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2 years ago

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A radioactive substance decays exponentially: The mass at time t is m(t) = m(0)ekt, where m(0) is the initial mass and k is a ne

gative constant. The mean life M of an atom in the substance is M = −k [infinity] tekt dt. 0 For a radioactive isotope, the value of k is −0.000192. Find the mean life of the isotope's atom. (Round your answer to one decimal place.) yr

1 answer:

velikii [3]2 years ago

3 0

Answer:

M = 1/0.000121 = 8264.5 years

Step-by-step explanation:

M = − k ∫∞₀ teᵏᵗdt

To obtain this mean life, we'll use integration by parts to integrate the function ∫ teᵏᵗdt

∫udv = uv - ∫ vdu

u = t

du/dt = 1

du = dt

∫ dv = ∫ eᵏᵗdt

v = eᵏᵗ/k

∫udv = ∫ teᵏᵗdt

uv = teᵏᵗ/k

∫ vdu = eᵏᵗ/k

∫ teᵏᵗdt = (teᵏᵗ/k) - ∫eᵏᵗ/k

But, ∫eᵏᵗ/k = (1/k) ∫eᵏᵗ = (1/k²) eᵏᵗ = eᵏᵗ/k²

∫ teᵏᵗdt = (teᵏᵗ/k) - eᵏᵗ/k²

The rest of the calculation is done on paper in the image attached to this question

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Celia uses 80% of her paycheck to pay her bills. Her bills total $272. How much is Celia paycheck?

Vanyuwa [196]

Step-by-step explanation:

It is 340 dollars in total, I do this in an ubcenventional way of dividing 272 by 80 which gave you 3.4 which is equal to one percent and then multiply that by 20 which is 68 and add that to the other 80 percent of her bill to get 340 dollars

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2 years ago

Consider two people being randomly selected. (For simplicity, ignore leap years.)

inna [77]

Answer:

$(a) = \frac{144}{133225} \\\\(b) = \frac{1}{365}$

Step-by-step explanation:

Part (a) the probability that two people have a birthday on the 9th of any month.

Neglecting leap year, there are 365 days in a year.

There are 12 possible 9th in months that make a year calendar.

If two people have birthday on 9th; P(1st person) and P(2nd person).

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Part (b) the probability that two people have a birthday on the same day of the same month

P(2 people selected have birthday on the same day of same month) + P(2 people selected not having birthday on same day of same month) = 1

P(2 people selected not having birthday on same day of same month):

$= \frac{365}{365} X \frac{364}{365} =\frac{364}{365}$

P(2 people selected have birthday on the same day of same month) $= 1-\frac{364}{365} \\\\= \frac{1}{365}$

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2 years ago

The probability that a randomly selected 2 2-year-old male garter snake garter snake will live to be 3 3 years old is 0.98861 0

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Answer:

a. $Probability = 0.97735$

b. $Probability = 0.92294$

c. $P(At\ Least\ One) = 1$

No, it is not unusual if at least 1 lives up to 3.

Step-by-step explanation:

Given

Represent the probability that a 2 year old snake will live to 3 with P(Live);

$P(Live) = 0.98861$

Solving (a): Probability that two selected will live to 3 years.

Both snakes have a chance of 0.98861 to live up to 3 years.

So, the required probability is:

$Probability = P(Live)\ and\ P(Live)$

$Probability = 0.98861 * 0.98861$

$Probability = 0.9773497321$

$Probability = 0.97735$ <em>--- Approximated</em>

Solving (b): Probability that seven selected will live to 3 years.

All 7 snakes have a chance of 0.98861 to live up to 3 years.

So, the required probability is:

$Probability = P(Live)^n$

Where $n = 7$

$Probability = 0.98861^7$

$Probability = 0.92294324145$

$Probability = 0.92294$ <em>--- Approximated</em>

Solving (c): Probability that at least one of seven selected will not live to 3 years.

In probabilities, the following relationship exist:

$P(At\ Least\ One) = 1 - P(None).$

So, first we need to calculate the probability that none of the 7 lived up to 3.

If the probability that one lived up to 3 years is 0.98861, then the probability than one do not live up to 3 years is 1 - 0.98861

This gives:

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The probability that none of the 7 lives up to 3 is:

$P(None) = P(Not\ Live)^7$

$P(None) = 0.01139^7$

Substitute this value for P(None) in

$P(At\ Least\ One) = 1 - P(None).$

$P(At\ Least\ One) = 1 - 0.01139^7$

$P(At\ Least\ One) = 0.99999999999997513055642436060443621$

$P(At\ Least\ One) = 1$ ---- Approximated

No, it is not unusual if at least 1 lives up to 3.

This is so because the above results, which is 1 shows that it is very likely for at least one of the seven to live up to 3 years

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2 years ago

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sp2606 [1]

Answer: The answer is $\textup{M}+\textup{H}\leq 10~~\textup{and}4\textup{M}+7\textup{H}\leq 90.$

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Since there cannot be more than 10 dances all together, so fisrt inequality can be written as

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Thus, the pair of inequalities is

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2 years ago

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VARVARA [1.3K]

Answer:

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Step-by-step explanation:

Given inequality : $4x + 0.9 \geq 0.1$

We are supposed to find Which of the following is equivalent to the inequality

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2 years ago