Mathematics, 24.02.2020 00:47, eliza35

The half-life of a radioactive substance is 50 years. If you start with some amount of this substance, what fraction will remain in 200 years? What fraction will remain in 300 years?

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The fraction will remain in 200 years is 1/16 and the fraction will remain in 300 years is 1/64 and this can be determined by using the formula of half-life.

Given :

The half-life of a radioactive substance is 50 years.

The Half-life of any substance is given by the formula:

$\rm A = A_02^{\frac{-t}{h}}$ --- (1)

where $\rm A_0$ is the initial quantity of the substance, A is the quantity of the substance remaining, t is the time elapsed and h is the half-life of the substance.

Now put the value of $\rm A_0$ , t and h in the equation (1).

$\rm A = 1\times 2^{\frac{-200}{50}}$

$\rm A = 1\times 2^{-4}$

$\rm A = \dfrac{1}{16}$

The fraction will remain in 200 years = 1/16

Now, for t = 300 the quantity of the substance remaining is:

$\rm A = 1\times 2^{\frac{-300}{50}}$

$\rm A = 1\times 2^{-6}$

$\rm A = \dfrac{1}{64}$

The fraction will remain in 300 years = 1/64

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answered: Kdub4evaa

The fraction will remain in 200 years is $\frac{1}{16}$ and in 300 years is $\frac{1}{64}$ .

Step-by-step explanation:

Given:

The half-life of a radioactive substance is 50 years.

Now, to find the fraction will remain in 200 years and fraction will remain in 300 years.

So, to get the fraction in 200 years we put formula:

Let the amount initial be 1.

Ao = 1 (amount initial).

t = 200 years (time).

h = 50 (half-life of a substance)

A = remaining amount after time.

$A=Ao2^\frac{-t}{h}$

$A=Ao\times 2^\frac{-t}{h}\\$

$A=1\times 2^\frac{-200}{50}$

$A=1\times 2^{-4}$

$A=1\times \frac{1}{2^4}$

$A=1\times \frac{1}{16}$

$A=\frac{1}{16}$

The fraction will remain in 200 years = $\frac{1}{16} .$

Now, to get the fraction will remain in 300 years we put same formula but the time will change:

t = 300 years (time).

$A=Ao\times 2^\frac{-t}{h} \\\\A=1\times 2^\frac{-300}{50} \\\\A=1\times 2^{-6} \\\\A=1\times \frac{1}{2^6} \\\\A=\frac{1}{64}$

The fraction will remain in 300 years = $\frac{1}{64} .$

Therefore, the fraction will remain in 200 years is $\frac{1}{16}$ and in 300 years is $\frac{1}{64}$ .

answered: Guest

. + λrh() e 1 1 2 4 4 4 4 4 3 4 4 4 4 4 + th−1(− s(x)2 dx ω + λrh−1() e 2 1 2 4 4

step-by-step explanation:

answered: Guest

answer: a video game has three difficulty levels: easy, medium, and expert. the game's scoring works in such a way that a player has to score a fixed number of points to advance to the next level, and then the scoring starts over. the total score needed to clear all three difficulty levels is 7,500. two times the number of points needed to clear the easy level is 500 more than the number of points needed to clear the medium level. twice the number of points needed to clear the medium difficulty level is 1,500 more than the number of points needed to clear the expert level. if x, y, and z represent the easy, medium, and expert levels, respectively, and a is the coefficient matrix of the system of equations modeling this situation, identify the inverse matrix, a-1, and the solution matrix, x, of the system of equations.

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The half-life of a radioactive substance is 50 years. If you start with some amount of this substanc...