for such experiment, you do it with care and to acquire and determine to put experience in it
Explanation:
because without you been or using experience the experiment will not correct
Answer:
static int checkSymbol(char ch)
{
switch (ch)
{
case '+':
case '-':
return 1;
case '*':
case '/':
return 2;
case '^':
return 3;
}
return -1;
}
static String convertInfixToPostfix(String expression)
{
String calculation = new String("");
Stack<Character> operands = new Stack<>();
Stack<Character> operators = new Stack<>();
for (int i = 0; i<expression.length(); ++i)
{
char c = expression.charAt(i);
if (Character.isLetterOrDigit(c))
operands.push(c);
else if (c == '(')
operators.push(c);
else if (c == ')')
{
while (!operators.isEmpty() && operators.peek() != '(')
operands.push(operators.pop());
if (!operators.isEmpty() && operators.peek() != '(')
return NULL;
else
operators.pop();
}
else
{
while (!operators.isEmpty() && checkSymbol(c) <= checkSymbol(operators.peek()))
operands.push(operators.pop());
operators.push(c);
}
}
while (!operators.isEmpty())
operands.push(operators.pop());
while (!operands.isEmpty())
calculation+=operands.pop();
calculation=calculation.reverse();
return calculation;
}
Explanation:
- Create the checkSymbol function to see what symbol is being passed to the stack.
- Create the convertInfixToPostfix function that keeps track of the operands and the operators stack.
- Use conditional statements to check whether the character being passed is a letter, digit, symbol or a bracket.
- While the operators is not empty, keep pushing the character to the operators stack.
- At last reverse and return the calculation which has all the results.
Answer:
Storage Size
Explanation:
the answer to which is better 8 Mbps or 2 MBps? Is 2 MBps (which is 16 Mbps). Using a less common, but more clear notation: which is better 8 Mbit/s or 2 MBps? Answer: 2 MBps, since that is 16 Mbit/s. (Marketing people use this confusion to their advantage if you're not sure which is intended, ask.)
Source http://cs.sru.edu/~mullins/cpsc100book/module02_introduction/module02-05_introduction.html
(if you want to read the full artical.)
Answer:
- import math
-
- def standard_deviation(aList):
- sum = 0
- for x in aList:
- sum += x
-
- mean = sum / float(len(aList))
-
- sumDe = 0
-
- for x in aList:
- sumDe += (x - mean) * (x - mean)
-
- variance = sumDe / float(len(aList))
- SD = math.sqrt(variance)
-
- return SD
-
- print(standard_deviation([3,6, 7, 9, 12, 17]))
Explanation:
The solution code is written in Python 3.
Firstly, we need to import math module (Line 1).
Next, create a function standard_deviation that takes one input parameter, which is a list (Line 3). In the function, calculate the mean for the value in the input list (Line 4-8). Next, use the mean to calculate the variance (Line 10-15). Next, use sqrt method from math module to get the square root of variance and this will result in standard deviation (Line 16). At last, return the standard deviation (Line 18).
We can test the function using a sample list (Line 20) and we shall get 4.509249752822894
If we pass an empty list, a ZeroDivisionError exception will be raised.
