Answer: Concurrent programs are the programs that execute at the same point of time. They are simultaneous in nature with other concurrent programs.They are executed with the help of threads to achieve the concurrency without the issue of scheduling. They are consider long running programs.
The program that has the low execution time compared with all other programs gets executed first.If the case of no progress is seen then another thread is executed.This type of execution gives the non- deterministic situation in which the possibility of leading completion of any particular program cannot be determined.
Answer:
B
Explanation:
When you initialize an instance of FunEvent(tags, year) and assign it to bc. The instance variables in this case are: self.tags = ["g", "ml"] and self.year = 2022. But then you alter tags, which will also change self.tags, since self.tags is a reference to the list you passed in as an argument. This is not the case when you do year=2023 because, first of all, integers are not mutable, and also because even if somehow integers were mutable, you're not changing the object in-place, you're simply changing the where the "variable" is pointing to. So for example if you did tags = ["g", "ml", "bc"] instead of tags.append("bc"), it would also not change the value of the instance variable "tags", because you wouldn't be changing the object in-place. So when you print(bc), the instance variables will be ["g", "ml", "bc"] and 2022. When you try to print an object, it call try to convert it into a string using the __str__ magic method. In this case it will return a string formatted as "Event(tags={self.tags}, year={self.year}) which will output "Event(tags=['g', 'ml', 'bc'], year=2022)" So the correct answer is B
Numeric Keypad is the answer
let's say they catch-up to each other after t hour.
so
in t hour distance travelled by Renee
d = speed × time = 50t
in t hour distance travelled by Kim
d = speed × time = 60(t-1) + 0×1 = 60(t-1)
Note: here kim didn't covered any distance in first hour and in rest t-1 hour it travelled all distance
now.as distance travelled by both is same so
60(t-1) = 50t
60t -60 = 50t
adding 60 both sides
60t = 50t+60
subtracting 50t both sides
60t -50t = 50t+60 - 50t
10t = 60
dividing by 10 both sides
t = 60/10 = 6
so in 6 hour both will catch-up to each other
