A company is deploying smartphones for its mobile salesforce. These devices are for personal and business use but are owned by the company. Sales personnel will save new customer data via a custom application developed for the company. This application will integrate with the contact information stored in the smartphones and will populate new customer records onto it. The customer application's data is encrypted at rest, and the application's connection to the back office system is considered secure. The Chief Information Security Officer (CISO) has concerns that customer contact information may be accidentally leaked due to the limited security capabilities of the devices and the planned controls. Which of the following will be the MOST efficient security control to implement to lower this risk?

A. Implement a mobile data loss agent on the devices to prevent any user manipulation with the contact information.
B. Restrict screen capture features on the devices when using the custom application and the contact information.
C. Restrict contact information storage dataflow so it is only shared with the customer application.
D. Require complex passwords for authentication when accessing the contact information.

Answer:

A race condition is actually possible in this scenario.

Explanation:

A race condition is an undesirable situation that occurs when a device or system attempts to perform two or more operations at the same time. In order to prevent the race condition from occurring the operations must be done in the proper sequence to be done correctly.

Here is an example of the race condition:  Let's assume the Current Balance of the shared bank account is $600.  The husband's program calls the withdraw($100) function and accesses the Current Balance of $600.  But right before it stores the new Current Balance, the wife's program calls function deposit($100) function and accesses the Current Balance of $600.  At this point, the husband's program changes the shared variable Current Balance to $500, followed by the wife's program changing the Current Balance variable to $700.  Based on calling both withdraw($100) and deposit($100) functions, the Current Balance should again be $600, but it is $700. Alternatively, the  Current Balance could also have been $500.

To prevent the race condition from occurring, the solution will be to create a method to "Lock" on the Current Balance variable.  Each method must "Lock" the variable before accessing it and changing it.  If a method determines that the variable is "Locked," then it must wait before accessing it.

Answer:

We will use hierarchical concepts,superclass and manager interference and use of object oriented programing in python and java and we need to create and call functions as well to solve that problem.

Answer:

Answer is in the attached screenshot!

Explanation:

Iterate through the given input strings, create a student class with the given name, assign the output array to the value of the class created from the given input string. Finally, return the array.

Answer:

Explanation:

K- is the working procedure:

It takes n no. of predefined cluster as input and data points.

It also randomly initiate n centers of the clusters.

In this case the initial centers are given.

Steps you can follow

Step 1. Find distance of each data points from each centers.

Step 2. Assign each data point to the cluster with whose center is nearest to this data point.

Step 3. After assigning all data points calculate center of the cluster by taking mean of data points in cluster.

repeat above steps until the center in previous iteration and next iteration become same.

A1(4,8), A2(2, 4), A3(1, 7), B1(5, 4), B2(5,7), B3(6, 6), C1(3, 7), C2(7,8)

Centers are X1=A1, X2=B1, X3=C1

A1 will be assigned to cluster1, B1 will be assigned to cluster2 ,C1 will be assigned to cluster3.

Go through the attachment for the solution.

The k-means algorithm is used to cluster data points into partitions based on their attributes, using the Euclidean distance function. After the first round of execution, the cluster centers are A1(4, 8), B1(5, 4), and C1(3, 7). The final points in the three clusters after the algorithm stops may include A1(4, 8), A2(2, 4), A3(1, 7), B1(5, 4), B2(5, 7), B3(6, 6), C1(3, 7), and C2(7, 8).

The k-means algorithm is used to cluster data points into partitions based on their attributes. In this case, we are clustering points in a two-dimensional space (x, y) into three clusters using the Euclidean distance function. The k-means algorithm involves initializing the cluster centers, iteratively recalculating the centroids, and reassigning the data points to the new centroids. The algorithm continues until convergence or a certain number of iterations.

(a) After the first round of execution, the three cluster centers are A1(4, 8), B1(5, 4), and C1(3, 7).

(b) The final points in the three clusters after the algorithm stops might be A1(4, 8), A2(2, 4), A3(1, 7) in one cluster, B1(5, 4), B2(5, 7), B3(6, 6) in another cluster, and C1(3, 7), C2(7, 8) in the third cluster. The exact assignment of points to clusters may vary depending on the distance calculations and convergence criteria used in the algorithm.

Answer:

See explaination

Explanation:

The program code

PastPresentFuture.java

import java.util.*;

import java.time.LocalDate;

public class PastPresentFuture

{

public static void main(String args[])

{

/*Initializing an array with days of month in a sequence, february with 28 days - it would be changed afterwards if the year entered is a leap year.*/

int daysOfmonth[]={31,28,31,30,31,30,31,31,30,31,30,31};

LocalDate today=LocalDate.now();

int mo,da,yr;

int todayMo,todayDa,todayYr;

//storing the values of today's day,month & year in 3 variables

todayMo=today.getMonthValue();

todayDa=today.getDayOfMonth();

todayYr=today.getYear();

//taking inputs of day,month & year from console which would be used later for checking

Scanner input=new Scanner(System.in);

System.out.println(":::Program to find whether a date is in past/present/future:::");

System.out.print("Enter month::");

mo=input.nextInt();

System.out.print("Enter day::");

da=input.nextInt();

System.out.print("Enter year::");

yr=input.nextInt();

/*finding whether the entered year is leap year or not based on standard logic and making the days of february to 29 if it is a leap year.*/

if(yr%100==0)

{

if(yr%400==0)

daysOfmonth[1]++;

}

else

{

if(yr%4==0)

daysOfmonth[1]++;

}

System.out.print("The date you entered -- ");

/*checking if the month value is between 1 to 12 and day entered is with in the boundary limits of the respective month using the array initialized above.

if the date entered is invalid, we are informing the user about the same and stopping the program.*/

if(!(mo>=1 && mo<=12 && da>=0 && da<=daysOfmonth[mo-1]))

{

System.out.println("invalid !!");

System.exit(0);

}

//logic to print the 4 statements as per the question

if(yr!=todayYr)

System.out.println(yr+" not this year.");

else

if(mo<todayMo)

System.out.println("month "+mo+" in an earlier month this year.");

else

if(mo>todayMo)

System.out.println("month "+mo+" in a later month this year.");

else

System.out.println("month "+mo+" is this month.");

}

}

Answer:

is_isbn = True

total = 0

number = input("Enter the ISBN: ")

if len(number) != 10:

   is_isbn = False

for i in range(9):

   if 0 <= int(number[i]) <= 9:

       total += int(number[i]) * (10 - i)

   else:

       is_isbn = False

if number[9] != 'X' and int(number[9]) > 9:

   is_isbn = False

if number[9] == 'X':

   total += 10

else:

   total += int(number[9])

if total % 11 != 0:

   is_isbn = False

if is_isbn:

   print("It is a valid ISBN")

else:

   print("It is NOT a valid ISBN")

Explanation:

Initialize the variables, total and is_isbn - keeps the value for the number if it is valid or not

Ask the user for the ISBN number

If number's length is less than 10, it is not valid, set is_isbn to false

Then, make calculation to verify the ISBN

Check the last digit, if it is not "X" and greater than 10, it is not valid, set is_isbn to false

Again, check the last digit, if it is "X", add 10 to the total, otherwise add its value

Check if total can divide 11 with no remainder. If not, it is not valid, set is_isbn to false

Finally, depending on the is_isbn variable, print the result

Answer:

SELECT order_id, order_date,

DATEADD(DAY,5,order_date) AS approx_ship_date,

ship_date,

DATEDIFF(DAY,ship_date,DATEADD(DAY,2,order_date)) AS days_to_ship

FROM Orders

WHERE YEAR(order_date) = 2018 AND MONTH(order_date) = 3                            

Explanation:

The first line of the SQL statement is a SELECT statement which selects    order_id, order_date and ship_date columns from Orders table.

The DATEADD() is used to add date and here it is used to add 5 days to order_date column and the resultant column is named as approx_ship_date using ALIAS.

DATEDIFF() function is used to return the difference between two dates and here it shows number of days between order_date and ship_date columns.

WHERE clause is used to retrieve orders from March 2018. YEAR function represents the year of order_date which is set as 2018 to retrieve the orders for 2018. MONTH function represents the month of order_date which is set to 3 which means March in order to retrieve the orders for March.

Final answer:

To retrieve order details with calculated shipping dates and filter for March 2018 orders from an Orders table, use an SQL SELECT statement with DATE_ADD for approx_ship_date, DATEDIFF for days_to_ship, and a WHERE clause for the date range.

Explanation:

The question involves writing an SQL SELECT statement to retrieve specific columns from an Orders table, including calculated columns for approx_ship_date and days_to_ship, and filtering the results for orders made in March 2018. Here's how you can do it:

SELECT
 order_id,
 order_date,
 DATE_ADD(order_date, INTERVAL 5 DAY) AS approx_ship_date,
 ship_date,
 DATEDIFF(ship_date, order_date) AS days_to_ship
FROM
 Orders
WHERE
 order_date BETWEEN '2018-03-01' AND '2018-03-31';

This SELECT statement fetches the order_id, order_date, and ship_date directly from the Orders table. It also calculates approx_ship_date by adding 5 days to the order_date and calculates days_to_ship as the difference between ship_date and order_date. The WHERE clause filters records to include only those with order_dates in March 2018.

Answer:

import java.util.Scanner;

public class num8 {

   public static void main(String[] args) {

       Scanner in = new Scanner(System.in);

       System.out.println("Enter first number");

       int X = in.nextInt();

       System.out.println("Enter second number");

       int Y = in.nextInt();

       int Z;

       if(X <= Y){

           Z = 0;

       }

       else if(X >= Y){

           Z = 1;

       }

   }

}

Explanation:

Answer:

#The program sets Z to 0 if the X <= Y. otherwise sets Z to 1 , if the X > Y.

clear Z; # sets Z to 0

clear U; # sets U to 0

clear V; # sets v to 0

clear T; # sets T to 0

while Y not 0 # repeat until Y times

# process to decrement Y by 1

while Y not 0

incr U;

decr Y;

decr U;

while U not 0

incr Y;

decr U;

#process to decrement X by 1

while X not 0

incr V

decr X;

while V not 0

decr V;

while V not 0

incr T;

decr V;

while T not 0

incr X

decr T

#end of main while loop

#after running the main while loop, X will be positive and non - zero number, # #if X > Y. otherwise X and Y will be zero.

#process to increment Z by 1, if X is positive and non - zero number

while X not 0

while X not 0

decr X;

incr Z;

#end of while

Explanation:

The program is totakes in as input two variables X and Y.

The main function of the program is to return or place a 0 in the variable Z if the variable X is less than or equal to Y, and place a 1 in the variable Z if the variable X is greater than Y.

When these conditions are met, end program.

To check for a matching pair of nuts and bolts in O(n) time, perform a linear scan using two indices since the arrays are sorted. Increment the indices appropriately and return true if a match is found before either index exceeds n, otherwise return false.

To solve the problem of finding if there is a matching pair of nuts and bolts in an O(n) time complexity, you can employ a simple linear scan. Since the arrays NUTS and BOLTS are both sorted in increasing order, you can iterate over them simultaneously to check for a match.

Algorithm:

This algorithm is guaranteed to run in O(n) since each list is traversed at most once.

Answer:

We will use script in oracle and SQl(Structured Query Language) and use of algorithms like cipher for encryption to insert a new flight into the flights table for the flight number 165 and that Cipher will show the desired message.

Answer:

#include <stdio.h>

int main()

{

//Store temp in Fahrenheit

int temp_fahrenheit;

//store temp in Celsius

float temp_celsius;

printf("\nTemperature conversion from Fahrenheit to Celsius is given below: \n\n");

printf("\n%10s\t%12s\n\n", "Fahrenheit", "Celsius");

//For loop to convert

temp_fahrenheit=0;

while(temp_fahrenheit <= 212)

{

temp_celsius = 5.0 / 9.0 *(temp_fahrenheit - 32);

printf("%10d\t%12.3f\n", temp_fahrenheit, temp_celsius);

temp_fahrenheit++;

}

return 0;

}

Explanation:

The program above used a while loop for its implementation.

It is a temperature conversion program that converts integer Fahrenheit temperatures from 0 to 212 degrees to floating-point Celsius temperatures with 3 digits of precision.

This can be calculated or the calculation was Perform using the formula celsius = 5.0 / 9.0 * ( fahrenheit - 32 ).

The expected output was later printed in two right-justified columns of 10 characters each, and the Celsius temperatures was preceded by a sign for both positive and negative values.

Final answer:

The provided transaction log represents updates made to the 'Part' table in a database, recording the details of each transaction and the changes made to specific rows.

Explanation:

The transaction log provided represents a series of updates made to the 'Part' table in a database. Each transaction is recorded with a unique 'TRL_ID' and 'TRX_ID'. The 'OPERATION' column indicates the type of operation performed (e.g., START, UPDATE, COMMIT). The 'TABLE' column specifies the table that was updated, and the 'ROW ID' column represents the specific row that was modified. The 'BEFORE VALUE' and 'AFTER VALUE' columns show the original and final values of the updated attribute.

Answer:

Check the explanation

Explanation:

# include <iostream>

#include<iomanip>

using namespace std;

int main ()

{double start,balance,rate,deposit=0,withdraw=0,amt,sum,totInterest=0,interest;

int i;

cout<<setprecision(2)<<fixed;

cout<<"Enter starting balance: ";

cin>>balance;

start=balance;

cout<<"Enter annual interest rate: ";

cin>>rate;

rate/=12.;

for(i=1;i<=3;i++)

  {cout<<"For month "<<i<<endl;

  sum=balance;

   cout<<"Enter total amount deposit: ";

   cin>>amt;

   while(amt<0)

       {cout<<"Must not be negative-retry\n";

       cout<<"Enter total amount deposit: ";

       cin>>amt;

       }

   deposit+=amt;

   balance+=amt;

   cout<<"Enter total amount withdrawn: ";

   cin>>amt;

   while(amt<0||amt>balance)

       {cout<<"Must not be negative, or greater than balance ($"<<balance<<")-retry\n";

       cout<<"Enter total amount withdrawn: ";

       cin>>amt;

       }

   withdraw+=amt;

   balance-=amt;

   interest=(sum+balance)/2.*rate;

   totInterest+=interest;

   balance+=interest;

}

cout<<"Starting balance at the beginning of the three month period: $"<<start<<endl;

cout<<"total deposits made during the three months: $"<<deposit<<endl;

cout<<"total withdrawals made during the three months: $"<<withdraw<<endl;

cout<<"total interest posted to the account during the three months $"<<totInterest<<endl;

cout<<"final balance: $"<<balance<<endl;

system("pause");

return 0;

}

Kindly check the output in the attached image below.

Answer:

seeexplaination

Explanation:

Code(header file):

#ifndef THROTTLESHIFTLVEL_H

#define THROTTLESHIFTLVEL_H

class throttle

{

int position;

public:

throttle()

{

position = 0;

}

// specify the number of shift levels when creating objects

throttle(int level)

{

position = level;

}

void shut_off()

{

position = 0;

}

void shift(int amount)

{

if(amount >= 0)

position = position + amount;

}

double flow() const

{

return position;

}

bool is_on() const

{

if (position > 0)

return true;

else

false;

}

};

#endif

Please kindly check attachment for screenshot

Final answer:

To create a new throttle ADT with a customizable number of shift levels, define the class with an additional parameter in the constructor. Provide methods for shifting the throttle to a specific level and releasing it.

Explanation:

The modified definition of the throttle class to create a new throttle ADT, which allows the user to specify the number of shift levels, can be written as follows:

class Throttle:

   def __init__(self, shift_levels):
       self.shift_levels = shift_levels

   def shift(self, level):
       # shift the throttle to the specified level
       pass

   def release(self):
       # release the throttle
       pass

Answer:

See attached solution

Explanation:

Answer:

Check the explanation

Explanation:

using System;

using System.IO;

public class Program

{

  public static void Main()

  {

      string[] lines;

      Console.WriteLine("enter name of the file");

      String fileName=Console.ReadLine();

      Console.WriteLine("enter path of the file");

      String filePath=Console.ReadLine();

      using (StreamReader streamReader = File.OpenText(filePath))

      {

          String text = streamReader.ReadToEnd();

          lines = File.ReadAllLines(filePath);

      }

      Console.WriteLine("Enter path of the file to write");

      String outputFilePath = Console.ReadLine();

      bool fileExist = File.Exists(outputFilePath);

      if (fileExist)

      {      

                  overWriteOrChangeFileName(outputFilePath);          

      }

      else

      {

          File.Create(outputFilePath);

          File.WriteAllLines(outputFilePath, lines);

      }

      Console.Write("Input file Path"+filePath);

      Console.Write("output file path"+outputFilePath);

      void overWriteOrChangeFileName(String filePathtoCheck){

          Console.WriteLine("File exists. Do you want to overwrite the file or do you want to change the location and file name of the new file.Press Y to overwrite and press C to change the location and filename");

          String userInput=Console.ReadLine();

          if(userInput.Equals("Y")){

File.WriteAllLines(outputFilePath, lines);

          }else{

              Console.WriteLine("enter new path of the file including filename");

              String NewFileName=Console.ReadLine();

              outputFilePath=NewFileName;

              if(File.Exists(outputFilePath)){

                  overWriteOrChangeFileName(outputFilePath);                  

              }

              else

      {

          File.Create(outputFilePath);

          File.WriteAllLines(outputFilePath, lines);

      }

          }

      }

  }

}

Answer:

see explaination and attachment

Explanation:

We can say that a Gantt chart is a visual view of tasks scheduled over time. Gantt charts are used for planning projects of all sizes and they are a useful way of showing what work is scheduled to be done on a specific day.

see attachment for the detailed step by step solution.

Answer:

='[Finance.xlsx]Fall2018'!$D$20

Explanation:

The formula that will correctly links to cell D20 in the Fall 2018 worksheet within the Finance workbook is given as:

='[Finance.xlsx]Fall2018'!$D$20

This will perform the operation and produce the required output.

Answer:

see explaination

Explanation:

Java code

//Header file section

import java.util.Scanner;

import java.io.*;

//main class

public class SalesTestDemo

{

//main method

public static void main(String[] args) throws IOException

{

String inFile;

String line;

double total = 0;

Scanner scn = new Scanner(System.in);

//Read input file name

System.out.print("Enter input file Name: ");

inFile = scn.nextLine();

FileReader fr = new FileReader(new File(inFile));

BufferedReader br = new BufferedReader(fr);

System.out.println("Name \t\tService_Sold \tAmount \tEvent Date");

System.out.println("=====================================================");

line = br.readLine();

//Each line contains the following, separated by semicolons:

//The name of the client, the service sold

//(such as Dinner, Conference, Lodging, and so on)

while(line != null)

{

String temp[] = line.split(";");

for(int i = 0; i < temp.length; i++)

{

System.out.print(temp[i]+"\t");

if(i == 1)

System.out.print("\t");

}

//Calculate total amount for each service category

total += Double.parseDouble(temp[2]);

System.out.println();

line = br.readLine();

}

//Display total amount

System.out.println("\nTotal amount for each service category: "+total);

}

}

inputSale.txt:

Peter;Dinner;1500;30/03/2016

Bill;Conference;100.00;29/03/2016

Scott;Lodging;1200;29/03/2016

Output:

Enter input file Name: inputSale.txt

Name Service_Sold Amount Event Date

=====================================================

Peter Dinner 1500 30/03/2016

Bill Conference 100.00 29/03/2016

Scott Lodging 1200 29/03/2016

Total amount for each service category: 2800.0

Answer:

String date = "21/05/2020";

String dayStr = date.substring(0,2);

int day = Integer.parseInt(dayStr);

System.out.println(day);

Explanation:

Create a variable called date which holds the current date

Create a variable called dayStr. Initialize it to the day part of the date using the substring method

Create a variable called day. Parse the dayStr and assign it to the day

Print the day

Answer:

Explanation:

Let G denote an unambiguous Grammar capable of producing simple mathematical expressions, involving operators +,-,*,/. These operators are left associative (which ensures left to right evaluation). S is the start symbol of the Grammar i.e. the production starts from S. n denotes a number and is a terminal i.e. we can't produce anything further from n. Then, the solution is as follows :

S → S + T |S - T | S

T→T | F | T*F|F

F → n

Here, S, T and F are variables. Note that /,* have been given precedence over +,-.

An unambiguous grammar for basic mathematical expressions without parentheses can be created using operator precedence. It defines a set of production rules that ensure expressions are evaluated following the order of operations.

An unambiguous grammar for basic mathematical expressions without parentheses can be created using the concept of operator precedence. We can define a grammar where the multiplication and division operators have higher precedence than the addition and subtraction operators. This way, expressions will be evaluated following the order of operations.

For example, the grammar could be defined as follows:

This grammar allows for the creation of unambiguous mathematical expressions using the basic operators +, -, *, and /, without the need for parentheses. The evaluation of these expressions will follow the order of operations, with * and / taking precedence over + and -.

Answer:

Check the explanation

Explanation:

The code will be

name=input('Enter file name:')

ls=open(name,'r').read().split('\n')

count=0;sums=0

for i in range(ls):

sums+=int(i)

count+=1

print('There were {} numbers in the file'.format(count))

print('The average is {}'.format(sums/count))

The output is

Answer:

Explanation:

CHECK THE ANSWER IN THE ATTACHMENT CORNNER

Answer:

see explaination please

Explanation:

a. Install a faster CPU. Answer:NO

Optional: a faster CPU reduces the CPU utilizationfurther since the CPU will spend more time waiting for a process toenter in the ready queue.

b. Install a bigger paging disk. Answer:NO

Optional: the size of the paging disk does not affect theamount of memory that is needed to reduce the pagefaults.

c. Increase the degree of multiprogramming Answer:NO

Since each process would have fewer frames available andthe page fault rate would increase

d. Decrease the degree of multiprogramming.Answer: YES

Optional: by suspending some of the processes, the otherprocesses will have more frames in order to bring their pages inthem, hence reducing the page faults.

e. Install more main memory. Answer:Likely

Optional: more pages can remain resident and do notrequire paging to or from the disks.

f. Install a faster hard disk or multiplecontrollers with multiple hard disks. Answer:Likely

(Optional)

g. Add prepaging to the page-fetchalgorithms.Answer: Likely

(Optional.)

h. Increase the page size. Answer:NO

Since each process would have fewer frames available and the page fault rate would increase

Answer:

Kindly note that, you're to replace "at" with shift 2 as the brainly text editor can't take the symbol

Explanation:

Copiable code:

# Define the class Interval.

class Interval(object):

   # Define the initializer.

   def __init__(self, a, b):

       # Set the minimum value as a and the maximum

       # value as b.

       self.a = min(a, b)

       self.b = max(a, b)

   # Define the function to return the length

   # of the interval.

   "at"property

   def length(self):

       return (self.b - self.a)

   # Define the function to return the center of

   # the interval.

   "at"property

   def center(self):

       # Divide the first and the last value by

       # 2 to compute the center.

       return (self.a + self.b)/2

   # Define the function to check if the two intervals

   # intersect or not.

   def intersects(self, other):

       # Compute the absolute difference between the

       # centers of the two intervals and compare it with

       # the sum of the half lengths of the intervals

       # to check if they intersect or not.

       return abs(self.center - other.center) <= (self.length/2 + other.length/2)

The complete program to test the above class is as follows:

Note: No output is generated by the code since all the test cases have been passed successfully.

Code to copy:

# Import the required packages to test the class.

from nose.tools import assert_equal, assert_false, assert_true

# Define the class Interval.

class Interval(object):

   # Define the initializer.

   def __init__(self, a, b):

       # Set the minimum value as a and the maximum

       # value as b.

       self.a = min(a, b)

       self.b = max(a, b)

   # Define the function to return the length

   # of the interval.

   "at"property

   def length(self):

       return (self.b - self.a)

   # Define the function to return the center of

   # the interval.

   "at"property

   def center(self):

       # Divide the first and the last value by

       # 2 to compute the center.

       return (self.a + self.b)/2

   # Define the function to check if the two intervals

   # intersect or not.

   def intersects(self, other):

       # Compute the absolute difference between the

       # centers of the two intervals and compare it with

       # the sum of the half lengths of the intervals

       # to check if they intersect or not.

       return abs(self.center - other.center) <= (self.length/2 + other.length/2)

# Create different intervals to check the working

# of the above class using assert test cases.

i = Interval(3, 5)

assert_equal(i.length, 2)

i = Interval(3, 5)

assert_equal(i.center, 4)

i = Interval(2, 5)

j = Interval(6, 7)

assert_false(i.intersects(j))

assert_false(j.intersects(i))

k = Interval(4, 6)

assert_true(i.intersects(k))

assert_true(k.intersects(i))

m = Interval(0, 8)

assert_true(i.intersects(m))

assert_true(m.intersects(i))

Answer:

According to IEEE-754 standard,equivalent  decimal of 32 bit floating point number is 78.1025390625

Explanation:

To convert the 32 bit floating point number from hexadecimal to decimal, following steps will be completed.

1. Convert this hexadecimal value "C29C3480" into binary number. Each digit in hexadecimal will convert into 4 bits of binary.

There

C= 1100

2 = 0010

9 =  1001

C = 1100

3 = 0011

4= 0100

8 = 1000

0 = 0000

So the number will be converted as given below:

1100 0010 1001 1100 0011 0100 1000 0000

2. Arrange the bits in order

1   = Sign bit = 32nd bit                 

10000101 = Exponent bits = From 31st to 23rd bits

00111000011010010000000 = Fraction bits= From 22nd to 0th bit

As the 32nd  bit, which is sign bit, is 1, so given number is negative.

3. Convert the Exponent bits into decimal values

10000101 = 1x2⁷+0x2⁶+0x2⁵+0x2⁴+0x2³+1x2²+0x2¹+1x2⁰

               = 1x2⁷ + 1x2² + 1x2⁰

               = 128 + 4 + 1 =  133

4. Calculate Exponent Bias

      exponent bias = 2ⁿ⁻¹ - 1

there n is total number of bits in exponent value of 32 bit number, Now in this question n=8, because there are total 8 bits in exponent.

 so,       exponent bias = 2⁸⁻¹ - 1 = 2⁷-1= 128-1= 127

         e = decimal value of exponent - exponent bias

          e  =     133 - 127 = 6

5.  Convert Fraction part into Decimal

 Mantissa=0x2⁻¹+0x2⁻²+1x2⁻³+1x2⁻⁴+1x2⁻⁵+0x2⁻⁶+0x2⁻⁷+0x2⁻⁸+0x2⁻⁹+1x2⁻¹⁰+1x2⁻¹¹+0x2⁻¹²+1x2⁻¹³+0x2⁻¹⁴+0x2⁻¹⁵+1x2⁻¹⁶+0x2⁻¹⁷+0x2⁻¹⁸+0x2⁻¹⁹+0x2⁻²⁰+0x2⁻²¹+0x2⁻²²+0x2⁻²³

              = 1x2⁻³+1x2⁻⁴+1x2⁻⁵+1x2⁻¹⁰+1x2⁻¹¹+1x2⁻¹³+1x2⁻¹⁶

              =  0.125 + 0.0625 + 0.03125 + 0.0009765625 + 0.00048828125 +                     0.00012207031 + 0.00001525878

Mantissa = 0.22035217284

6. Convert the number into decimal

Decimal number = (-1)^s x (1+Mantissa) x 2^e

there

s is the sign bit which is 1,

and

e =6.

so,

decimal number = (-1)^1 x (1+0.22035217284 ) x 2⁶

                           = -1 x 1.22035217284 x 64

                           = -78.1025390625

Answer:

This question is incomplete, here's the complete question:

1. Willow Brook National Bank operates a drive-up teller window that allows customers to complete bank transactions without getting out of their cars. On weekday mornings, arrivals to the drive-up teller window occur at random, with an arrival rate of 24 customers per hour or 0.4 customers per minute. 3. Use the single-server drive-up bank teller operation referred to in Problems 1 to determine the following operating characteristics for the system: a.) The probability that no customers are in the system. b.) The average number of customers waiting. c.) The average number of customers in the system. d.) The average time a customer spends waiting. e.) The average time a customer spends in the system. f.) The probability that arriving customers will have to wait for service.

Explanation:

Arrival rate \lambda = 24 customers per hour or 0.4 customers per minute

Service rate \mu​ = 36 customers per hour or 0.6 customers per minute (from problem 1)

a.) The probability that no customers are in the system , P0 = 1 - \lambda / \mu

= 1 - (24/36) = 1/3 = 0.3333

b.) The average number of customers waiting

Lq = \lambda^2 / [\mu(\mu - \lambda)] = 242 / [36 * (36 - 24)] = 1.33

c.) The average number of customers in the system.

L = Lq + \lambda / \mu = 1.33 + (24/36) = 2

d.) The average time a customer spends waiting.

Wq = \lambda / [\mu(\mu - \lambda)] = 24 / [36 * (36 - 24)] = 0.0555 hr = 3.33 min

e.) The average time a customer spends in the system

W = Wq + 1/\mu = 0.0555 + (1/36) = 0.0833 hr = 5 min

f.) The probability that arriving customers will have to wait for service.

= 1 - P0 = 1 - 0.3333 = 0.6667

Answer:

def str_replace(int_list, index):

   if int_list[index] < 7:

       int_list[index] = "small"

   elif int_list[index] > 7:

       int_list[index] = "large"

   return int_list

Explanation:

I believe you forgot to mention the comparison value. I checked if the integer at the given index is smaller/greater than 7. If that is not your number to check, you may just change the value.

Create a function called def str_replace that takes two parameters, an integer list and an index

Check if the number in given index is smaller than 7. If it is replace the number with "small".

Check if the number in given index is greater than 7. If it is replace the number with "large".

Return the list

The str_replace function replaces an integer in a list with 'small' or 'large' depending on the value. It demonstrates basic list manipulation and conditional logic in programming.

The str_replace function is designed to take a list of integers and an index. The function will check the value at the given index, and if the integer is less than 5, it will be replaced with the string "small"; if the integer is 5 or greater, it will be replaced with the string "large". This is a basic operation in programming, often related to the concept of list manipulation and conditional statements. Here's a step-by-step example in Python:

Answer:

Check the explanation

Explanation:

If raw_input() shows some error in Python v3 then call input() instead of raw_input().

def menu():

              option = 'n'

              while option not in ['r','p','d','h','l','q']:

                             print("Menu of choices:")

                             print("\t(r)ead employee data")

                             print("\t(p)rint employee payroll")

                             print("\t(d)isplay an employee by name")

                             print("\tfind (h)ighest paid employee")

                             print("\tfind (l)owest paid employee")

                             print("\t(q)uit")

                             option = input("Please enter your choice: ")

                             if option not in ['r','p','d','h','q']:

                                                                          print("Invalid choice. Please try again\n")

                             return option

def readEmployees(names,wages,hours):

              fileRead = 0

              try:

                             filename = input("Enter the file name: ")

                             f = open(filename,"r")

                             for line in f:

                                            values = line.split("\t")

                                            names.append(values[0])

                                            wages.append(float(values[1]))

                                            i=0

                                            total = 0

                                            while i < 5:

                                                           total = total + float(values[2+i])

                                                           i = i+1

                                            hours.append(total)

                             fileRead = 1

                             f.close()

              except ValueError:

                             print("Bad data in file " + filename)

              except IOError:

                             print("Error reading file " + filename)

              return fileRead

def printPayroll(names,wages,hours):

              print("")

              print("Name"+"    "+"Hours"+"    "+"Pay")

              print("----"+"    "+"-----"+"    "+"---")

              i = 0

              sortedNames = names

              sortedNames.sort()

              totalpay = 0

              for name in sortedNames:

                             i = names.index(name)

                             pay = hours[i] * wages[i]

                             name = names[i]

                             totalpay = totalpay + pay

                             name = repr(name).ljust(10)

                             name = name.replace("'","")

                             print( name + repr(hours[i]).ljust(9) + repr(pay).ljust(6))

                             i = i + 1

              print("\nTotal payroll = $" + str(totalpay))

def displayName(names,wages,hours):

              name = ""

              i=0

              size = len(names)

              name = input("Enter the employee's name: ")

              while i < size:

                             if names[i].lower() == name.lower():

                                            wage = wages[i]

                                            hour = hours[i]

                                            break

                             i = i +1

              if i == size:

                             print("Employee not found")

              else:

                             print(name + " worked " + str(hour) + " hours at $" + str(wage) + " per hour, and earned $" + str(wage * hour) + "\n")

def showHighLow(names,wages,hours, ind):

              highestIndex = 0

              if ind == 'h':

                             i =1

                             size = len(names)

                             wage = 0;

                             while i < size:

                                            wage = hours[i] * wages[i]

                                            if wage > (hours[highestIndex] * wages[highestIndex]):

                                                           highestIndex = i

                                            i = i + 1

                             print(names[highestIndex] + " earned $" + str(hours[highestIndex] * wages[highestIndex]) + "\n")

              elif ind == 'l':

                             lowestIndex = 0;

                             i =1

                             size = len(names)

                             wage = 0;

                             while i < size:

                                                                          wage = hours[i] * wages[i]

                                                                          if wage < (hours[lowestIndex] * wages[lowestIndex]):

                                                                                                                                      lowestIndex = i

                                                                          i = i + 1

                             print(names[lowestIndex] + " earned $" + str(hours[lowestIndex] * wages[lowestIndex]) + "\n")

def main():

              option = 'n'

              names    = list()

              wages    = list()

              hours    = list()

              fileRead = 0

              while option != 'q':

                             option = menu()

                             if option == 'r':

                                            fileRead = readEmployees(names,wages,hours)

                             elif option == 'p':

                                            if fileRead == 0:

                                                           print("Employee data has not been read.")

                                                           print("Please read the file before making this choice")

                                            else:

                                                           printPayroll(names,wages,hours)

                             elif option == 'd':

                                            if fileRead == 0:

                                                           print("Employee data has not been read.")

                                                           print("Please read the file before making this choice")

                                            else:

                                                           displayName(names,wages,hours)

                             elif option == 'h':

                                            if fileRead == 0:

                                                           print("Employee data has not been read.")

                                                           print("Please read the file before making this choice")

                                            else:

                                                           showHighLow(names,wages,hours,'h')

                             elif option == 'l':

                                            if fileRead == 0:

                                                           print("Employee data has not been read.")

                                                           print("Please read the file before making this choice")

                                            else:

                                                           showHighLow(names,wages,hours,'l')

                             elif option == 'q':

                                            print("GoodBye")

if __name__ == "__main__":

              main()

Kindly check the OUTPUT in the attached image below.