Answer:
Water holing is the correct answer to this question.
Explanation:
Waterholing:-
It is a kind of attack in which the attacker detects the sites that the targets of the group frequently access and then afflicts the sites with the ransomware. Which afflicts selected representatives of the target group.
The watering hole assault is a data breach wherein the individual attempts to infiltrate a particular demographic of end-users by harming sites reported to be visited by team members. The aim is to compromise a specific target data and gain network access at the perpetrator's place of work.
A university wants to schedule the classrooms for final exams. The attributes are given below:
course : course id, name, and department
section : section id, enrollment, and dependent as a weak entity set on course
room : room number, capacity, and building
exam: course id, section id, room number, and time.
1) Determine the relationships among these entity sets. Clearly specify the arity, the attributes, and the mapping cardinality of these relationships.
2) Construct an E-R diagram to for the scheduling system.
Answer:
See explaination
Explanation:
Clearly the entities are as follows:-
Course
Section
Room
Test
The relationship among entities are as follows:-
Course has Section
Test is conducted for Course
Test is conducted in Section
Test is conducted in Room
Attributes of each entities are as follows:-
Course (CourseID, Name, Department)
Section (SectionID, Enrollment)
Room (RoomNumber, Capacity, Building)
Test (Time)
Section is a week entity as, there may be same sections for different courses, therefore section uses the primary key of course entity as foreign key.
Also entity Test is dependent upon the entities Room,Section and Course, therefore primary keys of these entities will be used as foreign key in the Test entity.
Check attachment for the ER diagram
the smallest unit of time in music called?
Answer:
Ready to help ☺️
Explanation:
A tatum is a feature of music that has been defined as the smallest time interval between notes in a rhythmic phrase.
Answer:
A tatum bc is a feature of music that has been variously defined as the smallest time interval between successive notes in a rhythmic phrase "the shortest durational value
Write a program with total change amount as an integer input that outputs 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, the output is: no change Ex: If the input is: 45 the output is: 1 quarter 2 dimes
Answer:.
// Program is written in C++.
// Comments are used for explanatory purposes
// Program starts here..
#include<iostream>
using namespace std;
int main()
{
// Declare Variables
int amount, dollar, quarter, dime, nickel, penny;
// Prompt user for input
cout<<"Amount: ";
cin>>amount;
// Check if input is less than 1
if(amount<=0)
{
cout<<"No Change";
}
else
{
// Convert amount to various coins
dollar = amount/100;
amount = amount%100;
quarter = amount/25;
amount = amount%25;
dime = amount/10;
amount = amount%10;
nickel = amount/5;
penny = amount%5;
// Print results
if(dollar>=1)
{
if(dollar == 1)
{
cout<<dollar<<" dollar\n";
}
else
{
cout<<dollar<<" dollars\n";
}
}
if(quarter>=1)
{
if(quarter== 1)
{
cout<<quarter<<" quarter\n";
}
else
{
cout<<quarter<<" quarters\n";
}
}
if(dime>=1)
{
if(dime == 1)
{
cout<<dime<<" dime\n";
}
else
{
cout<<dime<<" dimes\n";
}
}
if(nickel>=1)
{
if(nickel == 1)
{
cout<<nickel<<" nickel\n";
}
else
{
cout<<nickel<<" nickels\n";
}
}
if(penny>=1)
{
if(penny == 1)
{
cout<<penny<<" penny\n";
}
else
{
cout<<penny<<" pennies\n";
}
}
}
return 0;
}
#define DIRECTN 100
#define INDIRECT1 20
#define INDIRECT2 5
#define PTRBLOCKS 200
typedef struct {
filename[MAXFILELEN];
attributesType attributes; // file attributes
uint32 reference_count; // Number of hard links
uint64 size; // size of file
uint64 direct[DIRECTN]; // direct data blocks
uint64 indirect[INDIRECT1]; // single indirect blocks
uint64 indirect2[INDIRECT2]; // double indirect
} InodeType;
Single and double indirect inodes have the following structure:
typedef struct
{
uint64 block_ptr[PTRBLOCKS];
}
IndirectNodeType;
Required:
Assuming a block size of 0x1000 bytes, write pseudocode to return the block number associated with an offset of N bytes into the file.
Answer:
WOW! that does not look easy!
Explanation:
I wish i could help but i have no idea how to do that lol
6. Why did he choose to install the window not totally plumb?
Answer:
Because then it would break
Explanation:
You achieve this by obtaining correct measurements. When measuring a window, plumb refers to the vertical planes, and level refers to the horizontal planes. So he did not install the window totally plumb
Algorithmic Complexity: what is the asymptotic complexity (Big-O) of each code section? Identify the critical section of each.\ Line 1: for (int i=0; i<532; i++) { f(n) = O( ) Line 2: for (int j=1; j
Answer:
Check the explanation
Explanation:
1) f(n) = O( 1 ), since the loops runs a constant number of times independent of any input size
there is no critical section in the code, as a critical section is some part of code which is shared by multiple threads or even processes to modify any shared variable.This code does not contain any variable which can be shared.
2) f(n) = O( log n! ), the outer loop runs for n times, and the inner loop runs log k times when i = k,ie the total number of print will be – log 1 + log2 +log3 +log4+…...+ log n = log (1 . 2 . 3 . 4 . ……. . n ) =log n!
there is no critical section in the code, as a critical section is some part of code which is shared by multiple threads or even processes to modify any shared variable.This code does not contain any variable which can be shared.
Note : Log (m *n) = Log m + Log n : this is property of logarithm
3) f(n) = [tex]O( n^2 )[/tex], since both outer and inner loop runs n times hence , the total iterations of print statement will be : n +n+n+…+n
for n times, this makes the complexity – n * n = n2
there is no critical section in the code, as a critical section is some part of code which is shared by multiple threads or even processes to modify any shared variable.This code does not contain any variable which can be shared.
Write a functionvector merge(vector a, vector b)that merges two vectors, alternating elements from both vectors. If one vector isshorter than the other, then alternate as long as you can and then append the remaining elements from the longer vector. For example, if a is 1 4 9 16and b is9 7 4 9 11then merge returns the vector1 9 4 7 9 4 16 9 1118. Write a predicate function bool same_elements(vector a, vector b)that checks whether two vectors have the same elements in some order, with the same multiplicities. For example, 1 4 9 16 9 7 4 9 11 and 11 1 4 9 16 9 7 4 9 would be considered identical, but1 4 9 16 9 7 4 9 11 and 11 11 7 9 16 4 1 would not. You will probably need one or more helper functions.19. What is the difference between the size and capacity of a vector
Answer:
see explaination for code
Explanation:
CODE
#include <iostream>
#include <vector>
using namespace std;
vector<int> merge(vector<int> a, vector<int> b) {
vector<int> result;
int k = 0;
int i = 0, j = 0;
while (i < a.size() && j < b.size()) {
if (k % 2 == 0) {
result.push_back(a[i ++]);
} else {
result.push_back(b[j ++]);
}
k ++;
}
while (i < a.size()) {
result.push_back(a[i ++]);
}
while(j < b.size()) {
result.push_back(b[j ++]);
}
return result;
}
int main() {
vector<int> a{1, 4, 9, 16};
vector<int> b{9, 7, 4, 9, 11};
vector<int> result = merge(a, b);
for (int i=0; i<result.size(); i++) {
cout << result[i] << " ";
}
}
A cloud provider is deploying a new SaaS product comprised of a cloud service. As part of the deployment, the cloud provider wants to publish a service level agreement (SLA) that provides an availability rating based on its estimated availability over the next 12 months. First, the cloud provider estimates that, based on historical data of the cloud environment, there is a 25% chance that the physical server hosting the cloud service will crash and that such a crash would take 2 days before the cloud service could be restored. It is further estimated that, over the course of a 12 month period, there will be various attacks on the cloud service, resulting in a total of 24 hours of downtime. Based on these estimates, what is the availability rating of the cloud service that should be published in the SLA?
Answer:
99.6
Explanation:
The cloud provider provides the certain modules of the cloud service like infrastructure as a service , software as a service ,etc to other companies .The cloud provider implements a new SaaS service that includes a cloud platform also the cloud provider intends to disclose the Service Level Agreement that is score based on its approximate accessibility during the next 12 months.
The Cloud providers predict that, depending on past cloud system data, it is a 25 % risk that a physical server holding the cloud platform will fail therefore such a failure will require 2 days to recover the cloud service so 99.6 is the Cloud storage accessibility ranking, to be released in the SLA.
Reggie receives a phone call from a computer user at his company who says that his computer would not turn on this morning. After asking questions, Reggie realizes that the computer seems to turn on because the user can hear fans spinning and the indicator light comes on and flashes during the boot. The Num Lock key is on, but there is no video on the monitor. Reggie also learns that the computer has a new video card. The night before the computer was working fine, according to the user. Reggie tests the video cord and the monitor. When Reggie reattaches the video cable to the video card port, the computer appears to be working. What might be the problem?
Answer:
The Integrated video may not be working properly well or functioning well.
Explanation:
Solution:
In this case, it is known that the computer was working fine a night before the problem.
If the video cable is connected to a wrong video port, then the computer was unable to work previous night.the integrated video cord has not seated because most times it occurs, due to the heat sink in the card. after restoring the video cable to the video card port, it works.
If the video was disabled in BIOS/UEFI, it does not show video, even after restoring the cable.
Create a Binary Expressions Tree Class and create a menu driven programyour program should be able to read multiple expressions from a file and create expression trees for each expression, one at a timethe expression in the file must be in "math" notation, for example x+y*a/b.display the preorder traversal of a binary tree as a sequence of strings each separated by a tabdisplay the postorder traversal of a binary tree in the same form as aboveWrite a function to display the inorder traversal of a binary tree and place a (before each subtree and a )after each subtree. Don’t display anything for an empty subtree. For example, the expression tree should would be represented as ( (x) + ( ( (y)*(a) )/(b) ) )
Answer:
Explanation:
Program:
#include<iostream>
#include <bits/stdc++.h>
using namespace std;
//check for operator
bool isOperator(char c)
{
switch(c)
{
case '+': case '-': case '/': case '*': case '^':
return true;
}
return false;
}
//Converter class
class Converter
{
private:
string str;
public:
//constructor
Converter(string s):str(s){}
//convert from infix to postfix expression
string toPostFix(string str)
{
stack <char> as;
int i, pre1, pre2;
string result="";
as.push('(');
str = str + ")";
for (i = 0; i < str.size(); i++)
{
char ch = str[i];
if(ch==' ') continue;
if (ch == '(')
as.push(ch);
else if (ch == ')')
{
while (as.size() != 0 && as.top() != '('){
result = result + as.top() + " ";
as.pop();
}
as.pop();
}
else if(isOperator(ch))
{
while (as.size() != 0 && as.top() != '(')
{
pre1 = precedence(ch);
pre2 = precedence(as.top());
if (pre2 >= pre1){
result = result + as.top() + " ";
as.pop();
}
else break;
}
as.push(ch);
}
else
{
result = result + ch;
}
}
while(as.size() != 0 && as.top() != '(') {
result += as.top() + " ";
as.pop();
}
return result;
}
//return the precedence of an operator
int precedence(char ch)
{
int choice = 0;
switch (ch) {
case '+':
choice = 0;
break;
case '-':
choice = 0;
break;
case '*':
choice = 1;
break;
case '/':
choice = 1;
break;
case '^':
choice = 2;
default:
choice = -999;
}
return choice;
}
};
//Node class
class Node
{
public:
string element;
Node *leftChild;
Node *rightChild;
//constructors
Node (string s):element(s),leftChild(nullptr),rightChild(nullptr) {}
Node (string s, Node* l, Node* r):element(s),leftChild(l),rightChild(r) {}
};
//ExpressionTree class
class ExpressionTree
{
public:
//expression tree construction
Node* covert(string postfix)
{
stack <Node*> stk;
Node *t = nullptr;
for(int i=0; i<postfix.size(); i++)
{
if(postfix[i]==' ') continue;
string s(1, postfix[i]);
t = new Node(s);
if(!isOperator(postfix[i]))
{
stk.push(t);
}
else
{
Node *r = nullptr, *l = nullptr;
if(!stk.empty()){
r = stk.top();
stk.pop();
}
if(!stk.empty()){
l = stk.top();
stk.pop();
}
t->leftChild = l;
t->rightChild = r;
stk.push(t);
}
}
return stk.top();
}
//inorder traversal
void infix(Node *root)
{
if(root!=nullptr)
{
cout<< "(";
infix(root->leftChild);
cout<<root->element;
infix(root->rightChild);
cout<<")";
}
}
//postorder traversal
void postfix(Node *root)
{
if(root!=nullptr)
{
postfix(root->leftChild);
postfix(root->rightChild);
cout << root->element << " ";
}
}
//preorder traversal
void prefix(Node *root)
{
if(root!=nullptr)
{
cout<< root->element << " ";
prefix(root->leftChild);
prefix(root->rightChild);
}
}
};
//main method
int main()
{
string infix;
cout<<"Enter the expression: ";
cin >> infix;
Converter conv(infix);
string postfix = conv.toPostFix(infix);
cout<<"Postfix Expression: " << postfix<<endl;
if(postfix == "")
{
cout<<"Invalid expression";
return 1;
}
ExpressionTree etree;
Node *root = etree.covert(postfix);
cout<<"Infix: ";
etree.infix(root);
cout<<endl;
cout<<"Prefix: ";
etree.prefix(root);
cout<<endl;
cout<< "Postfix: ";
etree.postfix(root);
cout<<endl;
return 0;
}
Evaluati urmatoarele expresii
5+2*(x+4)/3, unde x are valoare 18
7/ 2*2+4*(5+7*3)>18
2<=x AND x<=7 , unde x are valoare 23
50 %10*5=
31250/ 5/5*2=
Answer:
A) 22 ⅓
B) 111>18
C) There is an error in the expression
D) 25
E) 62500
Question:
Evaluate the following expressions
A) 5 + 2 * (x + 4) / 3, where x has a value of 18
B) 7/2 * 2 + 4 * (5 + 7 * 3) & gt; 18
C) 2 <= x AND x<= 7, where x has value 23
D) 50% 10 * 5 =
F) 31250/5/5 * 2 =
Explanation:
A) 5 + 2 * (x + 4) / 3
x = 18
First we would insert the value of x
5 + 2 * (x + 4) / 3
5 + 2(18 + 8) / 3
Then we would evaluate the expression by applying BODMAS : This stands for Bracket, Of, Division, Multiplication, addition and subtraction.
= 5 + 2(26) / 3
= 5 + 52/3
= 5 + 17 ⅓
= 22 ⅓
B) 7/2 * 2 + 4 * (5 + 7 * 3) > 18
we would evaluate the expression by applying BODMAS : This stands for Bracket, Of, Division, Multiplication, addition and subtraction.
7/2 * 2 + 4 * (5 + 7 * 3) >18
= 7/2 × 2 + 4× (5 + 7 × 3)>18
= (7×2)/2 + 4× (5+21) >18
= 14/2 + 4(26) >18
= 7 + 104 >18
= 111>18
C) 2 <= x AND x<= 7, where x has value 23
There is an error in the expression
D) 50% of 10 * 5
we would evaluate the expression by applying BODMAS : This stands for Bracket, Of, Division, Multiplication, addition and subtraction.
The 'of' expression means multiplication
= 50% × 10×5
= 50% × 50
50% = 50/100
=50/100 × 50
= 1/2 × 50
= 25
F) 31250/5/5 * 2
The expression has no demarcation. Depending on how it is broken up, we would arrive at different answers. Let's consider:
31250/(5/5 × 2)
Apply BODMAS
= 31250/[5/(5 × 2)]
= 31250/(5/10)
= 31250/(1/2)
Multiply by the inverse of 1/2 = 2/1
= 31250 × (2/1)
= 62500
Create a macro named mReadInt that reads a 16- or 32-bit signed integer from standard input and returns the value in an argument. Use conditional operators to allow the macro to adapt to the size of the desired result. Write a program that tests the macro, passing it operands of various sizes.
Answer:
;Macro mReadInt definition, which take two parameters
;one is the variable to save the number and other is the length
;of the number to read (2 for 16 bit and 4 for 32 bit) .
%macro mReadInt 2
mov eax,%2
cmp eax, "4"
je read2
cmp eax, "2"
je read1
read1:
mReadInt16 %1
cmp eax, "2"
je exitm
read2:
mReadInt32 %1
exitm:
xor eax, eax
%endmacro
;macro to read the 16 bit number, parameter is number variable
%macro mReadInt16 1
mov eax, 3
mov ebx, 2
mov ecx, %1
mov edx, 5
int 80h
%endmacro
;macro to read the 32 bit number, parameter is number variable
%macro mReadInt32 1
mov eax, 3
mov ebx, 2
mov ecx, %1
mov edx, 5
int 80h
%endmacro
;program to test the macro.
;data section, defining the user messages and lenths
section .data
userMsg db 'Please enter the 32 bit number: '
lenUserMsg equ $-userMsg
userMsg1 db 'Please enter the 16 bit number: '
lenUserMsg1 equ $-userMsg1
dispMsg db 'You have entered: '
lenDispMsg equ $-dispMsg
;.bss section to declare variables
section .bss
;num to read 32 bit number and num1 to rad 16-bit number
num resb 5
num1 resb 3
;.text section
section .text
;program start instruction
global _start
_start:
;Displaying the message to enter 32bit number
mov eax, 4
mov ebx, 1
mov ecx, userMsg
mov edx, lenUserMsg
int 80h
;calling the micro to read the number
mReadInt num, 4
;Printing the display message
mov eax, 4
mov ebx, 1
mov ecx, dispMsg
mov edx, lenDispMsg
int 80h
;Printing the 32-bit number
mov eax, 4
mov ebx, 1
mov ecx, num
mov edx, 4
int 80h
;displaying message to enter the 16 bit number
mov eax, 4
mov ebx, 1
mov ecx, userMsg1
mov edx, lenUserMsg1
int 80h
;macro call to read 16 bit number and to assign that number to num1
;mReadInt num1,2
;calling the display mesage function
mov eax, 4
mov ebx, 1
mov ecx, dispMsg
mov edx, lenDispMsg
int 80h
;Displaying the 16-bit number
mov eax, 4
mov ebx, 1
mov ecx, num1
mov edx, 2
int 80h
;exit from the loop
mov eax, 1
mov ebx, 0
int 80h
Explanation:
For an assembly code/language that has the conditions given in the question, the program that tests the macro, passing it operands of various sizes is given below;
;Macro mReadInt definition, which take two parameters
;one is the variable to save the number and other is the length
;of the number to read (2 for 16 bit and 4 for 32 bit) .
%macro mReadInt 2
mov eax,%2
cmp eax, "4"
je read2
cmp eax, "2"
je read1
read1:
mReadInt16 %1
cmp eax, "2"
je exitm
read2:
mReadInt32 %1
exitm:
xor eax, eax
%endmacro
;macro to read the 16 bit number, parameter is number variable
%macro mReadInt16 1
mov eax, 3
mov ebx, 2
mov ecx, %1
mov edx, 5
int 80h
%endmacro
;macro to read the 32 bit number, parameter is number variable
%macro mReadInt32 1
mov eax, 3
mov ebx, 2
mov ecx, %1
mov edx, 5
int 80h
%endmacro
;program to test the macro.
;data section, defining the user messages and lenths
section .data
userMsg db 'Please enter the 32 bit number: '
lenUserMsg equ $-userMsg
userMsg1 db 'Please enter the 16 bit number: '
lenUserMsg1 equ $-userMsg1
dispMsg db 'You have entered: '
lenDispMsg equ $-dispMsg
;.bss section to declare variables
section .bss
;num to read 32 bit number and num1 to rad 16-bit number
num resb 5
num1 resb 3
;.text section
section .text
;program start instruction
global _start
_start:
;Displaying the message to enter 32bit number
mov eax, 4
mov ebx, 1
mov ecx, userMsg
mov edx, lenUserMsg
int 80h
;calling the micro to read the number
mReadInt num, 4
;Printing the display message
mov eax, 4
mov ebx, 1
mov ecx, dispMsg
mov edx, lenDispMsg
int 80h
;Printing the 32-bit number
mov eax, 4
mov ebx, 1
mov ecx, num
mov edx, 4
int 80h
;displaying message to enter the 16 bit number
mov eax, 4
mov ebx, 1
mov ecx, userMsg1
mov edx, lenUserMsg1
int 80h
;macro call to read 16 bit number and to assign that number to num1
;mReadInt num1,2
;calling the display mesage function
mov eax, 4
mov ebx, 1
mov ecx, dispMsg
mov edx, lenDispMsg
int 80h
;Displaying the 16-bit number
mov eax, 4
mov ebx, 1
mov ecx, num1
mov edx, 2
int 80h
;exit from the loop
mov eax, 1
mov ebx, 0
int 80h
Write a statement that calls the recursive method backwardsAlphabet() with parameter startingLetter.
import java.util.Scanner; public class RecursiveCalls { public static void backwardsAlphabet(char currLetter) { if (currLetter == 'a') { System.out.println(currLetter); } else { System.out.print(currLetter + " "); backwardsAlphabet((char)(currLetter - 1)); } } public static void main (String [] args) { Scanner scnr = new Scanner(System.in); char startingLetter; startingLetter = scnr.next().charAt(0); /* Your solution goes here */ } }
Answer:
Following are the code to method calling
backwardsAlphabet(startingLetter); //calling method backwardsAlphabet
Output:
please find the attachment.
Explanation:
Working of program:
In the given java code, a class "RecursiveCalls" is declared, inside the class, a method that is "backwardsAlphabet" is defined, this method accepts a char parameter that is "currLetter". In this method a conditional statement is used, if the block it will check input parameter value is 'a', then it will print value, otherwise, it will go to else section in this block it will use the recursive function that prints it's before value. In the main method, first, we create the scanner class object then defined a char variable "startingLetter", in this we input from the user and pass its value into the method that is "backwardsAlphabet".How can u refer to additional information while giving a presentation
Answer:
There are various ways: Handing out papers/fliers to people, or presenting slides.
Write a program whose input is a character and a string, and whose output indicates the number of times the character appears in the string.
Ex: If the input is:
n Monday
the output is:
1
Ex: If the input is:
z Today is Monday
the output is:
0
Ex: If the input is:
n It's a sunny day
the output is:
2
Case matters.
Ex: If the input is:
n Nobody
the output is:
0
n is different than N.
This is what i have so far.
#include
#include
using namespace std;
int main() {
char userInput;
string userStr;
int numCount;
cin >> userInput;
cin >> userStr;
while (numCount == 0) {
cout << numCount << endl;
numCount = userStr.find(userInput);
}
return 0;
}
Which of the following should be the first page of a report?
O Title page
Introduction
O Table of contents
Terms of reference
Answer:
Title page should be the first page of a report.
hope it helps!
In this question, we give two implementations for the function: def intersection_list(lst1, lst2) This function is given two lists of integers lst1 and lst2. When called, it will create and return a list containing all the elements that appear in both lists. For example, the call: intersection_list([3, 9, 2, 7, 1], [4, 1, 8, 2])could create and return the list [2, 1]. Note: You may assume that each list does not contain duplicate items. a) Give an implementation for intersection_list with the best worst-case runtime. b) Give an implementation for intersection_list with the best average-case runtime.
Answer:
see explaination
Explanation:
a)Worst Case-time complexity=O(n)
def intersection_list(lst1, lst2):
lst3 = [value for value in lst1 if value in lst2]
return lst3
lst1 = []
lst2 = []
n1 = int(input("Enter number of elements for list1 : "))
for i in range(0, n1):
ele = int(input())
lst1.append(ele) # adding the element
n2 = int(input("Enter number of elements for list2 : "))
for i in range(0, n2):
ele = int(input())
lst2.append(ele) # adding the element
print(intersection_list(lst1, lst2))
b)Average case-time complexity=O(min(len(lst1), len(lst2))
def intersection_list(lst1, lst2):
return list(set(lst1) & set(lst2))
lst1 = []
lst2 = []
n1 = int(input("Enter number of elements for list1 : "))
for i in range(0, n1):
ele = int(input())
lst1.append(ele)
n2 = int(input("Enter number of elements for list2 : "))
for i in range(0, n2):
ele = int(input())
lst2.append(ele)
print(intersection_list(lst1, lst2))
Donnell backed up the information on his computer every week on a flash drive. Before copying the files to the flash drive, he always ran a virus scan against the files to ensure that no viruses were being copied to the flash drive. He bought a new computer and inserted the flash drive so that he could transfer his files onto the new computer. He got a message on the new computer that the flash drive was corrupted and unreadable; the information on the flash drive cannot be retrieved. Assuming that the flash drive is not carrying a virus, which of the following does this situation reflect?
a. Compromise of the security of the information on the flash drive
b. Risk of a potential breach in the integrity of the data on the flash drive
c. Both of the above
d. Neither of the above.
Answer:
b. Risk of a potential breach in the integrity of the data on the flash drive
Explanation:
The corrupted or unreadable file error is an error message generated if you are unable to access the external hard drive connected to the system through the USB port. This error indicates that the files on the external hard drive are no longer accessible and cannot be opened.
There are several reasons that this error message can appear:
Viruses and Malware affecting the external hard drive .Physical damage to external hard drive or USB memory .Improper ejection of removable drives.Assume the class Student implements the Speaker interface from the textbook. Recall that this interface includes two abstract methods, speak( ) and announce(String str). A Student contains one instance data, String class Rank. Write the Student class so that it implements Speaker as follows. The speak method will output "I am a newbie here" if the Student is a "Freshman", "I like my school" if the Student is either a "Sophomore" or a "Junior", or "I can not wait to graduate" if the student is a "Senior". The announce method will output "I am a Student, here is what I have to say" followed by the String parameter on a separate line. Finally, the class Rank is initialized in the constructor. Only implement the constructor and the methods to implement the Speaker interface.
Answer:
Check the explanation
Explanation:
Based on the above information, this the following resulting code.
The solution has namely one interface named Speaker, one class called Student and one Main class called StudentMain.
The StudentMain class just initialize various instances of the Student class with various classRank values and then subsequently calling in the speak() methods of all.
announce() method is called on one such instance of Student class to demonstrate its functioning.
Comments are added at required place to better help in understanding the logic.
Speaker.java
public interface Speaker {
abstract void speak();
abstract void announce(String str);
}
Student.java
public class Student implements Speaker {
/*
* Write the Student class so that it implements Speaker as follows.
The speak method will output "I am a newbie here" if the Student is a "Freshman",
"I like my school" if the Student is either a "Sophomore" or a "Junior",
or "I can not wait to graduate" if the student is a "Senior".
The announce method will output "I am a Student, here is what I have to say" followed by the String parameter on a separate line.
*/
String classRank;
// Based on the requirement taking in classRank String as a constructor parameter
public Student(String classRank) {
super();
this.classRank = classRank;
}
// Using the switch-case to code the above requirment
public void speak() {
switch(classRank) {
case "Freshman":
System.out.println("I am a newbie here");
break;
// This case would output the same result for the 2 case conditions
case "Sophomore":
case "Junior":
System.out.println("I like my school");
break;
case "Senior":
System.out.println("I can not wait to graduate");
break;
default:
System.out.println("Unknown classRank inputted");
}
}
//Based on the requirement, first line is printed first after second which is
// the inputted String parameter
public void announce(String str) {
System.out.println("I am a Student, here is what I have to say");
System.out.println(str);
}
}
StudentMain.java
public class StudentMain {
public static void main(String[] args) {
// Initializing all the Objects with required classRank
Student stu1=new Student("Freshman");
Student stu2=new Student("Sophomore");
Student stu3=new Student("Junior");
Student stu4=new Student("Senior");
Student stu5=new Student("Freshman");
//Calling the speak methods of the all the above create objects
stu1.speak();
stu2.speak();
stu3.speak();
stu4.speak();
stu4.speak();
stu4.announce("Wish u all the best");
}
}
Following is the output generated from the code run.
Describe how DMA is
used to
transfer data
from peripherals.
Answer:
Think of DMA as a co-processor that is used to quickly transfer data between main memory and peripherals without the intervention of the CPU.
Write a program that uses the map template class to compute a histogram of positive numbers entered by the user. The map’s key should be the number that is entered, and the value should be a counter of the number of times the key has been entered so far. Use –1 as a sentinel value to signal the end of user input.512355321-1Then the program should output the followings:The number 3 occurs 2 timesThe number 5 occurs 3 timesThe number 12 occurs 1 timesThe
Answer:
See explaination
Explanation:
#include <iostream>
#include <map>
#include <string>
#include <algorithm>
#include <cstdlib>
#include <iomanip>
using namespace std;
int main()
{
map<int, int> histogram;
int a=0;
while(a>=0)
{
cout << " enter value of a" ;
cin >>a;
histogram[a]++;
}
map<int,int>::iterator it;
for ( it=histogram.begin() ; it != histogram.end(); it++ )
{
cout << (*it).first << " occurs " << setw(3) << (*it).second << (((*it).second == 1) ? " time." : " times.") <<endl;
}
return 0;
}
Design an application for Bob's E-Z Loans. The application accepts a client's loan amount and monthly payment amount. Output the customer's loan balance each month until the loan is paid off. b. Modify the Bob's E-Z Loans application so that after the payment is made each month, a finance charge of 1 percent is added to the balance.
Answer:
part (a).
The program in cpp is given below.
#include <stdio.h>
#include <iostream>
using namespace std;
int main()
{
//variables to hold balance and monthly payment amounts
double balance;
double payment;
//user enters balance and monthly payment amounts
std::cout << "Welcome to Bob's E-Z Loans application!" << std::endl;
std::cout << "Enter the balance amount: ";
cin>>balance;
std::cout << "Enter the monthly payment: ";
cin>>payment;
std::cout << "Loan balance: " <<" "<< "Monthly payment: "<< std::endl;
//balance amount and monthly payment computed
while(balance>0)
{
if(balance<payment)
{ payment = balance;}
else
{
std::cout << balance <<"\t\t\t"<< payment << std::endl;
balance = balance - payment;
}
}
return 0;
}
part (b).
The modified program from part (a), is given below.
#include <stdio.h>
#include <iostream>
using namespace std;
int main()
{
//variables to hold balance and monthly payment amounts
double balance;
double payment;
double charge=0.01;
//user enters balance and monthly payment amounts
std::cout << "Welcome to Bob's E-Z Loans application!" << std::endl;
std::cout << "Enter the balance amount: ";
cin>>balance;
std::cout << "Enter the monthly payment: ";
cin>>payment;
balance = balance +( balance*charge );
std::cout << "Loan balance with 1% finance charge: " <<" "<< "Monthly payment: "<< std::endl;
//balance amount and monthly payment computed
while(balance>payment)
{
std::cout << balance <<"\t\t\t\t\t"<< payment << std::endl;
balance = balance +( balance*charge );
balance = balance - payment;
}
if(balance<payment)
{ payment = balance;}
std::cout << balance <<"\t\t\t\t\t"<< payment << std::endl;
return 0;
}
Explanation:
1. The variables to hold the loan balance and the monthly payment are declared as double.
2. The program asks the user to enter the loan balance and monthly payment respectively which are stored in the declared variables.
3. Inside a while loop, the loan balance and monthly payment for each month is computed with and without finance charges in part (a) and part (b) respectively.
4. The computed values are displayed for each month till the loan balance becomes zero.
5. The output for both part (a) and part (b) are attached as images.
You work at a cheeseburger restaurant. Write a program that determines appropriate changes to a food order based on the user’s dietary restrictions. Prompt the user for their dietary restrictions: vegetarian, lactose intolerant, or none. Then using if statements and else statements, print the cook a message describing how they should modify the order. The following messages should be used: - If the user enters "lactose intolerant", say "No cheese." - If the user enters "vegetarian", say "Veggie burger." - If the user enters "none", say "No alterations."
Answer:
See explaination
Explanation:
dietary_restrictions = input("Any dietary restrictions?: ")
if dietary_restrictions=="lactose intolerant":
print("No cheese")
elif dietary_restrictions == "vegetarian":
print("Veggie burger")
else:
print("No alteration")
Computer A has an overall CPI of 1.3 and can be run at a clock rate of 600 MHz. Computer B has a CPI of 2.5 and can be run at a clock rate of 750 MHz. We have a particular program we wish to run. When compiled for computer A, this program has exactly 100,000 instructions. How many instructions would the program need to have when compiled for Computer B, in order for the two computers to have exactly the same execution time for this program
Answer:
Check the explanation
Explanation:
CPI means Clock cycle per Instruction
given Clock rate 600 MHz then clock time is Cー 1.67nSec clockrate 600M
Execution time is given by following Formula.
Execution Time(CPU time) = CPI*Instruction Count * clock time = [tex]\frac{CPI*Instruction Count}{ClockRate}[/tex]
a)
for system A CPU time is 1.3 * 100, 000 600 106
= 216.67 micro sec.
b)
for system B CPU time is [tex]=\frac{2.5*100,000}{750*10^6}[/tex]
= 333.33 micro sec
c) Since the system B is slower than system A, So the system A executes the given program in less time
Hence take CPU execution time of system B as CPU time of System A.
therefore
216.67 micro = =[tex]\frac{2.5*Instruction}{750*10^6}[/tex]
Instructions = 216.67*750/2.5
= 65001
hence 65001 instruction are needed for executing program By system B. to complete the program as fast as system A
The number of instructions that the program would need to have when compiled for Computer B is; 65000 instructions
What is the execution time?
Formula for Execution time is;
Execution time = (CPI × Instruction Count)/Clock Time
We are given;
CPI for computer A = 1.3
Instruction Count = 100000
Clock time = 600 MHz = 600 × 10⁶ Hz
Thus;
Execution time = (1.3 * 100000)/(600 × 10⁶)
Execution time(CPU Time) = 216.67 * 10⁻⁶ second
For CPU B;
CPU Time = (2.5 * 100000)/(750 × 10⁶)
CPU Time = 333.33 * 10⁻⁶ seconds
Thus, instructions for computer B for the two computers to have same execution time is;
216.67 * 10⁻⁶ = (2.5 * I)/(750 × 10⁶)
I = (216.67 * 10⁻⁶ * 750 × 10⁶)/2.5
I = 65000 instructions
Read more about programming instructions at; https://brainly.com/question/15683939
Trace the complete execution of the MergeSort algorithm when called on the array of integers, numbers, below. Show the resulting sub-arrays formed after each call to merge by enclosing them in { }. For example, if you originally had an array of 5 elements, a = {5,2,8,3,7}, the first call to merge would result with: {2, 5} 8, 3, 7 ← Note after the first call to merge, two arrays of size 1 have been merged into the sorted subarray {2,5} and the values 2 and 5 are sorted in array a You are to do this trace for the array, numbers, below. Be sure to show the resulting sub-arrays after each call to MergeSort. int[] numbers = {23, 14, 3, 56, 17, 8, 42, 18, 5};
Answer:
public class Main {
public static void merge(int[] arr, int l, int m, int r) {
int n1 = m - l + 1;
int n2 = r - m;
int[] L = new int[n1];
int[] R = new int[n2];
for (int i = 0; i < n1; ++i)
L[i] = arr[l + i];
for (int j = 0; j < n2; ++j)
R[j] = arr[m + 1 + j];
int i = 0, j = 0;
int k = l;
while (i < n1 && j < n2) {
if (L[i] <= R[j]) {
arr[k] = L[i];
i++;
} else {
arr[k] = R[j];
j++;
}
k++;
}
while (i < n1) {
arr[k] = L[i];
i++;
k++;
}
while (j < n2) {
arr[k] = R[j];
j++;
k++;
}
printArray(arr, l, r);
}
public static void sort(int[] arr, int l, int r) {
if (l < r) {
int m = (l + r) / 2;
sort(arr, l, m);
sort(arr, m + 1, r);
merge(arr, l, m, r);
}
}
static void printArray(int[] arr, int l, int r) {
System.out.print("{");
for (int i = l; i <= r; ++i)
System.out.print(arr[i] + " ");
System.out.println("}");
}
public static void main(String[] args) {
int[] arr = {23, 14, 3, 56, 17, 8, 42, 18, 5};
sort(arr, 0, arr.length - 1);
}
}
Explanation:
See answer
The relationship between the temperature of a fluid (t, in seconds), temperature (T, in degrees Celsius), is dependent upon the initial temperature of the liquid (T0, in degrees Celsius), the ambient temperature of the surroundings (TA, in degrees Celsius) and the cooling constant (k, in hertz); the relationship is given by: ???? ???? ???????? ???? ???????????? ???? ???????????? ???????????????? Ask the user the following questions: From a menu, choose fluid ABC, FGH, or MNO. Enter the initial fluid temperature, in units of degrees Celsius. Enter the time, in units of minutes. Enter the ambient air temperature, in units of degrees Celsius. Enter the following data into the program. The vector contains the cooling constant (k, units of hertz) corresponding to the menu entries. K Values = [0.01, 0.03, 0.02] Create a formatted output statement for the user in the Command Window similar to the following. The decimal places must match. ABC has temp 83.2 degrees Celsius after 3 minutes. In
Answer:
See explaination
Explanation:
clc;
clear all;
close all;
x=input(' choose abc or fgh or mno:','s');
to=input('enter intial fluid temperature in celcius:');
t1=input('enter time in minutes:');
ta=input('enter ambient temperature in celcius:');
abc=1;
fgh=2;
mno=3;
if x==1
k=0.01;
elseif x==2
k=0.03;
else
k=0.02;
end
t=ta+((to-ta)*exp((-k)*t1));
X = sprintf('%s has temp %f degrees celcius after %d minutes.',x,t,t1);
disp(X);
(a) What is the difference between a compare validator and a range validator?
(b) When would you choose to use one versus the other?
Answer:
Check the explanation
Explanation:
a) A compare validator: A compare validator can be described as a particular method that is utilized to control and also enabled to perform different types of validation tasks. It is being used to carry out as well as to execute a correct data type that determines the entered value is proper value into a form field or not. For instance when in an application registration it makes use of the confirm password after the first-password.
(b) Whenever it is compulsory to determine if the value that has been entered is the correct value into the form field then the usage of compare validator will be needed whereas a range validator is used to confirm the entered value is in a specific range.
Explanation:
a) A compare validator: A compare validator can be described as a particular method that is utilized to control and also enabled to perform different types of validation tasks. It is being used to carry out as well as to execute a correct data type that determines the entered value is proper value into a form field or not. For instance when in an application registration it makes use of the confirm password after the first-password.
(b) Whenever it is compulsory to determine if the value that has been entered is the correct value into the form field then the usage of compare validator will be needed whereas a range validator is used to confirm the entered value is in a specific range.
Help its simple but I don't know the answer!!!
If you have a -Apple store and itunes gift card-
can you use it for in app/game purchases?
Answer:
yes you can do that it's utter logic
Answer:
Yes.
Explanation:
Itunes gift cards do buy you games/movies/In app purchases/ect.
Zoom Vacuum, a family-owned manufacturer of high-end vacuums, has grown exponentially over the last few years. However, the company is having difficulty preparing for future growth. The only information system used at Zoom is an antiquated accounting system. The company has one manufacturing plant located in Iowa; and three warehouses, in Iowa, New Jersey, and Nevada. The Zoom sales force is national, and Zoom purchases about 25 percent of its vacuum parts and materials from a single overseas supplier. You have been hired to recommend the information systems Zoom should implement in order to maintain their competitive edge. However, there is not enough money for a full-blown, cross-functional enterprise application, and you will need to limit the first step to a single functional area or constituency. What will you choose, and why?
Answer:A TPS focusing on production and manufacturing to keep production costs low while maintaining quality, and for communicating with other possible vendors. The TPS would later be used to feed MIS and other higher level systems.
Explanation:
Implement the function get_stats The function get_stats calculates statistics on a sample of values. It does the following: its input parameter is a csv string a csv (comma separated values) string contains a list of numbers (the sample) return a tuple that has 3 values: (n, stdev, mean) tuple[0] is the number of items in the sample tuple[1] is the standard deviation of the sample tuple[2] is the mean of the sample
Answer:
Check the explanation
Explanation:
PYTHON CODE: (lesson.py)
import statistics
# function to find standard deviation, mean
def get_stats(csv_string):
# calling the function to clean the data
sample=clean(csv_string)
# finding deviation, mean
std_dev=statistics.stdev(sample)
mean=statistics.mean(sample)
# counting the element in sample
count=len(sample)
# return the results in a tuple
return (count, std_dev, mean)
# function to clean the csv string
def clean(csv_string):
# temporary list
t=[]
# splitting the string by comma
data=csv_string.split(',')
# looping item in data list
for item in data:
# removing space,single quote, double quote
item=item.strip()
item= item.replace("'",'')
item= item.replace('"','')
# if the item is valid
if item:
# converting into float
try:
t.append(float(item))
except:
continue
# returning the list of float
return t
if __name__=='__main__':
csv = "a, 1, '-2', 2.35, None,, 4, True"
print(clean(csv))
print(get_stats('1.0,2.0,3.0'))
