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Delvig [45]
3 years ago
8

If 3.0 grams of strontium-90 in a rock sample remained in 1989, approximately how many grams of strontium-90 were present in the

original rock sample in 1933?
Mathematics
1 answer:
kirill [66]3 years ago
5 0

Answer:

<em>11.29 grams </em><em>of strontium-90 were present in the original rock sample in 1933.</em>

Step-by-step explanation:

Strontium has a half-life of 28.8 years. Therefore,

1989 - 1933 = 56 years.

56 / 28.8 = 1.94 half-lives

Thus, the quantity of the radioisotope remaining will double for each half-life elapsed. moving backwards in time.

Therefore, moving backwards in time by 1.94 half-lives, the quantity remaining will double by 1.94 times.

Thus, the amount remaining in 1933 is

3.0 × (1.94)² = <em>11.29 grams</em>

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If an object is dropped from a height of 144 feet, the function h(t)=-16t^2+144 gives the height of the object after t seconds.
shtirl [24]

Answer:

A. 3s

Step-by-step explanation:

The height of the object after t seconds is given by:

h(t) = -16t^2 + 144

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It hits the ground after t seconds, and t is found when h(t) = 0. So

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3 years ago
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castortr0y [4]

Answer:

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Step-by-step explanation:

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The exponential probability distribution, with mean m, is described by the following equation:

f(x) = \mu e^{-\mu x}

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P(X \leq x) = 1 - e^{-\mu x}

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P(X \leq x) = 1 - e^{-\mu x}

P(X \leq 7) = 1 - e^{-0.1*7} = 0.5034

50.34% probability that the arrival time between customers will be 7 minutes or less.

B. What is the probability that the arrival time between customers will be between 3 and 7 minutes?

P(3 \leq X \leq 7) = P(X \leq 7) - P(X \leq 3)

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P(3 \leq X \leq 7) = P(X \leq 7) - P(X \leq 3) = 0.5034 - 0.2592 = 0.2442

24.42% probability that the arrival time between customers will be between 3 and 7 minutes

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3 years ago
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