Answer:
The answer is 'PC4 21300 CL9".
Explanation:
Faster memory advertising is used to provide customer memory, which is essential for brand actions. It can't be taken throughout ad coverage. Retired-laboratory tests revealed its access to ad-memory detection aspects of the effect of its grade through publicity recovery, that's why it uses the PC4 21300 CL9, it transfers the 170400 bits per second.
Answer:
Two systems are connected by a router. Both systems and the router have transmission rates of 1,000bps. Each link has a propagation delay of 10ms. Also, it takes router 2ms in order to process the packet (e.g. decide where to forward it). Suppose the first system wants to send a 10,000 bit packet to the second system. How long will it take before receiver system receives the entire packet.
Transmission time for first Router = 10,000 bits / 1000 bps = 10 seconds
Receiving time for seond route r= 10,000 bits / 1000 bps = 10 seconds
Propagation delay = 10ms = .01 seconds x 2 for two delays = .02 seconds
First router 2ms to process = .002 seconds
Add all the times together and we get 20.022 seconds which is the same as or 20 seconds and 22 ms
we translate the following statement given in terms of logarithms. 211 base x can be expressed into log 211 / log x while 152 base 8 can be expressed int o log 152 over log 8. In this case,
log 211 / log x = log 152 / log 8log x = 0.962x = 10^0.962 = 9.1632
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:
What does Bob [1] return?
What about Bob[-2]?
What about Bob[1:-1]?
How to get the length of Bob?
Explanation: