You just need a little bit of math to solve this. Add up the items and see which one would be equal to $55.00 or less.
A = $59.90 so this is not the answer
B = $60.94 so this is not the answer
C = $55.99 so this is not the answer
D = $50.97
D is the correct answer.
First three bytes of the ethernet address exists assigned to vendors to identify the equipment.
<h3>What is Ethernet address?</h3>
A media access control address stands for a unique identifier allocated to a network interface controller for usage as a network address in communications within a network segment. This use exists as standard in most IEEE 802 networking technologies, including Ethernet, Wi-Fi, and Bluetooth.
This address exists as the identifying mark that allows a networked computer to communicate with the Internet. The ethernet address exists how we can associate an “IP address” to an individual computer — without this– number, there's no form to access any servers, websites, email, etc.
The first three bytes (pairs of hexadecimal characters) of any unicast address include that vendor address component of the MAC address. The staying three bytes carry the serial number of that vendor's interface card. Dell, Inc. Apple, Inc.
Hence, first three bytes of the ethernet address exists assigned to vendors to identify the equipment.
To learn more about Ethernet address refer to:
brainly.com/question/7284219
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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:
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Explanation:
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