Answer:
1.) 25 ; 15 ; 15
2.) 50 ; 15 ; 50
Explanation:
In the first function written :
The variable val was initially decaled or assigned a value of 25 and that was what was printed first.
However, after the example function was written, the val variable was finally assiagned a value of 15 within the function. However, it was also declared that the global variable takes uonthe val value. Hence, the val variable initially assigned a value, of 25 changes to 15 globally.
For the second code :
From the top:
Val was assigned a value of 50 ;
Hence,
print(val) gives an output of 50
Within the function definition which prints the value of val that is assigned a value of 25 within the function.
Since tbe global variable isnt reset.
Printing Val again outputs 50;since ito is outside the function.
It would be a audio file -B.
Answer:
I think its option a. and d.
Explanation:
option a. is right I'm sure.
option b. and c. are doubtful for me.
option d. can be true. but unlike a job letter, resumes carry a deeper view of the employee's accomplishments and qualifications.
but I still hold with option a. and b. if it's wrong please do inform, so I can refer.
Answer: 83.17
Explanation:
By definition, the dB is an adimensional unit, used to simplify calculations when numbers are either too big or too small, specially in telecommunications.
It applies specifically to power, and it is defined as follows:
P (dB) = 10 log₁₀ P₁ / P₂
Usually P₂ is a reference, for instance, if P₂ = 1 mW, dB is called as dBm (dB referred to 1 mW), but it is always adimensional.
In our question, we know that we have a numerical ratio, that is expressed in dB as 19.2 dB.
Applying the dB definition, we can write the following:
10 log₁₀ X = 19.2 ⇒ log₁₀ X = 19.2 / 10 = 1.92
Solving the logarithmic equation, we can compute X as follows:
X = 10^1.92 = 83.17
X = 83.17
