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Andru [333]
3 years ago
14

A sample of radium has a weight of 1.5 mg and a half-life of approximately 6 years.

Mathematics
1 answer:
pogonyaev3 years ago
5 0

Answer:

0.75 mg

Step-by-step explanation:

From the question given above the following data were obtained:

Original amount (N₀) = 1.5 mg

Half-life (t₁/₂) = 6 years

Time (t) = 6 years

Amount remaining (N) =.?

Next, we shall determine the number of half-lives that has elapse. This can be obtained as follow:

Half-life (t₁/₂) = 6 years

Time (t) = 6 years

Number of half-lives (n) =?

n = t / t₁/₂

n = 6/6

n = 1

Finally, we shall determine the amount of the sample remaining after 6 years (i.e 1 half-life) as follow:

Original amount (N₀) = 1.5 mg

Half-life (t₁/₂) = 6 years

Number of half-lives (n) = 1

Amount remaining (N) =.?

N = 1/2ⁿ × N₀

N = 1/2¹ × 1.5

N = 1/2 × 1.5

N = 0.5 × 1.5

N = 0.75 mg

Thus, 0.75 mg of the sample is remaining.

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Answer:

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Step-by-step explanation:

We are given the following information in the question:

Mean, μ = 530

Standard Deviation, σ = 119

We are given that the distribution of math scores is a bell shaped distribution that is a normal distribution.

Formula:

z_{score} = \displaystyle\frac{x-\mu}{\sigma}

a) P(test scores is between 411 and 649)

P(411 \leq x \leq 649)\\\\= P(\displaystyle\frac{411 - 530}{119} \leq z \leq \displaystyle\frac{649-530}{119})\\\\= P(-1 \leq z \leq 1)\\\\= P(z \leq 1) - P(z < -1)\\= 0.841 - 0.159 = 0.682 = 68.2\%

b) P(scores is less than 411 or greater than 649)

P(x < 411 < x, x > 649)\\=1 = P(411 \leq x \leq 649)\\=1 - 0.682\\=0.318 = 31.8\%

c) P(score greater than 768)

P(x > 768)

P( x > 768) = P( z > \displaystyle\frac{768 - 530}{119}) = P(z > 2)

= 1 - P(z \leq 2)

Calculation the value from standard normal z table, we have,  

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nordsb [41]

Answer:

<u>A. V (2) = $ 391,114.50</u>

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Step-by-step explanation:

Given formula:

V(t) = 420,000 * (0.965)^t

A. What would be worth the car′s worth in 2 years?

V(t) = 420,000 * (0.965)^t

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325,000 = 420,000 * (0.965)^t

325,000/420,000 = (0.965)^t

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