The laptop has a built-in wireless adapter or the wireless adapter is physically installed on a computer and it does not appear in Network Connections. What is most likely the problem when it does not show in Network Connections

The largest numeric value that can be represented in 3 bytes is 224

Explanation:

Final answer:

The largest value with three bytes using ASCII is 127127127, but for practical numerical storage in binary notation, the largest value with three bytes is 16777215, represented as 11111111 11111111 11111111 in binary.

Explanation:

The largest numeric value that could be represented with three bytes using one ASCII pattern per byte depends on the ASCII printable characters range. ASCII characters are represented within the range of 0 to 127, therefore in three bytes, the largest decimal values you could represent would be 127127127. However, this is not typically how numeric values are stored.

When using binary notation, each byte consists of 8 bits, and the maximum value for each bit is 1. Therefore, the largest value in a single byte is 11111111, which is 255 in decimal. Hence, in three bytes, the largest decimal number that can be stored is 16777215 (which is 11111111 11111111 11111111 in binary).

To output the average and product of three floating-point numbers as rounded integers as well as floating-point numbers with two decimal points, we can use the Python string formatting expression. Given the inputs 10.3, 20.4, and 5.0, the average is rounded to 11 and the product is rounded to 1050 as integers. As floating-point numbers, formatted with two decimal places, the average is 11.90 and the product is 1050.60.

The average calculation is (10.3 + 20.4 + 5.0) / 3, which equals 11.9 when not rounded. For the integer part, we round 11.9 to 11. The product is 10.3 * 20.4 * 5.0, which equals 1052.6, rounded to 1050 as an integer. To format these as floating-point numbers using Python string formatting: '%0.2f' % 11.9 for the average and '%0.2f' % 1052.6 for the product, yielding 11.90 and 1050.60, respectively.

Answer:

.

Explanation:

Answer:

Option A.

Explanation:

It is a business association working to protect your online privacy. Opa members agree to establish a User privacy Policy that is readable and understandable. So, the option of choosing who sees the data  is the provision is not available in the policy because It has data types policy gathered and how data collected is protected.

Answer:

The correct answer to the following question will be "False".

Explanation:

Therefore, the given statement is false.

Answer:

C++.

Explanation:

int main() {

   const float package_A = 9.95;

   const float package_B = 14.95;

   const float package_C = 19.95;

   const int package_A_extra = 2;

   const float package_B = 1;

/////////////////////////////////////////////////////////////////////////////

   int user_package_choice;

   int hours_used;

   cout<<"Your package? Enter option number,"<<endl;

   cout<<"1. Package A"<<endl<<"2. Package B"<<"3. Package C";

   cout<<endl;

   cin<<user_package;

   cout<<endl;

   cout<<"Hours used?: ";

   cin<<hours_used;

/////////////////////////////////////////////////////////////////////////////

   cout<<endl;

   float total_amount;

   if (user_package_choice == 1) {

       total_amount = package_A + ((hours_used - 10) * package_A_extra));

       if (total_amount - (package_B + ((hours_used - 20) * package_B_extra)) > 0)

          cout<<"If you had opted for Package B, you would have saved $"<<total_amount - (package_B + (hours_used * package_B_extra))<<endl;

       if (total_amount - package_C > 0)

           cout<<"If you had opted for Package C, you would have saved $"<<total_amount - package_C;

   }

   else if (user_package_choice == 2) {

       total_amount = package_B + ((hours_used - 20) * package_B_extra);

        if (total_amount - package_C > 0)

           cout<<"If you had opted for Package C, you would have saved $"<<total_amount - package_C;

   }

   else {

       total_amount = package_C;

   }

   return 0;

}

Final answer:

A program is needed to compute a customer's monthly internet bill based on their subscription package and calculate potential savings for switching to other packages.

Explanation:

The task involves writing a program to calculate a customer's monthly bill based on the subscription package for internet access. There are three packages: Package A offers 10 hours for $9.95 with additional hours at $2 each, Package B provides 20 hours for $14.95 with additional hours at $1 each, and Package C offers unlimited access for $19.95. The program will also calculate potential savings for customers of Package A if they switch to Package B or C, and for Package B customers if they switch to Package C, provided switching would result in savings.

Answer:

The correct option is B: Old brick-and-mortar rules, regulations, and habits

Explanation:

Globalization has been made possible because of the internet. However, the promise of what is considered "borderless commerce" is limited due to the old brick and mortar rules, regulations, and habits. Brick and mortar is considered the traditional businesses offering products and services to customer in a rented or owned store and requires face-to-face business. Hence globalization via the internet is limited because companies have to sort out their building and stores, follow every law and regulation to opening offices and other habits practiced by these brick-and-mortar businesses.

Answer:

   public static void printTotalInches(double num_feet, double num_inches){

       double inches = 12*num_feet;

       System.out.println("Total value in feets is: "+(inches+num_inches));

   }

This function is written in Java Programming Language. Find a complete program with a call to the function in the explanation section

Explanation:

public class ANot {

   public static void main(String[] args) {

//Calling the function and passing the arguments

   printTotalInches(5,8);

   }

   public static void printTotalInches(double num_feet, double num_inches){

       double inches = 12*num_feet;

       System.out.println("Total value in feets is: "+(inches+num_inches));

   }

}

Answer:

def calc_total_inches(num_feet, num_inches):

   return num_feet*12+num_inches

Explanation:

Final answer:

The task requires a Python code using a for loop to count down from a specified starting number, printing 'Ready!', each number on a new line, and 'Run!' at the end.

Explanation:

To write code that counts down from a given number and prints 'Ready!', followed by each number, and ending with 'Run!', we can use a for loop. Here is a sample Python code snippet that accomplishes this:

firstNumber = 3
print('Ready!')
for num in range(firstNumber, 0, -1):
   print(num)
   print('\n')
print('Run!')

This code starts by printing 'Ready!', then it enters a for loop that counts down from firstNumber to 1, printing each number followed by a newline. Finally, it prints 'Run!' after exiting the loop. Remember to replace firstNumber with the actual starting number you wish to count down from.

Answer:

The Java code is given below with appropriate comments

Explanation:

import java.util.*;

class Main {

   static Set<String> set = new HashSet();

   static void printPerms(char ch[], int ind){

       //If end of string is reached, add it to set

       if(ind==ch.length){

           set.add(new String(ch));

           return;

       }

       for(int i=ind;i<ch.length;i++){

           //Only swap if lower case

           if((ch[i]>='a' && ch[i]<='z')&&((ch[ind]>='a' && ch[ind]<='z'))){

               char t  = ch[i];

               ch[i] = ch[ind];

               ch[ind] = t;

           }

           printPerms(ch,ind+1);

           if((ch[i]>='a' && ch[i]<='z')&&((ch[ind]>='a' && ch[ind]<='z'))){

               char t  = ch[i];

               ch[i] = ch[ind];

               ch[ind] = t;

           }

       }

   }

   public static void main(String[] args) {

       printPerms("aBbCc".toCharArray(),0);

       System.out.println(set);

   }

}

Final answer:

The question involves creating an algorithm to generate permutations of a given string without altering the positions of uppercase alphabets, a problem suitable for Computers and Technology at the college level.

Explanation:

The task requires generating all permutations of a given string while maintaining the original positions of the uppercase alphabets. This is typically a problem that can be approached algorithmically, falling under the category of Computers and Technology. To solve it, you would need to find permutations of the lowercase letters only and then insert them back into the string at the positions not occupied by uppercase letters.

One way to achieve this is to extract all lowercase letters and use a permutation algorithm, such as Heap's algorithm, to generate all permutations while the uppercase letters act as anchors, holding their respective positions fixed. As each permutation of the lowercase letters is generated, it's interlaced with the uppercase ones to construct a full-string permutation that meets the condition. This process is repeated until all permutations of the lowercase characters are exhausted.

Answer:

c. When a mobile device is turned on, it connects to the nearest available base station through a control channel

Explanation:

A network is the communication and interconnection between Computer devices for sharing of data resources. The connection of these devices could be wireless or wired.

Wired network connection is the use of cables to connect devices, while wireless connection requires no wire.

Wireless communication could be electromagnetic, microwave, cellular etc. Cellular is mainly used for cell phone communication.

The cell phone has a database of cell tower connections and automatically connects to the nearest one through a control channel, when it's turned on.

Answer:

Option C and Option D.

Explanation:

When the title indicates, a greedy algorithm often makes the decision which imply to be the best overall. Which obviously makes a locally-optimal choice throughout the expectation that such a option can contribute to an answer that is generally optimal.

so, Earliest finish time and latest start time are the greedy methods resulting in an effective solution for the problems of operation selection.

Answer:

There are four options for this question: A and D are correct.

A. By adding a mobile layout and assigning it to a profile.

B. By adding quick actions in the publisher section

C. By adding actions in the Salesforce1 action bar section.

D. By adding Visualforce page to the mobile cards section.

Explanation:

We can add a different view for the users instead of the standard web, we can use a mobil design and assign a different profile, addition, we can implement the visualforce page to create a new view, with this option we make a better and interesting user interface with interaction with different views.

Answer:

0.24 seg

Explanation:

We have 600 MB of data if we convert it to Gb we get 0.6Gb of data, the bandwidth is telling us that we can transfer 2.5 Gb of data per second if we assume that the data transfer is lineal we can apply the rule of three to solve the problem:

2.5 Gb -> 1 seg

0.6 Gb -> X

Applying the cross multiplication we get that X = 0.6 Gb* 1 seg / 2.5 Gb = 0.24 seg

The time required to transfer a 600-MB database over a DMA channel with a bandwidth of 2.5 GB/s is 0.24 seconds.

To calculate the transfer speed of a 600-MB database over a DMA channel with a bandwidth of 2.5 GB/s, you first need to convert the database size to the same unit as the bandwidth. In this case, 600 MB = 0.6 GB.

The time to transfer data over a channel is equivalent to the size of the data (in GB) divided by the bandwidth (in GB/s). Therefore, the time to transfer the 600-MB database would be 0.6 GB / 2.5 GB/s = 0.24 seconds.

Learn more about Data Transfer here:

https://brainly.com/question/35863200

#SPJ3

Final answer:

To solve the student's problem, input the current and last month's house prices, calculate the price change, compute the estimated mortgage, and correctly format the output presenting the figures.

Explanation:

To create a program that takes the current price and last month's price of a house as inputs, and outputs a summary of the price change and estimated monthly mortgage, follow these instructions:

Here's how the output should look like:

This house is $200000. The change is $-10000 since last month. The estimated monthly mortgage is $850.000000.

```python

current_price, last_month_price = map(int, input().split())

print(f'This house is ${current_price}. The change is ${last_month_price - current_price} since last month. The estimated monthly mortgage is ${(current_price * 0.051) / 12:.6f}.')

```

The program takes two integer inputs, representing the current house price and last month's price. It then uses a formatted string to output a summary, including the current price, the change since last month (calculated by subtracting the current price from last month's price), and the estimated monthly mortgage computed using the given formula. The use of f-strings in Python allows for concise and readable code, embedding expressions directly within the string.

The map function is employed to convert the input values into integers. The program calculates the change in price by subtracting the current price from last month's price. To ensure precise spacing and formatting, the f-string includes the required text and variable values in a structured manner. The estimated monthly mortgage is calculated using the provided formula and formatted as a floating-point value with six decimal places.

The code adheres to the assignment's emphasis on precision in spacing, punctuation, and newlines. By using f-strings and the specified formula, the program produces the required output with clarity and accuracy.

Answer:

def new_list_function(initial_list):

   new_list = initial_list*3

   return new_list

def main():

   initial_list = []

   while True:

     item = input("Enter value to be added to list: ")

     item = item.rstrip()

     if item.lower() == "exit" and "EXIT" and "Exit":

         break

     initial_list.append(item)

     new_list = new_list_function(initial_list)

   for item in new_list:

     print(item)

main()

Explanation:

Answer:

Scripts for businesses are always followed exactly.

Explanation:

A computer script is a list of commands that are executed by a certain program or scripting engine. Scripts may be used to automate processes on a local computer.

An ER Diagram can be created for an art museum scenario, including entities like Artwork, Artist, Show, ShowCity, ShowSchedule, Gallery, and Loan.

The ER Diagram for the given scenario can be created with the following entities:

The relationships between the entities can be represented as follows:

Answer:

#include<iostream>

using namespace std;

int main()

{int num1,num2;

cout<<"enter 2 numbers:"<<endl;

cinn>>num1;

cinn>>num2;

cout<<("value without swapping of num1",num1);

cout<<("value without swapping of num2",num2);

 num1 ^= num2;

   num2 ^= num1;

   num1 ^= num2;

cout<<("num1 after swapping",num1);

cout<<("num2 after swapping,num2);

return 0;

}

Answer:

import random

paswrd = input("Enter Pass: ")

while len(paswrd) < 8:

print('Password not long enough.')

paswrd = input("Enter Pass: ")

paswrd = paswrd.replace('e', '@')

paswrd = paswrd.replace('E', '@')

paswrd = paswrd.replace('s', '$')

paswrd = paswrd.replace('S', '$')

paswrd = paswrd.replace('t', '+')

paswrd = paswrd.replace('T', '+')

paswrd = paswrd.capitalize()

paswrd = paswrd + (str(random.randint(1,999)))

print(paswrd)

Explanation:

MOHR-COULOMB FAILURE CRITERIA:

In 1900, MOHR-COULOMB states Theory of Rupture in Materials which defines as “A material fails due to because of a critical combination of normal and shear stress, not from maximum normal or shear stress”. Failure Envelope is approached by a linear relationship.

If you can not understand the below symbols see the attachment below

 f   f ()  

Where:      f = Shear Stress on Failure Plane  

       ´= Normal Stress on Failure Plane

 See the graph in the attachment

For calculating the shear stress, when Normal stress, cohesion and angle of internal friction are given. Use this formula:   shear stress =  f  c   tan 

Where,  

• f  is Shear Stress on Failure Plane

• c  is Cohesion

•  is Normal Total Stress on Failure Plane

•  is Friction Angle

Answer:

Explanation:

In this case, the price of the RAM can rise because cost increase, assume the news of the Wall Street Journal, people with earning lower $80,000 they won't buy RAM because is too expensive, and people with over these earnings will buy less RAM, in addition, the incomes will fall, and the demand of RAM will fall too, but for the same reason the price could fall for the poor demand.

Answer:

C++ Program

Explanation:

#include <iostream>

#include <iomanip>

using namespace std;

void arrSelectSort(double *, int);

void showArrPtr(double *, int);

void showAverage(double, int);

int main()

{

 //dynamically allocate an array

 //acumulator

 //averge scores

 //number of test scores

 double *scores,                        

 total = 0.0;                  

//average;                              

int nScores;                      

   //Get how many test scores the users wants to enter.

cout << "How many test scores would you like to process?";

cin >> nScores;

//condition about the scores

scores = new double[nScores];

if (scores == NULL)

 return 0;

//Get the number of each test

cout << "Enter the test scores below.\n";

for (int count = 0; count < nScores; count++)

{

 cout << "Test score #" << (count + 1) << ": ";

 cin >> *(scores + count);

}

//total score operation

for (int count = 0; count < nScores; count++)

{

 total += *(scores + count);

}

showAverage(total, nScores);

//the array pointers

arrSelectSort(scores, nScores);

cout << "the ascending order is: \n";

showArrPtr(scores, nScores);

//free memory.

delete[] scores;

scores = 0;

return 0;

}

// bubble sort  

void arrSelectSort(double *array, int size)

{

int temp;

bool swap;

do

{

 swap = false;

 for (int count = 0; count < (size - 1); count++)

 {

  if (*(array + count) > *(array + count + 1))

  {

   temp = *(array + count);

   *(array + count) = *(array + count + 1);

   *(array + count + 1) = temp;

   swap = true;

  }

 }

} while (swap);

}

// sort function

void showArrPtr(double *array, int size)

{

for (int count = 0; count< size; count++)

 cout << *(array + count) << " ";

cout << endl;

}

// average function

void showAverage(double total, int nScores)

{

double average;

//average operation

average = total / nScores;

//Display the results.

cout << fixed << showpoint << setprecision(2);

cout << "Average Score: " << average << endl;

}

Answer:

"Classification" is the correct answer to this question.

Explanation:

Classification is primarily a method of data processing, it makes easy for information to be separated and divided into a different business or personal goals by the data set demands.  

Answer:

import java.util.Scanner;

public class TestClock {

   public static void main(String[] args) {

       Scanner in = new Scanner(System.in);

       System.out.println("Enter num 1");

       double num1 = in.nextDouble();

       System.out.println("Enter num 2");

       double num2 = in.nextDouble();

       System.out.println("enter \"s\" for addition\t \"d\" for difference\t \"p\" for the product\t and \"q\" for the quotient");

       char op = in.next().charAt(0);

       if(op =='s'||op=='S'){

           double sum = num1+num2;

           System.out.println("The sum is "+sum);

       }

       else   if(op =='d'||op=='D'){

           double diff = num1-num2;

           System.out.println("The difference is "+diff);

       }

       else if(op =='p'||op=='P'){

           double prod = num1*num2;

           System.out.println("The product is "+prod);

       }

       else if(op =='q'||op=='Q'){

           double quo = num1/num2;

           System.out.println("The Quotient is "+quo);

       }

   }

}

Explanation:

Answer:

The answer is explained below:

Explanation:

There are various modes in a computer that can be used as per the need.

The three modes provided in a compute are Standby, sleep and Hibernate.

Hibernate: Hibernate mode saves all the open windows and then shuts down the computer. It also saves more energy because the computer goes off completely. A computer needs more time to wake up from hibernation mode.

Sleep mode: Sleep mode means different things on different computers and operating systems. In windows it means standby. It the computer has been sleeping for  more than three hours then it hibernates automatically.  

Standby: Standby mode is for saving energy, it puts your computer into energy saving mode.

Shutdown: If you shut down a computer then all the software programs are closed and the computer turns off. The last program to be closed is operating system.

Fully awake mode means that the computer is running normally.

Answer:

The correct answer to the following question will be "True".

Explanation:

Therefore, the given statement is true.

Following are the responses to the given question:

Given instruction:

[tex]10101100100001010000000000010100[/tex]

For point A)

Sign extend and leap "shift left 2": The sign extend can be determined by taking the LSB 16 bits and expanding it to 32 bits.

Sign Extend

[tex]00000000000000000000000000010100[/tex]

A jump command "shift left 2" can be discovered by extracting the LSB 26 bit and shifting left by 2 bits.

Jump Shift Left 2

[tex]0010000101000000000001010000[/tex]

For point B)

ALU control units' inputs are as follows:

Its ALUop has two LSBs.

[tex]ALUop\\\\00[/tex]

The instructions are a 6 bit LSB instruction.

Instruction

[tex]010100[/tex]

For point C)

New PC address:

[tex]New PC Address \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ Path \\\\ PC+4 \ \ \ \ \ \ \ \ PC \to Add (PC +4) \to Branch MUX \to Jump MUX \to PC[/tex]

For point D)

Input values:

[tex]ALU \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ Add (PC+4) \ \ \ \ \ \ \ \ \ \ \ Add (Branch)[/tex]

[tex]In put \#1 \ \ \ In put \# 2 \ \ \ \ In put \# 1 \ \ \ In put \# 2 \ \ \ \ In put \# 1 \ \ \ In put \# 2\\\\[/tex]

[tex]-4 \ \ \ \ 20 \ \ \ \ \ \ PC \ \ \ \ 4 \ \ \ \ \ \ PC+4 \ \ \ \ 20*4[/tex]

For point E)

Calculating the inputs into the "Registers" units:

[tex]\text{Bits\ 25-21 specify the Read Reg#1} \\\\\text{Bits\ 20-16 specify the Read Reg#2}[/tex]

[tex]Read \ \ \ \ \ \ Read \\\\Reg\#1 \ \ \ \ \ \ Reg\#2 \ \ \ \ \ \ WriteReg \ \ \ \ \ \ WriteData\ \ \ \ \ \ RegWrite[/tex]

[tex]4 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 5 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ X \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ X \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 0[/tex]

Learn more:

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Answer:

The method definition to this question as follows:

Method definition:

//method GetData

       public static void GetData(out int order, out int amount)//defining method GetData  

       {  

           String val1, val2;  //defining String variable

           Write("Enter order number ");  //message

           val1 = ReadLine();  //input value by user

           Write("Enter quantity ");  //message

           val2 = ReadLine();  //input value by user

           order = Convert.ToInt32(val1);  //convert value in integer and hold in order variable  

           amount = Convert.ToInt32(val2);  //convert value in integer and hold in amount variable

       }

Explanation:

In the above C# method definition code a method  GetData() is defined in which the GetData() method is a static method, in this method parameter two integer argument "order and amount" is passed, inside a method, two string variable "val1 and val2" is declared that is used to take input by the user and convert the value into integer and store this value in the function parameter variable.

Final answer:

The question concerns fixing errors in a C# code snippet, specifically with the GetData method call, constant declaration, and output formatting. After corrections, the code should work as intended, calculating the total price of an order.

Explanation:

The question pertains to an issue in a C# program where there are mistakes that need to be resolved to ensure the program runs correctly. Particularly, the errors are in the GetData method call and in the calculation of the total cost.

Firstly, the GetData method is meant to accept an order number and quantity which are not being assigned before the method is called. Secondly, there is an error in referring to the constant PRICE_EACH when calculating the total, as the variable in the calculation is misspelled as 'PRICEEACH'. In addition, the syntax errors like incorrect use of semicolons and missing closing parentheses in the WriteLine methods need to be corrected.

Here is the corrected version of the Main method:

static void Main()
{
int orderNum = 0, quantity = 0;
double total;
const double PRICE_EACH = 3.99;
// Assume GetData correctly assigns values to orderNum and quantity
GetData(orderNum, quantity);
total = quantity * PRICE_EACH;
WriteLine("Order #{0}. Quantity ordered = {1}", orderNum, quantity);
WriteLine("Total is {0}", total.ToString("C"));
}

Note that the pseudo-code for the GetData method implies there is additional logic required to populate the orderNum and quantity variables which is not provided.

Answer:

360 msec

Explanation:

Data provided in the question:

Distance between the two hosts = 15,000 kilometer

= 1.5 × 10⁷ m    [ ∵ 1 km = 1000 m ]

Transmission rate, R = 5 Mbps = 5 × 10⁶ bits per second

Propagation speed over the link = 2.5 × 10⁸ meters/sec

Size of file to be sent = 1,500,000 bit

Now,

Propagation delay (dprop)

= Distance ÷ Speed

= ( 1.5 × 10⁷ ) ÷ (2.5 × 10⁸)

= 0.06 sec

= 60 msec           [ ∵ 1 sec = 1000 millisecond ]

Transmission delay (dtrans)

= Size of file ÷ Rate of transmission

= 1500000 ÷ ( 5 × 10⁶ )

= 0.3 sec

= 300 msec

Therefore,

Time required for transmitting the file

= Propagation delay + Transmission delay

= 60 msec + 300 msec

= 360 msec

Final answer:

The total time to send a 1,500,000-bit file from host A to host B over a distance of 15,000 kilometers at a rate of 5 Mbps is 0.36 seconds, which is the sum of the transmission time (0.3 seconds) and the propagation delay (0.06 seconds).

Explanation:

To calculate the time taken to send a 1,500,000-bit file from host A to host B separated by 15,000 kilometers at a rate of 5 Mbps, you need to consider both the transmission time and the propagation delay. The transmission time is the time required to push all the file's bits onto the wire and is given by the file size divided by the transmission rate. The second part is the propagation delay, which is the time it takes for the first bit to travel from the source to the destination.

The transmission time can be calculated as: Transmission Time = File size / Rate = 1,500,000 bits / (5 x 10⁶ bits/sec) = 0.3 seconds.

The propagation delay is the distance divided by the speed of light in the medium, which is: Propagation Delay = Distance / Speed of light = 15,000 km / (2.5 x 10⁸ m/s) = 0.06 seconds. Since the distance needs to be in meters, we convert it by multiplying by 1,000: Propagation Delay = 15,000,000 m / (2.5 x 10⁸ m/s) = 0.06 seconds.

The total time to send the file is the sum of transmission time and propagation delay: Total Time = Transmission Time + Propagation Delay = 0.3 s + 0.06 s = 0.36 seconds.