Answer:
C. E-mail
Explanation:
Electronic mail is the best choice to convey urgent and highly sensitive information
Answer:
Round-Robin DNS
Explanation:
According to my experience in Information Technology and Networking it can be said that based on the information provided the best solution would be to set up a Round-Robin DNS. This term refers to a technique used to balance the load on a server, where a client request is sent to each server one at a time, and then the system repeats the process from the top of the request list. This prevents the server from being drowned in a sea of simultaneous requests.
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Answer: 83.17
Explanation:
By definition, the dB is an adimensional unit, used to simplify calculations when numbers are either too big or too small, specially in telecommunications.
It applies specifically to power, and it is defined as follows:
P (dB) = 10 log₁₀ P₁ / P₂
Usually P₂ is a reference, for instance, if P₂ = 1 mW, dB is called as dBm (dB referred to 1 mW), but it is always adimensional.
In our question, we know that we have a numerical ratio, that is expressed in dB as 19.2 dB.
Applying the dB definition, we can write the following:
10 log₁₀ X = 19.2 ⇒ log₁₀ X = 19.2 / 10 = 1.92
Solving the logarithmic equation, we can compute X as follows:
X = 10^1.92 = 83.17
X = 83.17
Answer: Crossover
Explanation:
Based on the information given in the question, the type of cable that should be used is the crossover cable.
Crossover cable is used for a direct connection of computing devices. A crossover cable is typically used for the connection of computing devices that are of identical types.
Since the LAN port on a router is to be connected to the uplink port on a switch and the switch does not support auto-MDI, the crossover cable can be used.
Answer:
Required memory size is 16k x 8
16k = 24 x 210 = 214
Hence, No. of address lines = 14
No. of data lines = 8
a) Size of IC 1024 x 1
Total number of ICs required = 16k x 8 / 1024 x 1 = 16 x 8 = 128
b) Size of IC 2k x 4
Total number of ICs required = 16k x 8 / 2k x 4 = 8 x 2 = 16
c) Size of IC 1k x 8
Total number of ICs required = 16k x 8 / 1k x 8 = 16 x 1 = 16
Explanation:
For a, 10 address lines from A0 to A9 are used to select any one of the memory location out of 1024 memory locations present in a IC.
For b, 11 address lines from A0 to A10 are used to select any one of the memory location out of 2k=2048 memory locations present in a IC.
For c, 10 address lines from A0 to A9 are used to select any one of the memory location out of 1k=1024 memory locations present in a IC.
