Answer:
<u>The total time elapsed from the time a bit is created (from the original analog signal at Host A) until the bit is decoded (as part of the analog signal at Host B is </u><u>25.11 ms</u>
Explanation:
Host A first converts the analog signal to a digital 64kbps stream and then groups it into 56-byte packets. The time taken for this can be calculated as:
time taken 1= 
= (56 x 8) bits / 64 x 10³ bits/s
= 7 x 10⁻³s
time taken 1= 7 ms
The transmission rate of the packet from Host A to Host B is 4 Mbps. The time taken to transfer the packets can be calculated as:
time taken 2= (56 x 8) bits / 4 x 10⁶ bits/s
= 1.12 x 10⁻⁴ s
time taken 2= 112 μs
The propagation delay is 18 ms.
To calculate the total time elapsed, we need to add up all the time taken at each individual stage.
<u />
<u> = Time taken 1 + Time taken 2 + Propagation Delay</u>
= 7 ms + 112 μs + 18 ms
= 0.025112 s
= 25.11 ms
What is the expected total cost for one year at the local community college if Kirk lives at home?
Answer: $6845
What is the expected total cost for one year at the out-of-state school if Kirk lives on campus?
Answer: $30,566
By using cpu-z or the performance ran in taskmgr
Answer:
O(n^2)
Explanation:
The number of elements in the array X is proportional to the algorithm E runs time:
For one element (i=1) -> O(1)
For two elements (i=2) -> O(2)
.
.
.
For n elements (i=n) -> O(n)
If the array has n elements the algorithm D will call the algorithm E n times, so we have a maximum time of n times n, therefore the worst-case running time of D is O(n^2)