Ask question

Ask question

All categories

3 years ago

8

The amount of carbon-14 present in animal bones after t years is given by P(t)equalsUpper P 0 e Superscript negative 0.00012 t

. A bone has lost 19% of its carbon-14. How old is the bone?

1 answer:

Vera_Pavlovna [14]3 years ago

7 0

Answer:

age of bone is 1756 years

Step-by-step explanation:

The amount of carbon-14 present in animal bones after t years

$P(t)= P_0e^{-0.00012t}$

P(t) is the carbon present

19% has lost. so carbon present is 100-19 = 81% present

out of 100 81 is presents

so P0 is 100

P(t) is 81

$81= 100e^{-0.00012t}$

divide by 100 on both sides

$0.81= e^{-0.00012t}$

take ln on both sides

$ln(0.081)=-0.00012tln(e)$

$ln(0.081)=-0.00012t$

divide both sides by -0.00012

t=1756.0085

age of bone is 1756 years

You might be interested in

A common inhabitant of human intestines is the bacterium Escherichia coli. A cell of this bacterium in a nutrient-broth medium d

nikitadnepr [17]

Answer:

a) k=2.08 1/hour

b) The exponential growth model can be written as:

$P(t)=Ce^{kt}$

c) 977,435,644 cells

d) 2.033 billions cells per hour.

e) 2.81 hours.

Step-by-step explanation:

We have a model of exponential growth.

We know that the population duplicates every 20 minutes (t=0.33).

The initial population is P(t=0)=58.

The exponential growth model can be written as:

$P(t)=Ce^{kt}$

For t=0, we have:

$P(0)=Ce^0=C=58$

If we use the duplication time, we have:

$P(t+0.33)=2P(t)\\\\58e^{k(t+0.33)}=2\cdot58e^{kt}\\\\e^{0.33k}=2\\\\0.33k=ln(2)\\\\k=ln(2)/0.33=2.08$

Then, we have the model as:

$P(t)=58e^{2.08t}$

The relative growth rate (RGR) is defined, if P is the population and t the time, as:

$RGR=\dfrac{1}{P}\dfrac{dP}{dt}=k$

In this case, the RGR is k=2.08 1/h.

After 8 hours, we will have:

$P(8)=58e^{2.08\cdot8}=58e^{16.64}=58\cdot 16,852,338= 977,435,644$

The rate of growth can be calculated as dP/dt and is:

$dP/dt=58[2.08\cdot e^{2.08t}]=120.64e^2.08t=2.08P(t)$

For t=8, the rate of growth is:

$dP/dt(8)=2.08P(8)=2.08\cdot 977,435,644 = 2,033,066,140$

(2.033 billions cells per hour).

We can calculate when the population will reach 20,000 cells as:

$P(t)=20,000\\\\58e^{2.08t}=20,000\\\\e^{2.08t}=20,000/58\approx344.827\\\\2.08t=ln(344.827)\approx5.843\\\\t=5.843/2.08\approx2.81$

3 0

3 years ago

I need help with this question

Murrr4er [49]

The number of cows is given by

$14\cdot 2^{\frac{y}{5}}$

So, after $k$ years, the number of cows will be

$14\cdot 2^{\frac{y+k}{5}}$

We want this number to be twice as much as the original:

$14\cdot 2^{\frac{y+k}{5}} = 2(14\cdot 2^{\frac{y}{5}})$

First of all, we can cancel 14 from both sides:

$2^{\frac{y+k}{5}} = 2\cdot 2^{\frac{y}{5}}$

Finally, on the right hand side, we can use the exponent rule

$a^b\cdot a^c=a^{b+c}$

to get

$2^{\frac{y+k}{5}} = 2^{\frac{y}{5}+1}$

To solve this equation, we must impose that the two exponents are the same:

$\dfrac{y+k}{5} = \dfrac{y}{5}+1 \iff \dfrac{y+k}{5} = \dfrac{y+5}{5}$

And clearly this is true if and only if $k=5$ . So, it will take 5 years for the cow heard to double in number.

You can do the exact same steps to find the doubling time for the sheeps.

8 0

3 years ago

Can 1,1,2 be the length is a triangle

Sholpan [36]

Alright, so in order to find this, you will nee to use the Triangle Inequality Theorem.
We’ll make the values = to variables to make it easier. a=1 b=1 c=2
All of the inequities need to be true.
a+b>c
b+c>a
c+a>b
Now replace the variables.
1+1>2 2>2 (False)
1+2>1 3>1 (True)
1+2>1 3>1 (True)
Since one is false, 1,1, and 2 cannot make a triangle.

4 0

3 years ago

Which measurements are for a rectangle with an area of 16 square units?

alexandr1967 [171]

Answer:

2 Units wide , 6 units long

4 0

3 years ago

Help please ?????? I don't understand

Stells [14]

Answer:

It's going up by a difference of 70, & 2.

Step-by-step explanation:

4 0

3 years ago