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Sergio [31]
3 years ago
9

If we have a priority queue storing approximately million numbers, and the bottom level is "full", which of the following statem

ents is false? (lower numbers have higher priority) A) The height of the structure is approximately 20. B) The width of the lowest level of the structure is approximately 500,000. C) Nearly half of the 1000 largest values must be in the bottom level of the structure. D) Nearly half of the 1000 smallest values must be in the top half of the tree (if there are k levels in the tree, the "top half" is the top k/2 levels). E) Nearly half of the 1000 smallest values may be in the bottom level of the structure.
Computers and Technology
1 answer:
Anna007 [38]3 years ago
7 0

Answer:

The width of the lowest level of the structure is approximately 500,000 is a false statement.

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Answer:

Explanation:

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2 years ago
A personal phone directory contains room for first names and phone numbers for 30 people. Assign names and phone numbers for the
Fynjy0 [20]

Answer:

# include <iostream.h>

# include <stdio.h>

# include <string.h>

using namespace std;

class citizen

{      

    int i;

    public string name[30];

    public long int phonenumber[30];

    public void addindividual(string name1)

     {      

           If (i<=30)

            {     int flag=0;

                  for(int j=0; j<=i;j++)

                  {

                     if (strcmp(name[i], name1)

                      {

                          flag=1;

                      }

                     else

                     {  

                           flag=0;

                     }

                  }

                  If (flag)

                 {

                  if (i<30)

                  {

                  for(j=i+1;j<=30; j++)

                  {

                        cout<<"Enter the name:"; getchar(name[j]);

                        cout<<"Enter the phone number:"; cin>>phonenumber[j];

                         i++;

                   }

                  else  

                   {

                        cout<<"The person already exists";

                        exit();

                   }

              }

            else

             {

                 cout<<"array is full:";

                 exit();

             }

     }

}

Void main()

{

     string str;

    cout<<" Enter name:";

    getline(cin, str); ;

     citizen c1=new citizen();

     c1.addindividual(name1);

}

Explanation:

With a little more effort you can make the program allow the user to enter any number of details, but less than 30 overall. We have used here flag, and as a programmer we know why we use the Flag. It is used to check whether certain Boolean condition is fulfilled or not. Here, we are checking whether a given name is present in the array of names, and if it is not present, we add that to the list. And if the name is present, we print, it already exist.

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The organization wants you to avoid situations that make you choose between its overall benefit and your
iren [92.7K]
Personal gain is the answer :))
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Which of the following is the most common tool that Windows administrators use on the domain controller?
baherus [9]

Answer:

D. The Active Directory Users and Computers

Explanation:

The Active Directory Users and Computers (ADUC) is one of the many tools used to administer the Active Directory and it is the most common tool that Windows admins use on the domain controller. It provides most of the admins functions such as

i. resetting of password

ii. unlocking users

iii. delegating of permissions to users to manage group policy

iv. managing Active Directory objects - users, computers, contacts, groups - and their attributes.

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<em>Hope this helps!</em>

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Consider five wireless stations, A, B, C, D, and E.
klio [65]

Answer:

Check the explanation

Explanation:

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C’s packet will be seen by A, B and D, but not by E. Thus, D can send to E at the sametime..

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Even though B’s packet will not be seen by D, other nodes, e.g., E, or C, can’t send to D since the packets from these nodes will interfere with the packets from B at A. Therefore, other communications is not likely at the same time.

C) Whenever B is sending to C, what other communications are possible?

B’s packet will be seen by E, A and C, by not by D. therefore, E can send to D at the same point.

4 0
3 years ago
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