How do you do these questions?

Step-by-step explanation:

(a)

g(x) = d/dx [2/√π ∫₀ˣ e^(-t²) dt]

g(x) = 2/√π e^(-x²)

g(-x) = 2/√π e^(-(-x)²) = 2/√π e^(-x²)

g(-x) = g(x)

Therefore, g(x) is even.

(b)

erf(x) = 2/√π ∫₀ˣ e^(-t²) dt

erf(x) = ∫₀ˣ g(t) dt

erf(-x) = ∫₀⁻ˣ g(t) dt

erf(-x) = -∫₋ₓ⁰ g(t) dt

Since g(t) is even, it is symmetrical about x=0.

erf(-x) = -∫₀ˣ g(t) dt

erf(-x) = -erf(x)

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.