Answer:
The complete program is:
import java.util.Scanner;
class Main
{
public static void main (String [] args) {
int numObjects;
Scanner scnr = new Scanner(System.in);
numObjects = scnr.nextInt(); // Program will be tested with values: 15, 40.
System.out.println(numObjects);
}
}
Explanation:
The program is as mentioned above, And we can check with inputs 15, and 40. And it was found that correct answer is obtained on running the program.
Answer:
Option 2: a type of mini chart that users can insert into a worksheet
Explanation:
A graph or chart is used to show trends and changes in a dataset.
When we have to show trends or changes in MS Excel, there are a lot of options available. One of them is Sparklines. Sparklines are like mini charts that can be used to show data graphically.
Hence,
The correct answer is:
Option 2: a type of mini chart that users can insert into a worksheet
Answer:
The WPA2 shared key is incorrect is the correct answer.
Explanation:
The WPA2 shared key is incorrect because when the technician installs a new wireless thermostat for the purpose to control the temperature of the meeting room. Then, admin determines that the thermostat is not connecting to the control system through the internet and the admin authenticate that its parameter of receiving thermostat is associated with the AP. So, that's why the following option is correct.
colorfulness and depending on your teacher pick colors that will go together on the project
Since both arrays are already sorted, that means that the first int of one of the arrays will be smaller than all the ints that come after it in the same array. We also know that if the first int of arr1 is smaller than the first int of arr2, then by the same logic, the first int of arr1 is smaller than all the ints in arr2 since arr2 is also sorted.
public static int[] merge(int[] arr1, int[] arr2) {
int i = 0; //current index of arr1
int j = 0; //current index of arr2
int[] result = new int[arr1.length+arr2.length]
while(i < arr1.length && j < arr2.length) {
result[i+j] = Math.min(arr1[i], arr2[j]);
if(arr1[i] < arr2[j]) {
i++;
} else {
j++;
}
}
boolean isArr1 = i+1 < arr1.length;
for(int index = isArr1 ? i : j; index < isArr1 ? arr1.length : arr2.length; index++) {
result[i+j+index] = isArr1 ? arr1[index] : arr2[index]
}
return result;
}
So this implementation is kind of confusing, but it's the first way I thought to do it so I ran with it. There is probably an easier way, but that's the beauty of programming.
A quick explanation:
We first loop through the arrays comparing the first elements of each array, adding whichever is the smallest to the result array. Each time we do so, we increment the index value (i or j) for the array that had the smaller number. Now the next time we are comparing the NEXT element in that array to the PREVIOUS element of the other array. We do this until we reach the end of either arr1 or arr2 so that we don't get an out of bounds exception.
The second step in our method is to tack on the remaining integers to the resulting array. We need to do this because when we reach the end of one array, there will still be at least one more integer in the other array. The boolean isArr1 is telling us whether arr1 is the array with leftovers. If so, we loop through the remaining indices of arr1 and add them to the result. Otherwise, we do the same for arr2. All of this is done using ternary operations to determine which array to use, but if we wanted to we could split the code into two for loops using an if statement.
