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Answer:
True: In binary search algorithm, we follow the below steps sequentially:
Input: A sorted array B[1,2,...n] of n items and one item x to be searched.
Output: The index of x in B if exists in B, 0 otherwise.
- low=1
- high=n
- while( low < high )
- { mid=low + (high-low)/2
- if( B[mid]==x)
- {
- return(mid) //returns mid as the index of x
- }
- else
- {
- if( B[mid] < x) //takes only right half of the array
- {
- low=mid+1
- }
- else // takes only the left half of the array
- {
- high=mid-1
- }
- }
- }
- return( 0 )
Explanation:
For each iteration the line number 11 or line number 15 will be executed.
Both lines, cut the array size to half of it and takes as the input for next iteration.
All of the above. If that’s not an option if it’s multiple choice let me know. Everyone uses web browser/the internet no matter the career.
Answer:
Answered below
Explanation:
class TestScores {
double test1;
double test2;
double test 3;
public TestScores (double test1, double test2, double teat3){
this.test1= test1;
this.test2 = test2;
this.test3 = teat3;}
//mutator
public setTest1(double test1){
this.test1 = test1;
}
//accessor
public double getTest1(){
return test 1;
}
//Write same accessors and mutators for test2 and test3
public double getTestAverage(){
double sum = test1+test2+test3;
return sum / 3;
}
}
class TestRun{
public static void main (String [] args){
TestScores scores = new TestScores(50.5, 40.0, 80.7)
int average = scores.getTestAverage();
System.out.print(average);
}
