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FromTheMoon [43]
2 years ago
5

Referential integrity constraints are concerned with checking INSERT and UPDATE operations that affect the parent child relation

ships.
a) true
b) false
Computers and Technology
1 answer:
Jet001 [13]2 years ago
6 0

Answer:

a) true

Explanation:

In Computer programming, integrity constraints can be defined as a set of standard rules that ensures quality information and database are maintained.

Basically, there are four (4) types of integrity constraints and these are;

1. Key constraints.

2. Domain constraints.

3. Entity integrity constraints.

4. Referential integrity constraints.

Referential integrity is a property of data which states that each foreign key value must match a primary key value in another relation or the foreign key value must be null.

For instance, when a foreign key in Table A points to the primary key of Table B, according to the referential integrity constraints, all the value of the foreign key in Table A must be null or match the primary key in Table B.

Hence, the referential Integrity constraints ensures that the relationship between the data in a table is consistent and valid.

Hence, referential integrity constraints are concerned with checking INSERT and UPDATE operations that affect the parent child relationships.

<em>This ultimately implies that, referential Integrity are rules used in database management systems (DBMS) to ensure relationships between tables when records are changed is VALID (INSERT and UPDATE).</em>

<em>In a nutshell, it always ensures a primary key must have a matching foreign key or it becomes null. </em>

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Write an if-else statement that displays 'Speed is normal' if the speed variable is within the range of 24 to 56. If the speed v
kari74 [83]

Answer:

import java.util.Scanner;

public class Speed{

int speed;

public Speed(int speed){

this.speed = speed;

}

public void checkSpeed(){

if(speed >= 24 || speed <= 56){

System.out.println("Speed is normal");

}

else

System.out.println("Speed is abnormal");

}

public static void main(String...args){

Scanner input = new Scanner(System.in);

int userSpeed = 0;

System.out.println("Enter a speed: ");

userSpeed = input.nextInt();

Speed obj1 = new Speed(userSpeed)

obj1.checkSpeed();

}

Explanation:

4 0
2 years ago
Consider a short, 10-meter link, over which a sender can transmit at a rate of 150 bits/sec in both directions. Suppose that pac
Katarina [22]

Answer:

The Tp value 0.03 micro seconds as calculated in the explanation below is negligible. This would lead to a similar value of time delay for both persistent HTTP and non-persistent HTTP.

Thus, persistent HTTP is not faster than non-persistent HTTP with parallel downloads.

Explanation:

Given details are below:

Length of the link = 10 meters

Bandwidth = 150 bits/sec

Size of a data packet = 100,000 bits

Size of a control packet = 200 bits

Size of the downloaded object = 100Kbits

No. of referenced objects = 10

Ler Tp to be the propagation delay between the client and the server, dp be the propagation delay and dt be the transmission delay.

The formula below is used to calculate the total time delay for sending and receiving packets :

d = dp (propagation delay) + dt (transmission delay)

For Parallel downloads through parallel instances of non-persistent HTTP :

Bandwidth = 150 bits/sec

No. of referenced objects = 10

For each parallel download, the bandwith = 150/10

  = 15 bits/sec

10 independent connections are established, during parallel downloads,  and the objects are downloaded simultaneously on these networks. First, a request for the object was sent by a client . Then, the request was processed by the server and once the connection is set, the server sends the object in response.

Therefore, for parallel downloads, the total time required  is calculated as:

(200/150 + Tp + 200/150 + Tp + 200/150 + Tp + 100,000/150 + Tp) + (200/15 + Tp + 200/15 + Tp + 200/150 + Tp + 100,000/15 + Tp)

= ((200+200+200+100,00)/150 + 4Tp) + ((200+200+200+100,00)/15 + 4Tp)

= ((100,600)/150 + 4Tp) + ((100,600)/15 + 4Tp)

= (670 + 4Tp) + (6706 + 4Tp)

= 7377 + 8 Tp seconds

Thus, parallel instances of non-persistent HTTP makes sense in this case.

Let the speed of propogation  of the medium be 300*106 m/sec.

Then, Tp = 10/(300*106)

               = 0.03 micro seconds

The Tp value 0.03 micro seconds as calculated above is negligible. This would lead to a similar value of time delay for both persistent HTTP and non-persistent HTTP. Thus, persistent HTTP is not faster than non-persistent HTTP with parallel downloads.

4 0
2 years ago
Given an n-element array X, algorithm D calls algorithm E on each element X[i]. Algorithm E runs in O(i) time when it is called
krek1111 [17]

Answer:

O(n^2)

Explanation:

The number of elements in the array X is proportional to the algorithm E runs time:

For one element (i=1) -> O(1)

For two elements (i=2) -> O(2)

.

.

.

For n elements (i=n) -> O(n)

If the array has n elements the algorithm D will call the algorithm E n times, so we have a maximum time of n times n, therefore the worst-case running time of D is O(n^2)  

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