Answer:
1000BaseT
Explanation: Just took the quiz and got it right!
1000BaseT is specific to Ethernet
802.11ac is a type of WiFi connection (even though it uses Ethernet)
4G is what they use to provide cell phone service
Answer:
See Explanation
Explanation:
Given
The attached function
What the recursion does is that; it adds up individual digits from N to 0
Solving (a): Each output when N = 6
For N = 6.
The function returns the following values:
f(6) = 6
Then: 6 + f(5)
Then: 6 + 5 + f(4)
Then: 6 + 5 + 4 + f(3)
Then: 6 + 5 + 4 + 3 + f(2)
Then: 6 + 5 + 4 + 3 + 2 + f(1)
Then: 6 + 5 + 4 + 3 + 2 + 1 + f(0)
Then: 6 + 5 + 4 + 3 + 2 + 1 + 0 = 21
Solving (b): The output when N = 7
Using the same process in (a) above.
The output is 28
int sum = 0, n;
do {cin>>n; sum+=n;}while (n!=0);
cout<<sum;
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
I think the answer is C but I could be wrong
