Answer: Preamble
Explanation:
Preamble is a 7- octet pattern of alternating the 0's and 1's and it allow devices in the network to easily synchronized the receiver clock. It also provide the bit level synchronization.
It basically allow the start of frame delimiter to provide the byte synchronization and assign a new frame. The bits are transmit orderly from left to right in the frame.
- The subnet mask would be a 32-bit integer which is formed by assigning the host bits to all 0's and the networking bits to so many 1's.
- In this method, the subnetting separates the IP address between host and network addresses.
- The subnet mask covers an IP address with its 32-bit number, thus the term "mask".
Given:
Network IP address 
Subnet numbers 
Calculating the borrow bits:
a)
Calculating the dotted decimal value for the subnet mask:
b)
The additional bits which will be needed to mask the subnet mask that is = 4.
Learn more: brainly.com/question/2254014
Answer:
The correct answer to the following question will be "False".
Explanation:
- The Open Systems Interconnection model is a conceptualization that describes and vastly simplifies a telecommunication or computer system's communication features, regardless of its inner structure and technologies underlying them.
- This model aims is to direct manufacturers and creators so that they would modularize with the digital communication devices and computer programs they build, and to promote a consistent context that defines the roles of a network or telecom device.
Therefore, the given statement is false.
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.
