Answer:
C = 292 Mbps
Explanation:
Given:
- Signal Transmitted Power P = 250mW
- The noise in channel N = 10 uW
- The signal bandwidth W = 20 MHz
Find:
what is the maximum capacity of the channel?
Solution:
-The capacity of the channel is given by Shannon's Formula:
C = W*log_2 ( 1 + P/N)
- Plug the values in:
C = (20*10^6)*log_2 ( 1 + 250*10^-3/10)
C = (20*10^6)*log_2 (25001)
C = (20*10^6)*14.6096
C = 292 Mbps
It’s in Wolfsburg Germany
Answer:
as all the people should go near stratosphere
Answer:
2.5=1500/Whp=> Whp=600 kWh
delWgain=1500-600=900 kWh
Money saved= 900* 6tk*=5400 tk
Answer:
The code is attached.
Explanation:
I created a string s including 6 colors with spaces in between. Then I converted the string into a list x by using split() method. I used three different methods for removing elements from the list. These methods are remove(), pop() and del.
Then I used methods append(), insert() and extend() for adding elements to the list.
Finally I converted list into a string using join() and adding space in between the elements of the list.
