Answer :
Below is a detailed, step-by-step solution.
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1. A medicine has a half-life of 8 hours. If the initial dose is [tex]$2000$[/tex] milligrams, the amount remaining in the body after [tex]$t$[/tex] hours is given by
[tex]$$
A(t) = 2000 \left(\frac{1}{2}\right)^{t/8}.
$$[/tex]
(a) To find the amount remaining after [tex]$20$[/tex] hours, substitute [tex]$t = 20$[/tex]:
[tex]$$
A(20) = 2000 \left(\frac{1}{2}\right)^{20/8} = 2000 \left(\frac{1}{2}\right)^{2.5}.
$$[/tex]
Evaluating [tex]$\left(\frac{1}{2}\right)^{2.5}$[/tex] gives approximately [tex]$0.17678$[/tex], so
[tex]$$
A(20) \approx 2000 \times 0.17678 \approx 353.55 \text{ milligrams}.
$$[/tex]
(b) For [tex]$2$[/tex] days, note that [tex]$2$[/tex] days equal [tex]$48$[/tex] hours. Substitute [tex]$t = 48$[/tex]:
[tex]$$
A(48) = 2000 \left(\frac{1}{2}\right)^{48/8} = 2000 \left(\frac{1}{2}\right)^{6}.
$$[/tex]
Since [tex]$\left(\frac{1}{2}\right)^6 = 0.015625$[/tex], we have
[tex]$$
A(48) \approx 2000 \times 0.015625 = 31.25 \text{ milligrams}.
$$[/tex]
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2. The number of people infected with a virus doubles every three days. If the initial number of infected people is [tex]$100$[/tex], the number after [tex]$t$[/tex] days is given by
[tex]$$
P(t) = 100 \cdot 2^{t/3}.
$$[/tex]
(a) After [tex]$9$[/tex] days, substitute [tex]$t = 9$[/tex]:
[tex]$$
P(9) = 100 \cdot 2^{9/3} = 100 \cdot 2^{3}.
$$[/tex]
Since [tex]$2^3 = 8$[/tex], it follows that
[tex]$$
P(9) = 100 \cdot 8 = 800 \text{ people}.
$$[/tex]
(b) After [tex]$15$[/tex] days, substitute [tex]$t = 15$[/tex]:
[tex]$$
P(15) = 100 \cdot 2^{15/3} = 100 \cdot 2^{5}.
$$[/tex]
Since [tex]$2^5 = 32$[/tex], then
[tex]$$
P(15) = 100 \cdot 32 = 3200 \text{ people}.
$$[/tex]
________________________________
In summary:
1. (a) After 20 hours, approximately [tex]$353.55$[/tex] milligrams remain.
(b) After 2 days (48 hours), approximately [tex]$31.25$[/tex] milligrams remain.
2. (a) After 9 days, there will be [tex]$800$[/tex] infected people.
(b) After 15 days, there will be [tex]$3200$[/tex] infected people.
________________________________
1. A medicine has a half-life of 8 hours. If the initial dose is [tex]$2000$[/tex] milligrams, the amount remaining in the body after [tex]$t$[/tex] hours is given by
[tex]$$
A(t) = 2000 \left(\frac{1}{2}\right)^{t/8}.
$$[/tex]
(a) To find the amount remaining after [tex]$20$[/tex] hours, substitute [tex]$t = 20$[/tex]:
[tex]$$
A(20) = 2000 \left(\frac{1}{2}\right)^{20/8} = 2000 \left(\frac{1}{2}\right)^{2.5}.
$$[/tex]
Evaluating [tex]$\left(\frac{1}{2}\right)^{2.5}$[/tex] gives approximately [tex]$0.17678$[/tex], so
[tex]$$
A(20) \approx 2000 \times 0.17678 \approx 353.55 \text{ milligrams}.
$$[/tex]
(b) For [tex]$2$[/tex] days, note that [tex]$2$[/tex] days equal [tex]$48$[/tex] hours. Substitute [tex]$t = 48$[/tex]:
[tex]$$
A(48) = 2000 \left(\frac{1}{2}\right)^{48/8} = 2000 \left(\frac{1}{2}\right)^{6}.
$$[/tex]
Since [tex]$\left(\frac{1}{2}\right)^6 = 0.015625$[/tex], we have
[tex]$$
A(48) \approx 2000 \times 0.015625 = 31.25 \text{ milligrams}.
$$[/tex]
________________________________
2. The number of people infected with a virus doubles every three days. If the initial number of infected people is [tex]$100$[/tex], the number after [tex]$t$[/tex] days is given by
[tex]$$
P(t) = 100 \cdot 2^{t/3}.
$$[/tex]
(a) After [tex]$9$[/tex] days, substitute [tex]$t = 9$[/tex]:
[tex]$$
P(9) = 100 \cdot 2^{9/3} = 100 \cdot 2^{3}.
$$[/tex]
Since [tex]$2^3 = 8$[/tex], it follows that
[tex]$$
P(9) = 100 \cdot 8 = 800 \text{ people}.
$$[/tex]
(b) After [tex]$15$[/tex] days, substitute [tex]$t = 15$[/tex]:
[tex]$$
P(15) = 100 \cdot 2^{15/3} = 100 \cdot 2^{5}.
$$[/tex]
Since [tex]$2^5 = 32$[/tex], then
[tex]$$
P(15) = 100 \cdot 32 = 3200 \text{ people}.
$$[/tex]
________________________________
In summary:
1. (a) After 20 hours, approximately [tex]$353.55$[/tex] milligrams remain.
(b) After 2 days (48 hours), approximately [tex]$31.25$[/tex] milligrams remain.
2. (a) After 9 days, there will be [tex]$800$[/tex] infected people.
(b) After 15 days, there will be [tex]$3200$[/tex] infected people.
