(28 ⋅ 3−5 ⋅ 60)−2 ⋅ 3 to the power of negative 2 over 2 to the power of 3, whole to the power of 4 ⋅ 228

The answer would look like this:

(2^8 × 5^–5 × 19^0)–2 ×× 228 = (2^–16 × 5^10 × 19^0) × × 2^28 = (2^–16 × 5^10 × 1) × × 2^28 = (2^–16 × 5^10 × 1) × 5^–8 × 2^–12 × 2^28 = (2^–16 × 2^–12 × 2^28) ×(5^10 ×5^–8) = 2^–28 × 2^28× 5^2 = 2^0 × 25 = 25

a) one student is chosen from classroom a, then a student is chosen from classroom b.  

step-by-step explanation:

independent events are ones in which the probability of one event does not affect the probability of the second event.

one student is chosen from classroom a; this does not affect the chances of choosing a student from classroom b.   this makes these independent events.

when one card is chosen from a standard deck and then set aside before a second card is chosen, the probability of the second card being chosen changes.   this makes them dependent events.

when one student is chosen from classroom a and then a second student is chosen from classroom a, the probability of choosing the second student changes.   this makes them dependent events.

when a letter is chosen from the alphabet and then a second letter is chosen, the probability of choosing the second letter changes.   this makes them dependent events.