Answer:
absolute
Explanation:
Use absolute references when you want different formulas to refer to the same cell.
What are absolute references?When dragging rows and columns to replicate formulas in Microsoft Excel, absolute references are the cell references you wish to maintain constant. For instance, the value in cell J4 is necessary for each and every cell in column K.
The reference would go from J4 to J5, then J6, and so on if you dragged it along column K. References that are absolute don't alter when copied or filled. To maintain the consistency of a row and/or column, utilize an absolute reference.
By including a dollar sign, a formula designates an absolute reference. The column reference, the row reference, or both may come before it.
Therefore, when you want different formulas to refer to the same cell, use absolute references.
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Write a function wordcount() that takes the name of a text file as input and prints the number of occurrences of every word in the file. You function should be case-insensitive so 'Hello' and 'hello' are treated as the same word. You should ignore words of length 2 or less. Also, be sure to remove punctuation and digits.
>>>wordcount('frankenstein.txt')
artifice 1
resting 2
compact 1
service 3
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.
Which can be used to code a simple web page? CSS JavaScript Text editor Web browser
Answer:
option (C) Text editor is the correct answer in this question.
Explanation:Used to code a web page, an editor is required that is a basic requirement.The other options are secondary ,text editors are provided with operating system and software development .Text editor specialized html editors can offer convenience and added functionality.Test editors require of html and any other web technology like Java script,and server-side scripting languages.
Answer:
It is text editor!
Explanation:
Suppose that an engineer excitedly runs up to you and claims that they've implemented an algorithm that can sort n elements (e.g., numbers) in fewer than n steps. Give some thought as to why that's simply not possible and politely explain
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.
X = 1, Y = 3, and Z = 2.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.
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If distances are recorded as 4-bit numbers in a 500-router network, and distance vectors are exchanged 3 times/second, how much total bandwidth (in bps) is used by the distributed routing algorithm
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
"Simon Says" is a memory game where "Simon" outputs a sequence of 10 characters (R, G, B, Y) and the user must repeat the sequence. Create a for loop that compares the two strings. For each match, add one point to user_score. Upon a mismatch, end the game. Ex: The following patterns yield a user_score of 4: simonPattern: R, R, G, B, R, Y, Y, B, G, Y userPattern: R, R, G, B, B, R, Y, B, G, Y
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:
For the Python program below, will there be any output, and will the program terminate?
while True: while 1 > 0: break print("Got it!") break
a. Yes and no
b. No and no
c. Yes and yes
d. No and yes
e. Run-time error
7.8 LAB: Palindrome A palindrome is a word or a phrase that is the same when read both forward and backward. Examples are: "bob," "sees," or "never odd or even" (ignoring spaces). Write a program whose input is a word or phrase, and that outputs whether the input is a palindrome.
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))
7. When using find command in word we can search?
a. Characters
b. Formats
c. Symbols
d. All of the above
Answer:
When using find command in word we can search all of the above
Complete this truth Table. Write a program that you can enter from the keyboard, a 1 or 0 into three Boolean variables, A,B,C. Write an if statement that tests the condition ( (A or B ) and C ). Run the program 8 separate times, testing each of the following 8 combinations. Write the results below in the following truth table.
Module 9-Alternative Sequences and Complex logical Criteria v17 (1).pdf. Adobe Acrobat Reader DC
File Edit View Window Help
Home Tools Module 9 - Altema...X
A B C (A or B ) (A or B ) and C
0 0 0
0 0 1
0 1 0
0 1 1
1 0 0
1 0 1
1 1 0
1 1 1
Answer:
Following are the code to this question:
#include<stdio.h>//defining header file
int AND(int x,int y) //defining a method AND that hold two variable in its parameter
{
if(x==1 && y==1)//defining if block to check x and y value is equal to 1
{
return 1;//return value 1
}
else //defining else block
{
return 0;//return value 0
}
}
int OR(int x,int y)//defining method OR that hold two variable in its parameter
{
if(x==0&&y==0)//defining if block to check x and y value is equal to 1
{
return 0;//return value 0
}
else //defining else block
{
return 1;//return value 1
}
}
int main()//defining main method
{
int a,b,c;//defining integer variable
int k=1;//defining integer variable k that holds a value 1
while(k<=8)//defining while loop for 8 time input
{
printf("Please insert 3 numbers in (0 0r 1):\n ");//print message
scanf("%d%d%d", &a, &b, &c);//input value
k++;//increment the value of k by 1
printf("value: %d\n",AND(OR(a,b),c));
}
return 0;
}
Output:
please find the attachment.
Explanation:
In the above-given C language code two methods "AND and OR" is declared, holds two integer variable "x and y" in its parameters, inside the method a conditional it used that can be defined as follows:
In the AND method, inside a conditional statement, if block check x and y both value is same that is 1 it will return 1 or it will goto else block in this it will return value 0. In the OR method, inside a conditional statement, if block check x and y both value is the same, that is 0 it will return 0 or it will goto else block in this it will return value 1.In the main method, four integers "a,b,c, and k" is declared in which variable "a, b, c" is used in the loop for 8 times input values from the user and print method is used to call the method and prints its return values.Discuss the differences between dimensionality reduction based on aggregation and dimensionality reduction based on techniques such as PCA and SVD.
Answer:
Following are the difference to this question can be defined as follows:
Explanation:
In terms of projections of information into a reduced dimension group, we can consider the dimension structure of PCA and SVD technologies. In cases of a change in the length, based on consolidation, there's also a unit with dimensions. When we consider that the days have been aggregated by days or the trade of a commodity can be aggregated to both the area of the dimension, accumulation can be seen with a variance of dimension.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.
What is dimensionality?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.
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What are the links between the operating systems, the software, and hardware components in the network, firewall, and IDS that make up the network defense implementation of the banks' networks.
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.
Without data compression, and focusing only on the data itself without any overhead, what transmission rate is required to send 30 frames per second of gray-scale images with resolution 1280 x 1024?
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
NMCI is the name used for a large-scale effort to link Navy and Marine Corps computer systems on bases, boats, and in offices around the world. When completed, this internal WAN will use Internet technology to link soldiers in the field with support personnel on bases, etc. NMCI is an example of a(n):
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.
a software development management tool that easily integrates into his business’s enterprise software/information system
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.
Write a program to help a travelling sales person keep up with their daily mileage driven for business. In your main method, the program should first ask the user how many days of mileage they want to enter and then collect the user input for each day's miles, storing the values entered in an appropriately sized array.
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;
Recall that within the ArrayBoundedQueue the front variable and the rear variable hold the indices of the elements array where the current front and rear elements, respectively, of the queue are stored. Which of the following code sequences could be used to correctly enqueue element into the queue, assuming that enqueue is called on a non-full queue and that the code also correctly increments numElements?
a. numElements++; elements[rear) - element:
b. front++; elements(front) - element:
c. rear = (rear + 1) % elements.length; elements[rear) - element;
d. front = (front + 1) % elements.length; elements[front) - element;
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.
#Write a function called after_second that accepts two #arguments: a target string to search, and string to search #for. The function should return everything in the first #string *after* the *second* occurrence of the search term. #You can assume there will always be at least two #occurrences of the search term in the first string. # #For example: # after_second("11223344554321", "3") -> 44554321 # #The search term "3" appears at indices 4 and 5. So, this #returns everything from the index 6 to the end. # # after_second("heyyoheyhi!", "hey") -> hi! # #The search term "hey" appears at indices 0 and 5. The #search term itself is three characters. So, this returns #everything from the index 8 to the end. # #Hint: This may be more complicated than it looks! You'll #have to look at the length of the search string and #either modify the target string or take advantage of the #extra arguments you can pass to find(). #Write your function here!
Answer:
Following are the code to this question:
def after_second(s,sub):#defining a method a fter_second
first = s.find(sub)#defining a variable first that hold method find value
if first != -1:#defining if block to check first variable value not equal to -1 using slicing
s = s[len(sub)+first:]#defining s variable to calculate sub parameter length of parameter and use slicing
second = s.find(sub)#defining second variable to calculate find method value
if second != -1:#defining if block to calculate second variable slicing
return s[len(sub)+second:]#return s variable value
print(after_second("heyyoheyhi","hey"))#defining print method to call after_second method
print(after_second("11223344554321","3"))#defining print method to call after_second method
Output:
hi
44554321
Explanation:
In the above python code a method "after_second" is declared, that accepts two-variable "s, and sub" as the parameter inside the method a first variable is declared that uses the inbuilt method "find" to find the value and stores it value. In the next step, two if blocks are used, in which both if blocks use the slicing to checks its value is not equal to "-1".
In the first, if block the first variable is declared that uses the s variable to calculate subparameter length by using slicing and defines the second variable that finds its value and stores its value. In the next, if block the s variable is used to return its calculated value, and at the end of the print, the method is used to call the method by passing parameter value and prints its return value.What is the main advantage of using DHCP? A. Allows you to manually set IP addresses B. Allows usage of static IP addresses C. Leases IP addresses, removing the need to manually assign addresses D. Maps IP addresses to human readable URLs
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:
What are some of the common security weaknesses inherent in Unix or Linux based systems and what techniques can be used to harden these systems against an attack?
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.
A signal travels from point A to point B. At point A, the signal power is 100 W. At point B, the power is 90 W. What is the attenuation in decibels?
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)
The statistical report is typically used to set targets for the organization's expenditures for the year and then report actual costs on a monthly or quarterly basis.
A. True
B. False
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.
Create an application in Java that asks a user for a number of hours, days, weeks, and years. It then computes the equivalent number of minutes (ignoring leap years).
Answer:
//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
Sample Output: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
Explanation:
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.
Which functions are examples of logical test arguments used in formulas? Check all that apply. OR IF SUM COUNT NOT AND
Answer: Which functions are examples of logical test arguments used in formulas? Check all that apply.
Explanation: A.or B.if E.not F. And
This is correct just took it :)
Among the following functions, those that are the logical test arguments used in formulas are OR, IF, NOT and, AND.
What are logical test argument functions ?To compare several conditions or numerous sets of conditions, logical functions are utilized. By weighing the arguments, it determines if the answer is TRUE or FALSE.
These operations are used to determine the outcome and aid in choosing one of the available data. The contents of the cell are assessed using the appropriate logical condition depending on the necessity. The types of logical functions used in this tutorial include: OR, AND, IF, NOT and XOR.
These all functions are used in different conditions. Therefore, for the given options the logical testing arguments are OR, IF, NOT and, AND. They are very common functions in Excel sheet.
Find more on logical functions :
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coefficient of x in expansion (x+3)(x-1)
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].
#Write a function called "in_parentheses" that accepts a
#single argument, a string representing a sentence that
#contains some words in parentheses. Your function should
#return the contents of the parentheses.
#
#For example:
#
# in_parentheses("This is a sentence (words!)") -> "words!"
#
#If no text appears in parentheses, return an empty string.
#Note that there are several edge cases introduced by this:
#all of the following function calls would return an empty
#string:
#
# in_parentheses("No parentheses")
# in_parentheses("Open ( only")
# in_parentheses("Closed ) only")
# in_parentheses("Closed ) before ( open")
#
#You may assume, however, that there will not be multiple
#open or closed parentheses.
#Write your function here!
def in_parentheses(a_string):
import re
regex = re.compile(".*?\((.*?)\)")
result = re.findall(regex, a_string)
return(result)
#Below are some lines of code that will test your function.
#You can change the value of the variable(s) to test your
#function with different inputs.
#
#If your function works correctly, this will originally
#print (including the blank lines):
#words!
#
#as he is doing right now
#
#
#!
print(in_parentheses("This is a sentence (words!)."))
print(in_parentheses("No parentheses here!"))
print(in_parentheses("David tends to use parentheses a lot (as he is doing right now). It tends to be quite annoying."))
print(in_parentheses("Open ( only"))
print(in_parentheses("Closed ) only"))
print(in_parentheses("Closed ) before ( open"))
print(in_parentheses("That's a lot of test cases(!)"))
Answer:
import regex as re
def in_parentheses(a_string):
regeX = re.compile(".*?\((.*?)\)")
result = re.findall(regeX, a_string)
return str(result).replace("[","").replace("]","")
print("test 1: "+in_parentheses("Open ( only"))
print("test 2: "+in_parentheses("This is a sentence (words!)."))
Can someone help me out with this one? I'm not sure why my code is not working
Now, let's improve our steps() function to take one parameter
that represents the number of 'steps' to print. It should
then return a string that, when printed, shows output like
the following:
print(steps(3))
111
222
333
print(steps(6))
111
222
333
444
555
666
Specifically, it should start with 1, and show three of each
number from 1 to the inputted value, each on a separate
line. The first line should have no tabs in front, but each
subsequent line should have one more tab than the line
before it. You may assume that we will not call steps() with
a value greater than 9.
Hint: You'll want to use a loop, and you'll want to create
the string you're building before the loop starts, then add
to it with every iteration.
Write your function here
def steps(number):
i = 1
while i < number + 1:
string = str(i) * 3
string1 = str(string)
if i != 0:
string1 = (i * 4*' ' + "\b" + "\b" + "\b" + "\b") + string
elif i == 1:
string1 = string
print(string1)
i = i + 1
#The following two lines will test your code, but are not
#required for grading, so feel free to modify them.
print(steps(3))
print(steps(6)
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.
Multiply each element in origList with the corresponding value in offsetAmount. Print each product followed by a semicolon (no spaces). Ex: If the input is: 4 5 10 12 2 4 7 3 the output is: 8; 20;70; 36; 1 #include 2 3 int main(void) { 4 const int NUM_VALS = 4; 5 int origList[NUM_VALS]; 6 int offsetAmount [NUM_VALS]; 7 int i; 8 9 scanf("%d", &origList[0]); 10 scanf("%d", &origList[1]); 11 scanf("%d", &origList[2]); 12 scanf("%d", &origList[3]); 13 14 scanf("%d", &offsetAmount[0]); 15 scanf("%d", &offsetAmount[1]); 16 scanf("%d", &offsetAmount[2]); 17 scanf("%d", &offsetAmount[3]); 18 19 \* Your code goes here */ 20 21 printf("\n"); 22 23 return 0; 24
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
}
Assemble a Server computer based on your budget (state the amount in Ghana Cedis), discussing the type of components (giving their exact names, model numbers, types, cost, architecture, etc.) you would need and give reasons why you need them.
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.
Dell Poweredge R730 has been its pattern. There should be the reason whether these servers are efficient, available, flexible, and support the concept of even a virtual environment. Its same servers are easy to use, as well as the memory will be connected to this server is 8 TB.____ enable users to create and edit collaborative Web pages quickly and easily; they are intended to be modified by others and are especially appropriate for collaboration.
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.
If each sandbox has size 16 megabytes, how many applets can be sandboxed on a machine with a 32-bit address space
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.