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jok3333 [9.3K]
3 years ago
10

Nickel-63 is a radioactive substance with a half-life of about 100 years. An artifact had

Mathematics
2 answers:
Marina CMI [18]3 years ago
3 0

Answer:

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

asambeis [7]3 years ago
3 0

Answer:

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)

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A. Complete the chart based on the initial conditions:
Nata [24]

Answer:

a.

Month\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ Pop(whole \ ant)\\month1 => x_1=80\times0.94^1=1128\\month2=>x_2=1200\times0.94^2=1060\\month3=>x_3=1200\times0.94^3=996\\month4=>x_4=1200\times0.94^4=936

b.

week Number\ \ \ \ \ \ \ \ \ \ \ \ \ \ Mass(g)\\week1 => x_1=80\times1.1^1=88g\\week2=>x_2=80\times1.1^2=96.8g\\week3=>x_3=80\times1.1^3=106.48g\\week4=>x_4=80\times1.1^4=117.128g

Step-by-step explanation:

a. From the information provided, we can deduce that the population death's follows a Geometric sequence in the form (a,ar,ar^2,ar^3...) where a-first \ term and r-common \ ratio

#Since the population is reducing, r can is obtained as r=1-r=0.94

#The n^t^h term is obtained using the formula x_n=ar^(^n^-^1^), given a=1200

The number of ants alive after every month (in first 4 months)

Month\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ Pop(whole \ ant)\\month1 => x_1=80\times0.94^1=1128\\month2=>x_2=1200\times0.94^2=1060\\month3=>x_3=1200\times0.94^3=996\\month4=>x_4=1200\times0.94^4=936

The ant's alive after 4 months is obtained as the value of x_5

x_n=ar^(^n^-^1^)\\1-x_5=1-1200\times 0.94^4=936.89\\\approx 936

Hence, 936 ants are alive after 4 months.

b. As with the above question, the kitten population follows a geometric sequence: (a,ar,ar^2,ar^3...).

#Since it's a growing population , the common ration is the sum of 100% + the growth rate,

r=1.1 and a=80 and x_n=ar^(^n^-^1^)

The population after 4weeks will be:

week Number\ \ \ \ \ \ \ \ \ \ \ \ \ \ Mass(g)\\week1 => x_1=80\times1.1^1=88g\\week2=>x_2=80\times1.1^2=96.8g\\week3=>x_3=80\times1.1^3=106.48g\\week4=>x_4=80\times1.1^4=117.128g

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