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tigry1 [53]
1 year ago
8

William has an internet connection that does not allow him to make calls when connected to the Internet. What internet service c

an he use over his existing telephone line that will enable him to make calls and browse the Internet at the same time?
A. Dial-up
B. ISDN
C. Cable internet access
D. Satellite internet access
Computers and Technology
2 answers:
saul85 [17]1 year ago
6 0
D. Satellite internet access
MArishka [77]1 year ago
6 0

Answer:

<u><em>Dial-up, 95% sure.</em></u>

Explanation:

It was kinda hard at first because ISDN & DIAL-UP are some what the same. After looking in to it though...<em> it seems like ISDN are not really known for there Internet and telephone services.</em> <u><em>However DIAL-UP  is pretty good with both, and there fast! </em></u>So, I personally choose DIAL-UP.

<u><em>I hope this helps! Stay awesome. </em></u>

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Which data type change will require the app builder to perform the additional steps in order to retain existing functionalities?
Readme [11.4K]

Answer:

Option D is the correct option.

Explanation:

The following option is correct because the lead alteration from number to text and the number to text types of data will reconstruct the important number of the app builder to accomplish those steps which is extra to continue to have that functionality which is in the existence. That's why the app builder has to be reconstructed the types of data for the custom fields.

7 0
1 year ago
Write a program that has the following String variables: firstName, middleName, and lastName. Initialize these with your first,
Licemer1 [7]

Answer:

The programming language is not stated; However, I'll answer your question using C++ programming language.

Comments are used for explanatory purpose

Program starts here

#include<iostream>

#include <string>

using namespace std;

int main()

{

//Declare Variables

string firstName, middleName, lastName;

char firstInitial, middleInitial, lastInitial;

/*Initialize firstName, middleName and lastName (Replace values with your details)*/

firstName = "First Name";

middleName = "Middle Name";

lastName = "Last Name";

 

// Get Initials

firstInitial = firstName.at(0);

middleInitial = middleName.at(0);

lastInitial = lastName.at(0);

 

//Print Results

cout<<"Lastname: "<<lastName<<endl;

cout<<"Firstname: "<<firstName<<endl;

cout<<"Middlename: "<<middleName<<endl;

cout<<"Last Initial: "<<lastInitial<<endl;

cout<<"First Initial: "<<firstInitial<<endl;

cout<<"Middle Initial: "<<middleInitial<<endl;

return 0;

}

3 0
2 years ago
Produce a list named prime_truths which contains True for prime numbers and False for nonprime numbers in the range [2,100]. We
LenKa [72]

Answer:

  1. def is_prime(n):
  2.    for i in range(2, n):
  3.        if(n % i == 0):
  4.            return False  
  5.    return True  
  6. prime_truths = [is_prime(x) for x in range(2,101)]
  7. print(prime_truths)

Explanation:

The solution code is written in Python 3.

Presume there is a given function is_prime (Line 1 - 5) which will return True if the n is a prime number and return False if n is not prime.

Next, we can use the list comprehension to generate a list of True and False based on the prime status (Line 7). To do so, we use is_prime function as the expression in the comprehension list and use for loop to traverse through the number from 2 to 100. The every loop, one value x will be passed to is_prime and the function will return either true or false and add the result to prime_truth list.

After completion of loop within the comprehension list, we can print the generated prime_truths list (Line 8).

3 0
2 years ago
Develop an sec (single error correction) code for a 16-bit data word. generate the code for the data word 0101000000111001. show
Kipish [7]

Answer:

code = 010100000001101000101

Explanation:

Steps:

The inequality yields 2^{k} - 1 > = M+K, where M = 16. Therefore,

The second step will be to arrange the data bits and check the bits. This will be as follows:

Bit position              number              Check bits            Data Bits

21                                   10101

20                                  10100

The bits are checked up to bit position 1

Thus, the code is 010100000001101000101

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