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Rudiy27
1 year ago
7

A mobile device user has tried to install a new app numerous times without success. She has closed all unused apps, disabled liv

e wallpapers, and restarted the device in an attempt to get the app to install.
Which of the following steps should the user take next?
A. Back up all user data and perform a factory reset.
B. Check for sufficient available storage space.
C. Disable full device encryption.
D. Generate a new public key.
Computers and Technology
1 answer:
elena-14-01-66 [18.8K]1 year ago
6 0

The correct answer is B. Check for sufficient available storage space.  

Explanation:

One of the factors that can cause it is not possible to install a new app is insufficient storage as space in the mobile device is needed to save the app. This can be easily solved either by deleting apps and other content or expanding the memory available through a Micro SD card. Also, before attempting any major procedure such as a complete reset it is important to check the storage space availability. In this context, the user should check for sufficient available storage space (option B) because this might be the cause of the issue she is experiencing; also, this is a basic step before considering others such as performing a factory reset.

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1 make sure only you know the password

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Write a static generic method PairUtil.minmax that computes the minimum and maximum elements of an array of type T and returns a
Gnesinka [82]

Answer:

Explanation:

The following code is written in Java. It is hard to fully create the code without the rest of the needed code including the T class and the Measurable interface. Regardless the following code can be implemented if you have that code available.

 public static T minmax(ArrayList<T> mylist) {

       T min = new T();

       T max = new T();

       for (int x = 0; x < mylist.size(); x++) {

           if (mylist.get(x) > max) {

               max = mylist.get(x);

           } else if (mylist.get(x) < min) {

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5 0
1 year ago
Create a program named Auction that allows a user to enter an amount bid on an online auction item. Include three overloaded met
Olin [163]

Answer:

Explanation:

The following code is written in Java and creates the overloaded methods as requested. It has been tested and is working without bugs. The test cases are shown in the first red square within the main method and the output results are shown in the bottom red square...

class Auction {

   public static void main(String[] args) {

       bid(10);

       bid(10.00);

       bid("10 dollars");

       bid("$10");

       bid("150 bills");

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   public static void bid(int bid) {

       if(bid >= 10) {

           System.out.println("Bid Accepted");

       } else {

           System.out.println("Bid not high enough");

       }

   }

   public static void bid(double bid) {

       if(bid >= 10) {

           System.out.println("Bid Accepted");

       } else {

           System.out.println("Bid not high enough");

       }

   }

   public static void bid(String bid) {

       if (bid.charAt(0) == '$') {

           if (Integer.parseInt(bid.substring(1, bid.length())) > 0) {

               System.out.println("Bid Accepted");

               return;

           } else {

               System.out.println("Bid not in correct Format");

               return;

           }

       }

       int dollarStartingPoint = 0;

       for (int x = 0; x < bid.length(); x++) {

           if (bid.charAt(x) == 'd') {

               if (bid.substring(x, x + 7).equals("dollars")) {

                   dollarStartingPoint = x;

               } else {

                   break;

               }

           }

       }

       if (dollarStartingPoint > 1) {

           if (Integer.parseInt(bid.substring(0, dollarStartingPoint-1)) > 0) {

               System.out.println("Bid Accepted");

               return;

           } else {

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               return;

           }

       } else {

           System.out.println("Bid not in correct format");

           return;

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1 year ago
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Answer:

Following are the code in the Java Programming Language:

try{  //try block.

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}

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}

Explanation:

In the following code, we set two blocks in Java Programming Language of the exception handling which is try block or catch block.

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1 year ago
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Vinil7 [7]

Answer:

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3 0
1 year ago
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