At a time hours after taking a tablet, the rate at which a drug is being eliminated r(t)= 50 (e^-01t - e^-0.20t)is mg/hr. Assuming that all the drug is eventually eliminated, calculate the original dose.

2500 mg

Step-by-step explanation:

Since r(t) is the rate at which the drug is being eliminated, we integrate r(t) with t from 0 to ∞ to find the original dose of drug, m. Since all of the drug will be eliminated at time t = ∞.

Since r(t) =  50 (e^-01t - e^-0.20t)

m = ∫₀⁰⁰50 (e^-01t - e^-0.20t)

= 50∫₀⁰⁰(e^-01t - e^-0.20t)

= 50[∫₀⁰⁰e^-01t - ∫₀⁰⁰e^-0.20t]

= 50([e^-01t/-0.01]₀⁰⁰ - [e^-0.20t/-0.02]₀⁰⁰)

= 50(1/-0.01[e^-01(∞) - e^-01(0)] - {1/-0.02[e^-0.02(∞) - e^-0.02(0)]})

= 50(1/-0.01[e^-(∞) - e^-(0)] - {1/-0.02[e^-(∞) - e^-(0)]})

= 50(1/-0.01[0 - 1] - {1/-0.02[0 - 1]})

= 50(1/-0.01[- 1] - {1/-0.02[- 1]})

= 50(1/0.01 - 1/0.02)

= 50(100 - 50)

= 50(50)

= 2500 mg

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.

it is true

step-by-step explanation:

all the combinations add up i used   calculator