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Please answer the following questions using exponential and logarithmic models.
4) A wooden artifact from an archaeological dig contains 70 percent of the Carbon-14 that is present in living trees. To the nearest year, about how many years old is the artifact? (The half-life of Carbon-14 is 5730 years.) In years
5) A tumor is injected with 0.5 grams of Iodine-125, which has a decay rate of 1.15% per day. Write an exponential model representing the amount of Iodine-125 remaining in the tumor after t days. Then use the formula to find the amount of Iodine-125 that would remain in the tumor after 60 day
6) A doctor prescribes 100 milligrams of a therapeutic drug that decays by about 20% each hour. To the nearest hour, what is the half-life of the drug?
7) A tumor is injected with 0.4 grams of Iodine-125, which has a decay rate of 1.15% per day. To the nearest day, how long will it take for half of the Iodine-125 to decay? How many Days?
8) A tumor is injected with 0.4 grams of Iodine-125, which has a decay rate of 1.15% per day. Write an exponential model representing the number of grams f of Iodine-125 remaining in the tumor after t days.
Then use the formula to find the amount of Iodine-125 that would remain in the tumor after 30 days. (Round your answer to four decimal places.) In grams
9) The half-life of Erbium-171 is 7.5 hours. What is the hourly decay rate? Express the result to four decimal places.
Express the hourly decay rate as a percentage to two decimal places. In percent %
4. Let the initial amount of Carbon be N0
Amount of carbon after t years = N0 e-kt
Given, half life = 5730 years
So, N0/2 = N0 e-k(5730)
=> 1/2 = e-5730k
=> ln (e-5730k) = ln (1/2)
=> -5730k = ln(1/2)
=> k = 0.000121
Let the artifact be t years old
0.7N0 = N0 e-0.000121t
=> 0.7 = e-0.000121t
=> ln (e-0.000121t) = ln 0.7
=> -0.000121t = ln (0.7)
=> t = 2947.73 years
5. 1-day decay percent = 1.15% = 0.0115
For x percent decrease , decay factor g is given by
g = 1 - 0.0115 = 0.9885
Initial amount = 0.5 grams
Amount of Iodine after t days = Initial Amount * (1-day decay factor)t
=> A(t) = 0.5(0.9885)t
Amount of Iodine after 60 days = 0.5(0.9885)60 = 0.2498 gm
6. 1-hour decay percent = 20% = 0.2
For x percent decrease , decay factor g is given by
g = 1 - 0.2 = 0.8
Let the initial amount be No
Amount of drug after t hours = No (0.8)t
Let the half life be n hours
So, No/2 = No (0.8)n
=> (0.8)n = 0.5
=> ln (0.8)n = ln 0.5
=> n ln 0.8 = ln 0.5
=> n = 3 hours
So, the half life of the drug is 3 hours
Please answer the following questions using exponential and logarithmic models. 4) A wooden artif...
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