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mojhsa [17]
4 years ago
10

Assume the existence of a BankAccount class with a method, getAvailable that returns the amount of available funds in the accoun

t (as an integer), and a subclass, OverdraftedAccount, with two integer instance variables:
Computers and Technology
1 answer:
julia-pushkina [17]4 years ago
6 0

Answer: This is hard but I will research ok

Explanation:

You might be interested in
)Dynamically allocate an object of Banana, using the pointer variable daco.
zepelin [54]

Answer:

d) daco = new Banana;

Explanation:

Dynamically allocated variables have their memory allocated in the heap memory.We declare a dynamical variable like this:-

int *a=new int ;

It means a pointer a is created on the stack memory which hold the address of the block that hold the value of variable a in heap memory.

We already have the pointer daco. We just have to initialize with keyword new.

It will be like daco=new Banana; which matches the option d.

5 0
3 years ago
Programs that do the same thing every time they run are called...
Misha Larkins [42]

Answer:

predictable

Explanation:

because it repeat itself all time

6 0
3 years ago
__________ is a term used to indicate any unwanted event that takes place outside normal daily security operations. This type of
Delvig [45]
The answer is “a security event”
6 0
3 years ago
To find the area, Andrea tiled a rectangle and sketched her answer. Sketch Andrea's rectangle, and find the area. Show your mult
seraphim [82]

Answer:

Area = 6.25

Explanation:

Given

Length = 2.5

Width = 2.5

See attachment for rectangle

Required

Determine the area of the rectangle

Area is calculated as:

Area = Length * Width

This gives:

Area = 2.5 * 2.5

Area = 6.25

<em>The area of Andrea's sketch is 6.25</em>

8 0
3 years ago
Write a Python function check_subset() that takes in two Python lists, representing sets of integers, A and B. Your function sho
wlad13 [49]

Answer:

  1. def check_subset(l1, l2):
  2.    status = False
  3.    count = 0
  4.    if(len(l1) > len(l2)):
  5.        for x in l2:
  6.            for y in l1:
  7.                if x == y:
  8.                    count += 1
  9.        if(count == len(l2)):
  10.            return True  
  11.        else:
  12.            return False
  13.    else:
  14.        for x in l1:
  15.            for y in l2:
  16.                if x==y:
  17.                    count += 1
  18.        if(count == len(l1)):
  19.            return True  
  20.        else:
  21.            return False
  22. print(check_subset([1,4,6], [1,2,3,4,5,6]))
  23. print(check_subset([2,5,7,9,8], [7,8]))
  24. print(check_subset([1, 5, 7], [1,4,6,78,12]))

Explanation:

The key idea of this solution is to create a count variable to track the number of the elements in a shorter list whose value can be found in another longer list.

Firstly, we need to check which list is shorter (Line 4). If the list 2 is shorter, we need to traverse through the list 2 in an outer loop (Line 5) and then create another inner loop to traverse through the longer list 1 (Line 6).  If the current x value from list 2 is matched any value in list 1, increment the count variable by 1. After finishing the outer loop and inner loop, we shall be able to get the total count of elements in list 2 which can also be found in list 1. If the count is equal to the length of list 2, it means all elements in the list 2 are found in the list 1 and therefore it is a subset of list 1 and return true (Line 10-11) otherwise return false.

The similar process is applied to the situation where the list 1 is shorter than list 2 (Line 15-24)

If we test our function using three pairs of input lists (Line 26-28), we shall get the output as follows:

True

True

False

6 0
3 years ago
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