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prisoha [69]
2 years ago
10

You are working on a ticket for a user who cannot access the Internet from their PC. After troubleshooting, how do you confirm t

hat the reported problem is solved
Computers and Technology
1 answer:
ki77a [65]2 years ago
8 0
Could be anything?
Maybe you get an email from them on their PC, which means they have internet
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<span>He would click on the Test 1 column and press Sort, then click on the Test 2 column and press Sort. 
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Match each definition to the correct type of media.
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Answer:

1-2

2-3

3-1

Explanation:

I couldn't find a way to explain this.

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A technician wants to replace a failing power supply on a high-end gaming computer. which form factor should the technician be l
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A technician need to use the form factor or the  technician be looking for what we call EPS12V.

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The EPS connector is known to be used to supply power to the  motherboard CPU socket and the PCI is known to be one that functions to express connector.

Note that this is said to be a server standard (EPS12V) and it is also known to be one that can be said to be a desktop standard (ATX12V).

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1 year ago
what are some of the challenges that could arise from setting up a file management system on a computer
seropon [69]
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3 years ago
Suppose you have two arrays of ints, arr1 and arr2, each containing ints that are sorted in ascending order. Write a static meth
telo118 [61]
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.


4 0
3 years ago
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