nu it would not but if they call microsoft to check it then yez the history would show up for them but dont try to make it notice able so they dont have to call
Answer and Explanation:
Given that total number of records in a file = 100000
Each record consists of= 240 bytes
Given that B= 2400 bytes.
Then total blocks in file = 100000 records * 240 / 2400
= 10000 blocks.
Time that will take for exhaustive read
= s + r + b.btt
= 16 + 8.3 + (10000) * 0.8
= 8024.3 msec
Now as per given X be the number of records that are searched randomly and it takes more than exhaustive read time.
Hence, X (s + r + btt) > 8024.3
X (16+8.3+0.8) > 8024.3
X > 8024.3/25.1
So that, X > 319.69
Hence, atleast 320 random reads should be made to search the file
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
