Look at the following array definition. char str[10]; Assume that name is also a char array, and it holds a C-string. Write code that copies the contents of name to str if the C-string in name is not to big to fit in str.

Answer:A

Explanation:

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:

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:

Here is the C++ code for the program:

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:

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.

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.

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.

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.

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:

This 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:

This 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.

This 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:

This 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.

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.

Read more about expected value at; https://brainly.com/question/25926316

Answer:

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.

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.

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.

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.

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.

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.

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.

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.

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:

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.

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

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();

}

}

}

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.

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.

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 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.

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:

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.

Read related link on:

https://brainly.com/question/12769747

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.

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)

Answer:

"3, 2 and 1" is the correct answer for the above question.

Explanation:

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.

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.