Nickel-63 is a radioactive substance with a half-life of about 100 years. An artifact had
9.8 milligrams of nickel-63 when it was first measured. Write an equation to represent
the mass of nickel-63, in milligrams, as a function of:
a. tt, time in years
b. dd, time in days

see below

Step-by-step explanation:

The equation for half life is

n = no  e ^ (-kt)

Where no is the initial amount of a substance , k is the constant of decay and t is the time

no = 9.8

1/2 of that amount is 4.9 so n = 4.9 and t = 100 years

4.9 = 9.8 e^ (-k 100)

Divide each side by 9.8

1/2 = e ^ -100k

Take the natural log of each side

ln(1/2) = ln(e^(-100k))

ln(1/2) = -100k

Divide each side by -100

-ln(.5)/100 = k

Our equation in years is

n = 9.8  e ^ (ln.5)/100 t)

Approximating ln(.5)/100 =-.006931472

n = 9.8 e^(-.006931472 t)  when t is in years

Now changing to days

100 years = 100*365 days/year

36500 days

Substituting this in for t

4.9 = 9.8 e^ (-k 36500)

Take the natural log of each side

ln(1/2) = ln(e^(-36500k))

ln(1/2) = -36500k

Divide each side by -100

-ln(.5)/36500 = k

Our equation in years is

n = 9.8  e ^ (ln.5)/36500 d)

Approximating ln(.5)/365=-.00001899

n = 9.8 e^(-.00001899 d)  when d is in days

a) x = 9.8 × 0.5^(t/100)

b) x = 9.8 × 0.5^(d/36500)

Step-by-step explanation:

a) x = 9.8 × 0.5^(t/100)

100 years = 365×100 days

= 36500 days

b) x = 9.8 × 0.5^(d/36500)

no association

step-by-step explanation:

if you have eyes and you can use them efficiently you can see that it just looks like random dots like do a line of best fit and like they're so spread out it's such a weak association you might as well consider it nonexistent

1/6

step-by-step explanation: