Question 1:
4. it increases the number of employees
Question 2:
2. false
Question 3:
2. it causes great damage to the ozone layer
Question 4:
1. true
Answer:
Hi Riahroo! This is a good question on the concept of relational databases.
We can normalize the relations as follows:
Flight
(flightnumber (unique), flighttime, airline_id, departure_city, arrival_city, passenger_id, pilot_id, airplane_id)
has_one_and_belongs_to :airline
has_many :passengers
has_one :pilot
Itinerary(passenger_id, flight_id)
Belongs_to
Passenger_details
(passengername (unique), gender, date_of_birth)
has_many :flights
Pilot
(pilotname (unique), gender, date_of_birth)
has_many :flights
airline(airlinename)
airplane(planeID, type, seats))
Explanation:
To normalize a relation, we have to remove any redundancies from the relationships between database objects/tables and simplify the structure. This also means simplifying many-to-many relationships. In this question, we see there is a many-to-many relationship between flights and passengers. To resolve this we can introduce a join table which simplifies this relationship to a one-to-many between the objects.
Answer:
mathematica ,is a software program helps to solve maths equation ...
Explanation:
mark me as brainliest ❤️
I would go with c. Audio and visuals catch peoples attention
Answer:
a) 4 processes
b) 2 resources
c) R1: 2 instances
R2: 2 instances
d) R2
e) R1
f) R1
g) No resource
h) R1
i) R2
j) R2
k) No
l)No deadlock
Explanation:
You need to know that the resources that are required by the processes for completion are shown by the request edge and the resources allocated are shown by the allocation edge. And thus, we can find what resources are allocated to the process, and required for the completion accordingly. And here again, the cycle is created, and hence deadlock may or may not occur. However, we see that resources have multiple instances and get freed on time. And hence, deadlock does not occur.
Like,
P4 uses R2 and free one instance of R2.
P3 then uses one instance of R2 and free R2.
P3 then uses one instance of R3 and free R3.
P1 uses one instance of R1 and free R1 one instance
P1 then uses R2 and free R2.
P2 uses R1 and free R1.
Hence, all the processes are complete and deadlock does not occur.
