Answer:
Sequence of popped values: h,s,f.
State of stack (from top to bottom): m, d
Explanation:
Assuming that stack is initially empty. Suppose that p contains the popped values. The state of the stack is where the top and bottom are pointing to in the stack. The top of the stack is that end of the stack where the new value is entered and existing values is removed. The sequence works as following:
push(d) -> enters d to the Stack
Stack:
d ->top
push(h) -> enters h to the Stack
Stack:
h ->top
d ->bottom
pop() -> removes h from the Stack:
Stack:
d ->top
p: Suppose p contains popped values so first popped value entered to p is h
p = h
push(f) -> enters f to the Stack
Stack:
f ->top
d ->bottom
push(s) -> enters s to the Stack
Stack:
s ->top
f
d ->bottom
pop() -> removes s from the Stack:
Stack:
f ->top
d -> bottom
p = h, s
pop() -> removes f from the Stack:
Stack:
d ->top
p = h, s, f
push(m) -> enters m to the Stack:
Stack:
m ->top
d ->bottom
So looking at p the sequence of popped values is:
h, s, f
the final state of the stack:
m, d
end that is the top of the stack:
m
Complete Question:
A campus bookstore sells both types and in the last semester sold 56% laptops and 44% desktops. Reliability rates for the two types of machines are quite different, however. In the first year, 5% of desktops require service, while 15% of laptops have problems requiring service.
Given that a computer required service, what is the probability that it was a laptop?
Answer:
Probability = 0.084
Explanation:
Given
Laptops = 56%
Desktop = 44%
Service Required (Laptop) = 15%
Service Required (Desktop) = 5%
Required
Determine the probability that a selected computer is a laptop and it requires service.
The question tests our knowledge of probabilities using "and" condition.
What the question requires is that, we calculate the probability of selecting a LAPTOP that REQUIRES SERVICE
Note the capitalised words.
This will be calculated as follows:
Probability = P(Laptop) and P(Service Required (Laptop))
[Substitute values for P(Laptop) and P(Service Required (Laptop))]
Probability = 56% * 15%
[Convert to decimal]
Probability = 0.56 * 0.15
Probability = 0.084
<span>The manual assignment of an IP address to a
system is called static addressing. Since the assignment of IP has been entered
manually using ARP utility, the network administrators probably forgot to
remove the manually configured nodes with IP addresses. </span>
Answer:
<em>sorry</em><em> </em><em>I </em><em>don't</em><em> </em><em>know</em><em> </em><em> </em><em>I </em><em> </em><em>can't</em><em> </em><em>will</em><em> </em><em>your</em><em> </em><em>help</em>
