Answer:

Step-by-step explanation:
The probability density function is :

With 0 < x < 3
To be a valid probability density function :

Where a < x < b
And also
f(x) ≥ 0 for a < x < b
Applying this to the probability density function of the exercise :





We can verify by replacing ''c'' in the original probability density function and integrating :


Also, f(x) ≥ 0 for 0 < x < 3
Answer:
It should be the last one, She made a sign error when multiplying.