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
Answer:
Abolition of intermediaries (rent collectors under the pre-Independence land revenue system); Tenancy regulation (to improve the contractual terms including the security of tenure); A ceiling on landholdings (to redistributing surplus land to the landless);
Answer:
Hello your question is incomplete attached below is the missing part and answer
options :
Effect A
Effect B
Effect C
Effect D
Effect AB
Effect AC
Effect AD
Effect BC
Effect BD
Effect CD
Answer :
A = significant
B = significant
C = Non-significant
D = Non-significant
AB = Non-significant
AC = significant
AD = Non-significant
BC = Non-significant
BD = Non-significant
CD = Non-significant
Explanation:
The dependent variable here is Time
Effect of A = significant
Effect of B = significant
Effect of C = Non-significant
Effect of D = Non-significant
Effect of AB = Non-significant
Effect of AC = significant
Effect of AD = Non-significant
Effect of BC = Non-significant
Effect of BD = Non-significant
Effect of CD = Non-significant
Hi
Acetylene and propane
I hope this help you!