Answer:
The signal would have experienced aliasing.
Step-by-step explanation:
Given that:
the bandwidth of the signal 
= 36MHz
= 36 × 10⁶ Hz
The sampling frequency 
= 36 × 10⁶ Hz
Suppose the sampling frequency is equivalent to the bandwidth of the signal, then aliasing will occur.
Therefore, according to the Nyquist criteria;
Nyquist criteria posit that if the sampling frequency is more above twice the maximum frequency to be sampled, a repeating waveform can be accurately reconstructed.
∴
By Nyquist criteria, for perfect reconstruction of an original signal, i.e. the received signal without aliasing effect;
Then,
∴
The signal would have experienced aliasing.
I’m sorry I don’t understand is there more to the question?
Answer:
a = -2
Step-by-step explanation:
a + 8/3 = 2/3
a = 2/3 - 8/3
a = -6/3 = -2
B.
Let's simply look at each conjecture and determine if it's true or false.
A. 2n– 1 is odd if n is positive: Since n is an integer, 2n will
always be even. And an even number minus 1 is always odd. Doesn't matter
if n is positive or not. So this conjecture is true.
B. 2n– 1 is always even: Once again, 2n will always be even. So 2n-1 will always be odd. This conjecture is false.
C. 2n– 1 is odd if n is even: 2n is always even, so 2n-1 will always
be odd, regardless of what n is. So this conjecture is true.
D. 2n– 1 is always odd: 2n will always be even. So 2n-1 will always be odd. Once again, this conjecture is true.
Of the 4 conjectures above, only conjecture B is false. So the answer is B.
