Answer:
Explanation:
tttttttttttt
What is the name of the top-level parentless folder in a digital file system?
Answer:
ROOT
Explanation:
Correct me if I'm wrong but I'm pretty sure it's called a "Root Folder" meaning it is the very first folder in the list
Create a script that internally calls the Linux command 'ps aux' and saves the output to a file named unprocessed.txt (you do not need to append to an existing file, if a file already exists, simply overwrite it). The program should pause and display the following message to the screen: Press 'q' to exit or 'ctrl-c' to process the file and exit' If the user presses 'q', exit the program leaving the filename on the disk as 'unprocessed.txt'. If the user presses 'ctrl-c', then catch the signal (hint - covered in chapter 16) and rename the file to 'processed.txt' and then exit. You must build a signal handler to catch the ctrl-c signal and rename the file. g
Answer:
see explaination
Explanation:
SIG{INT} = sub {
`mv unprocessed.txt processed.txt`;
print "\n";
exit;
};
`ps aux > unprocessed.txt`;
print "Press 'q' to exit or 'ctrl-c' to process the file and exit:\n";
$inp = <>;
if ($inp == 'q')
{
exit;
}
Write a Java program that prompts the user to enter a sequence of non-negative numbers (0 or more), storing them in an ArrayList, and continues prompting the user for numbers until they enter a negative value. When they have finished entering non-negative numbers, your program should return the mode (most commonly entered) of the values entered.
Answer: provided in explanation segment
Explanation:
the code to carry out this program is thus;
import java.util.ArrayList;
import java.util.Scanner;
public class ArrayListMode {
public static int getMode(ArrayList<Integer>arr) {
int maxValue = 0, maxCount = 0;
for (int i = 0; i < arr.size(); ++i) {
int count = 0;
for (int j = 0; j < arr.size(); ++j) {
if (arr.get(j) == arr.get(i))
++count;
}
if (count > maxCount) {
maxCount = count;
maxValue = arr.get(i);
}
}
return maxValue;
}
public static void main(String args[]){
Scanner sc = new Scanner(System.in);
ArrayList<Integer>list= new ArrayList<Integer>();
int n;
System.out.println("Enter sequence of numbers (negative number to quit): ");
while(true) {
n=sc.nextInt();
if(n<=0)
break;
list.add(n);
}
System.out.println("Mode : "+getMode(list));
}
}
⇒ the code would produce Mode:6
cheers i hope this helps!!!!
(6 pts) Write a bash shell script called 08-numMajors that will do the following: i. Read data from a class enrollment file that will be specified on the command line. ii. If the file does not exist, is a directory, or there are more or less than one parameters provided, display an appropriate error/usage message and exit gracefully. iii. Display the number of a specified major who are taking a given class.
Answer:
Check the explanation
Explanation:
Script:
#!/usr/bin/ksh
#using awk to manipulated the input file
if [ $# != 1 ]
then
echo "Usage: 08-numMajors.sh <file-name>"
exit 1
fi
if [ -f $1 ]
then
awk '
BEGIN {
FS=","
i=0
flag=0
major[0]=""
output[0]=0
total=0
}
{
if ($3)
{
for (j=0; j<i; j++)
{
# printf("1-%s-%s\n", major[j], $3);
if(major[j] == $3)
{
flag=1
output[j] = output[j] + 1;
total++
}
}
if (flag == 0)
{
major[i]=$3
output[i] = output[i] + 1;
i++;
total++
}
else
{
flag=0
}
}
}
END {
for (j=0; j<i; j++)
{
if(output[j] > 1)
printf("There are %d `%s` majors\n", output[j], major[j]);
else
printf("There is %d `%s` major\n", output[j], major[j]);
}
printf("There are a total of %d students in this class\n", total);
}
' < $1
else
echo "Error: file $1 does not exist"
fi
Output:
USER 1> sh 08-numMajors.sh sample.txt
There are 4 ` Computer Information Systems` majors
There is 1 ` Marketing` major
There are 2 ` Computer Science` majors
There are a total of 7 students in this class
USER 1> sh 08-numMajors.sh sample.txtdf
Error: file sample.txtdf does not exist
USER 1> sh 08-numMajors.sh
Usage: 08-numMajors.sh <file-name>
USER 1> sh 08-numMajors.sh file fiel2
Usage: 08-numMajors.sh <file-name>
Write a function to reverse a given string. The parameter to the function is a string. Function should store the reverse of the given string in the same string array that was passed as parameter. Note you cannot use any other array or string. You are allowed to use a temporary character variable. Define the header of the function properly. The calling function (in main()) expects the argument to the reverse function will contain the reverse of the string after the reverse function is executed.
Answer:
Following are the program to this question:
#include <iostream> //defining header file
using namespace std;
void reverse (string a) //defining method reverse
{
for (int i=a.length()-1; i>=0; i--) //defining loop that reverse string value
{
cout << a[i]; //print reverse value
}
}
int main() //defining main method
{
string name; // defining string variable
cout << "Enter any value: "; //print message
getline(cin, name); //input value by using getline method
reverse(name); //calling reverse method
return 0;
}
Output:
Enter any value: ABCD
DCBA
Explanation:
In the above program code, a reverse method is declared, that accepts a string variable "a" in its arguments, inside the method a for loop is declared, that uses an integer variable "i", in which it counts string value length by using length method and reverse string value, and prints its value.
In the main method, a string variable "name" is declared, which uses the getline method. This method is the inbuilt method, which is used to input value from the user, and in this, we pass the input value in the reverse method and call it.Write a program that opens two text files and reads their contents into two separate queues. The program should then determine whether the files are identical by comparing the characters in the queues. When two nonidentical characters are encountered, the program should display a message indicating that the files are not the same. If both queues contain the same set of characters, a message should be displayed indicating that the files are identical.
Answer:
The program in cpp for the given scenario is shown below.
#include <stdio.h>
#include <iostream>
#include <fstream>
#include <queue>
using namespace std;
int main()
{
//object created of file stream
ofstream out, out1;
//file opened for writing
out.open("words.txt");
out1.open("word.txt");
//queues declared for two files
queue<string> first, second;
//string variables declared to read from file
string s1, s2;
//object created of file stream
ifstream one("words.txt");
ifstream two("word.txt");
//first file read into the que
while (getline(one, s1)) {
first.push(s1);
}
//second file read into the queue
while (getline(two, s2)) {
second.push(s2);
}
//both files compared
if(first!=second)
cout<<"files are not the same"<<endl;
else
cout<<"files are identical"<<endl;
//file closed
out.close();
return 0;
}
OUTPUT:
files are identical
Explanation:
1. The object of the file output stream are created for both the files.
2. The two files are opened using the objects created in step 1.
3. The objects of the file input stream are created for both the files.
4. The queues are declared for both the files.
5. Two string variables are declared.
6. Inside a while loop, the text from the first is read into the string variable. The value of this string variable is then inserted into the queue.
7. The loop continues till the string is read and end of file is not reached.
8. Inside another while loop, the text from the second file is read into the second string variable and this string is inserted into another queue.
9. The loop continues till the end of file is not reached.
10. Using if-else statement, both the queues are compared.
11. If any character in the first queue does not matches the corresponding character of the second queue, the message is displayed accordingly.
12. Alternatively, if the contents of both the queues match, the message is displayed accordingly.
13. In the given example program, the message is displayed as "files are identical." This is because both the files are empty and the respective queues are considered identical since both the queues are empty.
14. Since queues are used, the queue header file is included in the program.
15. An integer value 0 is returned to indicate successful execution of the program.
Bob is stationed as a spy in Cyberia for a week and wants to prove that he is alive every day of this week and has not been captured. He has chosen a secret random number, x, which he memorized and told to no one. But he did tell his boss the value y = H(H(H(H(H(H(H(x))))))), where H is a one-way cryptographic hash function. Unfortunately, he knows that the Cyberian Intelligence Agency (CIA) was able to listen in on that message; hence, they also know the value of y. Explain how he can send a single message every day that proves he is still alive and has not been captured. Your solution should not allow anyone to replay any previous message from Bob as a (false) proof he is still alive.
Bob can use a hash chain method with a cryptographic hash function to send a message daily, validating his aliveness without risk of replay attacks. Each day he reveals the previous hash in the chain, which can be verified by hashing to obtain the initially shared value.
In order for Bob to prove he is still alive and has not been captured, without the risk of replay attacks, a strategy called hash chain can be employed using the cryptographic hash function. Bob starts with his secret random number x and computes a hash of it, H(x). He then repeatedly applies the hash function to this result a number of times equal to the number of days he needs to send a message (in this case, 7 times since he is there for a week), resulting in y = H(H(H(H(H(H(H(x))))))).
Each day, Bob unveils the previous hash in the chain. For example, on the first day, he sends H(H(H(H(H(H(x)))))), which anyone can hash once to verify it leads to y, the value his boss already knows. On the second day, he sends H(H(H(H(H(x))))), and so on. This proves that each message can only have come from someone with knowledge of the previous day's secret -- presumably, Bob himself. Because hash functions are one-way, even if someone knows the value H(x), they cannot reverse-engineer to find x.
This method ensures that even if the Cyberian Intelligence Agency (CIA) intercepted a message, they could not fake a future message since they would not be able to produce the next hash in the sequence without knowing the current one.
2. Suppose a computer using direct mapped cache has 220 words of main memory and a cache of 32 blocks, where each cache block contains 16 words. a. How many blocks of main memory are there? b. What is the format of a memory address as seen by the cache, that is, what are the sizes of the tag, block, and word fields? c. To which cache block will the memory reference 0DB6316 map? d. In many computers the cache block size is in the range of 32 to 128 bytes. What would be the main advantages and disadvantages of making the size of the cache blocks larger or smaller within this range
Answer:
(a) The block size = 16 words (b) the format for the sizes of the tag, block, and word fields are stated below:
Tag Block Offset
11 bits 5 bits 4 bits
(c)The memory address 0DB63 will map to 22nd block and 3 byte of a block. (d) The main advantage of the cache is, when the cache block is made larger or bigger then, there are fewer misses. this occur when the data in the block is used.
One of the disadvantage is that if the data is not used before the cache block is removed from the cache, then it is no longer useful. here there is also a larger miss penalty.
Explanation:
Solution
(a)The Cache of 32 blocks and the memory is word addressable
The main memory size is not stated, so let's consider it as 2^20
The main memory size = 2^20
The block size = 16 words
The Number of blocks = main memory size/ block size = 2^20/16. = 2^16= 64 K words.
(b) The Main memory is 2^20, which says it is 1 M words size and it requires 20 bits
Now,
From which 32 cache block requires 2^5, 5 bits and block size means offset requires 2^4, 4 bits
Thus, the sizes of the tag, block, and word fields are stated below:
Tag Block Offset
11 bits 5 bits 4 bits
(c)The hexadecimal address is 0DB63 , its binary equivalent is
0000 1101 1011 0110 0011
The tag = ( first 11 bits) = 0000 1101 101
block = ( next 5 bits ) = 1 0110
offset = ( next 4 bits ) = 0011
Therefore, the memory address 0DB63 will map to 22nd block and 3 byte of a block.
(d) The main advantage of the cache is that, when the cache block is made larger or bigger then, there are fewer misses. this occur when the data in the block is used.
One of the disadvantage is that if the data is not used before the cache block is removed from the cache, then it is no longer useful. here there is also a larger miss penalty.
In this exercise we want to use computer and C++ knowledge to write the code correctly, so it is necessary to add the following to the informed code:
(a) The block size = 16 words
(b) Tag Block Offset
11 bits 5 bits 4 bits
(c)The memory address 0DB63 will map to 22nd block and 3 byte of a block.
(d) The main advantage of the cache is, when the cache block is made larger or bigger then, there are fewer misses. One of the disadvantage is that if the data is not used before the cache block is removed from the cache, then it is no longer useful.
So analyzing the first question about the code we find that:
(a)The Cache of 32 blocks and the memory is word addressable :
The main memory size; [tex]2^{20[/tex] The block size; [tex]16 \ words[/tex] The Number of blocks = main memory size/ block size: [tex]2^{20}/16 = 2^{16}= 64[/tex]
(b) Says it is 1 M words size and it requires 20 bits. Now, from which 32 cache block requires 2^5, 5 bits and block size means offset requires 2^4, 4 bits. Thus, the sizes of the tag, block, and word fields are stated below:
Tag Block Offset 11 bits 5 bits 4 bits
(c)The hexadecimal address is 0DB63 , its binary equivalent is
The tag = ( first 11 bits) = 0000 1101 101 block = ( next 5 bits ) = 1 0110 offset = ( next 4 bits ) = 0011
Therefore, the memory address 0DB63 will map to 22nd block and 3 byte of a block.
(d) The main favored position or circumstance of the cache happen that, when the cache block is created best or considerable therefore, there happen hardly any misses. this take place when the information in visible form in the block exist secondhand. One of the loss exist that if the data exist not secondhand before the cache block exist detached from the cache, then it exist not any more valuable. in this place there exist in addition to a best miss punishment.
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Let's now use our new calculator functions to do some calculations!
A. Add the square of 3 to the square root of 9 Save the result to a variable called calc_a
B. Subtract the division of 12 by 3 from the the multiplication of 2 and 4 Save the result to a variable called calc_b
C. Multiply the square root of 16 by the sum of 7 and 9 Save the result to a variable called calc_c
D. Divide the square of 7 by the square root of 49 Save the result to a variable called calc_d
Answer:
Ans1.
double calc_a;
calc_a=Math.pow(3.0,2.0)+Math.sqrt(9);
Ans2.
double calc_b;
calc_b=((12.0/3.0)-(2.0*4.0));
Ans 3.
double calc_c;
calc_c=(Math.sqrt(16.0)*(7.0+9.0));
Ans 4.
double calc_d;
calc_d=Math.pow(7.0,2.0)/Math.sqrt(49.0);
Explanation:
The expressions are done with Java in answer above.
To solve these calculations, we can use mathematical functions on a calculator.
Explanation:To solve these calculations, we can use the mathematical functions on a calculator.
Calculation A:
To add the square of 3 to the square root of 9, we can write it as follows:
calc_a = [tex]3^2 + sqrt(9)[/tex]
calc_a = 9 + 3
calc_a = 12
Calculation B:
To subtract the division of 12 by 3 from the multiplication of 2 and 4, we can write it as follows:
calc_b = (2 * 4) - (12 / 3)
calc_b = 8 - 4
calc_b = 4
Calculation C:
To multiply the square root of 16 by the sum of 7 and 9, we can write it as follows:
calc_c = sqrt(16) * (7 + 9)
calc_c = 4 * 16
calc_c = 64
Calculation D:
To divide the square of 7 by the square root of 49, we can write it as follows:
calc_d = [tex](7^2) / sqrt(49)[/tex]
calc_d = 49 / 7
calc_d = 7
Write the function setKthDigit(n, k, d) that takes three integers -- n, k, and d -- where n is a possibly-negative int, k is a non-negative int, and d is a non-negative single digit (between 0 and 9 inclusive). This function returns the number n with the kth digit replaced with d. Counting starts at 0 and goes right-to-left, so the 0th digit is the rightmost digit.
Function are collections of named code blocks, that are executed when called or evoked.
The setKthDigit function written in Python, where comments are used to explain each line is as follows
#This defines the function
def setKthDigit(n, k, d):
#If the number of digits in n is greater than k
if len(str(n)) > k:
#This splits the number into a list
myWord = list(str(n))
#This reverses the list
myWord.reverse()
#This updates the kth digit
myWord[k] = str(d)
#This reverses the list
myWord.reverse()
#This initializes string newStr
newStr = ''
#This following loop creates the the number to return, as string
for i in myWord:
newStr+=i
#If otherwise
else:
#This initializes string newStr
newStr = ''
#This following loop creates the the number to return, as string
for i in range(1+k-len(str(n))):
newStr+=str(d)
newStr+=str(n)
#This returns the updated number, as an integer
return int(newStr)
Read more about functions at:
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Final answer:
The function setKthDigit replaces the kth digit of an integer n with a digit d, considering the rightmost digit as the 0th. The implementation involves string manipulation and careful handling of the sign of the input number.
Explanation:
To implement the setKthDigit function in Python, you need to first understand how to manipulate individual digits of a number. This involves converting the number into a string to easily access and modify specific digits. Then, you will convert the string back to an integer to return the final result. Below is a Python code snippet that demonstrates how to accomplish this task:
def setKthDigit(n, k, d):
n_str = str(abs(n)) # Convert the absolute value of n to a string
if k >= len(n_str): # Check if k is within the bounds of the number
return n # Return n as is if k is out of bounds
modified_n_str = n_str[:-k-1] + str(d) + n_str[-k:] if k != 0 else n_str[:-1] + str(d)
result = int(modified_n_str)
return result if n >= 0 else -result # Keep the original sign of n
This function works by first converting the number n into a string to easily manipulate its digits. If the digit to be replaced is indeed in the bounds of the number, the string is modified by replacing the kth digit with d. The modified string is then converted back to an integer, taking care to preserve the original sign of n.
Which programming language will you use to create an App in code.org?
A. Java Script
B. C++
C. Pearl
Answer:
A. Java Script
Explanation:
java gives languages
Answer:
A. Java Script
Explanation:
That is the only one they teach and is easier than others!
Data related to the inventories of Costco Medical Supply are presented below: Surgical Equipment Surgical Supplies Rehab Equipment Rehab Supplies Selling price $ 272 $ 130 $ 350 $ 164 Cost 167 129 257 161 Costs to sell 28 18 19 10 In applying the lower of cost or net realizable value rule, the inventory of surgical supplies would be valued at:
Answer:
The following are the table attachment to this question.
Explanation:
In the given question a table is defined, in the column section, we assign "Particulars, Surgical Equipment, Surgical Supplies Rehab Equipment, Rehab Supplies, and total " and in the row section we assign values.
In the next step, to calculate the net realizable value, we subtract selling price from the cost to cell, and compare the value from the costs. If the value is smaller than then the net realizable value we write its value and at last total column we add all the net realizable value, which is equal to 690. In this table, the value of rehab supplies is $154.3.27 LAB: Exact change (FOR PYTHON PLEASE)
Write a program with total change amount as an integer input, and output the change using the fewest coins, one coin type per line. The coin types are Dollars, Quarters, Dimes, Nickels, and Pennies. Use singular and plural coin names as appropriate, like 1 Penny vs. 2 Pennies.
Ex: If the input is:
0
(or less than 0), the output is:
No change
Ex: If the input is:
45
the output is:
1 Quarter
2 Dimes
Answer:
amount = int(input())
#Check if input is less than 1
if amount<=0:
print("No change")
else: #If otherwise
#Convert amount to various coins
dollar = int(amount/100) #Convert to dollar
amount = amount % 100 #Get remainder after conversion
quarter = int(amount/25) #Convert to quarters
amount = amount % 25 #Get remainder after conversion
dime = int(amount/10) #Convert to dimes
amount = amount % 10 #Get remainder after conversion
nickel = int(amount/5) #Convert to nickel
penny = amount % 5 #Get remainder after conversion
#Print results
if dollar >= 1:
if dollar == 1:
print(str(dollar)+" Dollar")
else:
print(str(dollar)+" Dollars")
if quarter >= 1:
if quarter == 1:
print(str(quarter)+" Quarter")
else:
print(str(quarter)+" Quarters")
if dime >= 1:
if dime == 1:
print(str(dime)+" Dime")
else:
print(str(dime)+" Dimes")
if nickel >= 1:
if nickel == 1:
print(str(nickel)+" Nickel")
else:
print(str(nickel)+" Nickels")
if penny >= 1:
if penny == 1:
print(str(penny)+" Penny")
else:
print(str(penny)+" Pennies")
Explanation:
The first answer is almost right, but this should give you the complete answer with the proper starting point and correct capitalizations.
The program is an illustration of conditional statements.
Conditional statements are statements whose execution depends on its truth value.
The program in Python, where comments are used to explain each line is as follows:
#This gets input for the amount
amount = int(input("Enter Amount: "))
#This checks if the amount is less than 1
if amount<=0:
print("No Change")
else: #If otherwise
#The next lines convert amount to various coins
dollar = int(amount/100) #Convert to dollar
amount = amount % 100 #Get remainder after conversion
quarter = int(amount/25) #Convert to quarters
amount = amount % 25 #Get remainder after conversion
dime = int(amount/10) #Convert to dimes
amount = amount % 10 #Get remainder after conversion
nickel = int(amount/5) #Convert to nickel
penny = amount % 5 #Get remainder after conversion
#The following if statements print the change
#The prints dollars
if dollar >= 1:
if dollar == 1:
print(str(dollar)+" dollar")
else:
print(str(dollar)+" dollars")
#The prints quarters
if quarter >= 1:
if quarter == 1:
print(str(quarter)+" quarter")
else:
print(str(quarter)+" quarters")
#The prints dimes
if dime >= 1:
if dime == 1:
print(str(dime)+" dime")
else:
print(str(dime)+" dimes")
#The prints nickels
if nickel >= 1:
if nickel == 1:
print(str(nickel)+" nickel")
else:
print(str(nickel)+" nickels")
#The prints pennies
if penny >= 1:
if penny == 1:
print(str(penny)+" penny")
else:
print(str(penny)+" pennies")
Read more about similar programs at:
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Can you find the reason that the following pseudocode function does not return the value indicated in the comments? // The calcDiscountPrice function accepts an item’s price and // the discount percentage as arguments. It uses those // values to calculate and return the discounted price. Function Real calcDiscountPrice(Real price, Real percentage) // Calculate the discount. Declare Real discount = price * percentage // Subtract the discount from the price. Declare Real discountPrice = price – discount // Return the discount price. Return discount End Function
Answer:
see explaination
Explanation:
Function Real calcDiscountPrice(Real price, Real percentage) // Calculate the discount.
Declare Real discount= (price * percentage) / 100.0 // Subtract the discount from the price.
//dividing by 100.0 because of % concept
Declare Real discountPrice = price - discount // Return the discount price.
Return discount End Function
Final answer:
The calcDiscountPrice function's issue is the incorrect calculation of the discount as the percentage is not converted into its decimal form before multiplication.
Explanation:
The problem with the provided pseudocode in calcDiscountPrice function is that the calculation for the discount is incorrect because the percentage is not converted into its decimal form before the calculation. The correct method to calculate the discount should be taking the percentage in decimal (by dividing the percentage by 100) and then multiplying it by the price. Hence, the corrected line of code should be:
Declare Real discount = price * (percentage / 100)
By doing so, the function will correctly calculate the discount and subtract it from the original price to get the discounted price. If the percentage is not divided by 100, you are not calculating a percentage of the price; you are instead calculating a multiple of the price, which is not the intended behavior for a discount.
Write a program that uses cin to get the name of a file from the user. The file will be a plain text file that contains information about different objects of type "BibleBook". BibleBook will be defined as a class. The private member attributes are: string title; int numChapters; double amountRead; The title will hold the title of the book, for example, "Genesis", "Exodus", "Psalms", etc. numChapters will hold the number of chapters in that book. amountRead will be a decimal ranging from 0 to 1. If the value is 0, that means you have read none of that book. If the number of 0.50, that means you have read half of that book, etc. You will need to discover the public member functions that need to be written based on the description. The text file will be organize like so: Each line will hold the title of a book, then a space, then the number of chapters, then a space, then the amount read of that book. Here's an example: Genesis 50 0.98 Matthew 28 1.00 Your program will read in this information and store each line in it's own object, and the objects will be part of a vector. In other words, there will be a vector of BibleBooks. Your program will then iterate through the vector, displaying all of the information about each object. The program will also identify for which object the amount read is the lowest. Sample output: Genesis 50 0.98 Matthew 28 1 The book with the lowest amount read is Genesis.
Answer:
See explaination
Explanation:
#include<iostream>
#include<vector>
#include<string>
#include<iostream>
#include<fstream>
using namespace std;
class BibleBooks{
public:
BibleBooks(string bk_title, int chapters, double percentage){
title=bk_title;
numChapters=chapters;
amountRead=percentage;
}
string getTitle(){
return title;
}
int getChapters(){
return numChapters;
}
double getPercentage(){
return amountRead;
}
private:
string title;
int numChapters;
double amountRead;
};
int main(){
char filename[200];
cout<<"Enter filename: ";
cin.get(filename,199);
vector<BibleBooks> bibleBooks;
ifstream infile(filename);
if(infile.is_open()){
string title; int chapters; double percentage;
while(infile>>title>>chapters>>percentage){
BibleBooks bb = BibleBooks(title,chapters,percentage);
bibleBooks.push_back(bb);
}
infile.close();
}else{
cout<<"Unable to read data from file: "<<filename;
}
//Your program will then iterate through the vector,
// displaying all of the information about each object.
int index_lowest_read=0;
for(int i=0; i<bibleBooks.size();i++){
cout<<bibleBooks.at(i).getTitle()<<" "
<<bibleBooks.at(i).getChapters()<<" "
<<bibleBooks.at(i).getPercentage()<<endl;
if(bibleBooks.at(index_lowest_read).getPercentage()>
bibleBooks.at(index_lowest_read).getPercentage()){
index_lowest_read=i;
}
}
cout<<"The book with the lowest amount read is "
<<bibleBooks.at(index_lowest_read).getTitle()<<endl;
}
A company has offices in Honolulu, Seattle, Ogden, and Dublin, Ireland. There are transmission links between Honolulu and Seattle, Seattle and Ogden, and Ogden and Dublin. Seattle needs to communicate at 1 Gbps with each other site. Seattle and Dublin only need to communicate with each other at 1 Mbps. Ogden and Dublin need to communicate at 2 Gbps, and Ogden and Seattle need to communicate with each other at 10 Mbps. How much traffic will each transmission link have to carry? Show your work.
Answer:
See explaination
Explanation:
Looking at telecommunications network, a link is a communication channel that connects two or more devices for the purpose of data transmission. The link is likely to be a dedicated physical link or a virtual circuit that uses one or more physical links or shares a physical link with other telecommunications links.
Please check attachment for further solution.
In the function below, use a function from the random module to return a random integer between the given lower_bound and upper_bound, inclusive. Don't forget to import the random module (before the function declaration).
For example, return_random_int(3, 8) should random return a number from the set: 3, 4, 5, 6, 7, 8.
length.py
def return_random_int(lower_bound, upper_bound):
Answer:
import random
def return_random_int(lower_bound, upper_bound):
return random.randrange(lower_bound, upper_bound)
print(return_random_int(3, 8))
Explanation:
Import the random module
Create a function called return_random_int takes two parameters, lower_bound and upper_bound
Return a number between lower_bound and upper_bound using random.range()
Call the function with given inputs, and print the result
The number of hits on a certain website follows a Poisson distribution with a mean rate of 4 per minute.a. What is the probability that 5 messages are received in a given minute?b. What is the probability that 9 messages are received in 1.5 minutes?c. What is the probability that fewer than 3 messages are received in a period of 30 seconds?
Answer:
a) [tex]P(X=5) = 15.63 \%[/tex]
b) [tex]P(X=9) = 6.88 \%[/tex]
c) [tex]P(X<3) = 67.65 \%[/tex]
Explanation:
a) What is the probability that 5 messages are received in a given minute?
The Poisson distribution is given by
[tex]P(X=x) = e^{-\lambda}\frac{\lambda^{x}}{x!}[/tex]
Where x is the number of messages received in some time interval t.
mean rate = λ = 4 per minute
number of messages = x = 5
[tex]P(X=x) = e^{-\lambda}\frac{\lambda^{x}}{x!}\\\\P(X=5) = e^{-4}\frac{4^{5}}{5!}\\\\P(X=5) = 0.1563\\\\P(X=5) = 15.63 \%[/tex]
b) What is the probability that 9 messages are received in 1.5 minutes?
mean rate = λ = 4*1.5 = 6
number of messages = x = 9
[tex]P(X=x) = e^{-\lambda}\frac{\lambda^{x}}{x!}\\\\P(X=9) = e^{-6}\frac{6^{9}}{9!}\\\\P(X=9) = 0.0688\\\\P(X=9) = 6.88 \%[/tex]
c) What is the probability that fewer than 3 messages are received in a period of 30 seconds?
mean rate = λ = 0.5*4 = 2
number of messages = x < 3
[tex]P(X<3) = P(X = 0) + P(X = 1) + P(X = 2)\\\\P(X=0) = e^{-2}\frac{2^{0}}{0!} = 0.1353\\\\ P(X=1) = e^{-2}\frac{2^{1}}{1!} = 0.2706\\\\ P(X=2) = e^{-2}\frac{2^{2}}{2!} = 0.2706\\\\ P(X<3) =0.1353+0.2706+ 0.2706\\\\P(X<3) = 0.6765\\\\P(X<3) = 67.65 \%[/tex]
A) The probability that 5 messages are received in a given minute is; 15.629%
B) The probability that 9 messages are received in 1.5 minutes is; 6.884%
C) The probability that fewer than 3 messages are received in a period of 30 seconds is; 67.668%
We are told that the number of hits follows a poisson distribution with mean of 4 per minutes.
Formula for poisson distribution is;
P(X = x) = [tex]e^{-\lambda}(\frac{\lambda^{x}}{x!})[/tex]
where;
x is number of items in counting
λ is the mean rate
A) We want to find the probability that 5 messages are received in a given minute. Thus;
λ = 4 × 1 = 4
x = 5
Thus;
P(X = 5) = [tex]e^{-4}(\frac{4^{5}}{5!})[/tex]
P(X = 5) = 0.15629
P(X = 5) = 15.629%
B) We want to find the probability that 9 messages are received in 1.5 minutes. Thus;
λ = 4 × 1.5 = 6
x = 9
Thus;
P(X = 9) = [tex]e^{-6}(\frac{6^{9}}{9!})[/tex]
P(X = 9) = 0.06884
P(X = 9) = 6.884%
C) We want to find the probability that 3 messages are received in 30 seconds. 30 seconds is 0.5 minutes. Thus;
λ = 4 × 0.5 = 2
x = 3
Thus;
P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2)
From online Poisson distribution calculator, we have;
P(X < 3) = 0.13534 + 0.27067 + 0.27067
P(X < 3) = 0.67668
P(X < 3) = 67.668%
Read more about Poisson distribution at; https://brainly.com/question/7879375
You should process the tokens by taking the first letter of every fifth word,starting with the first word in the file. Convert these letters to uppercase andappend them to a StringBuilder object to form a word which will be printedto the console to display the secret message.
Answer:
See explaination
Explanation:
import java.io.File;
import java.io.IOException;
import java.util.Scanner;
import java.util.StringTokenizer;
public class SecretMessage {
public static void main(String[] args)throws IOException
{
File file = new File("secret.txt");
StringBuilder stringBuilder = new StringBuilder();
String str; char ch; int numberOfTokens = 1; // Changed the count to 1 as we already consider first workd as 1
if(file.exists())
{
Scanner inFile = new Scanner(file);
StringTokenizer line = new StringTokenizer(inFile.nextLine()); // Since the secret.txt file has only one line we dont need to loop through the file
ch = line.nextToken().toUpperCase().charAt(0); // Storing the first character of first word to string builder as mentioned in problem
stringBuilder = stringBuilder.append(ch);
while(line.hasMoreTokens()) { // Looping through each token of line read using Scanner.
str= line.nextToken();
numberOfTokens += 1; // Incrementing the numberOfTokens by one.
if(numberOfTokens == 5) { // Checking if it is the fifth word
ch = str.toUpperCase().charAt(0);
stringBuilder = stringBuilder.append(ch);
numberOfTokens =0;
}
}
System.out.println("----Secret Message----"+ stringBuilder);
}
}
}
b) Write a Boolean equation for this sentence showing your work: (Use the variables S,R,O only for the right side) Give two possible interpretations of the logic and show which one is most likely to be the correct interpretation given your knowledge of roads: Let V(S,R,O) = Very Slippery: {1.2} The Road will be very slippery if it snows or it rains and there is oil on the road .
Answer:
Check the explanation
Explanation:
Kindly check the attached image below to see the solution to the above question.
What are possible challenges for cyber bullying
Answer:
well, you will have your account deleted/ arrested because of cyber bullying
Explanation:
so don't!!! (hint hint XD)
Answer:
1. they falsely believes he or she cannot control what is posted on his or her social networking site or who sends messages to him or her.
2. they feel uncomfortable telling you what someone else wrote on your their wall because the comment is inappropriate. Therefore your they fears you may be disappointed in him or her.
3. they feels powerless to deal with cyberbullying
For the preceding simple implementation, this execution order would be nonideal for the input matrix; however, applying a loop interchange optimization would create a nonideal order for the output matrix. Because loop interchange is not sufficient to improve its performance, it must be blocked instead. (a) What should be the minimum size of the cache to take advantage of blocked execution
Answer:
hi your question lacks the necessary matrices attached to the answer is the complete question
1024 bytes
Explanation:
A) The minimum size of the cache to take advantage of blocked execution
The minimum size of the cache is approximately 1 kilo bytes
There are 128 elements( 64 * 2 ) in the preceding simple implementation and this because there are two matrices and every matrix contains 64 elements .
note: 8 bytes is been occupied by every element therefore the minimum size of the cache to take advantage of blocked execution
= number of elements * number of bytes
= 128 * 8 = 1024 bytes ≈ 1 kilobytes
Consider the class Complex (see problem 11.13, page 633-635). The class enables operations on so called complex numbers. a) Modify the class to enable input and output of complex number through the overloaded << and >> operators (you should modify or remove the print function from the class). b) Overload the multiplication * and the division / operators. c) Overload the
Answer:
See explaination
Explanation:
// Complex.h
#ifndef COMPLEX_H
#define COMPLEX_H
#include <iostream>
using namespace std;
class Complex
{
friend ostream &operator<<(ostream &, const Complex &);
friend istream &operator>>(istream &, Complex &);
private:
// Declare variables
double real;
double imaginary;
public:
// Constructor
Complex(double = 0.0, double = 0.0);
// Addition operator
Complex operator+(const Complex&) const;
// Subtraction operator
Complex operator-(const Complex&) const;
// Multiplication operator
Complex operator*(const Complex&) const;
// Equal operator
Complex& operator=(const Complex&);
bool operator==(const Complex&) const;
// Not Equal operator
bool operator!=(const Complex&) const;
};
#endif
// Complex.cpp
// Header file section
#include "Complex.h"
#include <iostream>
using namespace std;
// Constructor
Complex::Complex( double realPart, double imaginaryPart )
: real( realPart ),
imaginary( imaginaryPart )
{
// empty body
}
// Addition (+) operator
Complex Complex::operator+( const Complex &operd2 ) const
{
return Complex( real + operd2.real,
imaginary + operd2.imaginary );
}
// Subtraction (-) operator
Complex Complex::operator-( const Complex &operd2 ) const
{
return Complex( real - operd2.real,
imaginary - operd2.imaginary );
}
// Multiplication (*) operator
Complex Complex::operator*( const Complex &operd2 ) const
{
return Complex(
( real * operd2.real ) + ( imaginary * operd2.imaginary ),
( real * operd2.imaginary ) + ( imaginary * operd2.real ) );
}
// Overloaded = operator
Complex& Complex::operator=( const Complex &right )
{
real = right.real;
imaginary = right.imaginary;
return *this; // enables concatenation
} // end function operator=
bool Complex::operator==( const Complex &right ) const
{
return ( right.real == real ) && ( right.imaginary == imaginary )
? true : false;
}
bool Complex::operator!=( const Complex &right ) const
{
return !( *this == right );
} // end function operator!=
ostream& operator<<( ostream &output, const Complex &complex )
{
output << "(" << complex.real << ", " << complex.imaginary << ")";
return output;
}
istream& operator>>( istream &input, Complex &complex )
{
input.ignore();
input >> complex.real;
input.ignore( 2 );
input.ignore();
return input;
}
// ComplexMain.cpp
// Header file section
#include <iostream>
#include "Complex.h"
using namespace std;
// main method
int main()
{
// Declare complex numbers
Complex x, y(4.3, 8.2), z(3.3, 1.1), k;
cout << "Enter a complex number in the form: (a, b): ";
// Demonstrating overloaded
// Accept a complex number
cin >> k;
// Display complex numbers
cout << "x: " << x
<< "\ny: " << y
<< "\nz: " << z
<< "\nk: " << k << '\n';
// Operator overloading
// Adding two complex numbers
x = y + z;
cout << "\nx = y + z:\n"
<< x << " = "
<< y << " + " << z << '\n';
// Subtract two complex numbers
x = y - z;
cout << "\nx = y - z:\n"
<< x << " = " << y << " - " << z << '\n';
// Multiply two complex numbers
x = y * z;
cout << "\nx = y * z:\n"
<< x << " = " << y << " * " << z << "\n\n";
if (x != k)
{
cout << x << " != " << k << '\n';
}
cout << '\n';
x = k;
if (x == k)
{
cout << x << " == " << k << '\n';
}
return 0;
}
Scott wants to make sure his current power supply has enough power to run his new system.
How will Scott determine whether his current power supply has enough power to run the new system?
Answer:
The correct answer to the following question will be "Wattage ".
Explanation:
The quantity of power needed for the operation of such an electrical system seems to be a Wattage.
The wattage means every device has the highest usable wattage. Remember, furthermore, that perhaps the PSU extracts Electrical energy from either the socket of the panel, transforms it to any other Voltage level, as well as supplies it to your device.By that same Scott determines for certain if his current supply voltage has sufficient energy to function the new program or system.Write a method printShampooInstructions(), with int parameter numCycles, and void return type. If numCycles is less than 1, print "Too few.". If more than 4, print "Too many.". Else, print "N: Lather and rinse." numCycles times, where N is the cycle number, followed by "Done.". End with a newline. Example output with input 2: 1: Lather and rinse. 2: Lather and rinse. Done. Hint: Declare and use a loop variable. import java.util.Scanner; public class ShampooMethod { /* Your solution goes here */ public static void main (String [] args) { Scanner scnr = new Scanner(System.in); int userCycles; userCycles = scnr.nextInt(); printShampooInstructions(userCycles); } }
Answer:
import java.util.Scanner;
public class nu3 {
public static void main(String[] args) {
Scanner in = new Scanner(System.in);
System.out.println("Enter number of cycles");
int numCycles = in.nextInt();
//Call the method
printShampooInstructions(numCycles);
}
public static void printShampooInstructions(int numCycles){
if (numCycles<1){
System.out.println("Too few");
}
if (numCycles>4){
System.out.println("Too Many");
}
else {
for (int i = 1;i <= numCycles; i++) {
System.out.println( numCycles+ ": Lather and rinse");
numCycles--;
}
System.out.println(" Done");
}
}
}
Explanation:
This is solved using Java programming language
The method printShampooInstructions() is in bold in the answer section
I have also provided a complete program that request user to enter value for number of cycles, calls the method and passes that value to it
The logic here is using if .... else statements to handle the different possible values of Number of cycles.
In the Else Section a loop is used to decrementally print the number of cycles
A lamp outside a front door comes on automatically when it is dark, and when someone stands on the doormat outside the front door. A pressure sensor under the mat changes from OFF (0) to ON (1) when someone stands on the doormat. The light sensor is ON (1) when it is light and OFF (0) when it is dark. Design a program to show what would happen to the lamp. Account for all possible scenarios by determining whether the pressure sensor and light sensor are ON or OFF. (HINT: Ask the user "Is it dark?" and "Is someone standing on the doormat outside the front door?")
Answer:
See explaination
Explanation:
#include <iostream>
#include <string>
using namespace std;
int main()
{
string raptor_prompt_variable_zzyz;
?? standing;
?? dark;
raptor_prompt_variable_zzyz ="Is it dark?";
cout << raptor_prompt_variable_zzyz << endl;
cin >> DARK;
if (DARK=="Yes")
{
raptor_prompt_variable_zzyz ="Is someone standing on the doormat outside the front door?";
cout << raptor_prompt_variable_zzyz << endl;
cin >> STANDING;
if (STANDING=="Yes")
{
cout << "LAMP IS ON" << endl; }
else
{
cout << "LAMP IS OFF" << endl; }
}
else
{
cout << "LAMP IS OFF" << endl; }
return 0;
}
Here's a simple program in pseudo-code to demonstrate the behavior of the lamp based on the inputs from the light sensor and pressure sensor:
Ask user: "Is it dark?" (yes or no)
Read user's response and store it in variable 'is_dark'
Ask user: "Is someone standing on the doormat outside the front door?" (yes or no)
Read user's response and store it in variable 'is_someone_on_doormat'
if is_dark equals yes and is_someone_on_doormat equals yes:
Turn on the lamp
Display message: "Lamp turned on because it is dark and someone is standing on the doormat."
else if is_dark equals yes and is_someone_on_doormat equals no:
Keep the lamp off
Display message: "Lamp remains off because it is dark but no one is standing on the doormat."
else if is_dark equals no and is_someone_on_doormat equals yes:
Turn on the lamp
Display message: "Lamp turned on because someone is standing on the doormat."
else:
Keep the lamp off
Display message: "Lamp remains off because it is not dark and no one is standing on the doormat."
This program covers all possible scenarios based on the inputs from the light sensor and pressure sensor and determines whether the lamp should be turned on or off accordingly.
Write a function remove_italicized_text that takes one parameter, a string named sentence. The string might contain HTML tags for representing italicized text. An HTML tag is a kind of code that directs the web browser to format content in a particular way.
Answer:
See explaination
Explanation:
def remove_italicized_text(x):
ans = ""
n = len(x) # length of string
i = 0
while i < n:
if x[i:(i+3)] == "<i>": # check for opening italic html tag
i += 3
cur = ""
while i < n:
if x[i:(i+4)] == "</i>": # check for closing italic html tag
break
cur += x[i]
i += 1
i += 4
ans += cur # add the text between two tags into output string
ans += " " # adding a space between two next
ans += cur # add the text again
else:
ans += x[i]
i += 1
return ans
sentence = "<i>Stony Brook</i>"
print("sentence = " + sentence)
print("Return value: " + remove_italicized_text(sentence))
print("\n")
sentence = "I <i>love</i> SBU, yes I do!"
print("sentence = " + sentence)
print("Return value: " + remove_italicized_text(sentence))
print("\n")
sentence = "Hey <i>Wolfie</i>, he's so fine, we hug him <i>all the time</i>!"
print("sentence = " + sentence)
print("Return value: " + remove_italicized_text(sentence))
print("\n")
sentence = "This text has no italics tags."
print("sentence = " + sentence)
print("Return value: " + remove_italicized_text(sentence))
print("\n")
Create a procedure named FindLargest that receives two parameters: a pointer to a signed doubleword array, and a count of the array's length. The procedure must return the value of the largest array member in EAX. Use the PROC directive with a parameter list when declaring the procedure. Preserve all registers (except EAX) that are modified by the procedure. Write a test program that calls FindLargest and passes three different arrays of different lengths. Be sure to include negative values in your arrays. Create a PROTO declaration for FindLargest.
Final answer:
The question is about writing an Assembly language procedure to find the largest number in an array. The procedure should be named FindLargest and use the PROC directive and PROTO declaration. The student must implement it, considering register preservation and set up test cases with arrays containing negative numbers.
Explanation:
Create a Procedure in Assembly Language
The student's question involves creating a procedure named FindLargest, which will find the largest value in an array of signed integers. This is a task that would be typically given in an assembly language programming course.
The PROC directive is used to define a new procedure in assembly language, and a PROTO declaration is used to declare the procedure's interface before it’s actually implemented, allowing other parts of the program to call it properly.
Here is an example template for the FindLargest procedure, which must be adapted to the specific assembly language being used, such as x86:
FindLargest PROC, arrayPtr:PTR DWORD, count:DWORD
; Procedure code goes here
FindLargest ENDP
And the corresponding PROTO declaration might look like this:
FindLargest PROTO, arrayPtr:PTR DWORD, count:DWORD
The test program would include calls to FindLargest with different arrays, for example:
invoke FindLargest, ADDR array1, LENGTHOF array1
invoke FindLargest, ADDR array2, LENGTHOF array2
invoke FindLargest, ADDR array3, LENGTHOF array3
Note: The procedure's task is to return the largest value in EAX, and all registers except for EAX must be preserved to maintain the state of the calling environment.
To create the FindLargest procedure, define it with the PROC directive, include operations to identify the largest value, and create a PROTO declaration. Also, write a test program to call FindLargest with different arrays. This ensures the largest value is correctly found and returned.
To create a procedure named FindLargest that receives a pointer to a signed doubleword array and the count of the array's length, follow these steps:
Define the PROC directive with the parameter list:
FindLargest PROC ; Procedure declarationInitialize registers and perform the required operations to find the largest value:
FindLargest PROCCreate a PROTO declaration for FindLargest:
FindLargest PROTO :PTR DWORD, :DWORDWrite a test program to call FindLargest with three different arrays:
.data1. In a language with zero-based indexing, at what index will the 100th item in an array be?
2. Same as 1.1, but for a language with one-based indexing.
The 100th item in an array would be at index 99 in a zero-based indexing language and at index 100 in a one-based indexing language. Indexing helps in accessing elements within data structures like arrays.
In a language with zero-based indexing, such as Python or Java, the 100th item in an array would be at index 99. This is because the first item in an array is at index 0, and the index increases by 1 for each subsequent item. Therefore, the index can be found by subtracting one from the item number (i.e., 100 - 1 = 99).
For a language with one-based indexing, like MATLAB or R, the 100th item would be at index 100 because the indexing starts from 1. Here, the index of the 100th item is the same as the item number.
Understanding the concept of array indexing is vital as it helps in accessing the elements in the data structure efficiently. The for loop can be set up with different starting indices and conditions depending on the language's indexing convention. For example, you might initialize the index i with 0 or 1, and the loop could run while i < 101 for zero-based or i <= 100 for one-based indexing.
Given a base Plant class and a derived Flower class, complete main() to create an ArrayList called myGarden. The ArrayList should be able to store objects that belong to the Plant class or the Flower class. Create a method called printArrayList(), that uses the printInfo() methods defined in the respective classes and prints each element in myGarden. The program should read plants or flowers from input (ending with -1), adding each Plant or Flower to the myGarden ArrayList, and output each element in myGarden using the printInfo() method.
Final answer:
The question involves creating an ArrayList that stores objects of type Plant and Flower, reading these objects from input, and printing their information using a defined method in Java.
Explanation:
The student's question pertains to object-oriented programming in Java and involves working with classes, inheritance, and polymorphism within an ArrayList. To solve this, you'll need to create an ArrayList that can store objects of both the Plant and Flower classes. The Plant class is the base class, while the Flower class is the derived class. You will then implement a printArrayList() method to iterate through the myGarden ArrayList and call the printInfo() method on each object.
Since both Plant and Flower objects can be added to myGarden, the printInfo() method needs to be defined in both the Plant and Flower classes, possibly overriding the method in Flower if it requires different behavior. The process of reading input and adding plants or flowers would continue until encountering a '-1', which signifies the end of input. After populating the ArrayList, the printArrayList() method will be called to display the information about each plant or flower in the garden.
Final answer:
The question involves creating an ArrayList in Java to hold Plant and Flower objects, utilizing polymorphism to store and process them. A printArrayList() method is employed to iterate through the ArrayList and call the printInfo() method on each Plant and Flower object.
Explanation:
The question revolves around object-oriented programming in Java, specifically dealing with the concept of polymorphism and ArrayLists. To accomplish the task, you will need to create an ArrayList that can hold objects of the Plant class as well as objects of the derived Flower class. This is possible because of polymorphism, where a Flower 'is-a' Plant, and thus can be treated as one.
To provide you with a starting point, here's an example of a possible printArrayList() method:
void printArrayList(ArrayList myGarden) {The actual main() is not provided here but would involve initializing the ArrayList, looping to read inputs, creating Plant or Flower objects based on the input, adding them to the myGarden ArrayList, and then calling the printArrayList() method to display the information for each plant or flower in the garden. Notice that the printInfo() method is called for each element; this method is expected to be defined in both the Plant and Flower classes, leveraging polymorphism for dynamic method dispatch.