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
Answer:
Press "Ctrl, "Shift" and "=" on your keyboard to turn off superscript formatting.
Answer:
Replace the first blank with:
message = "Hello " + name + ". Your lucky number is " + str(number)
Replace the second blank with:
return message
Explanation:
The first blank needs to be filled with a variable; we can make use of any variable name as long as it follows the variable naming convention.
Having said that, I decided to make use of variable name "message", without the quotes
The next blank is meant to return the variable on the previous line;
Since the variable that was used is message, the next blank will be "return message", without the quotes
Solution:
initial = float(eval(input('Enter the monthly saving amount: ')))
x = (1 + 0.00417)
month_one = initial * x
month_two = (initial + month_one) * x
month_three = (initial + month_two) * x
month_four = (initial + month_three) * x
month_five = (initial + month_four) * x
month_six = (initial + month_five) * x
print('The sixth month value is: '+str(month_six))
Don't forget the saving amount, and initialize the balance with that amount. Inside the loop, work out and add the interest and then add the saving amount for the next month.
balance = 801
for month in range(6):
balance = balance * (1.00417)
print(balance)
Answer:
probly the wifi connection
Explanation:
