Final answer:
In an M/M/1 queueing system with an average arrival rate of 8 per second and an average service time of 1/10 second, the average number of tasks in the system, either waiting or being served, is 18/5.
Step-by-step explanation:
An M/M/1 queueing system represents a single-server queueing system where arrivals follow a Poisson distribution and service times follow an exponential distribution. In this case, the average arrival rate is 8 per second and the average service time is 1/10 second.
To find the average number of tasks in the system, we can use Little's Law which states that the average number of tasks in the system is equal to the arrival rate multiplied by the average time spent in the system. In this case, the average time spent in the system is the sum of the average waiting time and the average service time.
The average waiting time can be found using the formula: (average arrival rate) * (average waiting time) = (average arrival rate) / (average service rate). Therefore, the average waiting time is 1/8 seconds.
So, the average number of tasks in the system is: (average arrival rate) * (average waiting time + average service time) = 8 * (1/8 + 1/10) = 18/5.