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dlinn [17]
3 years ago
7

Please help thanks!!!

Mathematics
1 answer:
lora16 [44]3 years ago
3 0

probablity \: of \: \: six = 29 : 150

<h2><u>hope</u><u> </u><u>it</u><u> </u><u>helps</u><u> </u><u>you</u><u> </u><u>❣❣</u><u> </u></h2>

<h2><em><u>Mark</u></em><em><u> </u></em><em><u>me</u></em><em><u> </u></em><em><u>as</u></em><em><u> </u></em><em><u>brainliest</u></em></h2>

<em><u>in</u></em><em><u> </u></em><em><u>fraction</u></em><em><u>=</u></em><em><u>29</u></em><em><u>/</u></em><em><u>15</u></em><em><u>0</u></em>

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The radioactive isotope of lead, Pb-209, decays at a rate proportional to the amount present at time t and has a half-life of 3.
blagie [28]

Answer:

N(t) =N_o (\frac{1}{2})^{\frac{t}{t_{1/2}}}

Where t_{1/2}= 3.3 hr represent the half life and the intial amount would be N_o = 1

And we want to find the time in order to have a 95% of decay so we can set up the following equation:

0.05 = 1 (0.5)^{t/3.3}

If we apply natural log on both sides we got:

ln(0.05) = \frac{t}{3.3} ln (0.5)

And solving for t we got:

t= 3.3 *\frac{ln(0.05)}{ln(0.5)}= 14.26

So then would takes about 14.26 hours in order to have  95% of the lead to decay

Step-by-step explanation:

For this case we can define the variable of interest amount of Pb209 and for the half life would be given:

N(t) =N_o (\frac{1}{2})^{\frac{t}{t_{1/2}}}

Where t_{1/2}= 3.3 hr represent the half life and the intial amount would be N_o = 1

And we want to find the time in order to have a 95% of decay so we can set up the following equation:

0.05 = 1 (0.5)^{t/3.3}

If we apply natural log on both sides we got:

ln(0.05) = \frac{t}{3.3} ln (0.5)

And solving for t we got:

t= 3.3 *\frac{ln(0.05)}{ln(0.5)}= 14.26

So then would takes about 14.26 hours in order to have  95% of the lead to decay

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3 years ago
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Answer:

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You want it in y = mx + b  form

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Add y to both sides

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-5 + y = -3x

Add 5 to both sides

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\text{We have:}\\\\f(x)=\dfrac{a}{x^3}\to y=\dfrac{a}{x^3}\\\\\text{and the point}\ \left(2,\ \dfrac{5}{4}\right)\to x=2,\ y=\dfrac{5}{4}.\\\\\text{Substitute:}\\\\\dfrac{5}{4}=\dfrac{a}{2^3}\\\\\dfrac{5}{4}=\dfrac{a}{8}\qquad\text{multiply both sides by 8}\\\\a=\dfrac{5}{\not4_1}\cdot\not8^2\\\\\boxed{a=10}

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