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Naddika [18.5K]
3 years ago
7

Free wifi is typically offered on _ network

Computers and Technology
2 answers:
Marina CMI [18]3 years ago
8 0

Free Wi-Fi is typically offered on public network

Effectus [21]3 years ago
5 0

Answer:

It's public

!

Explanation:

You might be interested in
Exercise#3: Write a ch program that performs the following: Declare a two arrays called arrayA and arrayB that holds integer and
Serga [27]

Answer:

The code for this problem is completed here. Comments are added, go and learn how things work.

Explanation:

// Arrays.java

import java.util.Random;

import java.util.Scanner;

public class Arrays {

     // method to fill an array with random numbers between low and high

     static void fillArray(int array[], int low, int high) {

           Random random = new Random();

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

                 // generating a value between low and high inclusive, adding to

                 // array

                 array[i] = random.nextInt(high - low + 1) + low;

           }

     }

     // method to print elements in an array

     static void printArray(int array[]) {

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

                 // displaying elements separated by tab space

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

                 // printing a line break after every 5th element, or if this is the

                 // last element.

                 if ((i + 1) % 5 == 0 || i == array.length - 1) {

                       System.out.println();

                 }

           }

     }

     // method to count the number of elements in array greater than value

     static int count(int array[], int value) {

           // initializing count to 0

           int c = 0;

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

                 if (array[i] > value) {

                       // incrementing c

                       c++;

                 }

           }

           return c;

     }

     // returns true if two arrays are same, else false

     static boolean isSame(int array1[], int array2[]) {

           if (array1.length != array2.length) {

                 // lengths mismatch

                 return false;

           }

           // checking if elements are same in same order

           for (int i = 0; i < array1.length; i++) {

                 if (array1[i] != array2[i]) {

                       // mismatch found

                       return false;

                 }

           }

           // same

           return true;

     }

     // returns the average of elements in an array

     static double findAverage(int array[]) {

           double sum = 0;

           // summing values

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

                 sum += array[i];

           }

           // finding average

           double avg = (double) sum / array.length;

           return avg;

     }

     // method to count the number of values greater than average of an array

     static int aboveAverage(int array[]) {

           // finding average of passed array using findAverage method

           double avg = findAverage(array);

           // finding number of elements in array>avg using count method

           int c = count(array, (int) avg);

           return c;

     }

     public static void main(String[] args) {

           // setting up a Scanner

           Scanner scanner = new Scanner(System.in);

           // prompting and reading size of arrayA

           System.out.print("Enter size of arrayA: ");

           int n1 = scanner.nextInt();

           // initializing arrayA

           int arrayA[] = new int[n1];

           // doing the same for arrayB

           System.out.print("Enter size of arrayB: ");

           int n2 = scanner.nextInt();

           int arrayB[] = new int[n2];

           // asking and reading low and high values

           System.out.print("Enter low and high values for random numbers: ");

           int low = scanner.nextInt();

           int high = scanner.nextInt();

           // filling both arrays

           fillArray(arrayA, low, high);

           fillArray(arrayB, low, high);

           // printing both arrays

           System.out.println("arrayA:");

           printArray(arrayA);

           System.out.println("arrayB:");

           printArray(arrayB);

           // reading an integer

           System.out.print("Enter an integer value: ");

           int value = scanner.nextInt();

           // displaying the number of elements in each array greater than above

           // value

           System.out.println("The number of elements greater than " + value

                       + " in arrayA: " + count(arrayA, value));

           System.out.println("The number of elements greater than " + value

                       + " in arrayB: " + count(arrayB, value));

           System.out.println("Both arrays are same: " + isSame(arrayA, arrayB));

           // finding and displaying average of arrayA

           double avgA = findAverage(arrayA);

           System.out.println("Average of all values in arrayA: " + avgA);

           // finding and displaying number of elements in arrayB greater than the

           // average of arrayB

           System.out.println("The number of elements greater than "

                       + "the average of arrayB" + " in arrayB: "

                       + aboveAverage(arrayB));

     }

}

6 0
3 years ago
The support group at Universal Containers wants agents to capture different information for product support and inquiry cases. I
Cloud [144]

Answer:

Record types.

Support processes.

Page layouts.

3 0
3 years ago
What explains the discrepancy between the number of bytes you can
sergejj [24]

Answer:

This is because, the advertisers report the storage space in decimal while the computer reads the storage space in binary.

Explanation:

The advertisers report the storage space in decimal or base 10 because humans count in decimal, whereas the computer reports the storage space in binary or base 2.

Since the computer storage is in bytes and 8 bits equal 1 byte, It is easier to write a storage space of 1 kB as 1000 B but it is actually supposed to be 1024 B(2¹⁰). So, there is a discrepancy of 1024 B - 1000 B = 24 B.

As  we go higher, the discrepancy increases. For example, 1 MB is advertised as 1000 kB = 1000000 B but is actually supposed to be 1024 kB = 1024 × 1024 B = 1048576 B. So, there is a discrepancy of 1048576 B - 1000000 B = 48576 B.

So, the actual number of bytes on the storage device is actually less than that reported due to the different number systems in which they are reported in. This discrepancy is less in memory cards or flash drives though in which the stated value of storage capacity might be the actual storage size.

Note that the base 10 or decimal system was chosen by advertiser since this is what consumers understand.

8 0
2 years ago
My home PC has IP address 192.168.1.22 and connects to the Internet through a NAT router. Assume I am downloading a web page fro
alekssr [168]

Answer and Explanation:

There are three sections to the inquiry.  

1. what occurs inside the PC or host when a bundle is produced by an application.  

2.how it goes from one host that is a source to another host that is a goal with numerous switches sitting in the middle.  

3.what goes inside goal have when it gets a parcel on the system.  

how about we examine individually  

Handling bundle at source machine:  

The application creates a bundle to be sent on the system and sends it to the layer beneath.  

The following layer is known as a vehicle layer which oversees start to finish correspondence between two machines. The convention utilized can be TCP or UDP. What is the contrast between these two conventions is another subject by and large.  

When the parcel is framed at the vehicle layer, it is sent to the system layer which includes source and goal IP in the bundle. A most significant field which is included at IP or system layer is Time To Live (TTL) which is utilized by middle switches/changes to choose if a parcel should be sent or not. (How goal IP is found?)  

After the system layer, the parcel arrives at the information connection or MAC layer, where the source and goal MAC address of machines are included. We will perceive how these fields change between each two neighbors. (How goal MAC is found?)  

The information interface layer pushes this bundle to the physical layer where it is sent as a flood of "0" and "1" on the physical medium accessible.  

Preparing a bundle at switch:  

Steering  

There are three cases which may happen when the switch investigates the directing table for goal IP  

1. On the off chance that there is a section relating to goal IP, we get the interface name the parcel ought to be sent to.  

2. On the off chance that there is no immediate passage, at that point IP is changed over into organize IP utilizing the veil and afterward checked once more. It ought to be noticed that the longest prefix match to locate the best sending interface.  

3. In the case of nothing matches, at that point the switch just advances it to the default goal designed.  

Sending  

Epitome  

Preparing parcel at the goal have  

The parcel is gotten at arrange card, physical layer, which produces a hinder to CPU and CPU peruses bundle in,  

At the information connect layer, goal MAC address is verified whether the parcel is bound to this machine, If indeed, the bundle is sent up to the system layer.  

At the IP layer, parcel approval like checksum confirmation, and so on is done and afterward gave to the applicable vehicle layer.  

Transport later at that point gives it to the fitting port with the goal that it arrives at the right application.

5 0
3 years ago
Need Help ASAP!
jonny [76]

Answer:

Ask about next steps and the best way to communicate.

Send a follow-up email within 24 hours.

Explanation:

3 0
3 years ago
Read 2 more answers
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