Which can be used to code a simple web page? CSS JavaScript Text editor Web browser

Answer:

Enterprise software/system

Explanation:

Enterprise software which is also known as Enterprise Application Software (EAS) is computer software that its primary function is to meet the needs of an organization rather than that of an individual.

EAS or Enterprise System is the software development management tool that easily integrates into a business' enterprise software system.

Answer:

num_ones = amount_to_change % 5

Explanation:

It is given that the cashier has to change the money in the maximum number of 5 dollar bills and remaining as $1 bills.

We have to write a single statement of code to assign the value to variable num_ones to find the value of $1 bills given amount_to_change.

Let us solve this by taking example:

If amount_to_change = $19

Then total number of $5 bills that can be given is 3. And amount that can be given as $5 bills is [tex]\bold{3 \times 5 = \$15}[/tex]

So, the remaining amount i.e. $19 - $15 = $4 will be given as one dollar bills.

Out of these $19, 4 bills of $1 can be found using the Modulus (%) operator.

Modulus operator leaves the remainder i.e.

Output of p % q is the remainder when a number 'p' is divided by 'q'.

19 % 5 = 4

4 is the number of one dollar bills to be given.

So, single line of code for the number of one dollar bills can be written as:

num_ones = amount_to_change % 5

Let us try it for amount_to_change  = $30

We know that 6 number of $5 bills can be used to change the amount of $30 and no one dollar bill required.

num_ones = 30 % 5 = 0 (because 30 is completely divisible by 5 so remainder is 0)

So, the correct statement is:

num_ones = amount_to_change % 5

Answer:

A total of 6,000 bps of bandwidth is used by the distributed routing algorithm

Explanation:

This is a bandwidth requirement question.

We proceed as follows;

To calculate the total number of bits for a routing table, we use the following formula;

Routing table=Number of routers * length of cost

we are given the following parameters from the question;

Number of routers = 500

length of cost = 4 bits

Routing table = 500*4

=2000

Hence, a routing table is 2000 bits in length.

Now we proceed to calculate the bandwidth required on each line using the formula below;

Bandwidth = no.of seconds * no.of bits in routing table

Bandwidth required on each line = 3*2000

=6000

Answer:

Replace /* Your code goes here */  with

for(i =0; i<NUM_VALS; i++) {

    printf("%d", origList[i]*offsetAmount[i]);

printf(";");

}

Explanation:

The first line is an iteration statement iterates from 0 till the last element in origList and offsetAmount

for(i =0; i<NUM_VALS; i++) {

This line calculates and print the product of element in origList and its corresponding element in offsetAmount

    printf("%d", origList[i]*offsetAmount[i]);

This line prints a semicolon after the product has been calculated and printed

printf(";");

Iteration ends here

}

Answer:

False

Explanation:

Budget and Financial reports are used to set targets for organization's expenditure. These reports display cost and expenditures related to all assets, raw material, inventory of the organization. The variance is then calculated based on targeted figures and the actual expenses. The deviation from target is found by the variance and then actions are taken for it. Managers rely heavily on these reports. Statistical reports are non financial reports. These focus on customer satisfaction, employee performance, staff turnover rate and assets performance.

Answer:

In several organizations, the operating system (OS) functions at the core of the personal computers in use and the servers. The OS ought to be secure from external threats to its stored data and other resources.

Explanation:

In this regard, Intrusion detection systems play a critical role in protecting the sensitive data, along with software firewalls. Hardware firewalls protect the hardware elements in the network (such as storage drivers etc), the IDS and as well as the IPS.

Given that,

Size of each sandbox = 16 MB

Address space = 32- bit

We need to calculate the total size of address space

Using formula of total size

[tex]\text{total size of address space}=2^{32}[/tex]

[tex]\text{total size of address space}=4\times10^9[/tex]

[tex]\text{total size of address space}=4\ Gigabytes[/tex]

We need to calculate the total of applets

Using formula of total applets

[tex]\text{total applets}=\dfrac{\text{total size of address space}}{\text{size of one applet}}[/tex]

Put the value into the formula

[tex]\text{total applets}=\dfrac{4\ GB}{16\ MB}[/tex]

[tex]\text{total applets}=\dfrac{4\times1024\ MB}{16\ MB}[/tex]

[tex]\text{total applets}=256[/tex]

Hence,  The total of applets are 256.

Explanation:

bottom [20, 20, 17, 11, 8, 8, 3, 21 top the following stack is not sorted (the 15 is out of place), so passing it to your method should return a result of false: bottom [18, 12, 15, 6, 11 top an empty or one-element stack is considered to be sorted. when your method returns, the stack should be in the same state as when it was passed in. in other words, if your method modifies the stack, you must restore it before returning.

Answer:

c. rear = (rear + 1) % elements.length; elements[rear] = element;

Explanation:

In the above statement:

Name of the array is elements.

rear holds current index of elements array where current rear element of queue is stored. Front are rear are two open ends of the queue and the end from which the element is inserted into the queue is called rear.

element is the element that is required to enqueue into the queue

Enqueue basically mean to add an element to a queue.

Here it is assumed that the queue is not full. This means an element can be added to the queue.

It is also assumed that code also correctly increments numElements.

rear = (rear + 1) % elements.length; This statement adds 1 to the rear and takes the modulus of rear+1 to the length of the array elements[]. This statement specifies the new position of the rear.

Now that the new position of rear is found using the above statement. Next the element can be enqueued to that new rear position using the following statement:

elements[rear] = element; this statement sets the element at the rear-th (new position) index of elements[] array.

For example we have a queue of length 5 and there are already 4 elements inserted into this queue. We have to add a new element i.e. 6 to the queue. There are four elements in elements[] array and length of the array is 5 so this means the queue is not full. Lets say that rear = 3

elements.length = 5

rear = 3

Using above two statements we get.

rear = (rear + 1) % elements.length;

       = 3 + 1 % 5

       = 4%5

      =  4

This computes the new position of rear. So the new position of rear is the 4-th index of elements[]. Now next statement:

elements[rear] = element;

elements[4] = 6

This statement adds element 6 to the 4-th index of elements[] array.

Thus the above statement enqueues element (element 6 in above example) into the queue.

Answer:

I am writing a Python program. Let me know if you want the program in some other programming language.        

import string  #to use string related functions

def wordcount(filename):  # function that takes a text file name as parameter and returns the number of occurrences of every word in file

   file = open(filename, "r")  # open the file in read mode

   wc = dict()  # creates a dictionary

   for sentence in file:  # loop through each line of the file

       sentence = sentence.strip()  #returns the text, removing empty spaces

       sentence=sentence.lower() #converts each line to lowercase to avoid case sensitivity

       sentence = sentence.translate(sentence.maketrans("", "", string.punctuation))  #removes punctuation from every line of the text file

       words = sentence.split(" ")  # split the lines into a list of words

       for word in words:  #loops through each word of the file

           if len(word)>2:  #checks if the length of the word is greater than 2

               if word in wc:  # if the word is already in dictionary

                   wc[word] = wc[word] + 1  #if the word is already present in dict wc then add 1 to the count of that word

               else:  #if the word is not already present

                   wc[word] = 1  # word is added to the wc dict and assign 1 to the count of that word                

   for w in list(wc.keys()):  #prints the list of words and their number of occurrences

       print(w, wc[w])  #prints word: occurrences in key:value format of dict        

wordcount("file.txt") #calls wordcount method and passes name of the file to that method

Explanation:

The program has a function wordcount that takes the name of a text file (filename) as parameter.

open() method is used to open the file in read mode. "r" represents the mode and it means read mode. Then a dictionary is created and named as wc. The first for loop, iterates through each line (sentence) of the text file. strip() method is used to remove extra empty spaces or new line character from each sentence of the file, then each sentence is converted to lower case using lower() method to avoid case sensitivity. Now the words "hello" and "Hello" are treated as the same word.

sentence = sentence.translate(sentence.maketrans("", "", string.punctuation))  statement uses two methods i.e. maketrans() and translate(). maketrans() specifies the punctuation characters that are to be deleted from the sentences and returns a translation table. translate() method uses the table that maketrans() returns in order to replace a character to its mapped character and returns the lines of text file after performing these translations.

Next the split() method is used to break these sentences into a list of words. Second for loop iterates through each word of the text file. As its given to ignore words of length 2 or less, so an IF statement is used to check if the length of word is greater than 2. If this statement evaluates to true then next statement: if word in wc:   is executed which checks if the word is already present in dictionary. If this statement evaluates to true then 1 is added to the count of that word. If the word is not already present  then the word is added to the wc dictionary and 1 s assigned to the count of that word.

Next the words along with their occurrences is printed. The program and its output are attached as screenshot. Since the frankenstein.txt' is not provided so I am using my own text file.

Answer:

[tex]\boxed{2}[/tex]

Explanation:

[tex](x+3)(x-1)[/tex]

Expand the brackets.

[tex]x(x-1)+3(x-1)[/tex]

[tex]x^2-x+3x-3[/tex]

Combine like terms.

[tex]x^2+2x-3[/tex]

The coefficient of [tex]x[/tex] is [tex]2[/tex].

Answer:

user_score = 0  #user score is initialized to 0

simonPattern = 'RRGBRYYBGY'# stores the sequence of characters in simonPattern

userPattern  = input("Enter user pattern: ")  # prompts the user to enter user pattern and stores the pattern in userPattern

for x in range(len(simonPattern)):  #loops through entire length of simonPattern using x as index variable

   if userPattern[x] == simonPattern[x]:  # checks if the element at x-th index of userPattern matches the element at x-th index of simonPattern

       user_score = user_score + 1  # adds 1 to user score if above if condition evaluates to true

   else:  #if a mismatch occurs  

       break  #loop break if mismatch occurs

print("User Score:",user_score) #prints the computed user score

Explanation:

The program first declares a simonPattern string variable and assign it a sequence of characters. You can also prompt to enter the Simon pattern as

simonPattern = input("Enter Simon pattern:")

If you want to hard code the values of both simonPattern and userPattern then change the first two statements after user_score =0 statement as:

simonPattern = 'RRGBRYYBGY'

userPattern  = 'RRGBBRYBGY'

Now I will explain the working of the for loop.

for x in range(len(simonPattern)):

len method is used which returns the length of the string stored in simonPattern. For example if simonPattern = 'RRGBRYYBGY' then len returns 10

range method is used to generate a sequence of numbers for x. This means value of x starts from 0 and it  keeps incrementing to 1 until the length of the simonPattern is reached.

At first iteration:

Let suppose user enters the pattern: RRGBBRYBGY

x starts at index 0. Value of x=0 is less than the length of simonPattern i.e. 10 so the body of loop executes. The body of loop contains the following if statement:

if userPattern[x] == simonPattern[x]:

It matches both the string for the value of x.

if userPattern[0] == simonPattern[0]:

First character (at 0th index) of userPattern is R and in simonPattern is also R so this statement executes:  user_score = user_score + 1  which adds 1 to the user_score. So user_score=1

Next iteration:

x=1

if userPattern[1] == simonPattern[1]:

Second character (at 1st index) of userPattern is R and in simonPattern is also R so this statement executes:  user_score = user_score + 1  which adds 1 to the user_score. So user_score=2

Next iteration:

x=2

if userPattern[2] == simonPattern[2]:

Third character (at 12nd index) of userPattern is G and in simonPattern is also G so this statement executes:  user_score = user_score + 1  which adds 1 to the user_score. So user_score=3

Next iteration:

x=3

if userPattern[3] == simonPattern[3]:

Fourth character (at 3rd index) of userPattern is B and in simonPattern is also B so this statement executes:  user_score = user_score + 1  which adds 1 to the user_score. So user_score=4

Next iteration:

x=4

if userPattern[4] == simonPattern[4]:

Fifth character (at 4th index) of userPattern is B and in simonPattern is R so the if part does not execute and program moves to the else part which has a break statement which means the loop break. The next print(user_score) statement print the value of user_score. As user_score=4 computed in above iterations so output is:

User Score: 4

Answer:

The other answer was so close, but the _ was left out so the answer was wrong.

Written in Python:

user_score = 0

simon_pattern = input()

user_pattern  = input()

for x in range(len(simon_pattern)):

   if user_pattern[x] == simon_pattern[x]:

       user_score = user_score + 1

   else:

       break

print('User score:', user_score)

Explanation:

Answer:

When using find command in word we can search all of the above

Answer: DHCP (dynamic host configuration protocol)  is a protocol which automatically processes the configuring devices on IP networks.It allows them to to use network services like: NTP and any communication proto cal based on UDP or TCP. DHCP assigns an IP adress and other network configuration parameters to each device on a network so they can communicate with  other IP networks. it is a part of DDI solution.

Explanation:

Answer:

Add this statement at the end of your steps() function

return ""

This statement will not print None at the end of steps.

Explanation:

Here is the complete function with return statement added at the end:

def steps(number):  # function steps that takes number (number of steps to print) as parameter and prints the steps specified by number

   i = 1  #assigns 1 to i

   while i < number + 1:  

       string = str(i) * 3  #multiplies value of i by 3 after converting i to string

       string1 = str(string)  # stores the step pattern to string1

       if i != 0:  # if the value of i is not equal to 0

           string1 = (i * 4*' ' + "\b" + "\b" + "\b" + "\b") + string  #set the spaces between steps

       elif i == 1:  # if value of i is equal to 1

           string1 = string  #set the string to string1

       print(string1)  # print string1

       i = i + 1  # increments i at each iteration

   return "" #this will not print None at the end of the steps and just add a an empty line instead of None.

Screenshot of the corrected program along with its output is attached.

Answer:

The programming language is not stated;

The program written in C++ is as follows (See Explanation Section for detailed explanation);

#include<iostream>

using namespace std;

int main()

{

 int numdays;

 cout<<"Number of days of mileage: ";

 cin>>numdays;

 int miles[numdays];

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

{

 cout<<"Miles traveled on day "<<i+1<<": ";

 cin>>miles[i];

}

 int total = 0;

 for(int i=0;i<numdays;i++)

{

 total+=miles[i];

}

cout<<"Total mileage traveled: "<<total;

 return 0;

}

Explanation:

This line declares the number of days of mileage

 int numdays;

This line prompts user for days of mileage

 cout<<"Number of days of mileage: ";

This line accepts input from the traveler for days of mileage

 cin>>numdays;

This line creates an array

 int miles[numdays];

The italicized is an iteration that collects the user input for each day

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

{

 cout<<"Miles traveled on day "<<i+1<<": ";

 cin>>miles[i];

}

This line initializes variable to 0

 int total = 0;

The italicized is an iteration that adds up the mileage traveled by the traveler each day  

 for(int i=0;i<numdays;i++)

{

 total+=miles[i];

}

This line prints the total miles traveled

cout<<"Total mileage traveled: "<<total;

Answer:

word = input("Write a word: ").strip().lower()

without_space = word.replace(" ","")

condition = without_space == without_space[::-1]

print("Is %s palindrome?: %s"%(word,condition))

//import the Scanner class

import java.util.Scanner;

//Begin class definition

public class NumberOfMinutes{

    //Begin main method

    public static void main(String []args){

       //Create an object of the Scanner class

       Scanner input = new Scanner(System.in);

       //initialize a variable nm to hold the number of minutes

       int nm = 0;

       //Prompt the use to enter the number of hours

       System.out.println("Please enter the number of hours");

       //Receive the input using the Scanner object and

       //Store the entered number of hours in a variable nh

       int nh = input.nextInt();

       //Prompt the user to enter the number of days

       System.out.println("Please enter the number of days");

       //Receive the input using the Scanner object and

       //Store the entered number of days in a variable nd

       int nd = input.nextInt();

       //Prompt the user to enter the number of weeks

       System.out.println("Please enter the number of weeks");

       //Receive the input using the Scanner object and

       //Store the entered number of weeks in variable nw

       int nw = input.nextInt();

       //Prompt the user to enter the number of years

       System.out.println("Please enter the number of years");

       //Receive the input using the Scanner object and

       //Store the entered number of years in a variable ny

       int ny = input.nextInt();

       //Convert number of hours to minutes and

       //add the result to the nm variable

       nm += nh * 60;

       //Convert number of days to minutes and

       //add the result to the nm variable

       nm += nd * 24 * 60;

       //Convert number of weeks to minutes and

       //add the result to the nm variable

       nm += nw * 7 * 24 * 60;

       //Convert number of years to minutes and

       //add the result to the nm variable

       nm += ny * 52 * 7 * 24 * 60;

       //Display the number of minutes which is stored in nm

       System.out.println("The number of minutes is " + nm);

    }   //End main method

}  //End of class definition

Please enter the number of hours

>>12

Please enter the number of days

>>2

Please enter the number of weeks

>>4

Please enter the number of years

>>5

The number of minutes is 2664720

The code contains comments explaining every line of the program. Please go through the comments. The actual lines of executable code are written in bold face to distinguish them from comments.

A sample output has also been provided above. Also, a snapshot of the program file, showing the well-formatted code, has been attached to this response.

Answer:

The summary including its given problem is outlined in the following section on the interpretation.

Explanation:

That's not entirely feasible, since at least n similarities have to be made to order n quantities. Find the finest representation where the numbers of 1 to 10 have already been arranged.

⇒ 1 2 3 4 5 6 7 8 9 10

Let's say that we identify one figure as the key then compared it towards the numbers across the left. Whether the correct number is greater, therefore, left number, are doing nothing to switch the location elsewhere.

Because although the numbers have already been categorized 2 has always been compared to 1 which would be perfect, 3 becomes especially in comparison to 2 and so much more. This should essentially take 9 moves, or nearly O(n) moves.

If we switch that little bit already

⇒ 1 3 2 4 5 6 7 8 9 10

3 Is contrasted with 1. 2 will indeed be matched against 3 as well as 2. Since 2 has indeed been exchanged, it must, therefore, be matched with 1 as there might be a case whereby each number z exchanged is greater than the number Y as well as the quantity X < Y.

Only one adjustment expanded the steps which culminated in n+1.

It should be noted that it won't be possible because at least n similarities have to be made to order n quantities.

From the information given, it should be noted that it's not possible for the engineer to implement an algorithm that can sort n elements (e.g., numbers) in fewer than n steps.

This is because there are at least n similarities have to be made to order n quantities. In this case, the numbers of 1 to 10 have already been arranged. When a number is greater, it should be noted that the left number will do nothing to switch the location elsewhere.

Therefore, the information given by the engineer isn't feasible.

Learn more about algorithms on:

https://brainly.com/question/24953880

Answer:

315 Mbps

Explanation:

Given data :

30 frames per second

resolution = 1280 * 1024

what transmission rate is required can be calculated as

frames per second = rendered frames / number of seconds passed

30 = 30 / 1

Transmission rate = the rate at which the images are transmitted from source to destination =315 Mbps without data compression

Answer:

Wikis

Explanation:

Wikis can be defined as a database that was designed by users which gives users the opportunity to add, edit and delete collaborative web pages easily.

It lets users to manage contents easily and they are used to create static websites.

some wikis can be accessed by the public. A great example is Wikipedia.

Wikis allows publishing and sharing of documents to be done easily.

Answer:

Following are the answer to this question:

Explanation:

The server would be generally a powerful processor instead of a desktop pc. It must choose the Dell server browser to choose a server for industrial applications.

Answer:

The answer is "Intranet".

Explanation:

The intranet becomes an information exchange computer network, in which the resources collaborate with OS and other computing infrastructure within the organization. It is usually done without access from third parties and primarily uses for analysis of the data.

Here, the NMCI links the computer network of navy and maritime bodies on the bases of both the boats and the regional offices, that's why we can say that it is the example of the Internet.

Answer:

The answer is below

Explanation:

1.  Lack of password enforcement

2.  Outdated third party applications: this includes outdated software such as Apache, PHP, MySQL, OpenSSL, and VNC

3.  General lack of patch management for the OS

4.  General lack of system hardening:

5.  Lack of backups

Ways to hardened the Linux/Unix based computer system is do the following:

1.  Perform minimal Installation: this involves, installing softwares that are necessary, and limiting the number of admin users on the computer system.

2.  Carry out Code Auditing: this is a means of preventing security issues, by withing more secure softwares to prevent variable declarations, or otherwise unexpected logic.

3.  Ensure there is a Traffic filter firewall: this will helps to attacks from external users or hackers. Install, Web Applicatio firewall, which helps to filter out traffic.

4.  Carry out Softwares updates: a software update can help prevent or guide against security leaks in a software

5.  Install security updates only: this is done by filtering out the security-related programs and only upgrade packages which are referred to in the custom file.

Answer:

I am in the world

Explanation:

monkey is the very largest monkey in the world at Tilak was the captain of the school football team Sudheer and Tilak did not get along at first because he was not a friendly boy

Answer:

223 8s an order of algorithm faults refrénce fifio 14 suppose in 14 and 123 no need to take other numbers

Answer:

[tex]Attenuation = 0.458\ db[/tex]

Explanation:

Given

Power at point A = 100W

Power at point B = 90W

Required

Determine the attenuation in decibels

Attenuation is calculated using the following formula

[tex]Attenuation = 10Log_{10}\frac{P_s}{P_d}[/tex]

Where [tex]P_s = Power\ Inpu[/tex]t and [tex]P_d = Power\ outpu[/tex]t

[tex]P_s = 100W[/tex]

[tex]P_d = 90W[/tex]

Substitute these values in the given formula

[tex]Attenuation = 10Log_{10}\frac{P_s}{P_d}[/tex]

[tex]Attenuation = 10Log_{10}\frac{100}{90}[/tex]

[tex]Attenuation = 10 * 0.04575749056[/tex]

[tex]Attenuation = 0.4575749056[/tex]

[tex]Attenuation = 0.458\ db[/tex] (Approximated)

Answer:

Following are the difference to this question can be defined as follows:

Explanation:

Dimensionality reduction based on aggregation involves selection of important character and variable from a large variable and dimensionality reduction based on techniques with the use of PCA and SVD.

Dimensionality involves reducing features that have been written or identified or constructing less features from a pool of important features.

Principal components analysis(PCA) and Single value decomposition(SVD) is a form or analysis that can be used to reduce some character that are not important to an information and can also help to select important variables from among many variables.

Therefore, dimensionality reduction based on aggregation involves selection of important character and variable from a large variable and dimensionality reduction based on techniques with PCA and SVD.

Learn more on dimensional analysis here,

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Answer:

The program written in Java is as follows;

The program takes input for hours, days, weeks and months;

Then the inputs are converted to minutes individually and then summed up

For you to better understand the program,  made use of comments to explain difficult lines,

The program is as follows

import java.util.*;

public class Assignment{

public static void main(String []args){

//This line declares all variables

int hours, days, weeks, years;

Scanner input = new Scanner(System.in);

//This line prompts user for hours input

System.out.print("Hours: ");

//This line accepts input for hours

hours = input.nextInt();

//This line prompts user for days  input        

System.out.print("Days: ");

//This line accepts input for days

days = input.nextInt();

//This line prompts user for weeks input

System.out.print("Weeks: ");

//This line accepts input for weeks

weeks = input.nextInt();

//This line prompts user for years input

System.out.print("Years: ");

//This line accepts input for years

years = input.nextInt();

//This line converts hours to minute

int minute = 60 * hours + 1440 * days + 10080 * weeks + 525600 * years;

//This line prints the equivalent minute of user input

System.out.println(minute +" minutes");

}

}