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scZoUnD [109]
3 years ago
12

A mobile device you are troubleshooting is experiencing a sharp decrease in performance after an hour of operation. The user pow

ers the device off each evening, and when it is powered on, performance returns to normal. After an hour of operation, performance degrades. Which of the following is the best action to take to resolve this issue?a. Reset the device to factory defaults. b. Update the mobile device firmware. c. Close or disable all unused running apps. d. Adjust the brightness control.
Computers and Technology
1 answer:
HACTEHA [7]3 years ago
8 0

Answer:

c. Close or disable all unused running apps.

Explanation:

Background running apps uses the RAM (random access memory) of the device thereby leaving the memory very low which makes the performance to be slow. A mobile device with low memory e.g 512 megabyte will easily decrease in performance compared to a device with memory of 1 gigabyte.

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Which of the following is the best reason to use a macro?: *
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Answer:c)You want to automate a repetitive task to save time

Explanation: Macro is the program that is utilized when there is a requirement of a task or operation to be repeated in a automatic way.This helps in saving the time without commanding the same operation to take place manually.

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Suppose that you are asked to modify the Stack class to add a new operation max() that returns the current maximum of the stack
Advocard [28]

Answer:

Following are the code to the given points:

Explanation:

For point 8.1:

public String max()//defining a method max

{

   String maxVal=null;//defining a string variable that holds a value

   for(int x=0;x<n;x++)

   {

       if(maxVal==null || a[i].compareTo(maxVal)>0)//defining if blok to comare the value

       {

           maxVal=a[i];//holding value in maxVal variable

       }

   }

   return maxVal;//return maxVal variable value

}

For point 8.2:

public void push(String item)//defining a method push that accepts item value in a parameter

{

       a[n]=item;//defining an array to hold item value

       if(n==0 || item.compareTo(maxVals[n-1])>0)//use if to comare item value

       {

               maxVals[n]=item;//holding item value in maxVals variable

       }

       else

       {

               maxVals[n]=maxVals[n-1];//decreasing the maxVals value

       }

       n++;//incrementing n value

}

public String pop()//defining a method pop

{

       return a[--n];//use return value

}

public String max()//defining a method max

{

       return maxVals[n-1];//return max value

}

  • In the first point, the max method is declared that compares the string and returns its max value.
  • In the second point, the push, pop, and max method are declared that works with their respective names like insert, remove and find max and after that, they return its value.
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Answer:

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Explanation:

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