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REY [17]
3 years ago
5

Your internet is out, and devices connected by WiFi and ethernet will not connect. You check several websites, and all your pack

ets fail. You then ping a website with an IP address, and to your surprise, it's a success. Which would you do next? a) Open the DNS settings on your router and try using a different DNS b) Open the DHCP settings on your router and try using a different pool of private IP addresses c) Open the Port Forwarding settings on your router and try turning on UPnP d) Open the Network settings on your router and try adjusting the default gateway e) Open the Network settings on your router and try turning off your firewall
Computers and Technology
1 answer:
tiny-mole [99]3 years ago
6 0

Answer:

a) Open the DNS settings on your router and try using a different DNS.

Explanation:

The DNS or domain name service is a protocol that majorly assigns a URL string to an IP address. This is because the IP address as a number is easily forgotten.

In the question above, the operator could not access the websites with the URL name but can confirm connectivity to the site with its IP address. This means that the IP address can be used to access the websites directly, so, the DNS protocol is either not configured or is down.

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Modify the code below to do the following:
Kitty [74]

Answer:

The code is appropriately given below with comments for better understanding

Explanation:

#include <linux/init.h>

#include <linux/module.h>

#include <linux/kernel.h>

#include <linux/hash.h>

#include <linux/gcd.h>

#include <asm/param.h>

#include <linux/jiffies.h>

/* This function is called when the module is loaded. */

static int simple_init(void)

{

  printk(KERN_INFO "Loading Module\n");

  printk(KERN_INFO "These are the HZ: %d\n", HZ);

  printk(KERN_INFO "These are the jiffies: %lu\n", jiffies);

  printk(KERN_INFO "Golden Ratio is: %lu\n", GOLDEN_RATIO_PRIME);  

  return 0;

}

/* This function is called when the module is removed. */

static void simple_exit(void) {

  printk(KERN_INFO "Removing Module");

  unsigned long a = gcd(3300, 24);

  printk(KERN_INFO "Greatest Common Denominator between 3,300 and 24 is: %lu\n", a);

  printk(KERN_INFO "These are the jiffies: %lu\n", jiffies);

}

/* Macros for registering module entry and exit points. */

module_init( simple_init );

module_exit( simple_exit );

MODULE_LICENSE("GPL");

MODULE_DESCRIPTION("Simple Module");

MODULE_AUTHOR("SGG");

7 0
3 years ago
Which implementation has the worst time complexity?
Nata [24]

Answer:

China and Michigan hskbvd

8 0
3 years ago
Create an application for a library and name it FineForOverdueBooks. TheMain() method asks the user to input the number of books
andrew11 [14]

Answer:

//The Scanner class is imported which allow the program to receive user input

import java.util.Scanner;

//Class Solution is defined to hold problem solution

public class Solution {

   // The main method which signify the begining of program execution

   public static void main(String args[]) {

       // Scanner object 'scan' is defined to receive input from user keyboard

       Scanner scan = new Scanner(System.in);

       // A prompt is display asking the user to enter number of books

       System.out.println("Please enter number of books: ");

       // the user response is assigned to numberOfBook

       int numberOfBook = scan.nextInt();

       // A prompt is displayed asking the user to enter the number of days over due

       System.out.println("Please enter number of days over due: ");

       // the user response is assigned to numberOfDaysOverDue

       int numberOfDaysOverDue = scan.nextInt();

       //displayFine method is called with numberOfBook and numberOfDaysOverDue as arguments

       displayFine(numberOfBook, numberOfDaysOverDue);

   

   }

   

   //displayFine method is declared having two parameters

   public static void displayFine(int bookNumber, int daysOverDue){

       // fine for first seven days is 10cent which is converted to $0.10

       double firstSevenDay = 0.10;

       // fine for more than seven days is 20cent which is converted to $0.20

       double moreThanSevenDay = 0.20;

       // the fine to be paid is declared

       double fine = 0;

       // fine is calculated in the following block

       if(daysOverDue <= 7){

           //the fine if the over due days is less than or equal 7

           fine = bookNumber * daysOverDue * firstSevenDay;

       } else{

           // the extra days on top of the first seven days is calculated and assigned to extraDays

           int extraDays = daysOverDue - 7;

           //fine for first seven days is calculated

           double fineFirstSevenDays = bookNumber * 7 * firstSevenDay;

           // fine for the extradays is calculated

           double fineMoreThanSevenDays = bookNumber * extraDays * moreThanSevenDay;

           // the total fine is calculated by adding fine for first seven days and the extra days

           fine = fineFirstSevenDays + fineMoreThanSevenDays;

       }

       // The total fine is displayed to the user in a nice format.

       System.out.printf("The fine for " + bookNumber + " book(s) for " + daysOverDue + " day(s) is: $%.02f", fine);

   }

}

Explanation:

The program first import Scanner class to allow the program receive user input. Then the class Solution was defined and the main method was declared. In the main method, user is asked for number of books and days over due which are assigned to numberOfBook and numberOfDaysOverDue. The two variable are passed as arguments to the displayFine method.

Next, the displayFine method was defined and the fine for the first seven days is calculated first if the due days is less than or equal seven. Else, the fine is calculated for the first seven days and then the extra days.

The fine is finally displayed to the user.

4 0
3 years ago
I made a binary sentence if you answer you get 75 points heres the code: 01101001 01100110 00100000 01111001 01101111 01110101 0
butalik [34]

01001001 00100000 01100001 01100011 01110100 01110101 01100001 01101100 01101100 01111001 00100000 01110101 01110011 01100101 01100100 00100000 01100001 00100000 01000010 01101001 01101110 01100001 01110010 01111001 00100000 01100011 01101111 01100100 01100101 00100000 01000011 01101111 01101110 01110110 01100101 01110010 01110100 01100101 01110010 00100000 01101100 01101101 01100001 01101111

6 0
3 years ago
Read 2 more answers
let m be a positive integer with n bit binary representation an-1 an-2 ... a1a0 with an-1=1 what are the smallest and largest va
Ahat [919]

Answer:

Explanation:

From the given information:

a_{n-1} , a_{n-2}...a_o in binary is:

a_{n-1}\times 2^{n-1}  + a_{n-2}}\times 2^{n-2}+ ...+a_o

So, the largest number posses all a_{n-1} , a_{n-2}...a_o  nonzero, however, the smallest number has a_{n-2} , a_{n-3}...a_o all zero.

∴

The largest = 11111. . .1 in n times and the smallest = 1000. . .0 in n -1 times

i.e.

(11111111...1)_2 = ( 1 \times 2^{n-1} + 1\times 2^{n-2} + ... + 1 )_{10}

= \dfrac{1(2^n-1)}{2-1}

\mathbf{=2^n -1}

(1000...0)_2 = (1 \times 2^{n-1} + 0 \times 2^{n-2} + 0 \times 2^{n-3} + ... + 0)_{10}

\mathbf {= 2 ^{n-1}}

Hence, the smallest value is \mathbf{2^{n-1}} and the largest value is \mathbf{2^{n}-1}

3 0
2 years ago
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