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You are training to complete in a local 5K run. You record your time scores in minutes after completing seven practice runs. Wri

trapecia [35]

Answer:

Following are the program to this question:

#include <iostream>//defining header file

using namespace std;

float Avg(float sum, int n)//defining method Avg that accepts float parameter

{

float average=sum/n;//defining float variable average that holds average value

return average;//return average

}

int main()//defining main method

{

float ar[ ] = {24.1,24.5,24.4,23.8,28.2,29.1,27.4};//defining float array

float sum;//defining float variable sum

int n=7,i;//defining integer variable

for (i = 0; i <n; i++)//defining loop to count total of array

{

sum=sum+ar[i];//add value in sum variable

}

cout<<"The average is: "<<Avg(sum,n);//use print method to call and print method return value

return 0;

}

Output:

The average is: 25.9286

Explanation:

In the above code, the float method "Avg" is declared, that accepts two parameters, that is "sum and n", inside the method, a float variable average is declared that divides and returns its values.

In the main method a float array "ar" is declared that holds values, in the next line, float variable "sum" and integer variable "n, j" is used.

In for loop, the "i" variable is used to add array values in the sum variable and pass into cout to call the "Avg" method and print its return value.

3 0

2 years ago

___________ is an unsecured client server application / protocol that transfers files between two computers.

Alexus [3.1K]

It is known as the File Transfer Protocol or FTP. It is based on a customer server demonstrate design and uses isolate control and information associations between the customer and the server. FTP clients may conform themselves with an unmistakable content sign-in convention, regularly as a username and secret word, yet can interface namelessly if the server is designed to permit it.

6 0

3 years ago

Just answer in five-line. Question:

Eddi Din [679]

Answer:

It is a half-duplex serial transmission mode, and this can be provided through the RS-485 two wired electrical interface. And this can be connected to the ethernet if required, connect to the LAN, and finally transmitting the video footage to all the screens. And the two reasons are:

1. Noiseless

2. Allows around 32 receivers.

Explanation:

Please check the answer.

3 0

3 years ago

I need to create a method named "root positive". which will either print the square root of the number passed to it or if the nu

rjkz [21]

Answer:

Explanation:

The following code is written in Java. It is a method that calculates the square root of a number as requested. The method first checks with an IF statement if the parameter value is a positive number and then calculates the square root and prints it to the screen. Otherwise, it prints Number must not be negative. A test case has been provided in the main method and the output can be seen in the attached image below.

import java.util.Scanner;

class Brainly {

public static void main(String[] args) {

Scanner in = new Scanner(System.in);

System.out.print("Enter a number of type double to calculate square root:");

double num = in.nextDouble();

rootPositive(num);

}

public static void rootPositive(double num) {

if (num > 0) {

System.out.println(Math.sqrt(num));

} else {

System.out.println("Number must not be negative.");

}

5 0

3 years ago

Suppose we have two String objects and treat the characters in each string from beginning to end in the following way: With one

Simora [160]

Answer:

If all the character pairs match after processing both strings, one string in stack and the other in queue, then this means one string is the reverse of the other.

Explanation:

Lets take an example of two strings abc and cba which are reverse of each other.

string1 = abc

string2 = cba

Now push the characters of string1 in stack. Stack is a LIFO (last in first out) data structure which means the character pushed in the last in stack is popped first.

Push abc each character on a stack in the following order.

c

b

a

Now add each character of string2 in queue. Queue is a FIFO (first in first out) data structure which means the character inserted first is removed first.

Insert cba each character on a stack in the following order.

a b c

First c is added to queue then b and then a.

Now lets pop one character from the stack and remove one character from queue and compare each pair of characters of both the strings to each other.

First from stack c is popped as per LIFO and c is removed from queue as per FIFO. Then these two characters are compared. They both match

c=c. Next b is popped from stack and b is removed from queue and these characters match too. At the end a is popped from the stack and a is removed from queue and they both are compared. They too match which shows that string1 and string2 which are reverse of each other are matched.

6 0

3 years ago