Do each of the following in the main function:
a) Write the function prototype for function zero, which takes a double array big_nums and does not return a value.
b) Write the function call for the function in part a.
c) Write the function prototype for function add1AndSum, which takes an integer array one_small and returns an integer.
d) Write the function call for the function described in part c.

Answer:

ES LA  B

Explanation:

Answer:

Wireless resources management is communication to address they faster response time.

Explanation:

Wireless communication that provide bandwidth, and transmission reliable, towards development.

This communication is discovered to the date and have consumed physical layer  of available wireless resources.

Wireless communication management all this critical topic  and the management techniques, and limited device battery power.

Communication as the address channel loss, and multi path .

Communication are the latest resources allocation and new or next generation and technologies.

Wireless communication theoretical concepts in a manner, and the necessary mathematical tools and fundamental tools.

Wireless communication device industries up to date of the topic,readers are benefit from a basic.

Answer:

line graph

Explanation:

the change in weight is linear

Answer:

Line Graph

Explanation:

Linear Weight change

Answer:

Alphabetic character spacing based on the width of each letter in a font. ... Proportional spacing is commonly used for almost all text. In this encyclopedia, the default text is a proportional typeface, and tables are "monospaced," in which all characters have the same fixed width.

hope this answer helps u

pls mark me as brainlitest .-.

Answer:

Following are the description of the given nodes:

Explanation:

The Multiple Access frequency - division facilitates control by granting every node a fixed time slot. Also, every slot seems to be identical in width. Its whole channel bandwidth is often used in every channel node. Compared to FDMA it takes more time.  

Time slot=3 frames for all the above scenario

Time interval = number of frames in first slot Time to send 1 frame number of images  

                      [tex]= 8 \times 3 \times 1 \\\\ = 8 \times 3 \\\\= 24 \ seconds[/tex]

Answer:

The programming language is not stated; however, I'll answer using Python programming language (See attachment for proper format)

tuition = 10000

rate = 0.04

for i in range(1,15):

tuition = tuition + tuition * rate

if i <= 10:

print("Year "+str(i)+" tuition:",end=" ")

print(round(tuition,2))

if i == 14:

print("Tuition 4th year after:",end=" ")

print(round(tuition,2))

Explanation:

The first 2 lines initializes tuition and rate to 10000 and 0.04 respectively

tuition = 10000

rate = 0.04

The next line iterates from year 1 to year 14

for i in range(1,15):

This line calculates the tuition for each year

tuition = tuition + tuition * rate

The next 3 lines prints the tuition for year 1 to year 10

if i <= 10:

print("Year "+str(i)+" tuition:",end=" ")

print(round(tuition,2))

The next 3 lines prints the tuition at the 4th year after year 10 (i.e. year 14)

if i == 14:

print("Tuition 4th year after:",end=" ")

print(round(tuition,2))

Answer:

"Phishing" will be the correct approach.

Explanation:

Answer:

D

Explanation:

The value of i is incremented by one then assigned to d

Answer:

a) An ARRAY ( 3 )

B) relatively INSENSITIVE ( 7 )

c) a HYBRID ( 4 )

Explanation:

They both used an array based implementation in the standard form and are both relatively insensitive to the order of the data. The main similarity is that each one creates an environment in which a hybrid strategy works more efficiently.

Quadratic selection sort is the sorting system that sorts out the smallest value in an entire list and arrange the list in such a way that the smallest value comes first.

Answer:

Here are the methods:

a) count

def count(object, list):  

   counter = 0

   for i in list:

       if i == object:

           counter = counter + 1

   return counter

b) in : return True if the item is in the list

def include(object, list):  

   for i in list:

       if i == object:

           return True

   return False

c)  reverse

def reverse(list):

   first = 0            

   last = len(list)-1    

   while first<last:

       list[first] , list[last] = list[last] , list[first]

       first = first + 1

       last = last - 1

   return list

d) index: return -1 if the item is not in the list  

def index(object, list):

   for x in range(len(list)):

       if list[x] == object:

           return x

   return -1

e)  insert

def insert(object, index, list):

   return list[:index] + [object] + list[index:]

Explanation:

a)

The method count takes object and a list as parameters and returns the number of times that object appears in the list.

counter variable is used to count the number of times object appears in list.

Suppose

list = [1,1,2,1,3,4,4,5,6]

object = 1

The for loop i.e. for i in list iterates through each item in the list

if condition if i == object inside for loop checks if the item of list at i-th position is equal to the object.  So for the above example, this condition checks if 1 is present in the list. If this condition evaluates to true then the value of counter is incremented to 1 otherwise the loop keeps iterating through the list searching for the occurrence of object (1) in the list.

After the complete list is moved through, the return counter statement returns the number of times object (i.e. 1 in the example) occurred in the list ( i.e. [1,1,2,1,3,4,4,5,6]  ) As 1 appears thrice in the list so the output is 3.

b)  

The method include takes two parameters i.e. object and list as parameters and returns True if the object is present in the list otherwise returns False. Here i have not named the function as in because in is a reserved keyword in Python so i used include as method name.

For example if list = [1,2,3,4] and object = 3

for loop for i in list iterates through each item of the list and the if condition if i == object checks if the item at i-th position of the list is equal to the specified object. This means for the above example the loop iterates through each number in the list and checks if the number at i-th position in the list is equal to 3 (object). When this if condition evaluates to true, the method returns True as output otherwise returns False in output ( if 3 is not found in the list). As object 3 is present in the list so the output is True.

c) reverse method takes a list as parameter and returns the list in reverse order.

Suppose list = [1,2,3,4,5,6]

The function has two variables i.e. first that is the first item of the list and last which is the last item of the list. Value of first is initialized to 0 and value of last is initialized to len(list)-1 where len(list) = 6 and 6-1=5 so last=5 for the above example. These are basically used as index variables to point to the first and last items of list.

The while loop executes until the value of first exceeds that of last.

Inside the while loop the statement list[first] , list[last] = list[last] , list[first]  interchanges the values of elements of the list which are positioned at first and last. After each interchange the first is incremented to 1 and last is decremented to 1.

For example at first iteration:

first = 0

last = 5

list[0] , list[5] = list[5] , list[0]

This means in the list [1,2,3,4,5,6] The first element 1 is interchanged with last element 6. Then the value of first is incremented and first = 1, last = 4 to point at the elements 2 and 5 of the list and then exchanges them too.

This process goes on until while condition evaluates to false. When the loop breaks  statement return list returns the reversed list.

d) The method index takes object and list as parameters and returns the index of the object/item if it is found in the list otherwise returns -1

For example list = [1,2,4,5,6] and object = 3

for loop i.e.  for x in range(len(list)):  moves through each item of the list until the end of the list is reached. if statement  if list[x] == object:  checks if the x-th index of the list is equal to the object. If it is true returns the index position of the list where the object is found otherwise returns -1. For the above examples 3 is not in the list so the output is -1  

e) insert

The insert function takes as argument the object to be inserted, the index where the object is to be inserted and the list in which the object is to be inserted.

For example list = [0, 1, 2, 4, 5, 6] and object = 3 and index = 3

3 is to be inserted in list  [1,2,4,5] at index position 3 of the list. The statement:

return list[:index] + [object] + list[index:]

list[:index] is a sub list that contains items from start to the index position. For above example:

list[:index] = [0, 1, 2]

list[index:] is a sub list that contains items from index position to end of the list.

list[index:] = [4, 5, 6]

[object] = [3]

So above statement becomes:

[0, 1, 2] + [3] + [4, 5, 6]

So the output is:

[0, 1, 2, 3, 4, 5, 6]    

Explanation:

Refactoring consists of improving the internal structure of an existing program's source code, while preserving its external behavior. 

The penalties for ignoring the requirements for protecting classified information when using social networking services are the same when using other media and methods of dissemination.

The penalties are the same because it does not matter what the medium used, the effect is that classified information and confidential details have been left unguarded and exposed.

When in hold of such information, it is not okay to leave it unguarded or exposed. A person who is found guilty of disclosure of classified materials would face sanctions such as:

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Answer:

Soft copy like audio

Hard copy e.g printouts

Explanation:

Answer:

pie chart

Explanation:

pie chart shows percentage per item

Answer:

pie chart

Explanation:

Answer:

This function will return a character grade for every floating-point grade for every student in the class.

function grade( floatPoint ){

      let text;

      switch ( floatPoint ) {

        case floatPoint < 40.0 :

             text= " F ";

             break;

        case floatPoint == 40.0 :

             text= " P ";

             break;

        case floatPoint > 40.0 && floatPoint <= 50.0 :

             text=  " C ";

             break;

        case floatPoint > 50.0 && floatPoint < 70.0 :

             text=  " B ";

             break;

        case floatPoint >=70.0 && floatPoint <= 100.0 :

             text= " C ";

             break;

        default :

            text= "Floating point must be between 0.0 to 100.0"

      }

     return text;

}

Explanation:

This javascript function would return a character grade for floating-point numbers within the range of 0.0 to 100.0.

Answer:

Following are the code to this question:

f1=input('Input first file name: ')#defining f1 variable that input file 1

f2=input('Input second file name: ')#defining f2 variable that input file 2

file1=open(f1,'r')#defining file1 variable that opens first files by using open method  

file2=open(f2,'r')#defining file1 variable that opens second files by using open method  

d1=file1.readlines()#defining d1 variable that use readlines method to read first file data

d2=file2.readlines()#defining d2 variable that use readlines method to read second file data

if d1==d2:#defining if block that check file data

   print('Yes')#when value is matched it will print message yes  

   exit# use exit keyword for exit from if block

for j in range(0,min(len(d1),len(d2))):#defining for loop that stores the length of the file  

   if (d1[j]!=d2[j]):#defining if block that check value is not matched

       print('No')#print the message "NO"

       print("mismatch values: ",d1[j]," ",d2[j])#print file values

Output:

please find the attached file.

Explanation:

code description:

Answer:

first = input("enter first file name: ")

second = input("enter second file name: ")

file_one = open(first, 'r')

file_two = open(second, 'r')

if file_one.read() == file_two.read():

  print("Both files are the same")

else:

  print("Different files")

file_one.close()

file_two.close()

Answer:

the answer is 300 feet in the daytime

Answer:

Here is the JAVA code:

public class Main{

public static int solution(int[] A) { //method that takes non-empty array A consisting of 0s and 1s

           int N = A.length; // number of 0s and 1s in array A

           int r = 0; //result of adjacency

           for (int i = 0; i < N - 1; i++;   )   { // iterates through A

               if (A[i] == A[i + 1])  //if i-th element of A is equal to (i+1)th element

                   r = r + 1;   }     //add 1 to the count of r

           if (r == N-1)   //for test cases like {1,1}

           {return r-1; }

           int max = 0; //to store maximum possible adjacency

           for (int i = 0; i <N; i++)   { //iterates through array

               int c = 0;

               if (i > 0)    { //starts from 1 and  covering the last

                   if (A[i-1] != A[i]) //checks if i-1 element of A is not equal to ith element of A

                       c = c + 1;    //adds 1 to counter variable

                   else

                      c = c - 1;   } //decrements c by 1

               if (i < N - 1)     {//starting with 0

                   if (A[i] != A[i + 1])  //checks if ith element of A is not equal to i+1th element of A

                       c = c + 1; //adds 1 to counter variable

                   else

                       c = c - 1;   }      //decrements c by 1        

               max = Math.max(max,c);   }  //finds the maximum of max and c

           return r + max;         } //returns result + maximum result

        public static void main(String[] args) {

   int[] A = {1, 1, 0, 1, 0, 0}; //sample array to test the method

   System.out.println(solution(A));} } //calls the method passing array to it

Explanation:

The program works as follows:

A[] = {1, 1, 0, 1, 0, 0}

N = A.length

N = 6

The A has the following elements:

A[0] =  1

A[1] = 1

A[2] = 0

A[3] = 1

A[4] = 0

A[5] = 0

Program iterates through array A using for loop. Loop variable i is initialized to 0

if condition if (A[i] == A[i + 1]) checks

if (A[0] == A[0 + 1])

A[0] =  1

A[0 + 1] = A[1] = 1

They both are 1 so they are equal

r = r + 1;  

Since the above if condition is true so 1 is added to the value of r Hence

r = 1

At each iteration of the loop the if condition checks whether the adjacent elements of A are equal. If true then r is incremented to 1 otherwise not.

So after all the iterations value of r = 2

if (r == N-1) evaluates to false because r=2 and N-1 = 5

So program moves to the statement:

for (int i = 0; i <N; i++)

This loop iterates through the array A

if (i > 0) condition checks if value of i is greater than 0. This evaluates to false and program control moves to statement:

if (i < N - 1) which makes if(0<5) This evaluates to true and program control moves to statement

if (A[i] != A[i + 1])  which means:

if (A[0] != A[0 + 1]) -> if (A[0] != A[1])

We know that

A[0] =  1

A[1] = 1

So this evaluates to false and else part is executed:

value of c is decremented to 1. So c=-1

max = Math.max(max,c) statement returns the max of max and c

max = 0

c = -1

So max = 0

value of i is incremented to 1 so i = 1

At next step:

if (i < N - 1) which makes if(1<5) This evaluates to true and program control moves to statement

if (A[i] != A[i + 1]) which means:

if (A[1] != A[1 + 1]) -> if (A[1] != A[2])

A[1] = 1

A[2] = 0

So the statement evaluates to true and following statement is executed

c = c + 1; The value of c is incremented to 1. So

c = -1 + 1

Hence

Hence c= 0, max = 0 and i = 2

next step:

if (i < N - 1) which makes if(2<5) This evaluates to true and program control moves to statement

if (A[i] != A[i + 1]) which means:

if (A[2] != A[2 + 1]) -> if (A[2] != A[3])

A[2] = 0

A[3] = 1

So the statement evaluates to true and following statement is executed

c = c + 1; The value of c is incremented to 1. So

c = 0 + 1

c = 1

Hence

The statement max = Math.max(max,c) returns the max of max and c

max = 0

c = 1

So max = 1

Hence  c= 1, max = 1 and i = 3

next step:

if (i < N - 1) which makes if(3<5) This evaluates to true and program control moves to statement

if (A[i] != A[i + 1]) which means:

if (A[3] != A[3 + 1]) -> if (A[3] != A[4])

A[3] = 1

A[4] = 0

So the statement evaluates to true and following statement is executed

c = c + 1; The value of c is incremented to 1. So

c = 1 + 1

c = 2

Hence

The statement max = Math.max(max,c) returns the max of max and c

max = 1

c = 2

So max = 2

Hence c= 2, max = 2 i = 4

next step:

if (i < N - 1) which makes if(4<5) This evaluates to true and program control moves to statement

if (A[i] != A[i + 1])  which means:

if (A[4] != A[4+ 1]) -> if (A[4] != A[5])

A[4] = 0

A[5] = 0

So this evaluates to false and else part is executed:

value of c is decremented to 1. So c=1

max = Math.max(max,c) statement returns the max of max and c

max = 2

c = 1

So max = 2

value of i is incremented to 1 so i = 5

next step:

if (i < N - 1) which makes if(5<5) This evaluates to false

if (i > 0) evaluates to true so following statement executes:

if (A[i-1] != A[i])

if (A[5-1] != A[5])

if (A[4] != A[5])

A[4] = 0

A[5] = 0

This statement evaluates to false so else part executes and value of c is decremented to 1

Hence

max = 2

c = 0

So max = 2

value of i is incremented to 1 so i = 6

The loop breaks because i <N evaluates to false.

Program control moves to the statement:

return r + max;

r = 2

max = 2

r + max = 2+2 = 4

So the output of the above program is:

4

Answer:

The sternum or breastbone is a long flat bone located in the central part of the chest.

The sternum or breastbone is a long flat bone located in the central part of the chest.

In the middle of the chest, there is a long, flat bone known as the sternum or breastbone. It forms the front of the rib cage and is joined to the ribs by cartilage, assisting in the protection of the heart, lungs, and major blood arteries from harm. It is one of the longest and largest flat bones in the body, somewhat resembling a necktie. The manubrium, body, and xiphoid process are its three regions.

Therefore, the sternum or breastbone is a long flat bone located in the central part of the chest.

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Hope this is correct

HAVE A GOOD DAY!

Answer:

a

Explanation:

because they need to protect the organization's information

Answer:

def lesser_adjacent_num ( N ):

     N = int( N )

     while N == True:

            next = N - 1

            print if  next >= 0 ? next : break

Explanation:

The python function above called "lesser_adjacent_num" accepts an argument called "N" which is an integer. The function print the number "next" which is a number less than the N integer for every loop for the condition where N is not zero.

Answer:

H cannot update it because it is probably protected.

If you cannot click in the table of contents, it is probably because it is protected. To disable this protection, choose Tools > Options > OpenOffice.org Writer > Formatting Aids, and then select Enable in the Cursor in protected areas section. If you wish to edit the table of contents without enabling the cursor, you can access it from the Navigator.

Explanation:

. To update a table of contents when changes are made to the document:

Right-click anywhere in the TOC.

From the pop-up menu, choose Update Index/Table. Writer updates the table of contents to reflect the changes in the document.

You can also update the index from the Navigator by right-clicking on Indexes > Table of Contents1 and choosing Index > Update from the pop-up menu.

Answer:

Here is the JAVA program:

import java.io.*;

import java.util.*;

public class Main {

public static void main(String[] args) throws FileNotFoundException{ //the start of main() function body, it throws an exception that indicates a failed  attempt to open the file

Scanner input = new Scanner(new File("file.txt")); //creates a Scanner object and a File object to open and scan through the file.txt    

coinFlip(input);    } //calls coinFlip method

public static void coinFlip(Scanner input) { //coinFlip method that accepts as its parameter a string input holding a file name

while(input.hasNextLine()) { //iterates through the input file checking if there is another line in the input file

Scanner scan = new Scanner(input.nextLine()); //creates a Scanner object

int head = 0; // stores count of number of heads

int count = 0; //stores count of  total number of tokens

while(scan.hasNext()) { //iterates through the sequence checking if there is another sequence in the input file

String token= scan.next(); // checks and returns the next token

if (token.equalsIgnoreCase("H")||token.equalsIgnoreCase("Heads")) { //compares H or Heads with the tokens in file ignoring lower case and upper case differences

           head++;                } //if a token i.e. any form of heads in file matches with the H or Heads then add 1 to the number of heads

           count++;            } //increment to 1 to compute total number of counts

double result = Percentage(head, count); //calls Percentage method passing number of heads and total counts to compute the percentage of heads

System.out.println(head + " heads " + "(" + result +"%)"); // prints the number of heads

if(result >= 50.00) { //if the percentage is greater or equal to 50

System.out.println("You win!");} //displays this message if above if condition is true

else //if the percentage is less than 50

{System.out.println("You lose!");}  }    } //displays this message if above if condition is false

public static double Percentage(int h, int total) { //method to compute the percentage of heads

double p = (double)h/total* 100; // divide number of heads with the total count and multiply the result by 100 to compute percentage

return p;    } } //returns result

Explanation:

The program is well explained in the comments mentioned with each line of the above code. I will explain how the method coinFlip  works.

Method coinFlip accepts a string holding a file name as its parameter. It opens that file and reads its contents as a sequence of tokens. Then it reads and scans through each token and the if condition statement:

if (token.equalsIgnoreCase("H")||token.equalsIgnoreCase("Heads"))

checks if the each token in the sequence stored in the file is equal to the H or Heads regardless of the case of the token. For example if the first token in the sequence is H then this if condition evaluates to true. Then the head++ statement increments the count of head by 1. After scanning each token in the sequence the variable count is also increased to 1.

If the token of the sequence is HeAds then this if condition evaluates to true because the lower or upper case difference is ignored due to equalsIgnoreCase method. Each time a head is found in the sequence the variable head is incremented to 1.

However if the token in the sequence is Tails then this if condition evaluates to false. Then the value of head variable is not incremented to 1. Next the count variable is incremented to 1 because this variable value is always incremented to 1 each time a token is scanned because count returns the total number of tokens and head returns total number of heads in the tokens.

Percentage method is used to return the percentage of the number of heads in the sequence. It takes head and count as parameters (h and total). Computes the percentage by this formula h/total* 100. If the result of this is greater than or equal to 50 then the message  You win is displayed otherwise message You lose! is displayed in output.

Answer:

total_owls = 0

num_owls_A = int(input())

num_owls_B = int(input())

total_owls = num_owls_A + num_owls_B

print("Number of owls:", total_owls)

Explanation:

Initialize the  total_owls as 0

Ask the user to enter num_owls_A and num_owls_B, convert the input to int

Sum the num_owls_A and num_owls_B and set it to the total_owls

Print the total_owls

avg_owls=0.0

num_owls_zooA = int(input())

num_owls_zooB = int(input())

num_owls_zooC = int(input())

num_zoos = 3

avg_owls = (num_owls_zooA + num_owls_zooB + num_owls_zooC) / num_zoos

print('Average owls per zoo:', int(avg_owls))

There is a problem while you are trying the calculate the average owls per zoo.

As you may know, in order to calculate the average, you need to get the total number of owls in all the zoos, then divide it to the number of zoos. That is why you need to sum num_owls_zooA + num_owls_zooB + num_owls_zooC and divide the result by num_zoos

Be aware that the average is printed as integer value. That is why you need to cast the avg_owls to an int as int(avg_owls)

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Answer:

Here is the Python and C++ program.

Python program:

gallons_paint=0.0  #declare and initialize gallons_paint  

wall_area = float(input())  #prompts user to enter wall area

gallons_paint = wall_area / 350  #formula to compute gallons of paint needed to cover the given square feet of walls

print(gallons_paint) #display the result

C++ program:

#include <iostream> //to use input output functions

using namespace std; //to identify objects like cin cout

int main(){ //start of main() function body

   float wall_area,gallons_paint=0; //declare variables

   cin>>wall_area; //reads input wall area from user

   gallons_paint=wall_area/350; //formula to compute gallons of paint needed to cover the given square feet of walls

   cout<<gallons_paint;} //print the result

Explanation:

The program prompts the user to enter the area of wall and stores the input value in wall_area

Suppose the user enters 250 as wall area

The the formula gallons_paint=wall_area/350; works as follows:

gallons_paint = wall_area/350;

                         = 250/350

                         = 0.71428

gallons_paint  = 0.71428

So the output is:

0.71428

The screenshot of the program along with its output is attached.

Answer:

Find the explanation below.

Explanation:

Asymmetric encryption is a system of communication in technology between two parties where keys which are variable values coined by an algorithm are used to encrypt and decrypt the messages between the sender and receiver. The two types of keys used in this process are the public and private keys. The public keys can be used by anyone to encrypt messages which are then sent to the receiver who decrypts the message using his private keys. The private keys on the other hand are only shared by the person who first set up the key. In sending messages this way, the receiver's public key is gotten from a public directory, and the message encrypted by the sender. The receiver can then decrypt the message using his private keys. But if the message is encrypted using a private key, it is decrypted by the receiver with a public key.

The one-way nature of this style of communication ensures data integrity and confidentiality since users are not required to disclose their keys and only the two parties can access the message. The two end users also preserve the right to authenticate the keys. The transport layer security protocol makes use of asymmetric encryption.

Answer:

a)

Account account = new Account();

b)

Account account = new Account(firstName, age);

c)

account.Age = newAge;

d)

int count = Account.Count;

Explanation:

a)

new is the keyword which is used to create an object.

Account is the class name.

account is the name of an object of class Account.

account object is created to access the class Account.

Account() is the constructor of Account class. This is the default constructor and it has no parameters. Constructor has the same name as class. When a account object is created, constructor Account will be invoked.

b)

Account is the class name.

account is the name of an object of class Account.

Account() is the constructor of Account class. This is the parameterized constructor and it has two parameters i.e. firstName and age. Constructor has the same name as class. When a account object is created, constructor Account will be invoked.

c)

account.Age = newAge;

Here account is the object name. The Age property to access and update the private field of the Account class is set to the variable named newAge.

d)

This int count = Account.Count; statement gets the value of a public static field named Count that’s defined in the Account class, and stores the value in a new int variable named count.