Answer:
(a)Applications Time Stamp Events
(b)S=0.5(W-
)+
, where
≤W≤
+8.
Explanation:
Some applications assume that clocks always advance, so they could timestamp events under this assumption.
In our case we have the wrong timed clock, say W and the hardware clock H which is supposed to advance at a perfect rate.
We proceed to construct a software clock such that after 8 seconds we can replace the wrong timed clock with the software clock in good conditions.
Let us denote the software clock with S.
Then, S=c(W-
)+
where:
=The current Time(10:27:54) and;
c is to be found.
We already know that S=
+4 when W=
+8,
So:
S=c(W-
)+
+4=c(
+8-
)+
4=8c
c=0.5
We obtain the formula
S=0.5(W-
)+
, where
≤W≤
+8.
Answer:
12,000,000 bytes.
Explanation:
If we want to store an uncompressed binary image, assuming that we will use 256 different levels to represent all tones in a monochromatic image, each pixel will be stored as an 8-bit sample, i.e. one byte per pixel.
The complete image will have 4000 pixels wide by 3000 pixels height, so an entire image will need 3000*4000= 12,000,000 pixels.
As we will use 8 bits= 1 Byte per pixel, we will need to store 12,000, 000 bytes for a complete uncompressed monochromatic (white & black) image.
Answer:
A. add level
D. Delete level
E. Copy level
F. Move up or down
Explanation:
I just did it on edg. 2020
Answer:
D
Explanation:
In the lab, the Ethernet II detail of the provided packet capture file indicated that Wireshark had determined that the type of traffic carried in the next layer was Internet Protocol (IP).