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qwelly [4]
3 years ago
13

Element X is a radioactive isotope such that every 14 years, its mass decreases by half. Given that the initial mass of a sample

of Element X is 360 grams, how long would it be until the mass of the sample reached 100 grams, to the nearest tenth of a year?
Mathematics
1 answer:
Alik [6]3 years ago
3 0

Answer:

25.9 years

Step-by-step explanation:

The half life of an isotope is related to the actual amounts by the equation:

N = Ni*(1/2)^(t/HL)

where N and Ni are the final and initial (Ni) amounts, t is the time (in years for this problem) and HL is the half-life, in years.

In this problem, we can write:

100 = 360*(1/2)^(x/14)

I'll reduce this to:

1 = 3.6*(1/2)^(x/14)

This can be solved by taking the log of both sides.

x = 25.87 or 25.9 years

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42 is 250% of what number
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8 0
3 years ago
Geologists estimate the time since the most recent cooling of a mineral by counting the number of uranium fission tracks on the
Oliga [24]

Answer:

(a) The value of P (X = 7) is 0.1388.

(b) The value of P (X ≥ 3) is 0.9380.

(c) The value of P (2 < X < 7) is 0.5433.

(d) \mu_{X}=6

(e) \sigma_{X}=2.45

Step-by-step explanation:

Let <em>X</em> = number of uranium fission tracks on per cm² surface area of the mineral.

The average number of track per cm² surface area is, <em>λ</em> = 6.

The random variable <em>X</em> follows a Poisson distribution with parameter <em>λ</em> = 6.

The probability mass function of a Poisson distribution is:

P(X=x)=\frac{e^{-\lambda}\lambda^{x}}{x!};\ x=0, 1, 2, 3...

(a)

Compute the value of P (X = 7) as follows:

P(X=6)=\frac{e^{-6}(6)^{7}}{7!}=\frac{0.0025\times 279936}{5040}=0.1388

Thus, the value of P (X = 7) is 0.1388.

(b)

Compute the value of P (X ≥ 3) as follows:

P (X ≥ 3) = 1 - P (X < 3)

              = 1 - P (X = 0) - P (X = 1) - P (X = 2)

              =1-\frac{e^{-6}(6)^{0}}{0!}-\frac{e^{-6}(6)^{1}}{1!}-\frac{e^{-6}(6)^{2}}{2!}\\=1-0.00248-0.01487-0.04462\\=0.93803\\\approx0.9380

Thus, the value of P (X ≥ 3) is 0.9380.

(c)

Compute the value of P (2 < X < 7) as follows:

P (2 < X < 7) = P (X = 3) + P (X = 4) + P (X = 5) + P (X = 6)

                   =\frac{e^{-6}(6)^{3}}{3!}+\frac{e^{-6}(6)^{4}}{4!}+\frac{e^{-6}(6)^{5}}{5!}+\frac{e^{-6}(6)^{6}}{6!}\\=0.08924+0.13385+0.16062+0.16062\\=0.54433\\\approx0.5443

Thus, the value of P (2 < X < 7) is 0.5433.

(d)

The mean of the Poisson distribution is:

\mu_{X}=\lambda=6

(e)

The standard deviation of the Poisson distribution is:

\sigma_{X}=\sqrt{\sigma^{2}_{X}}=\sqrt{\lambda}=\sqrt{6}=2.4495\approx2.45

5 0
4 years ago
PLEASE HELP THANK YOU
Ostrovityanka [42]
Look on the graph
When x = -4, (-4, 1)
The solution is g(-4) = 1
8 0
3 years ago
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