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) {
min = mylist.get(x);
}
}
return (min, max);
}
Answer:
int count =0;
for(int i=0;i<10;i++)
{
if(myArray[i]>=0)
{
count++;
}
}
cout<<"Number of positive integers is "<<count<<endl;
Explanation:
The above written loop is for counting positive integers in the myArray[].
For counting we have taken a count integer initialized with 0.On iterating over the array if the element is greater than or equal to 0 we consider it as positive and increasing the count.At the end printing the count.
Answer:
The correct answer is option A
Explanation:
Solution
Recall that:
From the question stated,the following segments of code that should be used in replacing the /* missing code */ so that the value 20 will be printed is given below:
Android a = new Android(x);
a.setServoCount(y);
System.out.println(a.getServoCount());
The right option to be used here is A.
