For applying the filter you need to select the data on which you want to apply it so it will allow you to look at only data you select from the filter options.
Answer:
Written in C++
void number(int n){
if(n%2 == 0)
cout<<2 * n;
else
cout<<5 * n;
}
Explanation:
The programming language is not stated.
However, I answered using C++
This line defines the function as void
void number(int n){
This line checks if the number is even
if(n%2 == 0)
If yes, it doubles the number and prints the output
cout<<2 * n;
If otherwise,
else
It multiplies the number by 5 and prints the output
cout<<5 * n;
}
To call the function from main, use:
number(n);
Note than n must be declared as integer
See attachment
Answer:
e(a) = 0
e(b) = 10
e(c) = 110
e(d) = 1110
Explanation:
The Worst case will happen when f(a) > 2*f(b) ; f(b) > 2*f(c) ; f(c) > 2*f(d) ; f(d) > 2*f(e) and f(e) > 2*f(f).
Where f(x) is frequency of character x.
Lets consider the scenario when
f(a) = 0.555, f(b) = 0.25, f(c) = 0.12, f(d) = 0.05, f(e) = 0.02 and f(f) = 0.005
Please see attachment for image showing the steps of construction of Huffman tree:- see attachment
From the Huffman tree created, we can see that endcoding e() of each character are as follows:-
e(a) = 0
e(b) = 10
e(c) = 110
e(d) = 1110
e(e) = 11110
e(f) = 11111
So we can see that maximum length of encoding is 5 in this case.
Answer:
He retired from fashion and commercial photography in 1937. ... Two years after he retired from the Navy, Edward Steichen became the director of the Photography Department at the Museum of Modern Art in New York. There he created what has become the most famous photographic exhibition of all time, The Family of Man.
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