Local Area Network (LAN), Hope this helps
public class MyClass {
public static void printChar(char ch1, char ch2, int numberPerLine){
int i = 0;
for (char c = ch1; c <= ch2; c++){
while (i < numberPerLine){
System.out.print(c + " ");
i += 1;
}
System.out.println("");
i = 0;
}
}
public static void main(String args[]) {
printChar('a', 'z', 10);
}
}
So far, this works by printing letters. If you need me to modify the code, I will.
Answer:
connect the sender with the desired recipients.
Explanation:
Integrated Marketing Communication (IMC) is a process through which organizations create seamless branding and coordination of their marketing and communication objectives with its business goals and target audience or consumers. The communication tools used in IMC are both digital and traditional media such as billboards, search engine optimization, magazines, television, blog, radio, webinars etc.
The communication channel used in IMC must connect the sender with the desired recipients.
The receiver is any individual who is able to read, hear or see and process the message being sent or communicated in the IMC communication process. Any interference the IMC communication process is known as noise.
An organization can analyze and measure the effectiveness of the IMC communication process by considering market share, sales, and customer loyalty.
8. according to Google, 1 byte = 8 bits
Answer:
Probability Distribution={(A, 4/7), (B, 2/7), (C, 1/7)}
H(X)=5.4224 bits per symb
H(X|Y="not C")=0.54902 bits per symb
Explanation:
P(B)=2P(C)
P(A)=2P(B)
But
P(A)+P(B)+P(C)=1
4P(C)+2P(C)+P(C)=1
P(C)=1/7
Then
P(A)=4/7
P(B)=2/7
Probability Distribution={(A, 4/7), (B, 2/7), (C, 1/7)}
iii
If X={A,B,C}
and P(Xi)={4/7,2/7,1/7}
where Id =logarithm to base 2
Entropy, H(X)=-{P(A) Id P(A) +P(B) Id P(B) + P(C) Id P(C)}
=-{(1/7)Id1/7 +(2/7)Id(2/7) +(4/7)Id(4/7)}
=5.4224 bits per symb
if P(C) =0
P(A)=2P(B)
P(B)=1/3
P(A)=2/3
H(X|Y="not C")= -(1/3)Id(I/3) -(2/3)Id(2/3)
=0.54902 bits per symb
