Answer:
showProduct(int,double)
for example: showProduct(10,10.5) is the correct answer even showProduct(10,10.0) is also correct but showProduct(10.0,10.5) or showProduct(10,10) or showProduct(10.0,10) are wrong calls.
Explanation:
The code is
- <em>public static void showProduct (int num1, double num2){</em>
- <em> int product;</em>
- <em> product = num1*(int)num2;</em>
- <em> System.out.println("The product is "+product);</em>
- <em> }</em>
showProduct is function which asks for two arguments whenever it is called, first one is integer and second one is of type double which is nothing but decimal point numbers. Generally, in programming languages, 10 is treated as integer but 10.0 is treated as decimal point number, but in real life they are same.
If showProduct( 10,10.0) is called the output will be 'The product is 100'.
Strange fact is that, if you enter showProduct(10,10.5) the output will remain same as 'The product is 100'. This happens because in the 3rd line of code,which is <em>product=num1*(int)num2</em>, (int) is placed before num2 which makes num2 as of type integer, which means whatever the value of num2 two is given, numbers after decimal is erased and only the integer part is used there.
This is necessary in JAVA and many other programming languages as you <u>cannot</u><u> multiply two different datatypes</u> (here one is int and another is double). Either both of them should be of type int or both should be of type double.
Answer:
// Program is written in C++ Programming Language
// Comments are used for explanatory purpose
// Program starts here
#include<iostream>
using namespace std;
int main()
{
// Declare integer variable n which serves as the quotient.
int n;
// Prompt to enter any number
cout<<"Enter any integer number: ";
cin>>n;
// Check for divisors using the iteration below
for(int I = 1; I<= n; I++)
{
// Check if current digit is a valid divisor
if(n%I == 0)
{
// Print all divisors
cout<<I<<" ";
}
}
return 0;
}
Answer:
a) Yes
b) Yes
c) Yes
d) No
e) Yes
f) No
Explanation:
a) All single-bit errors are caught by Cyclic Redundancy Check (CRC) and it produces 100 % of error detection.
b) All double-bit errors for any reasonably long message are caught by Cyclic Redundancy Check (CRC) during the transmission of 1024 bit. It also produces 100 % of error detection.
c) 5 isolated bit errors are not caught by Cyclic Redundancy Check (CRC) during the transmission of 1024 bit since CRC may not be able to catch all even numbers of isolated bit errors so it is not even.
It produces nearly 100 % of error detection.
d) All even numbers of isolated bit errors may not be caught by Cyclic Redundancy Check (CRC) during the transmission of 1024 bit. It also produces 100 % of error detection.
e) All burst errors with burst lengths less than or equal to 32 are caught by Cyclic Redundancy Check (CRC) during the transmission of 1024 bit. It also produces 100 % of error detection.
f) A burst error with burst length greater than 32 may not be caught by Cyclic Redundancy Check (CRC) during the transmission of 1024 bit.
Cyclic Redundancy Check (CRC) does not detect the length of error burst which is greater than or equal to r bits.
You click on the three dots and click the person silhouette with the plus sign on it.
The answer to your question is a
