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Ne4ueva [31]
3 years ago
7

Terbium-160 has a half-life of about 72 days. After 396 days, about how many milligrams of a 220 mg sample will remain?

Mathematics
2 answers:
liberstina [14]3 years ago
8 0
C....................
garik1379 [7]3 years ago
3 0

Answer:

Option A = 5 mg  

Step-by-step explanation:

Given : Terbium-160 has a half-life of about 72 days.

To find : After 396 days, about how many milligrams of a 220 mg sample will remain?

Solution :

We have given the Terbium-160 has a half-life of about 72 days.  

We can represent the situation with an exponential function,

A_t = A_0(0.5)^{\frac{t}{n}}

Where,

A_t is the amount at any time t,

A_0=220 is the original amount,

n=72 is the half-life

t=365 number of days

Substituting all the values,

A_t =220(0.5)^{\frac{396}{72}}

A_t =220(0.5)^{5.5}

A_t =220(0.022)

A_t =4.84

Approximately 4.84=5 mg

Therefore, Option A is correct.  

After 396 days, there will only be 5 mg of Terbium-160.                        

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ivann1987 [24]

Answer:

The right answer is :

(a) 23,200

(b) 24,514

(c) 22,926

Step-by-step explanation:

According to the question,

P_1 = 12000

P_2 = 20000

P_{sat}=80000

(a)

We know that the arithmetic growth formula will be:

⇒ P=Pi+K\times t...(1)

here,

⇒ K=\frac{P_2-P_1}{\Delta t}

        =\frac{20000-12000}{25}

        =\frac{80000}{25}

        =320

On putting the values in equation (1), we get

⇒ P_{2020}=20000+320\times 10

             =23,200

(b)

The geometric growth formula will be:

⇒ P=ln(Pi)+K\times t

here,

⇒ K=\frac{lnP_2-lnP_1}{\Delta t}

         =\frac{ln(20000)-ln(12000)}{25}

By putting the values of general log, we get

hence,

⇒ P_f=ln(20000)+0.0204\times 10

         =10.107

P_{2020}=e^{10.107}

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(c)

⇒ P_f=P_{sat}-(P_{sat}-P_i)e^{-K\times t}

or,

⇒ K=-\frac{1}{\Delta t}ln(\frac{P_{sat}-P_2}{P_{sat}-P_1} )

from here, we get

        =0.005

hence,

⇒ P_{2020}=80000-(80000-20000)e^{-0.005\times 10}

             =80000-(60000)e^{-0.005\times 10}

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5 0
3 years ago
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lidiya [134]
<h2>The reciprocals of the fraction \dfrac{37}{74} =  \dfrac{74}{37} or 2.</h2>

Step-by-step explanation:

We have,

\dfrac{37}{74}

To find, the reciprocals of the fraction \dfrac{37}{74} = ?

∴ \dfrac{37}{74}

Dividing numerator and denominator by 37, we get

= \dfrac{1}{2}

= \dfrac{37}{74} or, \dfrac{1}{2}

We know that,

The reciprocals of the fraction \dfrac{x}{y} =\dfrac{y}{x}

The reciprocals of the fraction \dfrac{37}{74} = \dfrac{74}{37} or \dfrac{2}{1}

= \dfrac{74}{37} or 2

Thus, the reciprocals of the fraction \dfrac{37}{74} =  \dfrac{74}{37} or 2.

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navik [9.2K]
M/(30-15) = m/(30+15) + 1
m/15 = m/45 + 1
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Answer: Yes you are correct. The answer is choice A

============================================================

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So by symmetry, P(Z > -1) = 0.84 approximately as well.

We'll convert the z score z = -1 into its corresponding x score

z = (x-mu)/sigma

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x-180 = -15

x = -15+180

x = 165

We don't land on any of the answer choices listed, but we get fairly close to 165.2, which is choice A. So you are correct.

I have a feeling that the table you have is probably more accurate than the one I'm using, so it's possible that you'd land exactly on 165.2 when following the steps above.

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