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HACTEHA [7]
2 years ago
6

The half-life of gold-195m is approximately 30.5 seconds.

Mathematics
1 answer:
SCORPION-xisa [38]2 years ago
6 0

The amount of the 12 grams sample of gold-195m having a half-life of 30.5 s that would remain after 41 s is 4.726 g

<h3>How to determine the number of half-lives </h3>
  • Half-life (t½) = 30.5 s
  • Time (t) = 41 s
  • Number of half-lives (n) =?

n = t / t½

n = 41 / 30.5

n = 82/61

<h3>How to determine the amount remaining </h3>
  • Original amount (N₀) = 12 g
  • Number of half-lives (n) = 82/61
  • Amount remaining (N) =?

N = N₀ / 2ⁿ

N = 12 / 2^(82/61)

N = 4.726 g

Learn more about half life:

brainly.com/question/26374513

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A textile mill wishes to establish a control procedure on flaws in towels it manufactures. Using an inspection unit of 50 units,
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Answer:

•A c-chart is the appropriate control chart

• c' = 8.5

• Control limits, CL = 8.5

Lower control limits, LCL = 0

Upper control limits, UCL = 17.25

Step-by-step explanation:

A c chart is a quality control chart used for the number of flaws per unit.

Given:

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Total flaws = 850

We now have:

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For control limits, we have:

CL = c'

UCL = c' + 3√c'

LCL = c' - 3√c'

The CL stands for the normal control limit, while the UCL and LCL are the upper and lower control limits respectively

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CL = c'

CL = 8.5

UCL = 8.5 + 3√8.5

= 17.25

LCL = 8.5 - 3√8.5

= -0.25

A negative LCL tend to be 0. Therefore,

LCL = 0

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3 years ago
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Answer:

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I=0.13x+1025 A textbook salesperson's monthly income depends on the amount of his sales. His monthly income is a salary of $1025
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Answer:

l = 0.13x+1025

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