Problem 1 :
The number of rabbits in a certain population doubles every 40 days. If the population starts with 12 rabbits, what will the population of rabbits be 160 days from now?
Problem 2 :
The population of a western town doubles in size every 12 years. If the population of town is 8,000, what will the population be 18 years from now?
Problem 3 :
The half-life of carbon-14 is approximately 6000 years. How much of 800 g of this substance will remain after 30,000 years?
Problem 4 :
A certain radioactive substance has a half-life of 12 days. This means that every 12 days, half of the original amount of the substance decays. If there are 128 milligrams of the radioactive substance today, how many milligrams will be left after 48 days?
1. Answer :
Doubling-Time Growth Formula :
A = P(2)^{t/d}
Substitute.
P = 12
t = 160
d = 40
Then,
A = 12(2)^{160/40}
= 12(2)^{4}
= 12(16)
= 192
So, the population of rabbits after 160 days from now will be 192.
2. Answer :
Doubling-Time Growth Formula :
A = P(2)^{t/d}
Substitute.
P = 8000
t = 18
d = 12
Then,
A = 8000(2)^{18/12}
= 8000(2)^{1.5}
Use a calculator.
≈ 22,627
So, the population after 18 years from now will be about 22,627.
3. Answer :
Half-Life Decay Formula :
A = P(1/2)^{t/d}
Substitute.
P = 800
t = 30000
d = 6000
Then,
A = 800(1/2)^{30000/6000}
= 800(1/2)^{5}
= 800(0.5)^{5}
= 800(0.03125)
= 25
So, 25 g of carbon-14 will remain after 30,000 years.
4. Answer :
Half-Life Decay Formula :
A = P(1/2)^{t/d}
Substitute.
P = 128
t = 48
d = 12
Then,
A = 128(1/2)^{48/12}
= 128(0.5)^{4}
= 128(0.0625)
= 8
So, 8 milligrams of radio active substance will be left after 48 days.
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