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Alenkinab [10]
3 years ago
11

1) Prompt the user to enter two words and a number, storing each into separate variables. Then, output those three values on a s

ingle line separated by a space. (Submit for 1 point) Enter favorite color: yellow Enter pet's name: Daisy Enter a number: 6 You entered: yellow Daisy 6
Computers and Technology
1 answer:
masya89 [10]3 years ago
7 0

Answer:

import java.util.Scanner;

public class TestClock {

   public static void main(String[] args) {

 Scanner in = new Scanner (System.in);

       System.out.print("Enter favorite color:");

 String word1 = in.next();

       System.out.print("Enter pet's name:");

 String word2 = in.next();

       System.out.print("Enter a number:");

 int num = in.nextInt();

       System.out.println("you entered: "+word1+" "+word2+" "+num);

   }

}

Explanation:

Using Java Programming language

  1. Import the Scanner class
  2. create an object of the scanner class
  3. Prompt user to enter the values for the variables (word1, word2, num)
  4. Use String concatenation in System.out.println to display the output as required by the question.

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Sam works in a real estate office. An offer comes in on a house but the homeowners are on vacation. He needs the homeowners to s
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Explanation:

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Your help desk has informed you that they received an urgent call from the vice president last night requesting his logon ID and
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Social Engineering Attack

Explanation:

The exploitation of individuals in order to gain confedential information is a Social Engineering Attack. These attacks typically take advanage of one's emotions or they use inpersonaltion to steal data.

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Lmk if this helps! :)

7 0
3 years ago
Write a method called findNames that meets the following specs: It takes two arguments: a list of strings, allNames, and a strin
Anna11 [10]

Answer:

public class Solution {

   public static void main(String args[]) {

       String[] allNames = new String[]{"Bob Smith", "Elroy Jetson", "Christina Johnson", "Rachael Baker", "cHRis", "Chris Conly"};

       String searchString = "cHRis";

       

       findNames(allNames, searchString);

   }

   

   public static void findNames(String[] listOfName, String nameToFind){

       ArrayList<String> resultName = new ArrayList<String>();

       

       for(String name : listOfName){

           if (name.toLowerCase().contains(nameToFind.toLowerCase())){

               resultName.add(name);

           }

       }

       

       for(String result : resultName){

           System.out.println(result);

       }

   }

}

Explanation:

The class was created called Solution. The second line define the main function. Inside the main function; we initialized and assign an array called allNames to hold the list of all name. Then a String called searchString was also defined. The searchString is the string to search for in each element of allNames array. The two variables (allNames and searchString) are passed as argument to the findNames method when it is called.

The method findNames is defined and it accept two parameters, an array containing list of names and the name to search for.

Inside the findNames method, we initialized and assigned an ArrayList called resultName. The resultName variable is to hold list of element found that contain the searchString.

The first for-loop goes through the elements in the allNames array and compare it with the searchString. If any element is found containing the searchString; it is added to the resultName variable.

The second for-loop goes through the elements of the resultName array and output it. The output is empty if no element was found added to the resultName variable.

During comparison, the both string were converted to lower case before the comparison because same lowercase character does not equal same uppercase character. For instance 'A' is not same as 'a'.

8 0
3 years ago
Consider the following Stack operations:
maxonik [38]

Answer:

Sequence of popped values: h,s,f.

State of stack (from top to bottom): m, d

Explanation:

Assuming that stack is  initially empty. Suppose that p contains the popped values. The state of the stack is where the top and bottom are pointing to in the stack. The top of the stack is that end of the stack where the new value is entered and existing values is removed. The sequence works as following:

push(d) -> enters d to the Stack

Stack:  

d ->top

push(h) -> enters h to the Stack

Stack:

h ->top

d ->bottom

pop() -> removes h from the Stack:

Stack:

d ->top

p: Suppose p contains popped values so first popped value entered to p is h

p = h

push(f) -> enters f to the Stack

Stack:

f ->top

d ->bottom

push(s) -> enters s to the Stack

Stack:

s ->top

f

d ->bottom

pop() -> removes s from the Stack:

Stack:

f ->top

d -> bottom

p = h, s

pop() -> removes f from the Stack:

Stack:

d ->top

p = h, s, f

push(m) -> enters m to the Stack:

Stack:

m ->top

d ->bottom

So looking at p the sequence of popped values is:

h, s, f

the final state of the stack:

m, d

end that is the top of the stack:

m

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