Answer:
The answer is below
Explanation:
The the sql command to list the total sales by customer , month , and product, with subtotals by customer and by month and a grand total for all product sales is:
SELECT S.CUS_CODE, T.TM_MONTH, S.P_CODE,
SUM(S.SALE_UNITS*S.SALE_PRICE) AS "TOTSALES"
FROM DWDAYSALESFACT AS S INNER JOIN DWTIME AS T ON S.TM_ID =
T.TM_ID
GROUP BY S.CUS_CODE,T.TM_MONTH,S.P_CODE WITH ROLLUP;
Answer:
The amount of data you use while talking on a phone is a form of data.
Explanation:
Markers are an often underutilized tool in the Premiere Pro CC toolbox. They are little colored tabs you can apply to both your timeline and your source clips to indicate important points in time with a color or note.
Answer:
Following are the code to this question:
file= open('book.txt') #open file
li= {} #define an empty list
for lines in file: # use for loop Split file data into words
d= lines.lower().strip(' !?').split() #define variable d that Add it to map
for val in d: #define loop to store data
if val not in li: #check value is not in list
li[val] = 0 #define list and assign value in 0
li[val] = li[val] + 1 #Sort the book data and add its value
m = sorted(li.items(),key = lambda x : -x[1]) #sorted value into the m variable
print(m[:10]) #print value
Output:
please find the attachment.
Explanation:
In the given python code first, we open the file "book.txt", in next line, an empty list is defined, that uses the for loop which can be described as follows:
- In the for loop is used, that reads the file data, and defines a variable "d", that stores the values into the map.
-
In the next line another loop is used, that check file values, if values are the same type so, it adds values and writes it.
- In the last line, m variable is used, that sorts the values and use the slicing to print its value.
Given 1234
i=1
user num=4#assume positive
while (user-num>=i);
print(i)
i+=1
#include <iostream>
using namespace std;
int main()
{int userNum=0;
int i=0;
userNum=4; ##assume positive
i=1;
while (i <=userNum){
cout<<i>>" ";
i=i+1;
cout <<endl;
return0;
}