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pantera1 [17]
3 years ago
14

What is the mean, median, mode, and range of 8, 35, 10, 12, 14?

Mathematics
1 answer:
Ivan3 years ago
3 0

answer:

mean: 15.8

median: 12

mode: no mode

range: 27

step-by-step explanation:

  • first know how to find each of the above and what they are
  • mean = (avg) add all the numbers, then divide by the number of numbers
  • median = the middle number from increasing order
  • mode = the one that repeats the most
  • range = the difference between the largest and smallest number

8, 35, 10, 12, 14

<u>mean</u>

8 + 35 + 10 + 12 + 14 = 79

  • there are five numbers

79 / 5 = 15.8

<u>median</u>

  • put them from smallest to largest

8, 10, 12, 14, 35

  • 12 is the median OR the middle number

<u>mode</u>

  • no number repeat, so no mode

<u>range</u>

35 - 8 = 27

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The half-life of cesium-137 is 30 years. Suppose we have a 180-mg sample. (a) Find the mass that remains after t years. (b) How
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Answer:

a) Q(t) = 180e^{-0.023t}

b) 11.4mg of cesium-137 remains after 120 years.

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Step-by-step explanation:

The following equation is used to calculate the amount of cesium-137:

Q(t) = Q(0)e^{-rt}

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The half-life of cesium-137 is 30 years.

This means that Q(30) = 0.5Q(0). We apply this information to the equation to find the value of r.

Q(t) = Q(0)e^{-rt}

0.5Q(0) = Q(0)e^{-30r}

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Applying ln to both sides of the equality.

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So

Q(t) = Q(0)e^{-0.023t}

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Q(t) = 180e^{-0.023t}

(b) How much of the sample remains after 120 years?

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Q(t) = 180e^{-0.023t}

Q(120) = 180e^{-0.023*120}

Q(120) = 11.4

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(c) After how long will only 1 mg remain?

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\ln{e^{-0.023t}} = \ln{0.00556}

-0.023t = \ln{0.00556}

t = \frac{\ln{0.00556}}{-0.023}

t = 225.8

225.8 years.

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