Question

Answer ASAP for five stars and brainliest!

Answer 1

Part 1:

Given that the half-life of a radioactive kind of antimony is 60 days. The amount of substance left of a radioactive substance with a half life of $t_{ \frac{1}{2} }$ after time t, is given by:

$N(t)=N_0\left( \frac{1}{2} \right)^{ \frac{t}{t_{ \frac{1}{2} }} }$

where: $N_0$ is the original amount of the substance.

Thus, in 240 days, the amount that will be left of the antimony given that the original amount of the antimony is 960 grams is given by:

$N(t)=960\left( \frac{1}{2} \right)^{ \frac{240}{60} } \\ \\ =960\left( \frac{1}{2} \right)^4=960\left( \frac{1}{16} \right) \\ \\ =60\ grams$


Part 2:

Given that the half-life of a radioactive kind of tellurium is 70 minutes. The amount of substance left of a radioactive substance with a half life of $t_{ \frac{1}{2} }$ after time t, is given by:

$N(t)=N_0\left( \frac{1}{2} \right)^{ \frac{t}{t_{ \frac{1}{2} }} }$

where: $N_0$ is the original amount of the substance.

Thus, in 280 minutes, the amount that will be left of the tellurium given that the original amount of the antimony is 480 grams is given by:

$N(t)=480\left( \frac{1}{2} \right)^{ \frac{280}{70} } \\ \\ =480\left( \frac{1}{2} \right)^4=480\left( \frac{1}{16} \right) \\ \\ =30\ grams$


Part 3:

Given that a town's population is currently 10,000. If the population of the town doubles every 38 years, than in 76 years from now, the population of the town must have doubled twice, thus in 76 years from now, the population of the town is given by:

$10000(2)^2=10,000(4)=40,000$