For this assignment, you will create flowchart using Flow gorithm and Pseudocode for the following program example: Hunter Cell Phone Company charges customer's a basic rate of $5 per month to send text messages. Additional rates apply as such:The first 60 messages per month are included in the basic billAn additional 10 cents is charged for each text message after the 60th message and up to 200 messages.An additional 25 cents is charged for each text after the 200th messageFederal, state, and local taxes add a total of 12% to each billThe company wants a program that will accept customer's name and the number of text messages sent. The program should display the customer's name and the total end of the month bill before and after taxes are added.

Answer:

Explanation:

MIPS program which increments from 0 to 15 and display results in Decimal on the console

In this program the user defined procedures print_int and print_eot were used to print the integer values and new line characters(\n) respectively.the mechanisam of the loop is explaine in the comment section of the program.

     addi $s0, $0, 0

     addi $s1, $0, 15

print_int:

              li $v0, 1                # system call to print integer

              syscall                

              jr $ra                     # return

print_eol:                      # prints "\n"

            li $v0, 4            

              la $a0, linebrk      

              syscall              

              jr $ra                  # return

main: . . .          

              li $a0, 0                # print 0

              jal print_int         # print value in $a0

loop:   move $a0, $s0           # print loop count

       jal print_int        

       jal print_eol           # print "\n" character

       addi $s0, $s0, 1        # increment loop count by 1

       ble $s1, $s0, loop      # exit if $s1<$s0

beq $s0, $0, end

end:

MIPS progam to increment from 0 to 15 and display results in Hexadecimal on the console

this program is slightly differed from the previous program in this program the system call issued in print_int is implemented with a system call that prints numbers in hex.

addi $s0, $0, 15

     addi $s1, $0, 0

print_int:

   li      $v0,34                  # syscall number for "print hex"

   syscall                         # issue the syscall

              jr $ra                     # return

print_eol:                      # prints "\n"

            li $v0, 4            

              la $a0, linebrk      

              syscall              

              jr $ra                  # return

main: . . .          

              li $a0, 0                # print 0

              jal print_int         # print value in $a0

loop:   move $a0, $s0           # print loop count

       jal print_int        

       jal print_eol           # print "\n" character

       addi $s0, $s0, 1        # increment loop count by 1

       ble $s1, $s0, loop      # exit if $s0>$s1

beq $s0, $0, end

end:

Answer:

In dynamic scoping the current block is searched by the compiler and then all calling functions consecutively e.g. if a function a() calls a separately defined function b() then b() does have access to the local variables of a(). The visible variables with the name of the function in which it was defined are given below.

Explanation:

a. main calls fun1; fun1 calls fun2; fun2 calls fun3

Solution:

b. main calls fun1; fun1 calls fun3

Solution:

c. main calls fun2; fun2 calls fun3; fun3 calls fun1

Solution:

Here variables c, d and e of fun2 are not visible .

d. main calls fun1; fun1 calls fun3; fun3 calls fun2

Solution:

Answer:

A. 20 30

Explanation:

Given

Linked list: 10, 20, 30, 40, 50, 60

Required

The output of the following code segment

pCur = head->next;

cout << pCur->data << " ";

cout << pCur->next->data << endl;

A linked list operates by the use of nodes which begins from the head to the next node, to the next, till it reaches the last;

The first line of the code segment; "pCur = head->next; " shifts the node from the head to the next node

The head node is the node at index 0 and that is 10;

This means that the focus has been shifted to the next at index 1 and that is 20;

So, pCur = 20

The next line of the code segment; cout << pCur->data << " "; prints pCur and a blank space

i.e. "20 " [Take note of the blank space after 20]

The last line "cout << pCur->next->data << endl; " contains two instructions which are

1. pCur = next->data;

2. cout<<pCur->data;

(1) shifts focus to the next node after 20 ; This gives pCur = 30

(2) prints the value of pCur

Hence, the output of the code segment is 20 30

Explanation:

1. I grew up in a small community and my family owned a small business which involved manual labor. Although, having little income, my Dad sought to give us the best education he could afford from his little savings. With the small family savings I went to college where I studied Information Technology, after graduating I worked for a startup firm for two years before I decided to proceed with post graduate studies and I attained my Masters in UA. I met a friend who recommended I study in University of Cumberlands, I hesitated initially but was convinced after I researched the institution.

2. I'm particularly interested in the applications of information technology in labor intensive businesses and work because of my family background.

3. I have my own startup firm that designs work communication software for companies, however I intend to broaden my knowledge through this program into Information Technology based systems using machine learning algorithm.

4. I am committed to work and study, as some have called me a curious mind because of my passion towards learning.

5. I view obtaining a doctorate as not just a personal achievement but of lasting benefits to society as I use knowledge derived to improve the work experience of society.

I am confident that after my program I would be one the renowned contributors to the applications of information technology in labor intensive businesses.

Answer:

w = ["Algorithm", "Logic", "Filter", "Software", "Network", "Parameters", "Analyze", "Algorithm", "Functionality", "Viruses"]

for i in range (len(w)):

   print(w[i].upper())

Explanation:

every element needs to be upper cased

A loop that prints the words in uppercase letters is written below with the help of Python.

Simply said, a function is a "chunk" of code that you may reuse repeatedly so instead of having to write it out several times. Software developers can divide an issue into smaller, more manageable parts, which can each carry out a specific task, using functions.

A function, according to a technical definition, is a relationship between an amount of parameters and a set of potential outputs, where every other input is connected to precisely one output.

All the functions are to be written in uppercase:

w = ["Algorithm", "Logic", "Filter", "Software", "Network", "Parameters", "Analyze", "Algorithm", "Functionality", "Viruses"]

for i in range (len(w)):

  print(w[i].upper())

Learn more about function, Here:

https://brainly.com/question/29050409

#SPJ5

Answer:

Explanation:

I mean I would copy and paste it. I hope thats right

Answer:

A digital security risk is an action that could result in damage to a computer or similar device's hardware, software, data etc.

Answer:

Sorted Array A { } = { 1, 4, 23, 32, 34, 34, 67, 78, 89, 100 }

Explanation:

Binary tree is drawn given that the binary tree do not follow both minimum heap and maximum heap property, therefore, it is not a heap.

See attached picture.

Answer:

~CaptnCoderYankee

Don't forget to award brainlyest if I got it right!

Answer:

Network Operating System (NOS)

Explanation:

A network operating system (NOS) is an operating system that makes different computer devices connect to a common network in order to communicate and share resources with each other using a server. A network operating system can be used by printers, computers, file sever among others, and they are connected together using a local area network. This local area network which they are connected to works as the server.

The NOS also acts as a network security because it could be used as an access control or even user authentication.

There are two types of NOS

1) Peer to peer network operating system.

2) Client/server network operating system

A network operating system (NOS) is a server-based operating system oriented to computer networking and may include directory services, network management, network monitoring, network policies, user group management, network security, and other network-related functions.

Answer:

A. Make Laws

Explanation:

Montesquieu believed the ways to go about limiting the power of a monarch would be to split up power among 3 groups, Judicial, Legislative and Executive. The American Presidential system and Constitution used Montesquieu's writings as a foundation for their principles.

Answer:

33600

Explanation:

Let M = ABE16

and N = DF416

In hexadecimal representation: 10 = A, 11 = B, 12 = C, 13 = D, 14 = E, 15 = F

Step 1: Find the 15's complement of N

 F  F  F  F  F

- D  F  4  1  6

2  0  B  E  9

Step 2: Add 15's complement of N to M

  A  B  E  1  6

+ 2  0  B  E  9  

  C  C  9  F  F

Step 3: Find the 15's complement of CC9FF

 F  F   F  F  F

- C  C  9  F  F

 3   3  6  0  0

Therefore, ABE16 - DF416 using 15's complement is 33600

Answer:

Given

The above lines of code

Required

Rearrange.

The code is re-arrange d as follows;.

#include<iostream>

int main()

{

int userNum;

scanf("%d", &userNum);

if (userNum > 0)

{

printf("Positive.\n");

}

else

{

printf("Non-positive, converting to 1.\n");

userNum = 1;

printf("Final: %d\n", userNum);

}

return 0;

}

When rearranging lines of codes. one has to be mindful of the programming language, the syntax of the language and control structures in the code;

One should take note of the variable declarations and usage

See attachment for .cpp file

Answer:

#Declare variable to get the input

#list from the user

input_list = input()

#split the list into words by space

list = input_list.split()

#Begin for-loop to iterate each word

#to get the word frequency

for word in list:

#get the frequency for each word

frequency=list.count(word)

#print word and frequency

print(word,frequency)

Explanation:

This program that we are told to write will be done by using python programming language/ High-level programming language.

The question wants us to write a program in which the output will be as below;

hey 1

hi 2

Mark 1

hi 2

mark 1

Just by imputing hey hi mark hi mark.

Just copy and run the code below.

#Declare variable to get the input

#list from the user

input_list = input()

#split the list into words by space

list = input_list.split()

#Begin for-loop to iterate each word

#to get the word frequency

for word in list:

#get the frequency for each word

frequency=list.count(word)

#print word and frequency

print(word,frequency)

In this exercise we have to use the knowledge in computational language in python to write the following code:

We have the code can be found in the attached image.

So in an easier way we have that the code is

input_list = input()

list = input_list.split()

for word in list:

frequency=list.count(word)

print(word,frequency)

See more about python at brainly.com/question/18502436

Answer:

Explanation:

PROGRAM

main.c

#include <fcntl.h>

#include <stdio.h>

#include <stdlib.h>

#include <sys/stat.h>

#include <sys/types.h>

#include <unistd.h>

#include "myio.h"

int checkInt(char *arg);

int main(int argc, char *argv[]) {

  int doubles, i, ints, lines;

  char newline;

  FILE *out;

  int x, y, z;

  newline = '\n';

  if (argc != 5) {

     printf("Usage is x y z output_filename\n");

     return 0;

  }

  if (checkInt(argv[1]) != 0)

     return 0;

  ints = atoi(argv[1]);

  if (checkInt(argv[2]) != 0)

     return 0;

  doubles = atoi(argv[2]);

  if (checkInt(argv[3]) != 0)

     return 0;

  lines = atoi(argv[3]);

  out = fopen(argv[4], "a");

  if (out == NULL) {

     perror("File could not be opened");

     return 0;

  }

  for (x = 0; x < ints; x++) {

     int n = ReadInteger();

     printf("%d\n", n);

     fprintf(out, "%d\n", n);

  }

  for (y = 0; y < doubles; y++) {

     double d = ReadDouble();

     printf("%lf\n", d);

     fprintf(out, "%lf\n", d);

  }

  for (z = 0; z < lines; z++) {

     char *l = ReadLine();

     printf("%s\n", l);

     fprintf(out, "%s\n", l);

     free(l);

  }

  fclose(out);

  return 0;

}

int checkInt(char *arg) {

  int x;

  x = 0;

  while (arg[x] != '\0') {

     if (arg[x] > '9' || arg[x] < '0') {

        printf("Improper input. x, y, and z must be ints.\n");

        return -1;

     }

     x++;

  }

  return 0;

}

myio.c

#include <limits.h>

#include <stdio.h>

#include <stdlib.h>

#include <unistd.h>

char *ReadInput(int fd) {

  char buf[BUFSIZ];

  int i;

  char *input;

  int r, ret, x;

  i = 1;

  r = 0;

  ret = 1;

  input = calloc(BUFSIZ, sizeof(char));

  while (ret > 0) {

     ret = read(fd, &buf, BUFSIZ);

     for (x = 0; x < BUFSIZ; x++) {

        if (buf[x] == '\n' || buf[x] == EOF) {

           ret = -1;

           break;

        }

        input[x*i] = buf[x];

        r++;

     }

     i++;

     if (ret != -1)

        input = realloc(input, BUFSIZ*i);

  }

  if (r == 0)

     return NULL;

  input[r] = '\0';

  input = realloc(input, r+1);

  return(input);

}

int ReadInteger() {

  char *input;

  int go, num, x;

  go = 0;

  do {

     go = 0;

     printf("Input an integer\n");

     input = ReadInput(STDIN_FILENO);

     for (x = 0; x < INT_MAX; x++) {

        if (x == 0&& input[x] == '-')

           continue;

        if (input[x] == 0)

           break;

        else if (input[x]> '9' || input[x] < '0') {

           go = 1;

           printf("Improper input\n");

           break;

        }

    }

  } while (go == 1);

  num = atoi(input);

  free(input);

  return num;

}

double ReadDouble(void) {

  int dec, exp;

  char *input;

  int go;

  double num;

  int x;

  do {

     go = 0;

     dec = 0;

     exp = 0;

     printf("Input a double\n");

     input = ReadInput(STDIN_FILENO);

     for (x = 0; x < INT_MAX; x++) {

        if (x == 0&& input[x] == '-')

           continue;

        if (input[x] == 0)

           break;

        else if (input[x] == '.' && dec == 0)

           dec = 1;

        else if (x != 0&& (input[x] == 'e' || input[x] == 'E') && exp == 0) {

           dec = 1;

           exp = 1;

        }

        else if (input[x]> '9' || input[x] < '0') {

           go = 1;

           printf("Improper input\n");

           break;

        }

     }

  } while (go == 1);

  num = strtod(input, NULL);

  free(input);

  return num;

}

char *ReadLine(void) {

  printf("Input a line\n");

  return(ReadInput(STDIN_FILENO));

}

char *ReadLineFile(FILE *infile) {

  int fd;

  fd = fileno(infile);

  return(ReadInput(fd));

}

myio.h

#ifndef _myio_h

#define _myio_h

/*

* Function: ReadInteger

* Usage: i = ReadInteger();

* ------------------------

* ReadInteger reads a line of text from standard input and scans

* it as an integer. The integer value is returned. If an

* integer cannot be scanned or if more characters follow the

* number, the user is given a chance to retry.

*/

int ReadInteger(void);

/*

* Function: ReadDouble

* Usage: x = ReadDouble();

* ---------------------

* ReadDouble reads a line of text from standard input and scans

* it as a double. If the number cannot be scanned or if extra

* characters follow after the number ends, the user is given

* a chance to reenter the value.

*/

double ReadDouble(void);

/*

* Function: ReadLine

* Usage: s = ReadLine();

* ---------------------

* ReadLine reads a line of text from standard input and returns

* the line as a string. The newline character that terminates

* the input is not stored as part of the string.

*/

char *ReadLine(void);

/*

* Function: ReadLine

* Usage: s = ReadLine(infile);

* ----------------------------

* ReadLineFile reads a line of text from the input file and

* returns the line as a string. The newline character

* that terminates the input is not stored as part of the

* string. The ReadLine function returns NULL if infile

* is at the end-of-file position. Actually, above ReadLine();

* can simply be implemented as return(ReadLineFile(stdin)); */

char *ReadLineFile(FILE *infile);

Answer:

Explanation:

a.  

// Rewriting the given function without using any square brackets

double computeAverage(const double* scores, int nScores)

{

const double* ptr = scores;

double tot = 0;

int i;//declaring integer variable

for(i=0;i<nScores;i++)

{

//using an integer variable i to visit each double in the array

tot += *(ptr+i);

}

return tot/nScores;

}

b.

// Rewriting the given function without using any square brackets

const char* findTheChar(const char* str, char chr)

{

for(int k=0;(*(str+k))!='\0';k++)

{

if ((*(str+k)) == chr)

return (str+k);

}

return nullptr;

}

c.

//Now rewriting the function shown in part b so that it uses neither square brackets nor any integer variables

const char* findTheChar(const char* str, char chr)

{

while((*str)!='\0')

{

if ((*str) == chr)

return (str);

str++;

}

return nullptr;

}

Answer:

6

Explanation:

Answer:

4

Explanation:

It is between 3 and 13. Please answer some of my questions too! :)

Answer:

#main function start

def main():

#list to store names entered by user

name_list=[]

#variable to take input from user to keep rotating the loop and take input from user again and again

ch='y'

fname=""

lname=""

while ch=='y' or ch=='Y':#keep the loop taking input from user till user do not want to enter more

#take input from user

name=[]

fname=input("Enter first name: ")

lname=input("Enter last name: ")

name.append(fname)

name.append(lname)

#append the previously input first and last name in main list

name_list.append(name)

ch=input("Do you want to continue: (y/Y for yes)")

#main function end

#get file name from user

filename=input("Enter output file name")

#open file in write mode

file=open(filename,"w")

#keep loop add data in file till main list 'name_list' have

for i in name_list:

#Write data in file

file.write("%s, %s\r\n" % (i[0],i[1]))

#close the file

file.close()

#call main function

main()

Explanation:

input the above code and see output

Find the given attachments

Answer: you want to list all LC structers

Explanation:

Answer:

total = 0

count = 0

while(True):

   grade = float(input("Enter a grade: "))

   if grade < 0:

       break

   else:

       total += grade

       count += 1

average = total/count

print("The average is: " + str(average))

Explanation:

*The code is in Python.

Initialize the total and count as 0

Create a while loop that iterates until a specific condition is met inside the loop

Inside the loop, ask the user to enter a grade. If the grade is smaller than 0, stop the loop. Otherwise, add the grade to the total and increment the count by 1.

When the loop is done, calculate the average, divide the total by count, and print it

Filtering is a useful way to see only the data that you want displayed in Access databases.

Answer:

Given: K(x,y) denotes statement:

"x knows y"

D denote domain of all people.

Explanation:

1. Everybody knows somebody.

Solution:

∀x∈D ∃y∈D : K(x, y)

∀ means for all. Here it is used for Everybody.

∃ means there exists some. Here it represents Somebody.

∈ means belongs to . Both x and y belongs to the domain D of all people.

2. There is somebody that no one knows.

Solution:

∀x∈D ∃y∈D : ¬K(x, y)

∀ means for all. ∃ means there exists some. ∈ means belongs to both x and y belongs to the domain D of all people.¬ this is negation sign which means not K(x,y). So the negation of everybody knows somebody can be expressed as there is somebody that no one knows.

3. There is no one who knows everybody

Solution:

This can be represented in both the ways below.

∀ y∈D ∃ x∈D : K(x, y)

∀ means for all. ∃ means there exists some. ∈ means belongs to both x and y belongs to the domain D of all people.

∀x∈D ∀y∈D : ¬K(x, y)

∀ means for all. The negation shows that there is no one who knows everybody.

Answer:

The correct option is D

D) CVS

Explanation:

A CVS file saves a workbook as a comma-delimited text file for use on another windows operating system. It ensures that line breaks, tab characters and other characteristics are interpreted correctly.

A CVS file stands for Comma Separated Value file, which basically allows the data to be saved in tabular format. However they differ from other spreadsheet file types because you can only have a single sheet in a file. Moreover, you cannot save cell, column or row in it.

Answer:

C. Resource Pool share value.

Explanation:

The vSphere is a term used to describe the VMware’s virtualization cloud platform. The vSphere comprises of the following, vCenter Server, ESXi, Virtual Machine File System (VMFS) and vCenter Client.

If a vSphere Administrator receive complaints that a database Virtual Machine (VM) is experiencing performance issues and the Virtual Machine (VM) is a member of the high priority resource pool and the cluster has not experienced contention.

In order to address immediate performance concerns, the vSphere Administrator should check the Resource Pool share value.

VMware resource pool refers to the aggregated central processing unit and memory allocated to a Virtual Machine (VM) for flexible management of the resources.

Also, the vSphere Administrator should install VMware tools to check which processes are having high CPU usage by using vimtop, as well as checking if vCenter Server is swapping.

Answer:

automated

Explanation:

Basically a Business Impact Analysis (BIA) estimates and determines the effects of a business activity and process disturbances. These disruptions can be natural or electronic disasters. It also collects information which is used to establish recovery plan. It identifies the business vulnerabilities and works on the strategies in order to reduce such potential hazards. The BIA involves both manual and automated processes. BIA involves automated processes which include the automated software tools that enables the protection of the confidential information of the users and also generates automated reports about the critical business processes.

Answer: Provided in the explanation section

Explanation:

(a). BIND is a type of DNS server used on the internet, actually it is reported to be most widely used DNS server on the internet. Also BIND is the de-facto standard on Unix like operating systems and Linux. Thus it can be used to locate computers on a network using domain names instead of their IP addresses. BIND was originally programmed at the University of California, Berkeley in the 1980s, this was made possible by a grant from US based DARPA program.

Originally, BIND stood for Berkeley Internet name daemon, but nowadays it is called Berkeley Internet name domain.

(b). In some of the clients in windows and Windows servers we will install LDP. By the LDP.exe we can test tDNS, Active directory woking , even we can add or modify. LDP is used to test and verify the Active Directory objects. We will install in windows machine then open the software, then from the menu we will select connect and type the IP adress of the Active Directory then we will select or type the objects of Active directory. For Linux system we have to openldap and can search the AD.

From Linux or Unix we can use the nmap tool which can test the network connectivity between all the computers whether it is windows or Linux. By nmap we can test by ICMP protocol to verify the network connectivity. By nmap we can scan all the port which are required to open from client to the server. We can also use nslookup to test DNS server from the client. By nslookup we can check that client can connect the server.

Answer:

servers to be upgraded are : APPLICATION AND LOAD BALANCING SERVERS

servers still appropriate to use : DNS AND DHCP SERVERS

Explanation:

The recommendations to be made in line with what the new director of information is asking for includes :

1 ) For the servers to be upgraded : The servers that requires upgrades  includes the APPLICATION SERVER and LOAD BALANCING SERVER. this is because these two servers are critical to the growth/expansion of any business, and they handle large volume of data

Recommendations on the servers upgrade includes:

Hardware : 2.3 GHz Intel Xeon Gold 5118 12-Core

CPU chip set :  Socket: FCLGA3647, Type: NSBM

Speed : processor 3.2 GHz

caching:   > 100 Gb

2)  For servers that do not necessarily need to be upgraded : The servers that do not need immediate upgrade are DNS and DHCP

Answer:

getInfo(); ==

getSy.Info()

Explanation:

Get System Info