Answer:
dude what r u doin
Explanation:
A MySetOperations is a collection of functions, on which the following set operations are defined:
myIsEmpty(A): Returns T if the set A is empty (A = Φ), F otherwise.
myDisjoint(A,B): Returns T if intersection of the set A and set B is non-empty
(A â© B â Φ), F otherwise.
myIntersection(A,B): Returns the intersection of the set A and set B (A â© B).
myUnion(A,B): Returns the union of the set A and set B (A ⪠B).
Write a collection of MySetOperations in Python.
Note: For this problem, you can use only the standard flow Python's instructions and the set build-in operations: len, in, not in, and add.
Answer:
Explanation:
The objective here is to write a collection of MySetOperations in Python.
During the process of submitting this answer after computing the program ; I keep getting a response message that goes thus "It appears that your answer contains either a link or inappropriate words. Please correct and submit again!" . In the bid to curb such dilemma , i have created a word document for the collection of MySetOperations in Python.
The document can be seen in the attached file below.
p25: File Write and Read1) User enters a file name (such as "myMovies.txt").2) User enters the titles of 4 of their favorite moveis (use a loop!).3) Program Writes the 4 titles to a file, one per line, then closes the file (use a loop!).4) Program Reads the 4 titles from "myMovies.txt" stores them in a list and shows the list5) The program writes the titles in reverse order into a file "reverseOrder.txt"Sample Run:Please enter a file name: myMovies.txtPlease enter a movie title #1: movie1Please enter a movie title #2: movie2Please enter a movie title #3: movie3Please enter a movie title #4: movie4Writing the 4 movie titles to file 'myMovies.txt'Reading the 4 movie titles from file into a list: [movie1, movie2, movie3, movie4]Writing the 4 movie titles in revers to 'reverseOrder.txt'Note: Do not use reverse() , reversed()Content of myMovies.txt:movie1movie2movie3movie4Content of reverseOrder.txtmovie4movie3movie2movie1
Answer:
The program goes as follows
Comments are used to explain difficult lines
#include<iostream>
#include<fstream>
#include<sstream>
#include<string>
using namespace std;
int main()
{
//Declare String to accept file name
string nm;
//Prompt user for file name
cout<<"Enter File Name: ";
getline(cin,nm);
//Create File
ofstream Myfile(nm.c_str());
//Prompt user for 4 names of movies
string movie;
for(int i = 1;i<5;i++)
{
cout<<"Please enter a movie title #"<<i<<": ";
cin>>movie;
//Write to file, the names of each movies
Myfile<<movie<<endl;
}
Myfile.close(); // Close file
//Create an Array for four elements
string myArr[4];
//Prepare to read from file to array
ifstream file(nm.c_str());
//Open File
if(file.is_open())
{
for(int i =0;i<4;i++)
{
//Read each line of the file to array (line by line)
file>>myArr[i];
}
}
file.close(); // Close file
//Create a reverseOrder.txt file
nm = "reverseOrder.txt";
//Prepare to read into file
ofstream Myfile2(nm.c_str());
for(int j = 3;j>=0;j--)
{
//Read each line of the file to reverseOrder.txt (line by line)
Myfile2<<myArr[j]<<endl;
}
Myfile2.close(); //Close File
return 0;
}
See attached for .cpp program file
How the full address space of a processor is partitioned and how each partition is used?
Answer:
?
Explanation:
?
which of the following correctly declares an array:
Choose one answer .
a. int array
b.array(10);
c.int array(10);
d.array array(10);
Answer:
a.
Explanation:
The option that correctly declares an array is int array.
What is an array?This is a term that connote a regular system or arrangement. Here, numbers or letters are said to be arranged in rows and columns.
In computing, An array is a term that describe memory locations box having the same name. All data in an array is said to be of the same data type.
Learn more about array from
https://brainly.com/question/26104158
Identify the flaws / limitations in the following ConvertToNumber method:
public static bool ConvertToNumber(string str)
{
bool canConvert = false;
try
{
int n = Int16.Parse(str);
if (n != 0)
{
canConvert = true;
}
}
catch (Exception ex)
{
}
bool retval = false;
if (canConvert == true)
{
retval = true;
}
return retval;
}
Answer:
Flaws and limitations identified in this program includes;
1.There was a not necessary usage of variable retrieval. Would have made use of canConvert.
2. Looking at the program, one will notice numerous typos. One of which is the fact that in JAVA we make use of Boolean instead of bool.
3.We rather use Integer.parseInt in JAVA and not Int16, cant make use of Int16.
4. The exception cant be printed
5. JAVA makes use of checkConversion instead of convertNumber as used in the program.
6. It cant work for decimal numbers, 0 and big integers.
Explanation:
See Answer for the detailed explaination of the flaws and limitations identified in the program.
The four compass points can be abbreviated by single-letter strings as "N", "E", "S", and "W". Write a function turn_clockwise that takes one of these four compass points as its parameter, and returns the next compass point in the clockwise direction. Here are some tests that should pass:
test(turn_clockwise("N") == "E")
test(turn_clockwise("W") == "N")
You might ask "What if the argument to the function is some other value?" For all other
cases, the function should return the value None:
test(turn_clockwise(42) == None)
test(turn_clockwise("rubbish") == None)
Answer:
The program code is written in the explanation.
Explanation:
def turn_clockwise(d):
if d=="N":
return "E"
elif d=="E":
return "S"
elif d=="S":
return "W"
elif d=="W":
return "N"
else:
return "None"
Output:
AYUDAAAA!!!!! URGENTE!!!!!!1
¿para que sirve la "BIG DATA"?
Answer:
Big data is a field that treats ways to analyze, systematically extract information from, or otherwise deal with data sets that are too large or complex to be dealt with by traditional data-processing application software. ... Big data was originally associated with three key concepts: volume, variety, and velocity.
SPANISH TRANSLATION
Big data es un campo que trata formas de analizar, extraer sistemáticamente información de, o de otra manera tratar con conjuntos de datos que son demasiado grandes o complejos para ser manejados por el software tradicional de aplicación de procesamiento de datos. ... Big data se asoció originalmente con tres conceptos clave: volumen, variedad y velocidad.
Using the College Registration example from Section 6.7.3 as a starting point, do the following:
-Recode the logic sing CMP and conditional jump instructions (instead of the .IF and .ELSEIF directives).
-Perform range checking on the credits value; it cannot be less than 1 or greater than 30. If an invalid entry is discovered, display an appropriate error message.
-Prompt the user for the grade average and credits values.
Display a message that shows the outcome of the evaluation, such as "The student can register" or "The student cannot register".
6.7.3 example:
.data
TRUE = 1
FALSE = 0
gradeAverage WORD 275 ; test value
credits WORD 12 ; test value
OkToRegister BYTE ?
.code
main PROC
mov OkToRegister,FALSE
.IF gradeAverage > 350
mov OkToRegister,TRUE
.ELSEIF (gradeAverage > 250) && (credits <= 16)
mov OkToRegister,TRUE
.ELSEIF (credits <= 12)
mov OkToRegister,TRUE
.ENDIF
Answer:
Check the explanation
Explanation:
INCLUDE Irvine32.inc
TRUE = 1
FALSE = 0
.data
gradeAverage WORD ?
credits WORD ?
oKToRegister BYTE ?
str1 BYTE "Error: Credits must be between 1 and 30" , 0dh,0ah,0
main PROC
call CheckRegs
exit
main ENDP
CheckRegs PROC
push edx
mov OkToRegister,FALSE
; Check credits for valid range 1-30
cmp credits,1 ; credits < 1?
jb E1
cmp credits,30 ; credits > 30?
ja E1
jmp L1 ; credits are ok
; Display error message: credits out of range
E1:
mov edx,OFFSET str1
call WriteString
jmp L4
L1:
cmp gradeAverage,350 ; if gradeAverage > 350
jna L2
mov OkToRegister,TRUE ; OkToRegister = TRUE
jmp L4
L2:
cmp gradeAverage,250 ; elseif gradeAverage > 250
jna L3
cmp credits,16 ; && credits <= 16
jnbe L3
mov OkToRegister,TRUE ; OKToRegister = TRUE
jmp L4
L3:
cmp credits,12 ; elseif credits <= 12
ja L4
mov OkToRegister,TRUE ; OKToRegister = TRUE
L4:
pop edx ; endif
ret
CheckRegs ENDP
END main
Write a program that prompts the user to enter the area of the flat cardboard. The program then outputs the length and width of the cardboard and the length of the side of the square to be cut from the corner so that the resulting box is of maximum volume. Calculate your answer to three decimal places. Your program must contain a function that takes as input the length and width of the cardboard and returns the side of the square that should be cut to maximize the volume. The function also returns the maximum volume.
Answer:
A program in C++ was written to prompts the user to enter the area of the flat cardboard.
Explanation:
Solution:
The C++ code:
#include <iostream>
#include <iomanip>
#include <cmath>
using namespace std;
double min(double,double);
void max(double,double,double&,double&);
int main()
{double area,length,width=.001,vol,height,maxLen,mWidth,maxHeight,maxVolume=-1;
cout<<setprecision(3)<<fixed<<showpoint;
cout<<"Enter the area of the flat cardboard: ";
cin>>area;
while(width<=area)
{length=area/width;
max(length,width,vol,height);
if(vol>maxVolume)
{maxLen=length;
mWidth=width;
maxHeight=height;
maxVolume=vol;
}
width+=.001;
}
cout<<"dimensions of card to maximize the cardboard box which has a volume "
<<maxVolume<<endl;
cout<<"Length: "<<maxLen<<"\nWidth: "<<maxLen<<endl;
cout<<"dimensions of the cardboard box\n";
cout<<"Length: "<<maxLen-2*maxHeight<<"\nWidth: "
<<mWidth-2*maxHeight<<"\nHeight: "<<maxHeight<<endl;
return 0;
}
void max(double l,double w,double& max, double& maxside)
{double vol,ht;
maxside=min(l,w);
ht=.001;
max=-1;
while(maxside>ht*2)
{vol=(l-ht*2)*(w-ht*2)*ht;
if(vol>max)
{max=vol;
maxside=ht;
}
ht+=.001;
}
}
double min(double l,double w)
{if(l<w)
return l;
return w;
}
Note: Kindly find the output code below
/*
Output for the code:
Enter the area of the flat cardboard: 23
dimensions of card to maximize the cardboard box which has a volume 0.023
Length: 4.796
Width: 4.796
dimensions of the cardboard box
Length: 4.794
Width: 4.794
Height: 0.001
*/
Answer each of the following with a TRUE (T) or FALSE (F).
(1) Pipelining allows instructions to execute sequentially.
(2) If you can find the value x in the cache, you can also find it in the main memory too.
(3) Only the lw instruction can access the Data Memory component.
(4) The bias in single precision is 127, while in double precision it is 1024.
(5) In a cache mapping architecture, the valid bit is always on until data is found in that specific block.
(6) Memory speed generally gets faster the farther you move from the processor.
(7) A cache is a small high speed memory that stores a subset of the information that is in RAM.
(8) The hit rate of a cache mapping architecture is calculated by doing 1 - hit penalty.
(9) Cache is the fastest type of memory.
(10) Single-cycle and multi-cycle machines differ in CPI but all the other values are the same.
Answer:
t
Explanation:
In the RSA system, the receiver does as follows:1. Randomly select two large prime numbers p and q, which always must bekept secret.2. Select an integer number E, known as the public exponent, such that (p 1)and E have no common divisors, and (q 1) and E have no commondivisors.3. Determine the private exponent, D, such that (ED 1) is exactly divisible byboth (p 1) and (q 1). In other words, given E, we choose D such that theinteger remainder when ED is divided by (p 1)(q 1) is 1.4. Determine the product n = pq, known as public modulus.5. Release publicly the public key, which is the pair of numbers n and E, K =(n, E). Keep secret the private key, K = (n, D).The above events are mentioned in the correct order as they are performed whilewriting the algorithm.Select one:TrueFalse
Answer:
False.
Explanation:
The correct option is false that is to say the arrangement is not true. So, let us check one or two things about the RSA System.
The RSA system simply refer to as a cryptosystem that was first established in the year 1973 but was later developed in the year 1977 by Ron Rivest, Adi Shamir and Leonard Adleman.
The main use of the RSA system is for encryption of messages. It is known as a public key cryptography algorithm.
The way the algorithm should have been arranged is;
(1). Randomly select two large prime numbers p and q, which always must bekept secret.
(2). Determine the product n = pq, known as public modulus.
(3). Select an integer number E, known as the public exponent, such that (p 1)and E have no common divisors, and (q 1) and E have no commondivisors.
(4). Determine the private exponent, D, such that (ED 1) is exactly divisible by both (p 1) and (q 1). In other words, given E, we choose D such that the integer remainder when ED is divided by (p1)(q 1) is 1.
(5). Release publicly the public key, which is the pair of numbers n and E, K =(n, E). Keep secret the private key, K = (n, D).
Therefore, The way it should have been arranged in the question should have been;
(1) , (4), (2), (3 ) and ( 5).
"Dean wants a quick way to look up staff members by their Staff ID. In cell Q3, nest the existing VLOOKUP function in an IFERROR function. If the VLOOKUP function returns an error result, the text ""Invalid Staff ID"" should be displayed by the formula. (Hint: You can test that this formula is working by changing the value in cell Q2 to 0, but remember to set the value of cell Q2 back to 1036 when the testing is complete.)"
Answer:
ierror(VLOOKUP(Q2,CBFStaff[[Staff ID]:[Name]],2,FALSE), "Invalid Staff ID")
Explanation:
Let me try as much as I can to explain the concept or idea of iferror in vlookup.
iferror have a typically function and result like an if else statement, its syntax is IFERROR(value,value _ if _ error), this simply means that if the the error is equal to value, value is returned if not, the next argument is returned.
Having said that, feom the question we are given,
let's substitute the value with vlookup function and add an else argument, it will look exactly this way;
IFERROR(VLOOKUP(),"Invalid Staff ID")// now this will set the message if vlookup cannot find the.
On the other hand using the values given, we will have;
ierror(VLOOKUP(Q2,CBFStaff[[Staff ID]:[Name]],2,FALSE), "Invalid Staff ID")
Tamera and Rupert each applied for the same credit card through the same company. Tamera has a positive credit history. Rupert has a negative credit history.
Which compares their credit limits and likely interest rates?
Answer:
it is two by five credit limit
Answer:
Your answer is A Tamera’s credit limit is most likely higher than Rupert’s, and her interest rate is most likely lower.
Explanation:
I got it right on the test and I hope this answer helps
You are building a predictive solution based on web server log data. The data is collected in a comma-separated values (CSV) format that always includes the following fields: date: string time: string client_ip: string server_ip: string url_stem: string url_query: string client_bytes: integer server_bytes: integer You want to load the data into a DataFrame for analysis. You must load the data in the correct format while minimizing the processing overhead on the Spark cluster. What should you do? Load the data as lines of text into an RDD, then split the text based on a comma-delimiter and load the RDD into a DataFrame. Define a schema for the data, then read the data from the CSV file into a DataFrame using the schema. Read the data from the CSV file into a DataFrame, infering the schema. Convert the data to tab-delimited format, then read the data from the text file into a DataFrame, infering the schema.
Answer:
see explaination
Explanation:
The data is collected in a comma-separated values (CSV) format that always includes the following fields:
? date: string
? time: string
? client_ip: string
? server_ip: string
? url_stem: string
? url_query: string
? client_bytes: integer
? server_bytes: integer
What should you do?
a. Load the data as lines of text into an RDD, then split the text based on a comma-delimiter and load the RDD into DataFrame.
# import the module csv
import csv
import pandas as pd
# open the csv file
with open(r"C:\Users\uname\Downloads\abc.csv") as csv_file:
# read the csv file
csv_reader = csv.reader(csv_file, delimiter=',')
# now we can use this csv files into the pandas
df = pd.DataFrame([csv_reader], index=None)
df.head()
b. Define a schema for the data, then read the data from the CSV file into a DataFrame using the schema.
from pyspark.sql.types import *
from pyspark.sql import SparkSession
newschema = StructType([
StructField("date", DateType(),true),
StructField("time", DateType(),true),
StructField("client_ip", StringType(),true),
StructField("server_ip", StringType(),true),
StructField("url_stem", StringType(),true),
StructField("url_query", StringType(),true),
StructField("client_bytes", IntegerType(),true),
StructField("server_bytes", IntegerType(),true])
c. Read the data from the CSV file into a DataFrame, infering the schema.
abc_DF = spark.read.load('C:\Users\uname\Downloads\new_abc.csv', format="csv", header="true", sep=' ', schema=newSchema)
d. Convert the data to tab-delimited format, then read the data from the text file into a DataFrame, infering the schema.
Import pandas as pd
Df2 = pd.read_csv(‘new_abc.csv’,delimiter="\t")
print('Contents of Dataframe : ')
print(Df2)
Explain why you cannot the Apple OS install on a regular windows computer and vice versa, without the aid of a virtualization software explain Virtualbox, parallel desktop, etc. In your response, explain why it works with a virtualization software.
Answer:
The reason is due to proprietary design of the Operating System (OS) which require a virtualization software to blanket or "disguise" the hardware (processor) borderlines of the computer onto which it is to be installed.
Explanation:
An Apple system that has the RISC processor and system architecture which has an operating system made entirely for the Apple system architecture
If the above Apple OS is to be installed on a windows computer, then the procedure to setup up the OS has to be the same as that used when on an Apple system, hence, due to the different processors and components of both systems, a virtualization will be be needed to be provided by a Virtual box, Parallels desktop or other virtualization software.
The following table contains data about projects and the hours charged against them:
ProjectBilling:
ProjectNbr ProjectName EmployeeNbr EmployeeName JobClass HourlyRate HoursBilled
15 Evergreen 103 June Phillips SA-3 96.75 23.8
15 Evergreen 101 John Baker DD-5 105.00 19.4
15 Evergreen 105 Alice Miller DD-5 105.00 35.7
15 Evergreen 106 William Smithfield SA-2 62.50 12.6
18 Amber Wave 114 Anna Jones SA-1 55.00 23.8
18 Amber Wave 101 John Baker DD-5 105.00 24.6
18 Amber Wave 112 Brad White SA-3 96.75 45.3
22 Starlight 107 Terry Ray SA-2 62.50 32.4
22 Starlight 115 Peter Novak SA-3 96.75 44.0
22 Starlight 101 John Baker DD-5 105.00 64.7
22 Starlight 114 Anna Jones SA-1 55.00 48.4
22 Starlight 108 Susan Brown SA-2 62.50 23.6
(A) Convert the table to 3NF.
Answer:
The final tables in 3NF are as follows.
Project( ProjectNbr, ProjectName)
Employee( EmployeeNbr, EmployeeName, JobClass)
Job( JobClass, HourlyRate)
ProjectBilling( ProjectNbr, EmployeeNbr, HoursBilled)
Explanation:
The given table is given below.
ProjectBilling( ProjectNbr, ProjectName, EmployeeNbr, EmployeeName, JobClass, HourlyRate, HoursBilled)
ProjectNbr -> ProjectName
EmployeeNbr -> EmployeeName
JobClass -> HourlyRate
ProjectNbr, EmployeeNbr -> HoursBilled
1NF
1. All the fields in the given table contain only a single value. The table is in 1NF.
2NF
2. New tables are formed based on the given functional dependencies.
Project( ProjectNbr, ProjectName)
Employee( EmployeeNbr, EmployeeName)
Job( JobClass, HourlyRate)
ProjectBilling( ProjectNbr, EmployeeNbr, HoursBilled)
3. Every table is assigned a primary key which are as follows.
ProjectNbr is the primary key for Project table.
EmployeeNbr is the primary key for Project table.
JobClass is the primary key for Project table.
(ProjectNbr, EmployeeNbr) is the composite primary key for the ProjectBilling table.
4. The tables which are related to each other are linked via primary key and foreign key.
5. In the ProjectBilling table, the composite primary key, ProjectNbr, EmployeeNbr is composed of the primary keys of the Project and Employee tables, i.e., ProjectNbr and EmployeeNbr respectively.
6. Job table is related to Employee table. Hence, primary key of Job table, JobClass, is introduced as foreign key in Employee table.
Employee( EmployeeNbr, EmployeeName, JobClass)
7. Partial dependency arises when composite primary key exists and non-prime attributes (columns other than the primary key) depend on a part of the primary key, i.e., partial primary key.
In the ProjectBilling table, no partial dependency exists.
8. All the tables are in 2NF as given below.
Project( ProjectNbr, ProjectName)
Employee( EmployeeNbr, EmployeeName, JobClass)
Job( JobClass, HourlyRate)
ProjectBilling( ProjectNbr, EmployeeNbr, HoursBilled)
3NF
9. All the tables are in 2NF.
10. In every table, all non-prime attribute depend only on the primary key.
11. No transitive dependency exists, i.e., all non-prime attributes do not depend on other non-prime attributes.
12. Hence, all the conditions are satisfied and the tables are in 3NF.
Suppose the daytime processing load consists of 65% CPU activity and 35% disk activity. Your customers are complaining that the system is slow. After doing some research, you learn that you can upgrade your disks for $8,000 to make them 3 times as fast as they are currently. You have also learned that you can upgrade your CPU to make it 1.5 times faster for $6,000. a) Which would you choose to yield the best performance improvement for the least
amount of money?
b) Which option would you choose if you don’t care about the money, but want a
faster system?
c) What is the break-even point for the upgrades? That is, what price would be
charged for the CPU(or the disk--change only one) so the results was the same cost per 1% increase in both.
Answer:
The answer to this question can be described as follows:
Explanation:
Given data:
Performance of the CPU:
The Fastest Factor Fraction of Work:
[tex]f_1=65 \% \\\\=\frac{65}{100} \\\\ =0.65[/tex]
Current Feature Speedup:
[tex]K_1=[/tex] 1.5
CPU upgrade=6000
Disk activity:
The quickest part is the proportion of the work performed:
Current Feature Speedup:
[tex]k_2=3[/tex]
Disk upgrade=8000
System speedup formula:
[tex]s=\frac{1}{(1-f)+(\frac{f}{k})}[/tex]
Finding the CPU activity and disk activity by above formula:
CPU activity:
[tex]S_{CPU}=\frac{1}{(1-f_1)+(\frac{f_1}{k_1})} \\\\=\frac{1}{(1-0.65)+(\frac{0.65}{1.5})} \\\\=1.276 \% ...\rightarrow (1) \\[/tex]
Disk activity:
[tex]S_{DISK} = (\frac{1}{(1-f_2)+\frac{f_2}{k_2}}) \\\\ S_{DISK} = (\frac{1}{(1-0.35)+\frac{0.35}{3}}) \\\\ = -0.5\% .... \rightarrow (2)[/tex]
CPU:
Formula for CPU upgrade:
[tex]= \frac{CPU \ upgrade}{S_{CPU}}\\\\= \frac{\$ 6,000}{1.276}\\\\= 4702.19....(3)[/tex]
DISK:
Formula for DISK upgrade:
[tex]=\frac{Disk upgrade} {S_{DISK}}\\\\= \frac{\$ 8000}{-0.5 \% }\\\\= - 16000....(4)[/tex]
equation (3) and (4),
Thus, for the least money the CPU alternative is the best performance upgrade.
b)
From (3) and (4) result,
The disc choice is therefore the best choice for a quicker system if you ever don't care about the cost.
c)
The break-event point for the upgrades:
=4702.19 x-0.5
= -2351.095
From (2) and (3))
Therefore, when you pay the sum for disc upgrades, all is equal $ -2351.095
Let a and b be two vector. i.e. a and b are two vectors, of possibly different sizes, containing integers. Further assume that in both a and b the integers are sorted in ascending order.
1. Write a function:
vector merge( vector a, vector b)
that will merge the two vectors into one new one and return the merged vector.
By merge we mean that the resulting vector should have all the elements from a and b, and all its elements should be in ascending order.
For example:
a: 2,4,6,8
b: 1,3,7,10,13
the merge will be: 1,2,3,4,6,7,8,10,13
Do this in two ways. In way 1 you cannot use any sorting function. In way 2 you must.
Answer:
A C++ program was used in writing a function in merging ( vector a, vector b) or that will merge the two vectors into one new one and return the merged vector.
Explanation:
Solution
THE CODE:
#include <iostream>
#include <vector>
using namespace std;
vector<int> merge(vector<int> a, vector<int> b) {
vector<int> vec;
int i = 0, j = 0;
while(i < a.size() || j < b.size()) {
if(i >= a.size() || (j < b.size() && b[j] < a[i])) {
if(vec.empty() || vec[vec.size()-1] != b[j])
vec.push_back(b[j]);
j++;
} else {
if(vec.empty() || vec[vec.size()-1] != a[i])
vec.push_back(a[i]);
i++;
}
}
return vec;
}
int main() {
vector<int> v1 = {2, 4, 6, 8};
vector<int> v2 = {1, 3, 7, 10, 13};
vector<int> vec = merge(v1, v2);
for(int i = 0; i < vec.size(); ++i) {
cout << vec[i] << " ";
}
cout << endl;
return 0;
}
Double any element's value that is less than controlValue. Ex: If controlValue = 10, then dataPoints = {2, 12, 9, 20} becomes {4, 12, 18, 20}.
import java.util.Scanner; public class StudentScores { public static void main (String [] args) { Scanner scnr = new Scanner(System.in); final int NUM_POINTS = 4; int[] dataPoints = new int[NUM_POINTS]; int controlValue; int i; controlValue = scnr.nextInt(); for (i = 0; i < dataPoints.length; ++i) { dataPoints[i] = scnr.nextInt(); } for (i = 0; i < dataPoints.length; ++i) { System.out.print(dataPoints[i] + " "); } System.out.println(); } }
Answer:
import java.util.Scanner;
public class StudentScores
{
public static void main(String[] args) {
Scanner scnr = new Scanner(System.in);
final int NUM_POINTS = 4;
int[] dataPoints = new int[NUM_POINTS];
int controlValue;
int i;
controlValue = scnr.nextInt();
for (i = 0; i < dataPoints.length; ++i) {
dataPoints[i] = scnr.nextInt();
}
for (i = 0; i < dataPoints.length; ++i) {
System.out.print(dataPoints[i] + " ");
}
System.out.println();
for (i = 0; i < dataPoints.length; ++i) {
if(dataPoints[i] < controlValue){
dataPoints[i] = dataPoints[i] * 2;
}
System.out.print(dataPoints[i] + " ");
}
}
}
Explanation:
*Added parts highligted.
After getting the control value and values for the array, you printed them.
Create a for loop that iterates through the dataPoints. Inside the loop, check if a value in dataPoints is smaller than the contorolValue. If it is, multiply that value with 2 and assign it to the dataPoints array. Print the elements of the dataPoints
Answer:
import java.util.Scanner;
public class numm3 {
public static void main (String [] args) {
Scanner scnr = new Scanner(System.in);
final int NUM_POINTS = 4;
int[] dataPoints = new int[NUM_POINTS];
int controlValue;
int i;
System.out.println("Enter the control Variable");
controlValue = scnr.nextInt();
System.out.println("enter elements for the array");
for (i = 0; i < dataPoints.length; ++i) {
dataPoints[i] = scnr.nextInt();
}
for (i = 0; i < dataPoints.length; ++i) {
System.out.print(dataPoints[i] + " ");
}
System.out.println();
//Doubling elements Less than Control Variable
for (i = 0; i < dataPoints.length; ++i) {
if (dataPoints[i]<controlValue){
dataPoints[i] = dataPoints[i]*2;
}
System.out.print(dataPoints[i] + " ");
}
System.out.println(); } }
Explanation:
See the additional code to accomplish the task in bold
The trick is using an if statement inside of a for loop that checks the condition (dataPoints[i]<controlValue) If true, it multiplies the element by 2
What keyboard functions lets you delete words
Answer:Backspace
Explanation:
If you want to delete words you use the backspace button.
Answer:
Backspace
Explanation:
In devising a route to get water from a local lake to its central water storage tank, the city of Rocky Mount considered the various routes and capacities that might be taken between the lake and the storage tank. The city council members want to use a network model to determine if the city's pipe system has enough capacity to handle the demand.
1. They should use ____________.
A. Minimal Spanning Tree Method
B. Shortest Path Method
C. Maximal Flow Method
D. Any of the above will do
Answer:
C. Maximal Flow Method
Explanation:
When it comes to optimization theory, Maximal Flow Method or problems have to do with finding a possible or viable flow through a flow network that obtains the maximum potential flow rate. The maximum flow problem can be viewed as an exceptional case of more complex and complicated network flow problems, such as that of the circulation problem.
The code below uses the Space macro which simply displays the number of blank spaces specified by its argument. What is the first number printed to the screen after this code executes? (ignore the .0000 from Canvas) main PROC push 4 push 13 call rcrsn exit main ENDP rcrsn PROC push ebp mov ebp,esp mov eax,[ebp + 12] mov ebx,[ebp + 8] cmp eax,ebx jl recurse jmp quit recurse: inc eax push eax push ebx call rcrsn mov eax,[ebp + 12] call WriteDec Space 2 quit: pop ebp ret 8 rcrsn ENDP
Answer:
The code implements a recursive function by decrementing number by 1 from 10 to 4 recursively from the stack and add an extra space 2 using the macro function
Explanation:
Solution
The code implements a recursive function by decrementing number by 1 from 10 to 4 recursively from the stack and add an extra space 2 using the macro function mwritespace
The first number is printed after the code executes is 9
Code to be used:
Include Irvine32.inc
INCLUDE Macros.inc ;Include macros for space
.code
main proc
push 4 ;Push 4 to the stack
push 10 ;Push 10 to the stack
call rcrsn ;call the function
exit
;invoke ExitProcess,0 ;After return from the function
;call exit function
main ENDP ;end the main
rcrsn PROC ;define the function
push ebp ;push the ebp into stack
mov ebp,esp ;set ebp as frame pointer
mov eax,[ebp + 12] ;This is for second parameter
mov ebx,[ebp + 8] ;This is for first parameter
cmp eax,ebx ;compare the registers
jl recurse ;jump to another function
jmp quit ;call quit
recurse: ;Implement another function
inc eax ;increment count value
push eax ;push the eax value into stack
push ebx ;push ebx into stack
call rcrsn ;call above function
mov eax,[ebp + 12] ;This is for second parameter
call WriteDec ;write the value on the screen
mWritespace 2 ;Space macro print 2 spaces
;pause the screen
quit: ;Implement quit function
pop ebp ;pop the pointer value from the stack
ret 8 ;clean up the stack
rcrsn ENDP
end main
Create two files to submit:
ItemToPurchase.java - Class definition
ShoppingCartPrinter.java - Contains main() method
Build the ItemToPurchase class with the following specifications:
Private fields
String itemName - Initialized in default constructor to "none"
int itemPrice - Initialized in default constructor to 0
int itemQuantity - Initialized in default constructor to 0
Default constructor
Public member methods (mutators & accessors)
setName() & getName()
setPrice() & getPrice()
setQuantity() & getQuantity()
(2) In main(), prompt the user for two items and create two objects of the ItemToPurchase class. Before prompting for the second item, call scnr.nextLine(); to allow the user to input a new string.
Answer:
Explanation:
public class ItemToPurchase {
private String itemName ;
private int itemPrice;
private int itemQuantity;
public ItemToPurchase(){
itemName = "none";
itemPrice = 0;
itemQuantity = 0;
}
public String getName() {
return itemName;
}
public void setName(String itemName) {
this.itemName = itemName;
}
public int getPrice() {
return itemPrice;
}
public void setPrice(int itemPrice) {
this.itemPrice = itemPrice;
}
public int getQuantity() {
return itemQuantity;
}
public void setQuantity(int itemQuantity) {
this.itemQuantity = itemQuantity;
}
}
ShoppingCartPrinter.java
import java.util.Scanner;
public class ShoppingCartPrinter {
public static void main(String[] args) {
Scanner scan = new Scanner(System.in);
ItemToPurchase item1 = new ItemToPurchase();
ItemToPurchase item2 = new ItemToPurchase();
System.out.println("Item 1");
System.out.print("Enter the item name: ");
String name1 = scan.nextLine();
System.out.print("Enter the item price: ");
int price1 = scan.nextInt();
System.out.print("Enter the item quantity: ");
int quantity1 = scan.nextInt();
item1.setName(name1);
item1.setPrice(price1);
item1.setQuantity(quantity1);
scan.nextLine();
System.out.println("Item 2");
System.out.print("Enter the item name: ");
String name2 = scan.nextLine();
System.out.print("Enter the item price: ");
int price2 = scan.nextInt();
System.out.print("Enter the item quantity: ");
int quantity2 = scan.nextInt();
item2.setName(name2);
item2.setPrice(price2);
item2.setQuantity(quantity2);
System.out.println("TOTAL COST");
int item1Total = item1.getPrice() * item1.getQuantity();
int item2Total = item2.getPrice() * item2.getQuantity();
System.out.println(item1.getName()+" "+item1.getQuantity()+" for $"+item1.getPrice()+" = $"+item1Total);
System.out.println(item2.getName()+" "+item2.getQuantity()+" for $"+item2.getPrice()+" = $"+item2Total);
System.out.println();
System.out.println("Total: $"+(item1Total + item2Total));
ItemToPurchase.java
======================================================
package brainly;
//Class header definition for class ItemToPurchase
public class ItemToPurchase {
//Write the private fields
private String itemName;
private int itemPrice;
private int itemQuantity;
//Create the default constructor
//And initialize all the fields to their respective initial values
public ItemToPurchase(){
this.itemName = "none";
this.itemPrice = 0;
this.itemQuantity = 0;
}
//Create the accessor methods as follows:
public String getItemName() {
return itemName;
}
public void setItemName(String itemName) {
this.itemName = itemName;
}
public int getItemPrice() {
return itemPrice;
}
public void setItemPrice(int itemPrice) {
this.itemPrice = itemPrice;
}
public int getItemQuantity() {
return itemQuantity;
}
public void setItemQuantity(int itemQuantity) {
this.itemQuantity = itemQuantity;
}
} // End of class definition
=======================================================
ShoppingCartPrinter.java
=======================================================
package brainly;
//import the Scanner class
import java.util.Scanner;
//Class header definition for class ShoppingCartPrinter
public class ShoppingCartPrinter {
public static void main(String[] args) {
//Create an object of the Scanner class
Scanner scnr = new Scanner(System.in);
//prompt the user to enter the name of first item
System.out.println("Item 1: Enter the item name");
//Store the input in a String variable
String firstItemName = scnr.nextLine();
//Prompt the user to enter the price of the first item
System.out.println("Enter the item price");
//Store the input in an int variable
int firstItemPrice = scnr.nextInt();
//Prompt the user to enter the quantity of the first item
System.out.println("Enter the item quantity");
//Store the input in an int variable
int firstItemQuantity = scnr.nextInt();
//Call the scnr.nextLine() method to allow user enter the
// second item
scnr.nextLine();
//Prompt the user to enter the name of the second item
System.out.println("Item 2: Enter the item name");
//Store the input in a String variable
String secondItemName = scnr.nextLine();
//Prompt the user to enter the price of the second item
System.out.println("Enter the item price");
//Store the input in an int variable
int secondItemPrice = scnr.nextInt();
//Prompt the user to enter the quantity of the second item
System.out.println("Enter the item quantity");
//Store the input in an int variable
int secondItemQuantity = scnr.nextInt();
//Create the first object using the first item
ItemToPurchase item1 = new ItemToPurchase();
//Create the second object using the second item
ItemToPurchase item2 = new ItemToPurchase();
//Get the total cost of the first item
int totalcostofitem1 = firstItemPrice * firstItemQuantity;
//Get the total cost of the second item
int totalcostofitem2 = secondItemPrice * secondItemQuantity;
//Calculate total cost of the two items
int total = totalcostofitem1 + totalcostofitem2;
//Print out the result.
System.out.println("Total: $" + total);
}
}
=======================================================
Sample Output=======================================================
>> Item 1: Enter the item name
Chocolate Chips
>> Enter the item price
3
>> Enter the item quantity
1
>> Item 2: Enter the item name
Bottled Water
>> Enter the item price
1
>>Enter the item quantity
10
>>Total: $13
======================================================
Explanation:The above code has been written in Java and it contains comments explaining every line of the code. Please go through the comments carefully.
The actual lines of code are written in bold face to distinguish them from comments.
Also, in the sample output, the user inputs are written in bold face.
Consider the following C program: void fun(void) { int a, b, c; /* definition 1 */ ... while (...) { int b, c, d; /* definition 2 */ ... <------------------ 1 while (...) { int c, d, e; /* definition 3 */ ... <-------------- 2 } ... <------------------ 3 }... <---------------------- 4 } For each of the four marked points in this function, list each visible variable, along with the number of the definition statement that defines it.
Answer:
Check the explanation
Explanation:
1.
void func(void){
int a,b,c; /*definition 1*/
/////* a,b,c for definition 1 are visible */
//// d, e are not visible
while(...){
int b, c, d; /* definition 2*/
////*
a from definition 1 is visible
b, c, d from definition 2 are visible
*/ ///
while(...){
int c, d, e;/* definition 3*/
////*
a from definition 1 is visible
b from definition 2 is visible
c, d, e from definition 3 are visible
*/ ///
}
////*
a from definition 1 is visible
b, c, d from definition 2 are visible
e not visible
*////
}
/////* a,b,c for definition 1 are visible */
///// d, e are not visible
}
Write a C++ program that opens and reads file question2.txt. File question1.txt has 3 students and 3 grades of each. You program should read values from file and assign values to 4 variables: (string) name, (double) grade1, (double) grade2, (double) grade3. Then, it should calculate the average grade of each student with the precision of 2 (use setprecision(n) function from header file) and find maximum and minimum grade in the class. Finally, you should have a report on both: • console screen • Output file named output2.txt in the following output format: Max Grade max Min Grade min Adam ave_grade John ave_grade Bill ave_grade
Answer:
see explaination
Explanation:
#include <fstream>
#include <iostream>
#include <string>
//to set precision
#include <iomanip>
#include <bits/stdc++.h>
//structure of student is created.
struct Student
{
std::string name;
double grade1;
double grade2;
double grade3;
double Average;
// Constructor initializes semester scores to 0.
Student()
{
grade1 = 0;
grade2 = 0;
grade3 = 0;
Average = 0;
}
//Read from the file and calculate the average, min and max grade with name.
void read(std::fstream& fil)
{
std::string temp;
fil >> name;
fil >> temp;
grade1 = std::stod(temp);
fil >> temp;
grade2 = std::stod(temp);
fil >> temp;
grade3 = std::stod(temp);
Average = ((grade1 + grade2 + grade3) /3.0);
}
//write to file- name and average with precision 2.
void write(std::fstream& fil)
{
fil << name << " " << std::setprecision(2) << std::fixed << Average << std::endl;
}
//Logic to calculate the minimum grade of the class.
double min()
{
double min = grade1;
if(min > grade2)
{
min = grade2;
}
if(min > grade3)
{
min = grade3;
}
return min;
}
//Logic to calculate the maximum grade of the class.
double max()
{
double max = grade1;
if(max < grade2)
{
max = grade2;
}
if(max < grade3)
{
max = grade3;
}
return max;
}
//Print the name and average with precision 2 in the console.
void print()
{
std::cout << name << " " ;
std::cout << std::setprecision(2) << std::fixed << Average << std::endl;
}
};
//As per this question,students are set to 3.
Student students[3];
//main method.
int main()
{
// Read from the file question2.txt
std::fstream fil;
fil.open("question2.txt", std::ios::in);
for(int i = 0; i < 3; i++)
students[i].read(fil);
fil.close();
// finding the max grade of the class.
int max = students[0].max();
int temp = students[1].max();
if(max <temp)
max = temp;
temp = students[2].max();
if(max < temp)
max = temp;
// finding the min grade of the class.
int min = students[0].min();
temp = students[1].min();
if(min > temp)
min = temp;
temp = students[2].min();
if(min > temp)
min = temp;
// print the output in console.
std::cout << "Max Grade " << max << std::endl;
std::cout << "Min Grade " << min << std::endl;
for(int i = 0; i < 3; i++)
students[i].print();
// write to output file output2.txt .
fil.open("output2.txt", std::ios::out);
fil << "Max Grade " << max << std::endl;
fil << "Min Grade " << min << std::endl;
for(int i = 0; i < 3; i++)
students[i].write(fil);
fil.close();
return 0;
}
A contiguous subsequence of a list S is a subsequence made up of consecutive elements of S. For example, if S is 5,15,-30,10,-5,40,10 then 5,15,-30 is a contiguous subsequence but 5,15,40 is not.
1. Give a linear-time algorithm for the following task:
Input: A list of numbers, A(1), . . . , A(n).
Output: The contiguous subsequence of maximum sum (a subsequence of length zero has sum zero).
For the preceding example, the answer would be 10,-5,40,10 with a sum of 55.
Answer:
We made use of the dynamic programming method to solve a linear-time algorithm which is given below in the explanation section.
Explanation:
Solution
(1) By making use of the dynamic programming, we can solve the problem in linear time.
Now,
We consider a linear number of sub-problems, each of which can be solved using previously solved sub-problems in constant time, this giving a running time of O(n)
Let G[t] represent the sum of a maximum sum contiguous sub-sequence ending exactly at index t
Thus, given that:
G[t+1] = max{G[t] + A[t+1] ,A[t+1] } (for all 1<=t<= n-1)
Then,
G[0] = A[0].
Using the above recurrence relation, we can compute the sum of the optimal sub sequence for array A, which would just be the maximum over G[i] for 0 <= i<= n-1.
However, we are required to output the starting and ending indices of an optimal sub-sequence, we would use another array V where V[i] would store the starting index for a maximum sum contiguous sub sequence ending at index i.
Now the algorithm would be:
Create arrays G and V each of size n.
G[0] = A[0];
V[0] = 0;
max = G[0];
max_start = 0, max_end = 0;
For i going from 1 to n-1:
// We know that G[i] = max { G[i-1] + A[i], A[i] .
If ( G[i-1] > 0)
G[i] = G[i-1] + A[i];
V[i] = V[i-1];
Else
G[i] = A[i];
V[i] = i;
If ( G[i] > max)
max_start = V[i];
max_end = i;
max = G[i];
EndFor.
Output max_start and max_end.
The above algorithm takes O(n) time .
Write a BASH script to create a user account. The script should process two positional parameters: o First positional parameter is supposed to be a string with user name (e.g., user_name) o Second positional parameter is supposed to be a string with user password (e.g., user_password) The script should be able to process one option (use getopts command): o Option -f arg_to_opt_f that will make your script direct all its output messages to file -arg_to_opt_f
#!/bin/bash
usage() {
echo "Usage: $0 [ -f outfile ] name password" 1>&2
exit 1
}
while getopts "f:" o; do
case "${o}" in
f)
filename=${OPTARG}
;;
*)
usage
;;
esac
done
shift $((OPTIND-1))
name=$1
password=$2
if [ -z "$password" ] ; then
usage
fi
set -x
if [ -z "$filename" ] ; then
useradd -p `crypt $password` $name
else
useradd -p `crypt $password` $name > $filename
fi
mm
Donte needs to print specific sections in the workbook, as opposed to the entire workbook. Which feature should he use?
Page Setup
Backstage view
Print Area
Option
Mark this and retum
Save and Exit
Answer:
Print Area
Explanation:
First, I'll assume Donte is making use of a spreadsheet application (Microsoft Office Excel, to be precise).
From the given options, only "Print Area" fits the given description in the question.
The print area feature of the Excel software allows users to print all or selected workbook.
To print a selection section, Donte needs to follow the steps below.
Go to File -> Print.
At this stage, Donte will have the option to set the print features he needs.
Under settings, he needs to select "Print Selection"; this will enable him Print sections of the entire workbook.
Refer to attachment for further explanation
Answer:
C. print area
Explanation:
The purpose of this assignment is to determine a set of test cases knowing only the function’s specifications, reported hereafter:The program takes as input a 2D matrix with N rows. The program finds the column index of the first non-zero value for each row in the matrix A, and also determines if that non-zero value is positive or negative. It records the indices of the columns, in order, in the array W. It records whether each value was positive or negative, in order, in the array T (+1 for positive, -1 for negative). Assume that all rows have at least one column with a non-zero value.As this is black-box testing, you will not have access to the source so you must use what you have learned.You need to include a comment that describes how you chose your test cases.Students should identify the input equivalence partitions classes and select test cases on or just to one side of the boundary of equivalence classes.Students will gain 25 points for each input equivalence partition classes and valid test cases reported.Any language can be used, preferably C or C++ but it's not an issue if it's not in these two. Thanks!
Answer:
The program is written using PYTHON SCRIPT below;
N=int(input(" Enter number of Rows you want:"))
M=[] # this for storing the matrix
for i in range(N):
l=list(map(int,input("Enter the "+str(i+1)+" Row :").split()))
M.append(l)
print("The 2D Matrix is:\n")
for i in range(N):
print(end="\t")
print(M[i])
W=[] # to store the first non zero elemnt index
T=[] # to store that value is positive or negative
L=len(M[0])
for i in range(N):
for j in range(L):
if (M[i][j]==0):
continue
else:
W.append(j) # If the value is non zero append that postion to position list(W)
if(M[i][j]>0): #For checking it is positive or negative
T.append(+1)
else:
T.append(-1)
break
print()
print("The first Non Zero element List [W] : ",end="")
print(W)
print("Positive or Negative List [T] : ",end="")
print(T)
Explanation:
In order for the program to determine a set of test cases it takes in input of 2D matrix in an N numbet of rows.
It goes ahead to program and find the column index of the first non-zero value for each row in the matrix A, and also determines if that non-zero value is positive or negative. The If - Else conditions are met accordingly in running the program.
Suppose an 80286 microprocessor is in protected mode. How it is switched back
to real address mode operations.
Answer:
hardware-initiated reset
Explanation:
Once in protected mode, the 80286 is designed to remain there until it is reset by hardware.
__
External hardware can be designed so that software can cause such a reset. In the 1984 PC/AT, such hardware combined with code in the BIOS allowed real mode re-entry and returned execution control to the program that caused the reset.