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3 years ago

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The half-life of a radioactive isotope is the time it takes for a quantity of the isotope to be reduced to half its initial mass

. Starting with 220 grams of a radioactive isotope, how much will be left after 5 half-lives?

1 answer:

Setler [38]3 years ago

7 0

Answer:

$\boxed{6.875 \text{\: grams}}$

Step-by-step explanation:

Half-lives are how long it takes for half of a substance to decay. If you start with a certain amount and it decays for a certain amount of half-lives, you are dividing by $2^{x}$ , where <em>x</em> is equal to the amount of half-lives.

In order to solve this question, simply divide the starting value by 2 raised to the value of half-lives.

$\large\boxed{\frac{220}{2^{5}}=6.875\text{\: grams}}$

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3 years ago

Animal populations are not capable of unrestricted growth because of limited habitat and food supplies. Under such conditions th

trasher [3.6K]

Answer:

(a) 100 fishes

(b) t = 10: 483 fishes

t = 20: 999 fishes

t = 30: 1168 fishes

(c)

$P(\infty) = 1200$

Step-by-step explanation:

Given

$P(t) =\frac{d}{1+ke^-{ct}}$

$d = 1200\\k = 11\\c=0.2$

Solving (a): Fishes at t = 0

This gives:

$P(0) =\frac{1200}{1+11*e^-{0.2*0}}$

$P(0) =\frac{1200}{1+11*e^-{0}}$

$P(0) =\frac{1200}{1+11*1}$

$P(0) =\frac{1200}{1+11}$

$P(0) =\frac{1200}{12}$

$P(0) = 100$

Solving (a): Fishes at t = 10, 20, 30

$t = 10$

$P(10) =\frac{1200}{1+11*e^-{0.2*10}} =\frac{1200}{1+11*e^-{2}}\\\\P(10) =\frac{1200}{1+11*0.135}=\frac{1200}{2.485}\\\\P(10) =483$

$t = 20$

$P(20) =\frac{1200}{1+11*e^-{0.2*20}} =\frac{1200}{1+11*e^-{4}}\\\\P(20) =\frac{1200}{1+11*0.0183}=\frac{1200}{1.2013}\\\\P(20) =999$

$t = 30$

$P(30) =\frac{1200}{1+11*e^-{0.2*30}} =\frac{1200}{1+11*e^-{6}}\\\\P(30) =\frac{1200}{1+11*0.00247}=\frac{1200}{1.0273}\\\\P(30) =1168$

Solving (c): $\lim_{t \to \infty} P(t)$

In (b) above.

Notice that as t increases from 10 to 20 to 30, the values of $e^{-ct}$ decreases

This implies that:

${t \to \infty} = {e^{-ct} \to 0}$

So:

The value of P(t) for large values is:

$P(\infty) = \frac{1200}{1 + 11 * 0}$

$P(\infty) = \frac{1200}{1 + 0}$

$P(\infty) = \frac{1200}{1}$

$P(\infty) = 1200$

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3 years ago

It x, y, and z are positive intergers and 3x = 4y = 7z, then the least possible value of x + y + z is???

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3 years ago

Scott took out a 4-year car loan for $5,500. He paid back a total of $7,370. What interest rate did he pay for this loan?

elena-14-01-66 [18.8K]

Amount of car loan taken by Scott = $5500
Amount of loan paid back by Scott = $7370
Then
Amount of money paid as interest by Scott = (7370 - 5500) dollars
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