We can start by substituting each value from the given sets into the inequality and checking which ones make the inequality true:
A. {−2, −1, 3}
For n = −2, 3(n – 2) = 3(-2 – 2) = −12 and 2n – 3 = 2(-2) – 3 = −7, so −12 ≤ −7 is false.
For n = −1, 3(n – 2) = 3(-1 – 2) = −9 and 2n – 3 = 2(-1) – 3 = −5, so −9 ≤ −5 is false.
For n = 3, 3(n – 2) = 3(3 – 2) = 3 and 2n – 3 = 2(3) – 3 = 3, so 3 ≤ 3 is true.
B. {−2, −1, 4}
For n = −2, 3(n – 2) = 3(-2 – 2) = −12 and 2n – 3 = 2(-2) – 3 = −7, so −12 ≤ −7 is false.
For n = −1, 3(n – 2) = 3(-1 – 2) = −9 and 2n – 3 = 2(-1) – 3 = −5, so −9 ≤ −5 is false.
For n = 4, 3(n – 2) = 3(4 – 2) = 6 and 2n – 3 = 2(4) – 3 = 5, so 6 ≤ 5 is false.
C. {−1, 0, 3}
For n = −1, 3(n – 2) = 3(-1 – 2) = −9 and 2n – 3 = 2(-1) – 3 = −5, so −9 ≤ −5 is false.
For n = 0, 3(n – 2) = 3(0 – 2) = −6 and 2n – 3 = 2(0) – 3 = −3, so −6 ≤ −3 is false.
For n = 3, 3(n – 2) = 3(3 – 2) = 3 and 2n – 3 = 2(3) – 3 = 3, so 3 ≤ 3 is true.
D. {−1, 1, 4}
For n = −1, 3(n – 2) = 3(-1 – 2) = −9 and 2n – 3 = 2(-1) – 3 = −5, so −9 ≤ −5 is false.
For n = 1, 3(n – 2) = 3(1 – 2) = −3 and 2n – 3 = 2(1) – 3 = −1, so −3 ≤ −1 is false.
For n = 4, 3(n – 2) = 3(4 – 2) = 6 and 2n – 3 = 2(4) – 3 = 5, so 6 ≤ 5 is false.
E. {1, 0, 5}
For n = 1, 3(n – 2) = 3(1 – 2) = −3 and 2n – 3 = 2(1) – 3 = −1, so −3 ≤ −1 is false.
For n = 0, 3(n – 2) = 3(0 – 2) = −6 and 2n – 3 = 2(0) – 3 = −3, so −6 ≤ −3 is false.
For n = 5, 3(n – 2) = 3(5 – 2) = 9 and 2n – 3 = 2(5) – 3 = 7, so 9 ≤ 7 is false.
So, only for n = 3, the inequality 3(n – 2) ≤ 2n – 3 is true. Hence, the only set of values that makes the inequality true is C. {−1, 0, 3}.