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1 year ago

Database management systems are expected to handle binary relationships but not unary and ternary relationships.'

Computers and Technology

1 answer:

alekssr [168]1 year ago

6 0

Answer:

False

Database management systems are expected to handle binary, unary and ternary relationships.

Explanation:

Unary Relationship: It is a recursive relationship which includes one entity in a relationship which means that there is a relationship between the instances of the same entity. Primary key and foreign key are both the same here. For example a Person is married to only one Person. In this example there is a one-to-one relationship between the same entity i.e. Person.

Binary Relationships: It is a relationship involving two different entities. These two entities identified using two relations and another relation to show relationship between two entities and this relation holds primary keys of both entities and its own primary key is the combination of primary keys of the both relations of the two entities. For example Many Students can read a Book and many Books can be obtained by a Student.

Ternary Relationships: is a relationship involving three entities and can have three tables. For example a Supplier can supply a specific Part of many Mobiles. Or many Suppliers may supply several Parts of many Mobile models.

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Jonah’s friend reminded him to check his report against the rubric provided by their teacher. His friend sent him these instruct

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Answer:

Steps 2 and 3 needs to be switched.

Explanation:

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1 year ago

Suppose your company has decided that it needs to make certain busy servers faster. Processes in the workload spend 60% of their

vekshin1

Answer:

CPU need 50% much faster

disk need 100% much faster

Explanation:

given data

workload spend time CPU = 60%

workload spend time I/O = 40%

achieve overall system speedup = 25%

to find out

How much faster does CPU need and How much faster does the disk need

solution

we apply here Amdahl’s law for the overall speed of a computer that is express as

S = $\frac{1}{(1-f)+ \frac{f}{k} }$ .............................1

here f is fraction of work i.e 0.6 and S is overall speed i.e 100% + 25% = 125 % and k is speed up of component

so put all value in equation 1 we get

S = $\frac{1}{(1-f)+ \frac{f}{k} }$

1.25 = $\frac{1}{(1-0.6)+ \frac{0.6}{k} }$

solve we get

k = 1.5

so we can say CPU need 50% much faster

and

when f = 0.4 and S = 125 %

put the value in equation 1

S = $\frac{1}{(1-f)+ \frac{f}{k} }$

1.25 = $\frac{1}{(1-0.4)+ \frac{0.4}{k} }$

solve we get

k = 2

so here disk need 100% much faster

7 0

1 year ago

In this programming assignment, you will create a hierarchy of classes (see below) that inherit from the beverage class. The bas

harina [27]

Answer:

public class GetInfo{

Beverage[] beverages=new Beverage[100];

int i=0;

GetInfo(Beverage b){

beverages[i]=b;

i++;

}

public void Display(){

for(int i=0;i<beverages.length;i++)

cout<<beverage[i].tostring();

}

Explanation:

we are taking Beverages array to store all values and in constructor we are adding that to the list and Display() function prints the vale

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1 year ago

Pick the correct statements regarding cell references.

shusha [124]

Statement two and three is correct.

Statement 1 is incorrect. A relative reference changes when a formula is copied to another cell while Absolute references remain constant. However, it is safe to say that an absolute address can be preceded by a $ sign before both the row and the column values. It is designated by the addition of a dollar sign either before the column reference, the row reference, or both. Statement C is also correct. A mixed reference is a combination of relative and absolute reference and the formula (= A1 + $B$2) is an example of a mixed cell reference.

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1 year ago

Read 2 more answers

Get user input as a boolean and store it in the variable tallEnough. Also, get user input as a boolean and store it in the varia

frozen [14]

Answer:

I will code in JAVA.

import java.util.Scanner;

public class Main {

public static void main(String[] args) {

boolean tallEnough;

boolean oldEnough;

Scanner input = new Scanner(System.in);

tallEnough = input.nextBoolean();<em> //wait the input for tallEnough</em>

oldEnough = input.nextBoolean(); <em>//wait the input for OldEnough</em>

if(tallEnough && oldEnough){

System.out.print(true);

} else {

System.out.print(false);

}

Explanation:

First, to accept user inputs you have to import the class Scanner. Then declare both variables before allowing the user to set input values for both boolean variables.

In the if-else statement checks if both variables are true, then prints true. Another case prints always false.

8 0

1 year ago