Answer:
The answer is below
Explanation:
The theoretical capacity of a channel (C) is the maximum data rate at which data can be transmitted in a channel. It is measured in bits per second (bps) and is given by the formula:
C = bandwidth * (SNR dB)/3
a) Bandwidth = 20 KHz = 20000 Hz, SNR dB =40
C = 20000 * (40)/3 = 26666.7 bps
C ≅ 267 Kbps
b) Bandwidth = 200 KHz = 200000 Hz, SNR dB =40]
C = 200000 * (4)/3 = 26666.7 bps
C ≅ 267 Kbps
c) Bandwidth = 1 MHz = 1000000 Hz, SNR dB =20
C = 1000000 * (20)/3 = 6666666.7 bps
C ≅ 6667 Kbps
Answer:
D. port disabled; SNMP or syslog messages
Explanation:
9251.44 or 231286/25 or 9251(11/25)
0.36*6*200*21.44-10.64
becomes
9262.08-10.64
gives you the result
The answer is 326.
Range is found by subtracting the smallest number in the data set from the largest number.
Highest number: 419
Lowest number: 93
419-93
=326
(Next time make sure to post this under the Mathematics section.)