Answer:
Q. When evaluating mining results, does data mining and evaluating become an intuitive process?
“Data mining” is technically associated with analysis of very large data sets, to appreciate patterns,and attempt to look at cause to effect relationships. Most of the data is quantitative in nature, and many of the tools relate to that of analysis of numerical /quantitative data.
Explanation:
One of the rules of the game of quantitative analysis, is to allow the “data to do the talking”, Your intuition CANNOT replace the results of quantitative analysis: whether through data mining or humble pencil and paper calculation on the back of an envelope.
With experience you may perhaps see a lot of “counter intuitive results”, Where the final outcome does not make “common sense” - but that is what the data is saying.
It is useful NOT to allow emotions, opinions, to come in the way of any sort of quantitative data analysis.
If you can specify what sort of data you are analyzing a more precise answer can be attempted.
Explanation:
Kendall should report the email as scam and delete email instead of forwarding it. She should also run her virus protection software as these kind of emails on the Internet are mostly Fraud and can contain virus so the user should avoid them.
Answer:
sqrt(area)
Explanation:
- Here sqrt is a short form of square root.
- area is a variable name having he area of square.
- When the function is called, it will give the square root of the value stored in variable area.
As we have to find the length of the diagonal, we must knew that as all sides of the square are same in length so are the diagonals. This means that each of the four sides of the square and it diagonals are equal in length.
So a square has: length=breadth=diagonal
As Area= length*breadth
√area = length (as length = diagonal length)
So √area = diagonal length
Answer:
Please find the answer below
Explanation:
// Online C compiler to run C program online
#include <stdio.h>
int main() {
// Write C code here
//printf("Hello world");
int userNum;
int i;
int j;
scanf("%d", &userNum);
/* Your solution goes here */
for(i = 0; i<=userNum; i++){
for(j = 0; j <= i; j++){
printf(" ");
}
printf("%d\n", i);
}
return 0;
}
