A sample of radium has a weight of 100 mg and a half-life of approximately 6 years.
1. How much of the sample will remain after 6 years?
3 years?
1 year?
* Round all answers to the nearest 10th place.
2. Find a function f which models the amount of radium f(t), in mg,
remaining after t years.

1a ) N_6 = 50 mg

1b) N_3 = 70.7 mg

1c) N_1 = 89.1 mg

2) f(t) = 100e^(-0.1155t)

Step-by-step explanation:

Formula to get amount remaining is;

N_t = N_o × ½^(t/t_½)

We are given;

N_o = 100 mg

t_½ = 6 years

We have; N_t = 100(½^(t/6))

Thus;

1a) amount left after 6 years is;

N_6 = 100 × ½^(6/6)

N_6 = 50 mg

1b) amount left after 3 years is;

N_3 = 100 × ½^(3/6)

N_3 = 70.7 mg

1c) amount left after 1 year is;

N_1 = 100 × ½^(1/6)

N_1 = 89.1 mg

2) to find the function to model the amount left after t years, let's use the formula;

f(t) = f_o × e^(-rt)

Where r is decay constant

Thus;

f(t) = 100e^(-rt)

We recall that N_t = 100(½^(t/6))

Thus;

100(½^(t/6)) = 100e^(-rt)

Divide both sides by 100 to get;

(½^(t/6)) = e^(-rt)

In (½^(t/6)) = In e^(-rt)

Using power rule, we have;

(t/6) In ½ = -rt In e

-0.6931t/6 = - rt (because In e = 1)

t will cancel out to give;

r = 0.6931/6

r = 0.1155

Thus;

f(t) = 100e^(-0.1155t)

ab will be equal to

answer: all of the odd numbers were opened

the first person was the one that was opening all of the lockers. since one is odd, this causes all of the odd number lockers to be opened in the end.