For anyone reading this in the future, the correct answer is transmission. I just took the quiz
Answer:
<h3>Connector names are written below Picture vise:</h3>
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Explanation:
Following is the brie Illustration of the terms used:
It is the newest connector in the market with the reversible/symmetrical design. It can be adapted to work with the legacy connectors such as USB-A, USB-B, USB-C and Micro USB.
It is used mostly in the connections of electronic devices such as printers and smartphones. It has A to B connectors as well as micro USB connectors and mini USB connectors.
It is used for the connection o compact devices such as smartphone and mp3 players. They are further grouped into three categories: Micro A, Micro B and micro USB 3.
<h3>
I hope it will help you!</h3>
I think marketing and branding. i dont think this is 100% correct
Answer:
A wave that has been digitized can be played back as a wave over and over, and it will be the same every time. For that reason, digital signals are a very reliable way to record information—as long as the numbers in the digital signal don’t change, the information can be reproduced exactly over and over again.
Explanation:
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.
