The argument for the function would be answer "D".
Answer:
c.
Explanation:
I believe that in this scenario, the best option for this data would be a hash table using open addressing with 1,800 entries. Hash tables consume more memory than lists but it makes up for it with much faster response time speeds. This is because hash tables work on a key:value system therefore, the license plate can easily be grabbed from the database extremely quickly by just plugging in the plate number. Doing so will retrieve all of the saved information from that license plate. That is why hash tables have a constant time complexity of O(1)
Hey!
I believe the answer to 1. Should be A. Rule of Thirds.
Here you go,
Import java.util.scanner
public class SumOfMax {
public static double findMax(double num1, double num2) {
double maxVal = 0.0;
// Note: if-else statements need not be understood to
// complete this activity
if (num1 > num2) { // if num1 is greater than num2,
maxVal = num1; // then num1 is the maxVal.
}
else { // Otherwise,
maxVal = num2; // num2 is the maxVal.
}
return maxVal;
}
public static void main(String[] args) {
double numA = 5.0;
double numB = 10.0;
double numY = 3.0;
double numZ = 7.0;
double maxSum = 0.0;
/* Your solution goes here */
maxSum = findMax(numA, numB); // first call of findMax
maxSum = maxSum + findMax(numY, numZ); // second call
System.out.print("maxSum is: " + maxSum);
return;
}
}
/*
Output:
maxSum is: 17.0
*/