Answer:
Here is the fractional_part() function:
def fractional_part(numerator, denominator):
if denominator != 0:
return (numerator % denominator)/denominator
else:
return 0
Explanation:
I will explain the code line by line.
The first statement it the definition of function fractional_part() which takes two parameters i.e. numerator and denominator to return the fractional part of the division.
Next is an if statement which checks if the value of denominator is 0. If this is true then the function returns 0. If this condition evaluates to false which means that the value of denominator is not 0 then return (numerator % denominator)/denominator is executed. Now lets see how this statement works with the help of an example.
Lets say the value of numerator is 5 and denominator is 4. (numerator % denominator)/denominator will first compute the modulus of these two values. 5 % 4 is 1 because when 5 is divided by 4 , then the remainder is 1. Now this result is divided by denominator to get the fractional part. When 1 is divided by 4 the answer is 0.25. So this is how we get the fractional part which is 0.25.
The program with output is attached.
Answer:
Explanation:
System models are created and used in order for all of the individuals involved in the process to be able to see exactly how each module that is being created connects to the rest of the system. This allows every developer to understand each module's/part's function and develop it with the entire system in mind. This also makes it easier to implement newer features. If a model is not made then the developers may miss key features and when it is time to piece all of the components of the system together, the overall process can become a jumbled mess.
The item that is least likely to cause an exhaust emission test failure as of a result of excessive nox emissions is; poorflow through the egr system.
<h3 /><h3>What is an exhaust emission?</h3>
Exhaust emissions simply means substances emitted into the atmosphere from the exhaust discharge nozzle of an aircraft or aircraft engine.
Now, when the test fails as a result of excessive nox emissions, we can say that high NOx emissions can occur when an engine's air-fuel mixture is too lean. Thus, we can conclude that the reason for exhaust emission test failure as of a result of excessive nox emissions is due to poorflow through the egr system.
Read more about emissions at; brainly.com/question/14154063
Answer:
hmm i would try calling him. ask your mom or other adults where he is!
Explanation:
hope you get help soon!
<u>Explanation</u>:
(a)
![f=A+A B+A C\\=A[1+B+C]=A \quad[1+x=1]\\F=A](https://tex.z-dn.net/?f=f%3DA%2BA%20B%2BA%20C%5C%5C%3DA%5B1%2BB%2BC%5D%3DA%20%5Cquad%5B1%2Bx%3D1%5D%5C%5CF%3DA)
No gate is required to implement this function
(b)
Note: Refer the first image.
(c)
![\begin{aligned}f &=\overline{A+B}+A \bar{B}+B \bar{C} \\&=(\bar{A} \bar{B})+A \bar{B}+B \bar{C} \\&=\bar{B}[A+\bar{A}]+B \bar{C} \\& F=\bar{B}+B \bar{C} =\bar{B}+\bar{C}\end{aligned}](https://tex.z-dn.net/?f=%5Cbegin%7Baligned%7Df%20%26%3D%5Coverline%7BA%2BB%7D%2BA%20%5Cbar%7BB%7D%2BB%20%5Cbar%7BC%7D%20%5C%5C%26%3D%28%5Cbar%7BA%7D%20%5Cbar%7BB%7D%29%2BA%20%5Cbar%7BB%7D%2BB%20%5Cbar%7BC%7D%20%5C%5C%26%3D%5Cbar%7BB%7D%5BA%2B%5Cbar%7BA%7D%5D%2BB%20%5Cbar%7BC%7D%20%5C%5C%26%20F%3D%5Cbar%7BB%7D%2BB%20%5Cbar%7BC%7D%20%3D%5Cbar%7BB%7D%2B%5Cbar%7BC%7D%5Cend%7Baligned%7D)
Note: Refer the second image
(d)
![\begin{aligned}f=& A B \bar{c}+\overline{A+\bar{c}} \\=& A B \bar{c}+\bar{A} \bar{c}=\bar{A} B \bar{c}+\bar{A} c \\f=& \bar{A}[c+B \bar{c}] . \\& f=\bar{A} B+\bar{A} c=\bar{A}(B+c)\end{aligned}](https://tex.z-dn.net/?f=%5Cbegin%7Baligned%7Df%3D%26%20A%20B%20%5Cbar%7Bc%7D%2B%5Coverline%7BA%2B%5Cbar%7Bc%7D%7D%20%5C%5C%3D%26%20A%20B%20%5Cbar%7Bc%7D%2B%5Cbar%7BA%7D%20%5Cbar%7Bc%7D%3D%5Cbar%7BA%7D%20B%20%5Cbar%7Bc%7D%2B%5Cbar%7BA%7D%20c%20%5C%5Cf%3D%26%20%5Cbar%7BA%7D%5Bc%2BB%20%5Cbar%7Bc%7D%5D%20.%20%5C%5C%26%20f%3D%5Cbar%7BA%7D%20B%2B%5Cbar%7BA%7D%20c%3D%5Cbar%7BA%7D%28B%2Bc%29%5Cend%7Baligned%7D)
Note: Refer the third image
(e)
![\begin{aligned}f=& A \bar{B}+\bar{B} C+A \bar{B} \\&=\bar{B}[A+\bar{A}+c] \\&=\bar{B}[1+C]\end{aligned}](https://tex.z-dn.net/?f=%5Cbegin%7Baligned%7Df%3D%26%20A%20%5Cbar%7BB%7D%2B%5Cbar%7BB%7D%20C%2BA%20%5Cbar%7BB%7D%20%5C%5C%26%3D%5Cbar%7BB%7D%5BA%2B%5Cbar%7BA%7D%2Bc%5D%20%5C%5C%26%3D%5Cbar%7BB%7D%5B1%2BC%5D%5Cend%7Baligned%7D)
(f)
![\begin{aligned}f &=A B C+A B D+A B C \\&=A B[C+C]+A B D \\&=A B+A B D \\&=B[A+A D] \\&=B[A+D] \\\therefore & A=B[A+D]\end{aligned}](https://tex.z-dn.net/?f=%5Cbegin%7Baligned%7Df%20%26%3DA%20B%20C%2BA%20B%20D%2BA%20B%20C%20%5C%5C%26%3DA%20B%5BC%2BC%5D%2BA%20B%20D%20%5C%5C%26%3DA%20B%2BA%20B%20D%20%5C%5C%26%3DB%5BA%2BA%20D%5D%20%5C%5C%26%3DB%5BA%2BD%5D%20%5C%5C%5Ctherefore%20%26%20A%3DB%5BA%2BD%5D%5Cend%7Baligned%7D)
Note: Refer the fourth image
