Mathematics, 27.05.2021 04:00 bunng7387

Element X is a radioactive isotope such that every 69 years, its mass decreases by half. Given that the initial mass of a sample of Element X is 20 grams, how much of the element would remain after 30 years, to the nearest whole number?

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Answer from: Quest

what "following statement"?

step-by-step explanation:

Answer from: Quest

step-by-step explanation:

i found the life tables for the year 2012.

life table of males at age 30. probability of dying between ages x and x+1

total population:

29–30 . . . . . . . . . . . . 0.001030

30–31 . . . . . . . . . . . . 0.001061

males:

29–30 . . . . . . . . . . . . 0.001420

30–31 . . . . . . . . . . . . 0.001448

white population

29–30 . . . . . . . . . . . . 0.001005

30–31 . . . . . . . . . . . . 0.001037

white males

29–30 . . . . . . . . . . . . 0.001372

30–31 . . . . . . . . . . . . 0.001399

non hispanic white population

29–30 . . . . . . . . . . . . 0.001083

30–31 . . . . . . . . . . . . 0.001116

non hispanic white males

29–30 . . . . . . . . . . . . 0.001477

30–31 . . . . . . . . . . . . 0.001503

jacob, non hispanic white male. 30 years old.

probability that jacob lives to be 30 years old.

since the given probabilities are probabilities of dying, i think we simply subtract it from 1 to get the probability of living. we will use the probability of non hispanic white male.

probability of living to be 30 y.o.

29 - 30 : 1 - 0.001477 = 0.998523

30 - 31 : 1 - 0.001503 = 0.998497

Answer from: Quest

One shot. if 60% of 10 is 6, then what is 70%? we know it would be seven (since 60% is mathematically 60/100, or 6/10, then 70% is 70/100, or 7/10) but we can check anyway using ratios: 6 : 60% = x : 70%*inverse multiplicationx * 60% = 6 * 70%*divide both sides by 60%x = (6 * 70%) / 60%x = 7and since he has made 6 shots already, he would need just one more to get 7 shots or 70%. if the explanation is too confusing, comment and i will try to

Answer from: Quest

$3^{4x-5}=\left(\dfrac{1}{27}\right)^{2x+10}\\\\3^{4x-5}=\left(\dfrac{1}{3^3}\right)^{2x+10}\qquad\text{use}\ a^{-n}=\dfrac{1}{a^n}\\\\3^{4x-5}=\left(3^{-3}\right)^{2x+10}\qquad\text{use}\ (a^n)^m=a^{n\cdot m}\\\\3^{4x-5}=3^{-3(2x+10)}\qquad\text{use distributive property}\\\\3^{4x-5}=3^{(-3)(2x)+(-3)(10)}\\\\3^{4x-5}=3^{-6x-30}\iff4x-5=-6x-30\qquad\text{add 5 to both sides}\\\\4x=-6x-25\qquad\text{add 6x to both sides}\\\\10x=-25\qquad\text{divide both sides by 10}\\\\\boxed{x=-2.5}$

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Element X is a radioactive isotope such that every 69 years, its mass decreases by half. Given that...

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