Your question wasn't very clear, but I think I understand what you want. Additionally, you should really state what language you're working with. Here it is in C#, and shouldn't be too much of a hassle to translate in to other languages.
Console.Write("Enter payment: ");
float payment;
if (float.TryParse(Console.ReadLine(), out payment))
Console.WriteLine((Math.Floor(payment * 100) / 100) * 0.15, + " at 15% tip.");
else
Console.WriteLine("Invalid input.");
Answer
calculator calc;
Explanation
An object is an instance of a class. And a class is what defines or describes the behavior or the state of the object of its type. When a class is defined no memory is allocated until when an object is created memory is allocated.
Answer:
code = 010100000001101000101
Explanation:
Steps:
The inequality yields 
, where M = 16. Therefore,
The second step will be to arrange the data bits and check the bits. This will be as follows:
Bit position number Check bits Data Bits
21 10101
20 10100
The bits are checked up to bit position 1
Thus, the code is 010100000001101000101
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.
Answer:
C. 220.100.100.45 to 220.100.100.46
Explanation:
The Classless IP subnetting of 220.100.100.0 begins from the fourth octet of the IP address. To get 45 subnet mask, it uses 6 bits from the fourth octet, which approximately give 64 subnets, while the remaining 2 bits are used for host IP addressing.
The useable host IP addresses are gotten from the formula '
-2', with n=2 bits.
useable host IP addresses = 2^2 - 2 = 2 addresses per subnet.
While the 12th subnet is 12 x 2^2 = 44.
This means that the 12th subnet mask starts with 220.100.100.44 (as the network address) and ends with 220.100.100.47 as broadcast IP address, while '.45' and '.46' are the assignable addresses of the subnet.
