Answer:
The correct option is D = regionNumber
Explanation:
In this scenario we want to know customers who owe more than $1000 each, in each of 12 sales regions. And the customer data variables include name, zip-code, balanceDue and regionNumber; based on the customer data variables names and zip-code will not really affect our output. It is based on balanceDue that we increment the number of customer owing in a particular region (regionNumber).
Therefore, we would add 1 to an array element whose subscript would be represented by regionNumber since we are interested to know the number of customer owing in each of the 12 sales regions.
Answer:
D. 25x24x23x22x21.....3x2x1.
Explanation:
The ceasar cipher is a encryption technique, that uses a combination of information or keys to encrypt an information.
The 25 shift ceasar cipher has 25 different combinations, so the number of probable random cipher substitution is 25!,
That is = 25 × 24×23× 22×..... ×2×1.
What is the problem what am I supposed to do?
Answer:
You could search up almost anything you want
they have a lot of storage
you can find good information
they help you with a lot of things
Explanation:
also how they can process information in seconds
Answer:
If all the character pairs match after processing both strings, one string in stack and the other in queue, then this means one string is the reverse of the other.
Explanation:
Lets take an example of two strings abc and cba which are reverse of each other.
string1 = abc
string2 = cba
Now push the characters of string1 in stack. Stack is a LIFO (last in first out) data structure which means the character pushed in the last in stack is popped first.
Push abc each character on a stack in the following order.
c
b
a
Now add each character of string2 in queue. Queue is a FIFO (first in first out) data structure which means the character inserted first is removed first.
Insert cba each character on a stack in the following order.
a b c
First c is added to queue then b and then a.
Now lets pop one character from the stack and remove one character from queue and compare each pair of characters of both the strings to each other.
First from stack c is popped as per LIFO and c is removed from queue as per FIFO. Then these two characters are compared. They both match
c=c. Next b is popped from stack and b is removed from queue and these characters match too. At the end a is popped from the stack and a is removed from queue and they both are compared. They too match which shows that string1 and string2 which are reverse of each other are matched.
