Answer:
Following are the code in the C Programming Language.
//set integer datatype variable
int score;
//check condition is the score is in the range of 0 to 100
if(score > 0 && score < 100){
//print if condition is true
printf("Valid test scores");
}else{
//otherwise print the following string.
printf("test scores are Invalid");
}
Explanation:
<u>Following are the description of the code.</u>
In the following code that is written in the C Programming Language.
- Set an integer data type variable i.e., score.
- Then, set the if conditional statement to check the condition is the variable "score" is greater than 0 and less the 100.
- If the following statement is true then print "Valid test scores".
- Otherwise, it print "test scores are Invalid".
Answer:
b. Combatant Command
Explanation:
The Combatant commands are commands which are primarily authorized by the Secretary of Defense with approval from the President. They are then executed by the Combatant Commanders (CCDR's), who receive information from the higher authorities through the Chairman of the Joint Chiefs of Staff. The combatant commanders head the combatant commands and are usually four-star Generals. The combatant commands have a broad and continuing purpose.
They ensure synergy for the command and control of the United States military forces. They are formed on the basis of geography or function. The United States Defense Department has eleven (11) combatant commands which serve either a geographic or functional purpose.
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.
Answer:
Displayport uses a lower voltage than DVI and HDMI.
Explanation:
DisplayPort are cables and connector used to stream video, audio, usb or other kinds of data to the monitor screen of a computer. As defined, it can send video and audio signals on the same cable, over a long distance at a high speed.
The voltage requirement for DisplayPort is 3.3 volts while HDMI and DVI uses 5 volts.
