Using your knowledge of how an ATM is used, develop a set of use-cases that could serve as a basis for understanding the require
ments for an ATM system
1 answer:
Answer:
Use cases are known to be a set of instruction or processes between a User/Actor with the system to produce a desired input.
With the aid of a diagram, the set of use cases that are carried out in this ATM are given below:
Insert PIN
(1)Perform required transaction
(2)Withdrawal
(3)Deposit
(4)Transfer
(5)Change PIN
(6)Exit
Note: Kindly find an attached diagram of the Use case as part of the solution to process carried out at the ATM
Sources: The diagram of the Use case for ATM was researched and taken from Quizlet.
Explanation:
Solution
Use cases are normally a set of instruction or processes between a User/Actor with the system to produce a desired input.
A use case diagram or image is a graphical representation of all the use case or processes that connects or interact with the system
The use case diagram is a part of Unified Modelling Language also called the UML.
The set of use cases that are carried out in this ATM use case diagram to know the requirements of the ATM is shown below:
- Insert PIN
- Perform required transaction
- Withdrawal
- Deposit
- Transfer
- Change PIN
- Exit
Now both the customer/client and Bank are seen as Actors.
Actors are the ones or people that interface with the system.
You might be interested in
The answer & explanation for this question is given in the attachment below.
Answer: b) Area under the elastic portion of the stress-strain curve
Explanation:
By definition, resilience is the strain-energy density stored by the material when it is stressed to the proportional limit defined by Hooke's Law. Resilience is given by the following expression:
μ(r) = / 2E
μ(r) is the modulus of resilience
σ(pl) is the stress to the proportional limit
E is the elastic modulus
When you look at the stress-strain curve, the area under the elastic portion (up to the proportional limit) can be obtained by the area of a triangle with base equal to the strain (σ) and height equal to the stress (ε):
Ω = (b . h) / 2 = (σ(pl) . ε) / 2
Using Hooke's Law: σ = E . ε → ε = σ/E
Replacing the expression in the area equation:
Ω = (σ(pl) . ε) / 2 = / 2E = μ(r)