Write a C++ program that takes as input a list ( an array) of n integrers and finds the number of negative inetgers inthe list (array ).
Side Note: The problem did not come out ofthis book but the information that I study is from this book. And Iam not getting how to write the program.

Answer:

Normalization is a technique in which database are designed and organizes tables in such manner that reduces dependency and redundancy of the data.  It can divide larger tables into the smaller tables and used them using different relationships.

3rd form of normalization is defined as, the table whose non primary key field are dependent on the primary key only and have no dependence any other non key primary field in the tables.

4th form of normalization is defined as, is used in database normalization where they are a non trivial multi value dependency other than candidate key. It builds on the first,second and third normal forms and the Boyce Codd Normal Form.

Answer:

The correct answer is b. Asymmetric Distribution.

Explanation:

Asymmetric Distribution has to do with those larger workloads which are issued to IT resources with higher processing capacities. The Pay-Per-Use Monitor is where the billing system relies on. The Workload Prioritization is where workloads are prioritized according to their level. And the SLA monitor is a contract between a service provider and the customer. So, the most correct answer is b.Asymmetric Distribution.

Answer:

Yes, i agree with the given statement that the tremendous diversity of the source system is the primary reason for their complexity as, when we storing the large amount of the data and the data warehousing are based on the solution to implemented to the data are get transformed for matched the desired queries. There is also the duplicate entry elimination which increased the complexity of the data.

Answer:

J frame class has to be extended for the creation of a window

Explanation:

J frame needs to be extended to create a window, this is because this class in windows has a title, support buttons components, a border or we can say it has decorations for windows.  

The rest of the classes like J button and J text fields lie within the J frame and the Container class is a super class and provides only outlines to J Frame class.

Answer:

C - Technology is the use of computers and software to create economic efficiency

Explanation:

The economic idea of technology refers to everything that can assist in producing goods more efficiently (faster/cheaper/higher quality). Economists refer to technology as an innovative method of executing things, rather than referring to physical hardware.

Answer:

The Correct Answer is C

Explanation:

Answer:

Refer to the images for different S-R Latches.

There are 3 images attached 1 having S-R latch with 2 NAND gates 2nd having S-R Latch with 2 NOR gates and last having S-R latch with control input.

Difference between NAND gated S-R latch and NOR gated S-R latch is the output is reversed.

In latch with control input the latch will only work when the control input or Enable signal in our image is HIGH or 1.

Answer:

Physical Object and Radio Frequency Interference

Explanation:

Great question, it is always good to ask away and get rid of any doubts that you may be having.

There are two main factors that affect a networks performance when talking about Wireless Mesh Network. The two factors are Physical Object and Radio Frequency Interference.

Physical Objects can cause interference in mesh networks. Common Objects such as trees and houses are common causes of interference on wireless mesh networks. The best connection is gained with a clear sight from one node to another. Specific objects such as aluminum made objects can cause complete signal severance.

Radio Frequency Interference is the second main factor when dealing with mesh networks. If there are many signals on the same frequency, Information or packets can get lost in the process causing really bad connection.

I hope this answered your question. If you have any more questions feel free to ask away at Brainly.

Answer:The answer is (c).first.

Explanation:

In doubly linked list each node possesses the address of next node(except last node) because there is no node present after the last node .

and the address of previous node(except first node) because there is no node present before the first node.

So the conclusion is that the first node does not contain the address of previous node and the last node does not contain the address of the next node.

Answer:

a. values

Explanation:

The values() method returns an array of all values of an enumeration. This method is defined automatically by the java compiler for the enum data type.

For example:

enum Traffic_Signal {RED,YELLOW,GREEN};

for(Traffic_Signal t : Traffic_Signal.values()){

   System.out.println(t);

}

This code segment will print out all the valid values in the Traffic_Signal enumeration.

We power on the computer before we begin because a computer has a special programme in it called operating system which needs to be started for us to use the computer

Answer: so i can turn on and you could do what ever u want

Explanation:

Answer Explanation

There are many things to improve the security of business with the help of internet

Answer:

#include<iostream>

using namespace std;

//create the function which add two number

void addTwoNumber(int num_1,int num_2)

{

   int result = num_1 + num_2;  //adding

   cout<<"The output is:"<<result<<endl;  //display on the screen

}

//main function

int main(){

   //calling the function

   addTwoNumber(3,6);

   return 0;

}

Explanation:

First, include the library iostream for using the input/output instructions.

then, create the function which adds two numbers. Its return type is void, it means the function return nothing and the function takes two integer parameters.

then, use the addition operation '+' in the programming to add the numbers and store the result in the variable and display the result.

create the main function for testing the function.

call the function with two arguments 3 and 6.

then, the program copies the argument value into the define function parameters and then the program start executing the function.

Answer and explanation : The TCP/IP means TRANSMISSION CONTROL PROTOCOL AND INTERNET PROTOCOL It governs all the communication which are performed over network it has a set of protocol. It defines how different types of conversation are performed without any fault through a network

THERE ARE 5 TYPES OF LAYER IN TCP/IP MODEL

Answer:

Clock rate also known as clock speed is the rate at which microprocessor executes instructions.That means clock rate is directly proportional to the execution speed of an instruction.Microprocessor is the core of a computer a system it contains all the functions of a central processing unit.So if the clock rate is more  execution speed of an instruction is more if clock rate is less execution speed of an instruction is less.

Answer:

8 times

Explanation:

The while is execute again and again until the condition is TRUE.

In the question:

the value of i=2, when program enter the loop it check the condition when

2 <= 16, condition TRUE. it update the value i = 4.

Again the loop check condition, 4 <= 16, condition TRUE, i become 6 and so on...

4,6,8,10,12,14,16

when i = 16 loop condition is also TRUE. 16 <= 16. So, it again run the loop and i become 18.

then, the loop condition false and it exit the loop.

so, i run until 18

4,6,8,10,12,14,16,18

Therefore, the answer is 8 times.  

The statements that can be blocked in programming, similar to those depending on an if statement, depend on a(n) 'else'. An 'else' statement allows for the conditional execution of a block of code when the 'if' condition is not met.

Just as you can block statements that depend on an if statement, you can also block statements that depend on a(n) c. else. In programming, an else statement is used to execute a block of code when the condition in the if statement is not met. Both if and else statements are structured to handle Boolean expressions which evaluate to either true or false. These control flow statements form the building blocks of logical structures within code and enable conditional execution of code segments.

It's important to distinguish these structures from constants, operators, and statement constants. A constant holds a value that does not change throughout the execution of the program, whereas an operator is used to perform operations on variables and values. A statement constant, on the other hand, is a symbolic representation of a particular statement (or truth value) that remains unchanged within the logical framework of the discussion or argument. Using conditional statements like if-then-else allows programmers to control the flow of execution and make decisions within a program.

Answer:

Internet marketing typically is one-way and impersonal- False.

The following statement is false.

Answer:

#include <iostream>

using namespace std;

int main()

{

 int myArray[] = {-11,3,-8,30,-2,0,5,7,-100,44};

 int n = sizeof(myArray)/sizeof(myArray[0]);

 for(int i=0;i<n;i++){

     if(myArray[i]>=0){

         cout<<myArray[i]<<endl;

     }

 }

  return 0;

}

Explanation:

First include the library iostream in c++ programming for using the input/output function.

Then, create the main function and define the array with the elements which contain both positive element as well negative elements.

after, use the for loop for traversing the array and inside the loop take the conditional statement for check if element in the array is positive.. if it true then print the element on the screen with separate line.

Answer:

c

Explanation:

xxhdudhshshsudjdjd

Answer: The shell sort is based on insertion sort. Here the list of elements are divided into smaller sub list which are sorted based on insertion sort.

Its best case time complexity is O(n* logn) and worst case is O(n* log^2 n)

Explanation:

Shell sort is an inplace sorting here we begin by dividing the list into sublist and sorting the list with insertion sort. We create interval for dividing the list into sub list until we reach the smallest interval of 1.

The best case is O(n* logn).

Answer:

All of the given

Explanation:

Hence all the given options are examples  of standard creation committee.

Answer:

Digital electronics involves 2 states which are abstracted as 0/1 or true/false.

Explanation:

Digital electronics involves 2 states. For TTL logic this corresponds to 0 Volt (0) or 5 Volt (1). Analyzing further, a  digital waveform has a square shape  (not necessarily a perfect square) with 2 levels denoting the two states, namely, true(1) or false(0). So a 0 or 1 is not actually absolute but 0 corresponds to voltage level below a threshold voltage whereas 1 corresponds to voltage level above a threshold voltage.

Answer: True

Explanation:

Embedded system processor are more powerful than general purpose processor because the embedded  processors are designed to work in a particular machine for which they get the right amount of RAM and hardware however general purpose processors are to work in every machine which makes them little slow in some machine as they do not get the proper hardware and RAM and have to cope with much more demanding situations.

Answer:

A traffic controller is basically a system design to regulate traffic in a controlled manner with a set of traffic rules to follow so as to ensure public safety .

The main functions of a traffic light controller are listed below:

Answer:

Variables which are declared within a function are called Local variables.They are named local because they are used locally only means they cannot be used outside the function in which they are declared.These local variables exist in the memory till the function execution starts and are deleted when the execution of function ends.Hence they lose their value between calls to the function in which they are declared.

Local variables lose their values between function calls because they are scoped to the function block, meaning they are removed from memory after the function ends. They prevent naming conflicts and allow modular coding by shadowing any variables with the same name defined outside the function.

Local variables lose their values between calls to the function in which they are defined because they are scoped to the function block. This means that once a function call ends, the local variables that were defined within that function become inaccessible and are typically removed from memory, a process sometimes assisted by a garbage collector in certain programming languages.

Local variables are advantageous because they prevent naming conflicts and ensure that variables within a function do not affect variables with the same name outside the function, effectively shadowing them. This encapsulation allows for modular code, where functions can be understood and debugged in isolation without concerning other parts of the program.

In contrast to global variables, which persist throughout the runtime of the program, local variables defined within, for example, a block in C++ are known to be 'short-lived'. They encourage good programming practices by being limited to their respective scopes, thereby improving code clarity and potential memory efficiency.

C++ program that computes the area and perimeter of a specified shape

#include <iostream>

#include <cmath>

using namespace std;

void rectangle() //Defining function for rectangle

{ int h,w;

cout << "Enter height: ";

//taking input

cin >> h;

cout << "Enter width: ";

cin >> w;

cout << "The perimeter of the rectangle is " <<2*h+ 2*w << " and the area is " <<h*w << endl;  //printing output

}

void triangle()  //Defining function for triangle

{ int s1,s2,s3,h,w;

cout << "Side 1: ";  //Taking input

cin >> s1;

cout << "Side 2: ";

cin >> s2;

cout << "Side 3: ";

cin >> s3;

cout << "Enter the height: ";

cin >> h;

cout << "Enter the base length: ";

cin >> w;

cout << "The perimeter of the triangle is " <<s1+s2+s3 << " and the area is " <<(.5)*w*h << endl; //printing output

}

void circle()//Defining Function for the circle

{

const double p=3.14;

int w;

cout << "Enter the radius: "; //Taking input

cin >> w;

cout << "The perimeter of the circle is " << p*2*w << " and the area is " << p*w*w<< endl;  //printing output

}

int main()  //driver function

{

int s;

cout << "Enter the shape (1 for rectangle,2 for triangle, 3 for circle): ";

//Asking user for the shape

cin >> s;

switch(s)  //checking which shape it chooses

{

case 1:

rectangle();  //If user type 1 ,then calling rectangle function

break;

case 2:

triangle();   //If user type 2 ,then calling triangle function

break;

case 3:

circle();  //If user type 3,then calling circle function

break;

default:

cout <<"Enter valid choice for shape";  //If user type other than 1,2,3

}

return 0;  

}

Output

Enter the shape (1 for rectangle,2 for triangle, 3 for circle): 1

Enter height:2

Enter width: 3

The perimeter of the rectangle is 10 and the area is 6

Answer: Servers perform actions and replies back to clients when connected in the form of the back end of application being used.

Workstations on the other hand are systems where high performance work is performed for getting a high quality of output.

Explanation:

Examples of servers are the application servers or web server. Examples of workstations include graphics editing and audio editing workstations. If we compare in term so reliability i.e there response to failures or the frequencies of failures, it is found that workstations are more reliable in comparison to servers.

Answer:

First-fit, Best-fit, and Worst-fit these are the techniques for the memory allocation of process in the memory locations.

First -fit: Allocating the first memory location to the process which is just smaller than or equal.

Best-fit: Allocating the smallest memory location that is greater than or equal to the size of process.  

Worst-fit: Allocating the largest memory location in the storage to the processes.

These are the memory locations-100 k,600 k,400 k,500 k,300 k                    and size of processes-117 k,412 k,325 k and 510 k

For the first-fit,

We allocate 117 k to the 600 k memory location

Then, we allocate 412 k to the 500 k because others don't have the size.

Then,we allocate 325 k to the 400 k.

510 k will stop,as no memory locations can execute.

For the Best-fit,

We allocate 117 k to the 300 memory location

Then,we allocate 412 k to the 500 k.

Then,we allocate 325 k to the  400 k not 600 k according to best-fit.

Then we allocate 510 k to the 600 k.

For the worst fit,

We allocate 117 k to the 600 k memory location according to the worst-fit algorithm.

Then,we allocate 412 k to the 500 k.

Then,we allocate 325 k to the 400 k

We can' allocate 510 k in any memory location.

Explanation of First-fit, Best-fit, and Worst-fit algorithms in memory management.

In **First-fit** algorithm, the system allocates the first available partition that is big enough for the process. For the given partitions and processes, the 117K process would be placed in the 400K partition, the 412K process in the 600K partition, the 325K process in the 400K partition, and the 510K process in the 600K partition.

**Best-fit** algorithm assigns the smallest partition that is big enough for the process. The 117K process would go into the 400K partition, the 412K process in the 500K partition, the 325K process in the 400K partition, and the 510K process in the 600K partition.

**Worst-fit** algorithm allocates the largest available partition that can accommodate the process. In this scenario, the 117K process would be placed in the 600K partition, the 412K process in the 600K partition, the 325K process in the 400K partition, and the 510K process in the 600K partition.

Answer:

To check if the year comes under each 100th year, lets check if the remainder when dividing with 100 is 0 or not.

Similarly check for 400th year and multiple 0f 4. The following C program describes the function.

#include<stdio.h>

#include<stdbool.h>

bool is_leap_year(int year);

void main()

{

int y;

bool b;

printf("Enter the year in yyyy format: e.g. 1999 \n");

scanf("%d", &y);     // taking the input year in yyyy format.

b= is_leap_year(y);  //calling the function and returning the output to b

if(b==true)

{

 printf("Thae given year is a leap year \n");

}

else

{

 printf("The given year is not a leap year \n");

}

}

bool is_leap_year(int year)

{

if(year%100==0)   //every 100th year

{

 if(year%400==0)   //every 400th year

 {

  return true;

 }

 else

 {

  return false;

 }

}

if(year%4==0)  //is a multiple of 4

{

 return true;

}

else

{

 return false;

}

}

Explanation:

Output is given as image