Answer:
12.1 seconds
Explanation:
Step 1:
Assuming the following data :
N: Size of packet = 30 Mb
T: Transmission rate = 10 Mbps
C: Speed of light in copper = 2 * 10^8 m/s
L: Length of link 5000 km
Step 2:
Calculation of the Transmission delay and the Propagation Delay
Transmission delay = 
= 
= 3s
Propagation delay = 
= 
= 0.025s
Step 3:
Calculation of the end-to -end delay
The end-to-end delay = 4*(Transmission Delay+Propagation Delay) = 4*(3+0.025) = 12.1 seconds
The amount of data that can be stored in the storage unit.
Answer:(C) Anyone with access to the personal computers could log on to the mainframe.
Explanation:
It is no doubt that log on procedures are cumbersome so what most users do is that they store log on sequences in their computer so that they could use the main frame facilities whenever they access it the second time.
Here we have to identify the associated risk with this approach.
This process could lead to access of the computer system to users and thereby making all the data available in the system vulnerable. So option C is correct.
option (A) is incorrect as the back up option who also be affected when the system is accessible to anyone.
option (B) is incorrect because it has nothing to do anything with training of the personnel.
option (D) is incorrect because it one has nothing to do with the physical system what vulnerable reside is inside the system.
C. Thinksmart - thinksmart only sells products and softwares of their own. Thinksmart is a online selling company that sells moslty softwares on their domain. Think smart is not a popular cloud suite or an online office suite. So the answer to your question is C.
