Answer:
Theoretically Yes
Explanation:
The data given is linearly separable. So, the subset of the data will also be linearly separable. And it will pass for all training dataset. However, you should definitely never expect such thing In any real-life problem because the data is noisy, for a bazilion of reasons, so no model is guaranteed to perform perfectly.
Answer:
sendObject(John_Doe);
Explanation:
The above code has been written in Java.
Since the calling class is the same that declares it, to invoke the method, simply call its name with its argument(s) in a pair parentheses. The name of the method is "sendObject" and its argument is a reference to an object of type Customer saved in a variable called "John_Doe". This can be written as follows:
sendObject(John_Doe);
Hope this helps!
My idea on creating a new and fresh Gaming experience around viewing occasions is that one should incorporate emotions into games that people around around viewing occasions can feel.
<h3>What is a gaming experience?</h3>
Gaming is known to be that experience or one can say the origin of identity and also the community that is made for gamers.
Note that the experience is one that goes far more than the game itself, and as such, My idea on creating a new and fresh Gaming experience around viewing occasions is that one should incorporate emotions into games that people around around viewing occasions can feel.
Learn more about Gaming experience from
brainly.com/question/27355039
#SPJ1
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:
The motherboard must be configured to accept a particular type of memory.
Explanation:
The motherboard must be configured to accept a particular type of memory, especially when dealing with multi-channel memory. On motherboard, different modes can be configured to run simultaneously . For motherboard to take advantage of the triple memory channels on the board. You can also check the manual that came with the motherboard to know how to configure it.
