Answer: i dont see the following sorry cant help ya
Explanation:
You have a specific number of words typed a minute and you need to know the science involved in making it
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
Answer:
3 bits
Explanation:
Capacity of main memory=16 Bytes=24
The number of address bits= 4 bits.
The size of the word= 1 Byte=20
The word bits=0.
Number of lines =4
Number of sets required=21
The sets bits is =1
The number of offset bits=20=0
Number of tag bits= total number of address bits - (word bits + offset bits + set bits)
= 4 - 0 -0- 1
= 3 bits
