println("Number of B's: " + B);System.out.println("Number of C's: " + C);System.out.println("Number of D's: " + D);System.out.println("Number of F's: " + F);System.out.println("Highest score: " + max);System.out.println("Lowest score: " + min);System.out.println("Average: " + average);}}

A web page that allows interaction from the user​

Answers

Answer:

Its A Dynamic Web Page

Explanation:

Which of the following statements is not true?A. A structured chart is a sequential representation of program designB. the Real-Time system is a particular case of a on-line-systemC. Batch totals are not incorporated while designing real-time applicationsD. 4GLs are used for application proto typingE. None of the above

Answers

Answer:

A structured chart is a sequential representation of program design

Explanation:

It is true that a real-time system which is a term to describe an operating system working in relation to real-time is actually a form of the online system. Hence, option B is not correct.

It is also true that Batch totals are not incorporated while designing real-time applications because Batch Data processing is carried out in a separate manner and at a time when the computer is free. Thus Option C is not Correct

It is also true that 4GL which stands for Forth generation programming language is used for application prototyping. Again Option D is not Correct.

However, a structured chart is not a sequential representation of program design, but rather a break down to the infinitesimal module in the program design. Hence, option A is the correct answer.

Bargain Bob's auto dealership sells vehicles. He sells Chrysler, Jeep, and Dodge brand vehicles. He tracks customer and manufacturer. Bob tracks only the main owner—all co-owners and co-signers are recorded elsewhere.Based on the description above, what is the maximum cardinality between each instance of "Customer" and "Vehicle?"

Answers

Based on the given description in the question, the maximum cardinality assignment of the customer and vehicle is; one customer to one vehicle mapping

Cardinal Mapping

Cardinality is simply defined as the mapping of entities or group of entities to a certain cardinal value. This means that cardinality seeks to highlight the relationship between two objects such as eggs in a crate or shoes on a rack.

Now, in this question, we see that the car dealer sells one car to a customer and tracks the customer and manufacturer for that particular vehicle. This is simply 1 to 1 mapping and as such the maximum cardinality assignment of the customer and vehicle is one (1) to one (1) mapping.

Read more on Mapping at; brainly.com/question/1625866

Answer:

Customer(1) - (1) Vehicle.

Explanation:

Cardinality is the mapping of entities or group of entities to a cardinal value. It tries to show the relationship between two objects like a cups in a shelf or plates in racks.

The car dealer in the question, sells one car to a customer and keep or prioritise the record of the main owner of the acquired vehicle. So the maximum cardinality assignment of the customer and vehicle is one (1) to one (1) mapping.

Write a function named minMax() that accepts three integers arguments from the keyboard and finds the smallest and largest integers. Include the function minMax() in a working program. Make sure your function is called from main().Test the function by passing various combinations of three integers to it.

Answers

Answer:

public class Main

{

public static void main(String[] args) {

 minMax(1, 2, 3);

 minMax(100, 25, 33);

 minMax(11, 222, 37);

}

public static void minMax(int n1, int n2, int n3){

    int max, min;

    if(n1 >= n2 && n1 >= n3){

        max = n1;

    }

    else if(n2 >= n1 && n2 >= n3){

        max = n2;

    }

    else{

        max = n3;

    }

   

    if(n1 <= n2 && n1 <= n3){

        min = n1;

    }

    else if(n2 <= n1 && n2 <= n3){

        min = n2;

    }

    else{

        min = n3;

    }

    System.out.println("The max is " + max + "\nThe min is " + min);    

}

}

Explanation:

*The code is in Java.

Create a function named minMax() that takes three integers, n1, n2 and n3

Inside the function:

Declare the min and max

Check if n1 is greater than or equal to n2 and n3. If it is set it as max. If not, check if n2 is greater than or equal to n1 and n3. If it is set it as max. Otherwise, set n3 as max

Check if n1 is smaller than or equal to n2 and n3. If it is set it as min. If not, check if n2 is smaller than or equal to n1 and n3. If it is set it as min. Otherwise, set n3 as min

Print the max and min

Inside the main:

Call the minMax() with different combinations

Given the following: int funcOne(int n) { n *= 2; return n; } int funcTwo(int &n) { n *= 10; return n; } What will the following code output? int n = 30; n = funcOne(funcTwo(n)); cout << "num1 = " << n << endl; Group of answer choices Error num1 = 60 num 1 = 30 num1 = 600 num1 = 300

Answers

Answer:

num1 = 600

Explanation:

Given

The attached code snippet

Required

The output of n = funcOne(funcTwo(n));  when n = 30

The inner function is first executed, i.e.

funcTwo(n)

When n = 30, we have:

funcTwo(30)

This returns the product of n and 10 i.e. 30 * 10 = 300

So:

funcTwo(30) = 300

n = funcOne(funcTwo(n)); becomes: n = funcOne(300);

This passes 300 to funcOne; So, we have:

n = funcOne(300);

This returns the product of n and 2 i.e. 300 * 2 = 600

Hence, the output is 600

A database is composed of several parts known as database ____

Answers

fragments.....is the answer...
Hope it Helpz..!!
Other Questions
Quicksort reaches the worst-case time complexity when:Partition is not implemented in placePicking the largest one of input or partitioned data as pivot valueMedian of input or partitioned data is expensive to calculateData is sorted already and always pick the median as pivotChoose the incorrect statement:When the median is always picked as pivot in input/partitioned data, then quicksort achieves the best-case time complexity.Mergesort has O(N(log(N)) time complexity for its worst case, average case and best caseInsertionsort reaches its best-case time complexity O(N log(N)) when the input data is pre-sortedQuicksort is practically fast and frequently used sorting algorithm.Choose the incorrect statement:In the lower bound analysis by using decision tree, each branch uses one comparison to narrow down possible casesIn the lower bound analysis by using decision tree, he number of required comparisons can be represented by height of decision treeA decision tree to sort N elements must have N^2 leavesO(N log(N)) lower bound means that comparison-based algorithm cannot achieve a time complexity better than O(N log(N))Choose the incorrect statement regarding time complexity of union-find operation:Inverse Ackermann function does not depend on N and is a constant factor.When we use arbitrary union and simple find for union-find operation, the worst-case time complexity is O(MN) for a sequence of M operations and N elementsUnion-by-size and Union-by-rank both improve the time complexity to O(M log(N)) for a sequence of M operations and N elementsTo finish the entire equivalence class computation algorithm, we need to go over each pair of elements, so if we use union-by-rank with path compression for find operation, then the overall time complexity is O(N^2 log*N), where log*N denotes the inverse Ackermann function.Choose the incorrect statement regarding Dijstraâs algorithmDijstraâs algorithm is a greedy algorithmDijstraâs algorithm requires to dynamically update distance/costs/weights of paths.To begin with, Dijstraâs algorithm initializes all distance as INFDijstraâs algorithm can be implemented by heaps, leading to O(|E|+|V| log(|V|)) time complexity, where, particularly, log(|V|) is due to "insert" operation in heaps.