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Liono4ka [1.6K]

3 years ago

15

The amount, A, in milligrams, of radioactive material remaining in a container can be modeled by the exponential function A(t)=5

(0.5)^0.25t where T is time, in years. Based on this model how many years does it take for half of the original radioactive material to be left remaining?

1 answer:

Arturiano [62]3 years ago

3 0

Answer:

4 years

Step-by-step explanation:

The exponent of 0.5 is 1 when t=4. So, half the original amount will be remaining when t=4.

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3 years ago

The half-life of a radioactive element is 130 days, but your sample will not be useful to you after 80% of the radioactive nu

gtnhenbr [62]

Answer:

We can use the sample about 42 days.

Step-by-step explanation:

Decay Equation:

$\frac{dN}{dt}\propto -N$

$\Rightarrow \frac{dN}{dt} =-\lambda N$

$\Rightarrow \frac{dN}{N} =-\lambda dt$

Integrating both sides

$\int \frac{dN}{N} =\int\lambda dt$

$\Rightarrow ln|N|=-\lambda t+c$

When t=0, N= $N_0$ = initial amount

$\Rightarrow ln|N_0|=-\lambda .0+c$

$\Rightarrow c= ln|N_0|$

$\therefore ln|N|=-\lambda t+ln|N_0|$

$\Rightarrow ln|N|-ln|N_0|=-\lambda t$

$\Rightarrow ln|\frac{N}{N_0}|=-\lambda t$ .......(1)

$\frac{N}{N_0}=e^{-\lambda t}$ .........(2)

Logarithm:

$ln|\frac mn|= ln|m|-ln|n|$
$ln|ab|=ln|a|+ln|b|$

$ln|e^a|=a$

$ln|a|=b \Rightarrow a=e^b$
$ln|1|=0$

130 days is the half-life of the given radioactive element.

For half life,

$N=\frac12 N_0$ , $t=t_\frac12=130$ days.

we plug all values in equation (1)

$ln|\frac{\frac12N_0}{N_0}|=-\lambda \times 130$

$\rightarrow ln|\frac{\frac12}{1}|=-\lambda \times 130$

$\rightarrow ln|1|-ln|2|-ln|1|=-\lambda \times 130$

$\rightarrow -ln|2|=-\lambda \times 130$

$\rightarrow \lambda= \frac{-ln|2|}{-130}$

$\rightarrow \lambda= \frac{ln|2|}{130}$

We need to find the time when the sample remains 80% of its original.

$N=\frac{80}{100}N_0$

$\therefore ln|{\frac{\frac {80}{100}N_0}{N_0}|=-\frac{ln2}{130}t$

$\Rightarrow ln|{{\frac {80}{100}|=-\frac{ln2}{130}t$

$\Rightarrow ln|{{ {80}|-ln|{100}|=-\frac{ln2}{130}t$

$\Rightarrow t=\frac{ln|80|-ln|100|}{-\frac{ln|2|}{130}}$

$\Rightarrow t=\frac{(ln|80|-ln|100|)\times 130}{-{ln|2|}}$

$\Rightarrow t\approx 42$

We can use the sample about 42 days.

7 0

3 years ago

Let's say you take an exam worth 240 points and your score marked the 78th percentile of all the exam scores. Based on this you

gladu [14]

Answer:

28%

Step-by-step explanation:

78th percentile means 78% scored less than/equal to you.

100 - 78 = 22%

3 0

3 years ago

Mrs. Davis bought 6 pounds of grades for $18. At this rate, how much would 72 ounces of grapes have cost?

saveliy_v [14]

This the problem we are looking at: 6 pounds/18 = 72 ounces /?
To solve this, we have to make the unit of measurement the same. There are 16 ounces in 1 pound. So, there are 72 ounces in 4.5. To find that you just do 72 divided by 16. This is the new problem we are looking at: 6 pounds/18 = 4.5 pounds/?
To find ?, you must divided 6 divided by 4.5 to find the difference between the two to keep the same rate. That equals 1.33333333.
What is done to the numerator, you must do to the denominator to keep the same rate so now you divide 18 divided by 1.33333333 which equals 13.5

So for 72 ounces (4.5 pounds) it will cost $13.50

I hope this helps:)

4 0

3 years ago