Answer:
my_list = ["Rain fell from blue sky", "while I was having coffee,", "procrastinating"]
print(my_list[0])
print(my_list[1])
print(my_list[2])
Explanation:
Using Python Programming language:
Since the given list contains three string elements, the indexes are from 0-2
The print function can then be used to print elements are specific indexes as given above
Present a detailed data model for your project scenario. You can create your data model using Microsoft Visio 2010, which you will have access to through Lab and Microsoft Excel, which comes with Microsoft Office. Other tools may be used as long as the output is legible and conforms to standard format. (i.e., I will not be able to grade the data model if I cannot tell what it is supposed to be!). Your data model should include a minimum of an ERD and metadata chart (data dictionary).
Answer:
THE FARMERS PRICE CENTER
The farmer's price center is about helping farmers acquire farm inputs easily
and conveniently. Through relevant branches and supportive statt, the farmers
price center provides a fresh perspective on the day to day hardships that farmers
Tace. The tarmer's price center is run by a practicing tarnmer with tirst hand
information as he holds a bachelors degree in agriculture. His unique perspective
allows farmers to automatically connect with his point of view. The farmer's center
is equally informal as it is also a mentor or a friend. Besides selling farm inputs,
the center guides tarmers overcome their challenges.
The tarmer's price center addresses the tarming and tarming management
sector. The people who purchase from famers price center are either practicing
tarmers or want to start a venture in agriculture. They have a natural drive to create
agricultural products and seek out professional advice. The ideal customer for the
farmer's price center would be the new farmer or investor who wants to start an
agricultural venture or manage their firm better. The farmer's price center provides
a comprehensive resource that answers the questions farmers have while at the
same time giving thema step by step way to succeed. Mostly, farmers do not have
The farrmers price center will be getting 5,000 unique customers per month
and be ranked as a pace setter and a trusted mentor and be ranked as a leader in
agricultural products and advice site. To achieve these long term goals, the
Tarmer s price center needs to apply imbound marketing te chniques to get tound.
Part or this strategy will be to contract sales representatives to market the varlous
products to the retail shops across the countryside. These sales representatives are
to be paid a retainer and commission upon hitting their sales target. The (products
are transported by our fleet of trucks to the customer's premises. Payment shall be
made on cheque at the end of the month according to the number of products
delivered. Each sales representative shall prepare a report on their daily sales with
accordance to their specitic routes.
ne raners price center is all about neiping ramers tnve n tne cndotic
world of business filled with scrupulous products. The farmer's price center will
achieve upwards of 5,000 unique customers and S10,000 in revenue per month
within the next two years.
The class shown below called is called CSVReader. Use this class exactly as it is written (specifically don’t add any instance variables) except add the code for the methods readFile, numberOfRows, numberOfFields, and field. Not that this class stores the data in a 2-dimensional ArrayList of String. You can assume that the fields in a CSV file are separated by commas. You will find it useful to have a Scanner object that reads the file line by line, and a second Scanner object that processes each line.
import java.util.Scanner;
import java.util.ArrayList;
import java.io.File;
import java.io.FileNotFoundException;
public class CSVReader
{
ArrayList> fields;
public CSVReader()
{
fields = new ArrayList>();
}
public void readFile(String filename) throws FileNotFoundException
{
File inputFile = new File (filename);
String word;
String line;
int index = 0;
Scanner in = new Scanner(inputFile);
while(in.hasNextLine())
{
line = in.nextLine();
Scanner lineScanner = new Scanner(line);
lineScanner.useDelimiter(",");
fields.add(new ArrayList());
while(lineScanner.hasNext())
{
word = lineScanner.next();
fields.get(index).add(word);
}
lineScanner.close();
index++;
}
in.close();
}
public int numberOfRows()
{
return fields.size();
}
public int numberOfFields(int row)
{
if(row < 0) {return 0;}
if(row >= fields.size()){return 0;}
return fields.get(row).size();
}
String field(int row, int column)
{
if( row >= 0 && row < fields.size()){
if(column >= 0 && column < fields.size())
{
return fields.get(row).get(column);
}
else return "";
}
else return "";
}
}
Note: Your program should ignore the first row of data (which is the column headers) even though your CSVReader class will read and store these values.
• You can assume that the names of the realtors in Realty.csv consist of a single-word first name followed by one blank followed by a single-word last name.
• The MLS number should be left justified in a field 8 characters wide
• The name should be written as Lastname, Firstname left justified in a field 20 characters wide
• The street address should be left justified in a field 20 characters wide
• The city should be left-justified in a field 12 characters wide
• The state should be left justified in a field 3 characters wide
• The zip should be left justified in a field 6 characters wide
• The price should be right justified in a field 12 characters wide, preceded by a dollar sign, with decimal separators and 2 digits of precision.
• The square footage, number of bedrooms, number of baths, and price per square foot should look like the example, with price per square foot right justified.
Include the summary information at the bottom.
M5678 Smith, Jane 606 Cardinal St. Maryville TN 37803 $230,000.00 1800 SF 4 beds 2 baths price/sf: $127.78
M3499 McCormick, Lance 418 Scenic Dr. Knoxville TN 37919 $649,000.00 3900 SF 4 beds 4 baths price/sf: $166.41
M2345 Smith, Jane 814 St. Andrew St. La Crosse WI 54601 $99,000.00 1100 SF 2 beds 1 baths price/sf: $90.00
M1265 Finley, Heather 517 Avon St. La Crosse WI 54601 $110,000.00 1200 SF 3 beds 1 baths price/SF: $91.67
M8690 Burmeister, Georgia 728 Alice Bell Rd Knoxville TN 37917 $169,000.00 2100 SF 3 beds 2 baths price/sf: $80.48
M8356 Nichols, Roger 119 Cherokee Blvd Knoxville TN 37919 $999,999.00 3800 SF 5 beds 4 baths price/sf: $263.16
M8211 Jones, Steven 619 Derby St Chattanooga TN 37404 $549,000.00 2800 SF 4 beds 3 baths price/sf: $196.07
M8044 Smith, Harold 872 La Crosse St. La Crosse WI 54603 $210,000.00 2250 SF 3 beds 2 baths price/sf: $93.33
M4789 O'Connor, Judy 4000 McArthur Ave Chattanooga TN 37404 $179,000.00 1675 SF 3 beds 3 baths price/sf: $106.87
M9000 Smith, Alice 1901 3rd St. Chattanooga TN 37411 $158,000.00 2400 SF 3 beds 3 baths price/sf: $65.83
There is a total of 10 homes on the market with an average price of $335,299.90
Using the class provided above make a Tester method that prints out the file provided Realty.csv as well as making sure other files like Realty.csv can also be used in the Tester made us the Class provided. Make it to where any CSV file can be used by hardcoding it in but for this question only focus on the file shown
CSV File contents:
MLS Realtor name Street Address City State Zip Price square footage number of bedrooms number of bathrooms
M5678 Jane Smith 606 Cardinal St. Maryville TN 37803 230000 1800 4 2
M3499 Lance McCormick 418 Scenic Dr. Knoxville TN 37919 649000 3900 4 4
M2345 Jane Smith 814 St. Andrew St. La Crosse WI 54601 99000 1100 2 1
M1265 Heather Finley 517 Avon St. La Crosse WI 54601 110000 1200 3 1
M8690 Georgia Burmeister 728 Alice Bell Rd Knoxville TN 37917 169000 2100 3 2
M8356 Roger Nichols 119 Cherokee Blvd Knoxville TN 37919 999999 3800 5 4
M8211 Steven Jones 619 Derby St Chattanooga TN 37404 549000 2800 4 3
M8044 Harold Smith 872 La Crosse St. La Crosse WI 54603 210000 2250 3 2
M4789 Judy O'Connor 4000 McArthur Ave Chattanooga TN 37404 179000 1675 3 3
M9000 Alice Smith 1901 3rd St. Chattanooga TN 37411 158000 2400 3 3
Answer:
import java.io.File;
import java.io.FileNotFoundException;
import java.util.ArrayList;
import java.util.Scanner;
public class CSVReader {
ArrayList fields;
public CSVReader() {
fields = new ArrayList();
}
public void readFile(String filename) throws FileNotFoundException {
File inputFile = new File(filename);
String word;
String line;
int index = 0;
int ind2 = 0;
int valPrice = 0, valSqft = 0, valSum = 0;
Scanner in = new Scanner(inputFile);
while (in.hasNextLine()) {
line = in.nextLine();
Scanner lineScanner = new Scanner(line);
lineScanner.useDelimiter(",");
fields.add(new ArrayList());
while (lineScanner.hasNext()) {
word = lineScanner.next();
if (index != 0) {
if (ind2 == 2) {
System.out.print(", " + word);
} else if (ind2 == 8) {
System.out.printf(" $" + Integer.valueOf(word) / 1000);
if (Integer.valueOf(word) % 1000 > 0) {
System.out.printf(",%.2f", (float) Integer.valueOf(word) % 1000);
} else {
System.out.print(",000.00");
}
valPrice = Integer.valueOf(word);
valSum += Integer.valueOf(word);
} else if (ind2 == 9) {
System.out.print(" SF " + word);
valSqft = Integer.valueOf(word);
} else if (ind2 == 11) {
System.out.print(" beds " + word);
System.out.printf(" baths price/sf: $ %.2f\n", (float) valPrice / valSqft);
} else if (ind2 > 0) {
System.out.print(" " + word);
} else if (ind2 == 0) {
System.out.print(word);
}
}
((ArrayList) fields.get(index)).add(word);
ind2++;
}
lineScanner.close();
index++;
ind2 = 0;
valPrice = 0;
valSqft = 0;
}
System.out.printf(
"There is a total of " + (numberOfRows() - 1) + " homes on the market with an average price of $"+(valSum / (numberOfRows() - 1))/1000+",%.2f",
(float) (valSum / (numberOfRows() - 1))%1000);
in.close();
}
public int numberOfRows() {
return fields.size();
}
public int numberOfFields(int row) {
if (row < 0) {
return 0;
}
if (row >= fields.size()) {
return 0;
}
return ((ArrayList) fields.get(row)).size();
}
String field(int row, int column) {
if (row >= 0 && row < fields.size()) {
if (column >= 0 && column < fields.size()) {
return (String) ((ArrayList) fields.get(row)).get(column);
} else
return "";
} else
return "";
}
public static void main(String[] args) throws FileNotFoundException {
CSVReader test = new CSVReader();
test.readFile("R.csv");
}
}
Explanation:
Given an unsorted std::vector and a number n, what is the worst-case time complexity for finding the pair of integers whose sum is closest to n, using no additional memory? For example, given the vector(12, 3, 17, 5, 7} and n = 13, we would get the pair(5, 7).
A.Θ(log n)
B.Θ(n)
C.Θ(n log n)
D.Θ(n2)
E.0(29
Answer:A
Explanation:
An issue that many organizations are currently struggling with is the presence and ease of using external data storage locations, many of which are free to use. Why is this an issue
Answer:
Restricted company information and other assets stored in external sites means that information may be at risk as it is outside the control of the organization.
Explanation:
I dont know if that made sense but i hope it did.
2. Feet to Inches One foot equals 12 inches. Design a function named feetToInches that accepts a number of feet as an argument, and returns the number of inches in that many feet. Use the function in a program that prompts the user to enter a number of feet and then displays the number of inches in that many feet.
Final answer:
To create a function that converts feet to inches, you multiply the number of feet by 12, which is the unit equivalence. For conversions, you can also set up proportions to find the resulting number of inches. Understanding these conversions is crucial in various measurement situations.
Explanation:
Designing a Function to Convert Feet to Inches
To design a function that converts feet to inches, you simply need to multiply the number of feet by 12 because one foot is equivalent to 12 inches. This process of conversion uses the unit equivalence to go from a larger unit to a smaller unit of measure, which involves multiplication. For example, to convert 5 feet to inches, the calculation would be 5 feet multiplied by 12 inches per foot, which equals 60 inches.
Using Proportions for Conversion
To use proportions, you can set up a proportion such as 1 foot is to 12 inches as x feet is to the unknown number of inches. You would then cross multiply to solve for the unknown number of inches. This method is also helpful when dealing with fractional feet and ensures accuracy in conversion.
Conversion factors and the understanding of customary units of length, such as inches and feet, are essential for comparing and converting measurements effectively. Being familiar with these methods is useful in various practical scenarios, such as estimating the height or length of objects.
Write a class named Employee that holds the following data about an employee in attributes: name, ID number, department, and job title.
Once you have written the class, write a python program that creates 3 Employee objects to hold the following data:
Name: Susan Meyers, ID number: 47899, Department: Accounting, Job Title: Vice President
Name: Mark Jones, ID Number: 39119, Department: IT, Job Title: Programmer
Name: Jon Rogers, ID Number: 81774, Department: Manufacturing, Job Title: Engineer
The prgram should store this data in the 3 objects and then display the data for each employee on the screen.
The Python program defines an `Employee` class with attributes for name, ID number, department, and job title. It then creates three `Employee` objects with specific data and displays the details for each employee.
Here's a Python implementation of the `Employee` class and a program that creates three `Employee` objects with the specified data:
```python
class Employee:
def __init__(self, name, id_number, department, job_title):
self.name = name
self.id_number = id_number
self.d-epartment = department
self.job_title = job_title
# Creating three Employee objects
employee1 = Employee("Susan Meyers", 47899, "Accounting", "Vice President")
employee2 = Employee("Mark Jones", 39119, "IT", "Programmer")
employee3 = Employee("Jon Rogers", 81774, "Manufacturing", "Engineer")
# Displaying data for each employee
print("Employee 1:")
print("Name:", employee1.name)
print("ID Number:", employee1.id_number)
print("Department:", employee1.d-epartment)
print("Job Title:", employee1.job_title)
print()
print("Employee 2:")
print("Name:", employee2.name)
print("ID Number:", employee2.id_number)
print("Department:", employee2.d-epartment)
print("Job Title:", employee2.job_title)
print()
print("Employee 3:")
print("Name:", employee3.name)
print("ID Number:", employee3.id_number)
print("Department:", employee3.d-epartment)
print("Job Title:", employee3.job_title)
```
This program defines a `Employee` class with the specified attributes and then creates three instances of this class, initializing them with the given data. Finally, it prints out the details of each employee.
1. (15%) Consider a computer system with three users: Alice, Bob, and Cyndy. Alice owns the file a, and Bob and Cyndy can read it. Bob owns the file b, and Cyndy can read and write the file b, but Alice can only read it. Cyndy owns the file c, but neither Alice and Bob can read or write it. If a user owns a file, he/she can also execute the file.(a) Create the ACM (access control matrix) of the system.(b) Show the ACL of the ACM and the CL of the ACM.(c) Cyndy gives Alice permission to read c, and Alice removes Bob's ability to read a. Show the ACM after these changes.
Answer:
answered in Image and explanation is given below.
Explanation:
CL is euqal to user permission on every file
ACL is equal to how permissions are defined for one life for example File a, read, write, execute.
What is the output of the following program?
public class testmeth
{
static int i = 1;
public static void main(String args[])
{
System.out.println(i+" , ");
m(i);
System.out.println(i);
}
public void m(int i)
{
i += 2;
}
}
a) 1 , 3
b) 3 , 1
c) 1 , 1
d) 1 , 0
e) none of the above.
Answe:
none of the above
Explanation: non-static method m(i) cannot be cannot be referenced from a static context
Explanation:
The output of the program is 1 , 1. The parameter i inside the method m does not affect the original variable i in the testmeth class.
Explanation:The output of the program is 1 , 1.
In the program, the variable i is initially assigned the value of 1. The System.out.println(i+" , ") statement will print the value of i, which is 1, followed by a comma and a space. Then, the m(i) method is called.
However, the m(int i) method does not modify the value of the i variable in the testmeth class. In Java, primitive data types such as int are passed by value, so any modifications made to the parameter i inside the method will not affect the original variable i. Therefore, the value of i remains 1 and the last System.out.println(i) statement will print 1.
A fair six-sided die is rolled repeatedly and independently. Let An be the event of rolling n sixes in 6n rolls, and let Bn be the event of rolling n or more sixes in 6n rolls. (a) Does P(An) and P(Bn) change with n? (b) Use a computer to investigate what happens to P(An) and P(Bn) as n becomes very large.
Answer:
R code:
n=1:100*1000
p1=round(dbinom(n,6*n,1/6),4)
p2=round(1-pbinom(n-1,6*n,1/6),4)
levels=factor(c("n","P(A_n)","P(B_n)"))
X=cbind(n,p1,p2)
colnames(X)=c(expression(n),expression(P(A_n)),expression(P(B_n)))
Output:
n P(A_n) P(B_n)
[1,] 1000 0.0138 0.5054
[2,] 2000 0.0098 0.5038
[3,] 3000 0.0080 0.5031
[4,] 4000 0.0069 0.5027
[5,] 5000 0.0062 0.5024
[6,] 6000 0.0056 0.5022
[7,] 7000 0.0052 0.5020
[8,] 8000 0.0049 0.5019
[9,] 9000 0.0046 0.5018
[10,] 10000 0.0044 0.5017
[11,] 11000 0.0042 0.5016
[12,] 12000 0.0040 0.5016
[13,] 13000 0.0038 0.5015
[14,] 14000 0.0037 0.5014
[15,] 15000 0.0036 0.5014
[16,] 16000 0.0035 0.5013
[17,] 17000 0.0034 0.5013
[18,] 18000 0.0033 0.5013
[19,] 19000 0.0032 0.5012
[20,] 20000 0.0031 0.5012
[21,] 21000 0.0030 0.5012
[22,] 22000 0.0029 0.5011
[23,] 23000 0.0029 0.5011
[24,] 24000 0.0028 0.5011
[25,] 25000 0.0028 0.5011
[26,] 26000 0.0027 0.5011
[27,] 27000 0.0027 0.5010
[28,] 28000 0.0026 0.5010
[29,] 29000 0.0026 0.5010
[30,] 30000 0.0025 0.5010
[31,] 31000 0.0025 0.5010
[32,] 32000 0.0024 0.5010
[33,] 33000 0.0024 0.5009
[34,] 34000 0.0024 0.5009
[35,] 35000 0.0023 0.5009
[36,] 36000 0.0023 0.5009
[37,] 37000 0.0023 0.5009
[38,] 38000 0.0022 0.5009
[39,] 39000 0.0022 0.5009
[40,] 40000 0.0022 0.5008
[41,] 41000 0.0022 0.5008
[42,] 42000 0.0021 0.5008
[43,] 43000 0.0021 0.5008
[44,] 44000 0.0021 0.5008
[45,] 45000 0.0021 0.5008
[46,] 46000 0.0020 0.5008
[47,] 47000 0.0020 0.5008
[48,] 48000 0.0020 0.5008
[49,] 49000 0.0020 0.5008
[50,] 50000 0.0020 0.5008
[51,] 51000 0.0019 0.5008
[52,] 52000 0.0019 0.5007
[53,] 53000 0.0019 0.5007
[54,] 54000 0.0019 0.5007
[55,] 55000 0.0019 0.5007
[56,] 56000 0.0018 0.5007
[57,] 57000 0.0018 0.5007
[58,] 58000 0.0018 0.5007
[59,] 59000 0.0018 0.5007
[60,] 60000 0.0018 0.5007
[61,] 61000 0.0018 0.5007
[62,] 62000 0.0018 0.5007
[63,] 63000 0.0017 0.5007
[64,] 64000 0.0017 0.5007
[65,] 65000 0.0017 0.5007
[66,] 66000 0.0017 0.5007
[67,] 67000 0.0017 0.5007
[68,] 68000 0.0017 0.5007
[69,] 69000 0.0017 0.5006
[70,] 70000 0.0017 0.5006
[71,] 71000 0.0016 0.5006
[72,] 72000 0.0016 0.5006
[73,] 73000 0.0016 0.5006
[74,] 74000 0.0016 0.5006
[75,] 75000 0.0016 0.5006
[76,] 76000 0.0016 0.5006
[77,] 77000 0.0016 0.5006
[78,] 78000 0.0016 0.5006
[79,] 79000 0.0016 0.5006
[80,] 80000 0.0015 0.5006
[81,] 81000 0.0015 0.5006
Here we observed that as n goes to infinity, both probabilities are constant and Plan) is almost zero and PB) is almost 0.5.
Explanation:
Final answer:
The probability of rolling exactly n sixes in 6n rolls (P(An)) and the probability of rolling n or more sixes in 6n rolls (P(Bn)) both change with n and would require computational methods to investigate as n becomes large.
Explanation:
The probability question at hand involves a fair six-sided die and two types of events, An and Bn. The event An is rolling exactly n sixes in 6n rolls, and the event Bn is rolling n or more sixes in 6n rolls.
a. The probability P(An) would change with n because it represents the probability of a very specific outcome (exactly n sixes), which depends on the number of trials (rolls). P(Bn) also changes with n because it accumulates probabilities of all cases where at least n sixes occur, up to all 6n being sixes.
b. To investigate what happens to P(An) and P(Bn) as n becomes very large, a computer simulation would be required. Typically, the probabilities can be expected to approach certain limits due to the law of large numbers. However, the exact behavior would depend on the specifics of the binomial probability distribution.
When dealing with large n, the central limit theorem comes into play, which would usually imply that the distribution of the average number of sixes approaches a normal distribution. However, obtaining exact probabilities would require computational methods.
An internet access provider (IAP) owns two servers. Each server has a 50% chance of being "down" independently of the other. Fortunately, only one server is necessary to allow the IAP to provide service to its customers, i.e., only one server is needed to keep the IAP’s system up. Suppose a customer tries to access the internet
Question Continuation
Suppose a customer tries to access the internet on 8 different occasions, which are sufficiently spaced apart in time, so that we may assume that the states of the system corresponding to these 8 occasions are independent. What is the probability that the customer will only be able to access the internet on 2 out of the 8 occasions?
Answer:
0.311
Explanation:
Given
p = success = 2/8 = ¼ --- Probability that both servers down at any one event
p + q = 1
q = failure = 1 - ¼ = ¾ --- Probability that at most one server down at any one event
The number of ways that the customer will only be able to access the internet on 2 out of the 8 occasions is solved by applying binomial distribution below;
(p + q) ^n where n = 8 and r = 2 is
nCr * p^r * q ^ (n-r)
This becomes
8C2 * (¼)² * (¾)^6
= 8!/(6!2!) * 1/16 * 729/4096
= 0.31146240234375
= 0.311 --- approximated
Final answer:
The probability that at least one server is operational and the IAP can provide service is 75%, calculated by considering the independent 50% chance of each server being up and subtracting the 25% probability that both servers are down simultaneously.
Explanation:
Considering the scenario provided, we can calculate the probability of the Internet Access Provider (IAP) being able to provide service to its customers even with only one functioning server. Each server has a 50% chance of being down, which equivalently means each has a 50% chance of being up and operational. Since the occurrence of one server being down is independent of the other, we find the probability of both servers being down simultaneously to calculate the likelihood the IAP cannot provide service.
The probability of both servers being down is 0.5 (probability of Server 1 being down) × 0.5 (probability of Server 2 being down), which equals 0.25 or 25%. Consequently, the probability of at least one server being up and thus the IAP being able to provide service is 1 - 0.25, which equals 0.75 or 75%. This probability indicates a relatively high chance that the IAP will have at least one operational server at any given time.
1. Rectangle Area The area of a rectangle is calculated according to the following formula: Area=Width×LengthArea=Width×Length Design a function that accepts a rectangle’s width and length as arguments and returns the rectangle’s area. Use the function in a program that prompts the user to enter the rectangle’s width and length, and then displays the rectangle’s area.
Answer:
The method is given below.
double area(double w, double l)
{
double area = w*l;
return area;
}
Explanation:
The cpp program using the above method is given below.
#include <iostream>
using namespace std;
double area(double w, double l)
{
double area = w*l;
return area;
}
int main() {
// variables to hold parameters of the rectangle
double width;
double length;
// user asked to enter parameters of the rectangle
cout<<"Enter the width of the rectangle: ";
cin>>width;
cout<<"Enter the length of the rectangle: ";
cin>>length;
// area of the rectangle returned by area() and displayed
cout<<"The area of the rectangle is "<<area(width, length)<<endl;
return 0;
}
OUTPUT
Enter the width of the rectangle: 12
Enter the length of the rectangle: 8.9
The area of the rectangle is 106.8
The program includes the following.
1. The area() method takes two double parameters as argument of the rectangle.
2. The area() method computes the area using the arguments and returns this value.
3. The area() method has return type of double.
4. The area() method has a double variable, area, which holds the product of the arguments.
5. The main() method has return type of integer and returns an integer value. This is done to assure successful execution of the program.
6. Inside main(), two double variables are declared.
7. User is prompted to enter the parameters of the rectangle which are stored in these variables.
8. The user input is passed to the area() method as parameters.
9. This method returns the computed value of the area of the rectangle.
10. The area is displayed to the console. A new line is inserted at the end of the output.
11. All the variables are declared with double datatype to accommodate both decimal and non-decimal values.
12. All the code is written inside the respective method. No classes are used.
An upper-layer packet is split into 10 frames, each of which has an 80% chance of arriving undamaged. If no error control is done by the data link protocol, how many times must the message be sent on average to get the entire thing through
Answer:
The expected or average no. of transmissions = 9.32
Explanation:
Chance of each frame to arrive undamaged = 80%
Total number of frames = 10
Probability of getting whole message to arrive undamaged = p = 0.80¹⁰
p = 0.1073
q = 1 - p = 1 - 0.1073 = 0.8927
where q is the probability of not getting whole message to arrive undamaged
The expected or average no. of transmissions = ∑ [tex]kp(1-p)^{k-1}[/tex]
where k is from 0 to ∞
The above series can be reduced to p/(1-q)²
The expected or average no. of transmissions = p/(1-q)²
The expected or average no. of transmissions = 0.1073/(1-0.8927)²
Therefore, the expected or average no. of transmissions = 9.32
The number of times that the message must be sent on average to get the entire thing through is; 10 times
The question did not tell us anything about datalink protocols.
Now, let us assume that there will be a successful transmission of the whole fragmented message with the single condition that all the frames arrive intact.
Thus;
P(successful transmission of the message) = P(all the frames arrive intact); p = 0.8¹⁰
Now, probability of unsuccessful transmission of the message;
q = 1 - p
Now, the Mean value of number of tries is simply the Expected value of k.
The formula for that is;
Σ[k][P(k)]
where k is from 1 to ∞
The binomial expansion of that formula gives us;
1p + 2qp + 3q²p + 4q³p .....
This can be reduced to;
p(1 + 2q + 3q² + 4q³ .....)
This can be further reduced to;
p/(1 - q)²
Recall that; q = 1 - p
Thus;
p/(1 - q²) = p/(1 - (1 - p))²
⇒ 1/p
Put 0.8¹⁰ for p to get;
1/0.8¹⁰ = 9.31
Thus, mean number of tries will be approximately 10.
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In python please:
Design a program that asks the user to enter a store’s sales for each day of the week. The amounts should be stored in a list. Use a loop to calculate the total sales for the week and display the result.
Answer:
declare a list and loop through to get the total sales
Explanation:
input_string = input("Enter sales: ")
userList = input_string.split()
print("user list is ", userList)
print("Calculating total of the sales")
sum = 0
for num in userList:
sum += int(num)
print("Total sales= ", sum)
Write programs with loops that compute:
(a) The sum of all even numbers between 2 and 100 (inclusive).
(b) The sum of all squares between 1 and 100 (inclusive).
(c) The sum of all odd numbers between a and b (inclusive), where a and b are inputs.
(d) Ask the user for a sequence of integer inputs and print: The smallest and largest of the inputs.
Answer:
The programs are written in MATLAB.
(a)
sum = 0;
for i = 2:100 %This for loop starts from 2 and ends at 100
sum = sum + i; %This line sums each values of the loop
end
disp(sum); %This command displays the sum at command window
(b)
sum = 0;
for i = 2:100
sum = sum + i^2; %This line sums the squares of each values in the loop
end
disp(sum);
(c)
sum = 0;
a = input('Please enter a: ');
b = input('Please enter b: ');
for i = a:b
if mod(i,2) == 1 %This command checks whether the value in the loop is odd or not
sum = sum + i; %This command sums each odd value in the loop
end
end
(d)
seq = input('Please enter a few numbers with square brackets [] around them: ');
small = seq(1); %Initialize small
large = seq(1); % Initialize large
for i = 1:length(seq) %Iterate over the sequence
if seq(i) < small
small = seq(i); %Decide if the value in the loop is smallest
elseif seq(i) > large
large = seq(i); %Decide if the value in the loop is largest
end
end
fprintf('The smallest of the inputs is %g,\nThe largest of the inputs is %g.\n',small,large);
To compute the sum of all even numbers between 2 and 100, you can iterate through each number in that range and check if it's even. Similarly, to compute the sum of all squares between 1 and 100, you can iterate through each number in that range and calculate its square. To find the sum of all odd numbers between two inputs, you can iterate through the range defined by the inputs and check if each number is odd. Finally, to find the smallest and largest numbers from a sequence of integer inputs, you can compare each input to the current smallest and largest numbers.
Explanation:To compute the sum of all even numbers between 2 and 100, you can use a loop that iterates through each number in that range. You can check if a number is even by using the modulus operator (%), which returns the remainder when divided by 2. If the remainder is 0, then the number is even and you can add it to the sum. Here's an example in Python:
sum = 0This will output the sum of all even numbers between 2 and 100, which is 2550.
To compute the sum of all squares between 1 and 100, you can use another loop that iterates through each number in that range. You can calculate the square of a number by multiplying it by itself. Here's an example in Python:
sum = 0This will output the sum of all squares between 1 and 100, which is 338350.
To compute the sum of all odd numbers between a and b (inclusive), where a and b are inputs, you can use a loop that iterates through each number in that range. You can check if a number is odd by using the modulus operator (%), which returns the remainder when divided by 2.
sum = 0This will prompt the user to enter values for a and b, and then output the sum of all odd numbers between a and b.
To find the smallest and largest numbers from a sequence of integer inputs, you can track the smallest and largest numbers by comparing each input to the current smallest and largest numbers. Here's an example in Python:
smallest = NoneThis will continuously prompt the user to enter integer inputs until they enter 'q' to quit, and then output the smallest and largest numbers from the input sequence.
"In about 100 words, discuss the technologies that Walmart’s trucking fleet might use to better manage their operations. Include a discussion of tracking technologies, GPS devices, and/or apps for use by truck drivers on their smartphones or tablet devices."
Answer:
Walmart’s trucking fleet might use real time GPS systems to better route them to their destinations. This would allow them to bypass traffic and accidents that are on the route. The RFID would also allow them to monitor their cargo and to ensure that all items are delivered to the proper locations. The real tracking via the GPS and would allow management to monitor their locations in real-time and make sure that divers are where they are supposed to be and notmaking any unscheduled stops. Management of Walmart would also be able to track when stores receive merchandise for tracking and accounting purposes, they would also be able to track how store sales are moving on certain products.
Write a program that reads a list of integers, and outputs those integers in reverse. The input begins with an integer indicating the number of integers that follow. For coding simplicity, follow each output integer by a space, including the last one. Assume that the list will always contain fewer than 20 integers.
Answer:
The program to this question can be described as follows:
Program:
#include<stdio.h> //defining header file
int main() //defining main method
{
const int total_number= 20; //defining integer constant variable
int a[total_number],i; //defining integer array and variable
printf("Enter total number, that you want to insert in list: "); //message
scanf("%d",&a[0]); //input value by user
printf("Input list: "); //message
for(i=1;i<=a[0];i++) // loop for input value
{
scanf("%d",&a[i]); //input value by user
}
printf("List in reverse order:\n");
for(i=a[0];i>0;i--) //loop for print list in reverse order
{
printf("%d ",a[i]); //print value
}
return 0;
}
Output:
Enter total number, that you want to insert in list: 5
Input list: 6
4
2
8
1
List in reverse order:
1 8 2 4 6
Explanation:
In the above C language program, three integer variable "total_number, a[], and i " is declared, in which total_number is a constant variable, that holds value 20, which is passed in an array, in array first element we input the total number of the list.
In the next step, an array variable is used, that uses for loop to input number in list. Then another for loop is declared that prints the array value in its reverse order, with a white space.Look at the following pseudocode module header: Module myModule (Integer a, Integer b, Integer c) Now look at the following call to myModule: Call myModule (3, 2, 1) When this executes, what value will be stored in a? What value will be stored in b? What value will be stored in c?
Answer:
"3, 2 and 1" is the correct answer for the above question.
Explanation:
The module or function is used to some task for which it is designed. This task is defined in the body of the module at the time of definition.Some functions or modules are parameterized based, which means that this type of function takes some value to process any task.The above-defined module also takes three values to process, a,b and c. this value is passed at the time of calling 3,2 and1.The three are passed as the first value which is assigned for the a variable because the a variable is the first argument of the function.The two are passed as the second value which is assigned for the b variable because the b variable is the second variable which is defined as the second argument of the function.The one is passed as the third value which is assigned for the c variable because the c variable is the third variable which is defined as the third argument of the function. Hence the answer is a=3, b=2 and c=1.____________________ material is information, graphics, images, or any other physical or electronic item that could have value as evidence in a legal proceeding, whether criminal or civil.
Answer:
Explanation:
Evidentiary material is information, graphics, images, or any other physical or electronic item that could have value as evidence in a legal proceeding, whether criminal or civil. Information of evidentiary value may be found on digital media such as CDs, (DVDs), floppy disks, thumb drives, hard drives, and memory expansion cards found in digital cameras and mobile phones.
Write a C program to calculate the minimum, maximum, and average of n grades entered by the user. The program uses a function int g(int n) to calculate the requirements above and also prints the grades in reverse order. The function should utilize the concepts of recursion and static storage class specifier.
Answer:
#include <stdio.h>
int p;
int g(int n)
{
static int minimum=101, maximum=-1, average=0;
if(n==0)
return -1;
else
{
printf("Enter course grade %d: ",p-n+1);
int grade;
scanf("%d",&grade);
g(n-1);
printf("%d\n",grade);
if(minimum>grade)
minimum=grade;
if(maximum<grade)
maximum=grade;
average=average+grade;
if(n==p)
{
printf("minimum= %d\n",minimum);
printf("maximum= %d\n",maximum);
printf("Average= %.2f\n",(double)average/(double)n);
}
return -1;
}
}
int main()
{
printf("Enter the number of grades:\n");
scanf("%d",&p);
g(p);
return 0;
}
Explanation:
Use a conditional statement to check to find the maximum and minimum grade.Check if n is equal to p, then display the values of maximum and minimum.Calculate the average by dividing the sum with the total numbers and then display its value.Inside the main function, get number of grades from user as an input.Lastly, call the function g and pass it the value of grade.Write an INSERT statement that adds a row to the InvoiceCopy table with the following values: VendorID: 32 InvoiceTotal: $434.58 TermsID: 2 InvoiceNumber: AX-014-027 PaymentTotal: $0.00 InvoiceDueDate: 11/8/06 InvoiceDate: 10/21/06 CreditTotal: $0.00 PaymentDate: null
_______ is a communications protocol used to send information over the web. HyperText Markup Language (HTML). URL (Uniform Resource Locator). Web 2.0 TCP/IP
TCP/IP is a communications protocol used to send information over the web.
Explanation:
TCP / IP, which stands for the Transmission Control Protocol / Internet Protocol, is a set of communication protocols used to connect internet network devices. The whole suite of the Internet Protocol — a set of rules and procedures — is commonly called TCP / IP.
The TCP / IP protocol suite acts as a layer of abstraction between web applications and the routing / switching fabric. TCP / IP defines how data is shared over the internet by supplying end-to-end communications that specify how the data should be split into packets, signed, distributed, routed and received at the destination.
8. Brad is a security manager and is preparing a presentation for his company's executives on the risks of using instant messaging (IM) and his reasons for wanting to prohibit its use on the company network. Which of the following should not be included in his presentation? a. Sensitive data and Files can be transferred from system to system over IM. b. Users can receive information-including malware-from an attacker posing as a legitimate sender. c. IM use can be stopped by simply blocking specific ports on the network firewalls. d. A security policy is needed specifying IM usage restrictions.
Answer:
Letter c is correct. IM use can be stopped by simply blocking specific ports on the network firewalls.
Explanation:
The letter c is the alternative that should not be included in Brad's presentation to executives at his company about the risks of using instant messaging (IM).
This alternative refers only to the existing way to stop the use of instant messaging by blocking specific ports on network firewalls, and does not offer a reason why the use should be prohibited in the company, unlike other alternatives, which present the risks inherent in use of instant messaging in the company, and would therefore be more plausible reasons for executives to consider.
Kelvin called the meeting to order. The first person to address the group was Susan Hamir, the network design consultant from Costly & Firehouse, LLC. She reviewed the critical points from the design report, going over its options and outlining the trade-offs in the design choices. When she finished, she sat down and Kelvin addressed the group again: "We need to break the logjam on this design issue. We have all the right people in this room to make the right choice for the company. Now here are the questions I want us to consider over the next three hours." Kelvin pressed the key on his PC to show a slide with a list of discussion questions on the projector screen.
a. What questions do you think Kelvin should have included on his slide to start the discussion?
b. If the questions were broken down into two categories, they would be cost versus maintaining high security while keeping flexibility. Which is more important for SLS?
Answer:
a. Some of the questions Kelvin should have included on his slide to start the discussion are:-
• Why are there differences in opinion on internet architecture?
• What is the level of security that needs to be implemented?
• How can it be achieved, and what is the cost of this implantation?
• What is the best design, loops and pitfalls identified in each design, and shortcomings with each design?
b. It is always better to invest in maintaining high security with flexibility. Investing in a great design helps not only establishing a better ground to begin with, but can avoid headaches for later, when there may be a security breach/threat. Investing in a better design may help to prevent future issues.
Explanation:
The questions that Kelvin should have included on his slide to start the discussion include:
What is the importance of internet architecture to the company?Are there any challenges in its implementation?Internet architecture refers to distinct networks that interact with a common protocol. The layers of internet architecture include the network layer, data link layer, and application layer.
It should be noted that cost, high security, and flexibility are important for SLS. Therefore, it's important to choose an architecture that addresses all of them.
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#Write a function called hide_and_seek. The function should #have no parameters and return no value; instead, when #called, it should just print the numbers from 1 through 10, #follow by the text "Ready or not, here I come!". Each #number and the message at the end should be on its own #line. #
Answer:
# hide_and_seek function is defined
def hide_and_seek():
# for loop from 1 to 10
# and print each number
for number in range(1, 11):
print(number)
# the last message is displayed on its own line
print("Ready or not, here I come!")
# the function is called
hide_and_seek()
Explanation:
The code is well commented. A sample image of the output when the code is executed is attached.
Final answer:
The hide_and_seek function is written in Python and prints numbers 1 through 10 on separate lines, followed by "Ready or not, here I come!" on its own line. The function uses a for loop and requires no parameters or return values.
Explanation:
Writing the hide_and_seek Function
To write a function called hide_and_seek that prints numbers from 1 through 10 followed by the message "Ready or not, here I come!", you can use the Python programming language. This function does not require any parameters to be passed and does not return any value. Instead, it performs an action by printing out a series of statements. Here is how you can define this function:
def hide_and_seek():
for i in range(1, 11):
print(i)
print("Ready or not, here I come!")
When you call this function using hide_and_seek(), it will execute a loop that prints each number from 1 to 10, each on a new line, and then it prints the message "Ready or not, here I come!" also on a new line.
Write a program trapezoid.cpp that prints an upside-down trapezoid of given width and height. However, if the input height is impossibly large for the given width, then the program should report,
Final answer:
The question pertains to writing a C++ program to print an upside-down trapezoid or report an error if the input height is impossible for the given width. An example code snippet demonstrates basic logic using loops to print the trapezoid shape. It includes error handling for when the input height is too large.
Explanation:
You asked how to write a program trapezoid.cpp that prints an upside-down trapezoid of given width and height, or reports an error if the height is impossibly large for the given width. The subject of this question falls under Computers and Technology, particularly involving programming skills. Let's assume you are going to use C++ for this task since the file extension is .cpp.
Here is an example of how you could approach this problem:
\#include <iostream>This example C++ program checks if the height is too large and if it's not, it proceeds to print the upside-down trapezoid.
This program prints an upside-down trapezoid of a given width and height using C++. It first checks if the given height is possible for the given width, and prints an error message if it is not.
Program to Print an Upside-Down Trapezoid
Here is a simple C++ program to print an upside-down trapezoid of given width and height. If the input height is too large compared to the width, the program should report that the input size is impossible.
Steps:
Read inputs for width and height.Check if the height is possible given the width.Print the trapezoid if possible, else report an error.Here is the C++ code for the program:
#include <iostream>Problem 2 A 10,000 rpm (revolutions per minute) harddisk can transfer data at a rate of 160 Mbps (megabits per second). The time to find a track is negligible. Determine the data amount for which it takes as long to transfer it off the disk as it takes to find the right sector.
Answer:
491.52 Kb
Explanation:
Number of revolution in a minute (= 60000 milliseconds) = 10000
Number of revolution in 1 millisecond = 1 / 6
So for a half revolution the time required = 3 milliseconds
Data Transfer = 160 X 1024 Kb/1000 s = 163.84 Kb/milliseconds
Data amount = 163.84 X 3 = 491.52 Kb
Explain how the principles underlying agile methods lead to the accelerated development and deployment of software. Why is it necessary to introduce some methods and documentation from plan-based approaches when scaling agile methods to larger projects that are developed by distributed development teams?
Answer:
Explanation:
Agile is several approaches to develop software, this method evolve depends on the team, clients and organization.
In this case, the development can elaborate efficient software because is not necessary to add all the functions in the first version, we can optimize depends on our needs and request about the program.
Agile software development is perfect to work with small teams, but there are introducing elements to scale the development process in this way easier.
Write an application that counts by five from 5 through 500 inclusive, and that starts a new line after every multiple of 50 (50, 100, 150, and so on).
Answer:
Explanation:
We must create a cycle for to count 5 by 5 then a with a conditional if, we are going to identify every multiple of 50 to start in a new line.
class Count{
public static void main(String[] args){
for(int i=5; i<=500; i+=5){
System.out.print(i + " ");
if(i%50==0)
System.out.println();
}
}
}
Choosing a High-Availability Solution You have been hired to set up the network and servers for a new company, using Windows Server 2012 R2. The company is investing in an application critical to the company’s business that all employees will use daily. This application uses a back-end database and is highly transaction oriented, so data is read and written frequently. To simplify maintenance of the front-end application, users will run it on remote desktop servers. The owner has explained that the back-end database must be highly available, and the remote desktop servers must be highly scalable to handle hundreds of simultaneous sessions. What high-availability technologies do you recommend for the back-end database server and the remote desktop servers? Explain your answer.
Answer:
We have following high availability solutions available in sql server:-
Logshiping:It is the process of automating backup of database and transaction logfiles on a server and then restoring on standby server.Through diagram below,logshipping can be understood properly.
Primary database is the database which needs to be available.It is available on primary server as the name says.Secondary database is available on secondary server where the all the transaction logfile will be restored to make the database in sync.This ca be chosen for high availability solution in OLTP environment.
It is very easy to setup.Just by using GUI,it can be established.
We have norecovery mode and stand by mode.Database on secondary server is not available if chosen no recovery mode.Stand by mode is chosen to use database for reading purpose when it is not being restored.
We can have multiple secondary server.On one server database can be used for reporting purpose and can be used for other purpose on another server.
It requires low maintance.
Mirroring:It is also a high availability solution which makes database available on secondary server.Below is the diagram showing mirroring:-
It increases availability of a database.
It increases data protection
It increases availability of database on production server during upgrades.
We have two modes high performance and high safety.In hgh performance mode, database mirroring session operates asynchronously and uses only the principal server and mirror server. The only form of role switching is forced service (with possible data loss).
In high safety mode,the database mirroring session operates synchronously and, optionally, uses a witness, as well as the principal server and mirror server.
Replication:It is also a high availability solution.It can copy database from one server and transfer the data on another server and keep the database in sync.We have different type of replication like snapshot,transaction,merge,peer to peer.Below diagram can be used to understand replication:-
In replication,we can transfer various database objects like table,views,sp's etc.Even we can transfer database object from sql server to oracle.
Transaction replication is used when data changes very frequently and we want database to be transferred immediately. Snapshot is used when database changes less frequently and it requires to be replicated periodically.Merge is used when changes on publisher as well as subscriber can be replicated.
Other than that,we have alwaysOn and clustering.In alwaysOn, we have primary replica and secondary replicas.Clustering setup is required in alwaysOn but storage is not required on SAN.It is really good,if we want to restrict one server to take backup and other for reporting purpose.
In clustering,we have common storage which is shared by all nodes in cluster.So if a node fails,other nodes are available and there would not be much impact as database files are available on separate server.
Explanation:
See attached picture.
Final answer:
High availability for the back-end database can be achieved using a Windows Server 2012 R2 clustering setup with Always On Availability Groups, regular SQL Server backups, and segregating database servers from web servers. For scalability of remote desktop servers, use Remote Desktop Services with load balancing. Secure your data transmissions with SSL and invest in SSDs for reliability.
Explanation:
To ensure high availability for the back-end database of your critical company application, consider implementing a cluster of servers using Windows Server 2012 R2 with the Always On Availability Groups feature. This will ensure that if one server fails, another can take over without losing data or transactional continuity. Additionally, SQL Server backups should be configured regularly and stored in a secure, off-site location to prevent data loss in case of catastrophic failures.
For the remote desktop servers that need to be highly scalable, consider utilizing Remote Desktop Services (RDS) with load balancing. This allows you to distribute the load across multiple servers, enabling the system to handle hundreds of remote sessions efficiently. Moreover, implementing Remote Desktop Gateway can enhance security by allowing remote connections only through authorized channels.
Concerning data management and security, segregate your database servers from web servers, even if they run the same OS, to reduce security risks. Make sure your database uses SSL connections to enhance security during data transmission and configure appropriate limitations on concurrent connections based on your expected user load.
Lastly, solid-state drives (SSDs) for your database and remote desktop servers could improve performance and reliability, thanks to their speed and reduced risk of mechanical failure. Make sure the drives are compatible with your system and have adequate backups in place to ensure continuous availability of services.
You want to use the motherboard with DDR3 to triple channel the RAM in order to increase performance. When you open your computer, you notice that four RAM slots are available. How can you use the increased performance of triple channel with four slots available?
Answer:
Check the motherboard documentation to find which modules are supported. Purchase additional modules that are the same as what is currently installed.Final answer:
Triple channel RAM requires three memory sticks and a compatible motherboard, typically with six slots. If your motherboard has four slots, it likely only supports dual-channel or single-channel configurations. You should use pairs of identical RAM modules to benefit from dual-channel mode.
Explanation:
To utilize the increased performance of triple channel RAM on a motherboard with DDR3 memory and four RAM slots, you need to understand that triple channel memory configurations require three memory sticks working in concert. Unfortunately, if your motherboard supports only dual-channel or single-channel modes with its four slots, you cannot set up a triple channel configuration. Instead, to maximize performance, you should fill either two or four slots with identical RAM modules (same capacity, speed, etc.) to take advantage of dual-channel capabilities.
If your motherboard actually supports triple channel memory (which is less common and typically associated with older Intel platforms like X58), it would often have six slots to accommodate the triple channel configuration. In such a case, three identical RAM sticks should be installed in the correct slots as indicated by the motherboard's manual to enable triple channel mode. Since your motherboard has only four slots, this indicates it likely does not support triple channel memory configuration.
In your scenario, with only four RAM slots, the best approach is to use pairs of identical RAM modules to enable dual-channel mode, which also offers a performance boost over single-channel, albeit less than what triple channel would provide.