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vfiekz [6]
3 years ago
10

Help a brotha out..................

Computers and Technology
1 answer:
Sveta_85 [38]3 years ago
3 0

Answer:

highlight a cell in column A; place your cursor on the border of the cell so it turns into a two sided arrow and drag

Explanation:

y

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How many pieces are there in a normal laptop??? Good luck and solve carefully
polet [3.4K]

What do you mean pieces? Like Laptop components? If so then motherboard,ram,cpu, and psu for a basic laptop so 4 I guess (5 if you want to include a gpu).

4 0
3 years ago
Write the SQL statements that define the relational schema (tables)for this database. Assume that person_id, play_id, birth_year
Sunny_sXe [5.5K]

Answer:

  • SQL statement that defines table for Actor

CREATE TABLE Actor(

person_id integer primary key,

name varchar2(40) not null,

birth_year integer check ((birth_year) <= 2019)

);

  • SQL statement that defines table for Play

CREATE TABLE Play(

play_id integer primary key,

title varchar2(60) not null,

author varchar2(60) not null,

year_written integer check ((year_written) <= 2019)

);

  • SQL statement that defines table for Role

CREATE TABLE Role (

person_id integer,

character_name varchar2(60) not null,

play_id integer,

constraint fk_person foreign key (person_id) references actor(person_id),

constraint fk_play foreign key (play_id) references play(play_id),

primary key (person_id, character_name, play_id)

);

Explanation:

Other information that were not added to the question are as below:

The following database contains information about three tables i.e. actors, plays, and roles they performed.

Actor (person_id, name, birth_year)

Play (play_id, title, author, year_written)

Role (person_id, character_name, play_id)

Where: Actor is a table of actors, their names, and the year they were born. Each actor has a unique person_id, which is a key.

Play is a table of plays, giving the title, author, and year written for each play. Each play has a unique play_id, which is a key.

Role records which actors have performed which roles (characters) in which plays.

Attributes person_id and play_id are foreign keys to Actor and Play respectively.

All three attributes make up the key since it is possible for a single actor to play more than one character in the same play

Further Explanation:

In SQL, in order to define relational schema (Tables) for a database, we use CREATE TABLE statement to create a new table in a database.  The column parameters specify the names of the columns of the table.  The datatype parameter specifies the type of data the column can hold (varchar, integer, date)

4 0
3 years ago
The IntList class contains code for an integer list class. Study it; notice that the only things you can do are: create a list o
torisob [31]

Explanation:

public class Int_List

{

protected int[] list;

protected int numEle = 0;

 

public Int_List( int size )

{

list = new int[size];

public void add( int value )

{

if ( numEle == list.length )

{

System.out.println( "List is full" );

}

else

{

list[numEle] = value;

numEle++;

}

}

public String toString()

{

String returnStr = "";

for ( int x = 0; x < numEle; x++ )

{

returnStr += x + ": " + list[x] + "\n";

}

return returnStr;

}

}

public class Run_List_Test

{

public static void main( String[] args )

{

 

Int_List myList = new Int_List( 7 );

myList.add( 102 );

myList.add( 51 );

myList.add( 202 );

myList.add( 27 );

System.out.println( myList );

}

}

Note: Use appropriate keyword when you override "tostring" method

8 0
3 years ago
What options are available for storing backups, physically?
RUDIKE [14]

The options are available for storing backups, physically are:

  • In both on site and off site,, a person can backup data to a given system that is located on-site, or the backups can be sent to any  remote system that is off-site.

<h3>What is a backup?</h3>

This is known to be a device that helps to save information or data temporarily or permanently.

Note that in the above, The options are available for storing backups, physically are:

  • In both on site and off site,, a person can backup data to a given system that is located on-site, or the backups can be sent to any  remote system that is off-site.

Learn more about backups from

brainly.com/question/17355457

#SPJ12

7 0
2 years ago
I have six nuts and six bolts. Exactly one nut goes with each bolt. The nuts are all different sizes, but it’s hard to compare t
juin [17]

Answer:

Explanation:

In order to arrange the corresponding nuts and bolts in order using quicksort algorithm, we need to first create two arrays , one for nuts and another for bolts namely nutsArr[6] and boltsArr[6]. Now, using one of the bolts as pivot, we can rearrange the nuts in the nuts array such that the nuts on left side of the element chosen (i.e, the ith element indexed as nutArr[i]) are smaller than the nut at ith position and nuts to the right side of nutsArr[i] are larger than the nut at position "I". We implement this strategy recursively to sort the nuts array. The reason that we need to use bolts for sorting nuts is that nuts are not comparable among themselves and bolts are not comparable among themselves(as mentioned in the question)

The pseudocode for the given problem goes as follows:

// method to quick sort the elements in the two arrays

quickSort(nutsArr[start...end], boltsArr[start...end]): if start < end: // choose a nut from nutsArr at random randElement = nutsArr[random(start, end+1)] // partition the boltsArr using the randElement random pivot pivot = partition(boltsArr[start...end], randElement) // partition nutsArr around the bolt at the pivot position partition(nutsArr[start...end], boltsArr[pivot]) // call quickSort by passing first partition quickSort(nutsArr[start...pivot-1], boltsArr[start...pivot-1]) // call quickSort by passing second partition quickSort(nutsArr[pivot+1...end], boltsArr[pivot+1...end])

// method to partition the array passed as parameter, it also takes pivot as parameter

partition(character array, integer start, integer end, character pivot)

{

       integer i = start;

loop from j = start to j < end

       {

check if array[j] < pivot

{

swap (array[i],array[j])

               increase i by 1;

           }

 else check if array[j] = pivot

{

               swap (array[end],array[j])

               decrease i by 1;

           }

       }

swap (array[i] , array[end])

       return partition index i;

}

7 0
3 years ago
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