5. At 2pm the number of bacteria in a colony was 100, by 4pm it was 4000. Write an exponential function to model the population y of bacteria x hours after 2pm.
6. Using the information in problem 5, how many bacteria were there at 7pm that day?
7. Kaplan Corporation bought a computer for $1500. It is expected to depreciate at a rate of 22% per year. What will the value of the computer be in 2 years?
8. Radioactive iodine is used to determine the health of the thyroid gland. It decays according to the equation y = ae-0.0856t, where t is in days. Find the half-life of this substance.
9. An anthropologist finds there is so little remaining Carbon-14 in a prehistoric bone that instruments cannot measure it. This means that there is less than 0.5% of the amount of Carbon-14 the bones would have contained when the person was alive. How long ago did the person die?
10. The Mendes family bought a new house 10 years ago for $120,000. The house is now worth $191,000. Assuming a steady rate of growth, what was the yearly rate of appreciation?
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5. To write an exponential function to model the population of bacteria, we need to find the growth rate.
We can use the equation for exponential growth, which is given by: y = ab^x, where y is the population of bacteria, x is the time in hours after 2pm, a is the initial population, and b is the growth rate.
To find the growth rate, we can use the given information at 2pm and 4pm.
At 2pm, the population was 100, which means a = 100.
At 4pm, the population was 4000, which means y = 4000. We need to find the value of b.
Using the equation, we can substitute the values: 4000 = 100 * b^2.
Simplifying, we get: b^2 = 4000/100 = 40.
Taking the square root of both sides, we find: b = √40 = 2√10.
Therefore, the exponential function to model the population of bacteria is: y = 100 * (2√10)^x.
6. To find the number of bacteria at 7pm, we need to substitute x = 7 (7 hours after 2pm) into the exponential function we found in question 5.
Substituting x = 7 into the function, we get: y = 100 * (2√10)^7.
Simplifying this expression will give us the number of bacteria at 7pm.
7. To find the value of the computer after 2 years, we need to use the formula for exponential depreciation.
The formula for exponential depreciation is: V = P(1 - r)^t, where V is the current value, P is the initial value, r is the depreciation rate, and t is the time in years.
In this case, the initial value is $1500, the depreciation rate is 22% (which can be written as 0.22), and the time is 2 years.
Substituting the values into the formula, we get: V = 1500(1 - 0.22)^2.
Simplifying this expression will give us the value of the computer after 2 years.
8. The half-life of a substance is the time it takes for half of the initial amount to decay.
We are given the decay equation: y = ae^(-0.0856t), where y is the remaining amount of radioactive iodine, a is the initial amount, and t is the time in days.
To find the half-life, we need to set y equal to half of the initial amount, and solve for t.
The half-life equation is: 0.5a = ae^(-0.0856t).
Cancelling out the 'a' on both sides and simplifying, we get: 0.5 = e^(-0.0856t).
To solve for t, we need to take the natural logarithm of both sides: ln(0.5) = ln(e^(-0.0856t)).
Using the properties of logarithms, we can simplify: ln(0.5) = -0.0856t.
Finally, solving for t, we get: t = ln(0.5) / -0.0856.
Therefore, the half-life of the substance is ln(0.5) / -0.0856.
9. To find out how long ago the person died, we need to use the concept of exponential decay.
According to the given information, there is less than 0.5% of the amount of Carbon-14 remaining in the bones. This means that the remaining amount, y, is less than 0.005 times the initial amount, a.
The decay equation for Carbon-14 is: y = ae^(-kt), where y is the remaining amount, a is the initial amount, k is the decay constant, and t is the time elapsed.
Since the remaining amount is less than 0.005 times the initial amount, we can write: y < 0.005a.
Substituting the decay equation, we get: ae^(-kt) < 0.005a.
Cancelling out the 'a' on both sides, we get: e^(-kt) < 0.005.
To solve for t, we need to take the natural logarithm of both sides: ln(e^(-kt)) < ln(0.005).
Using the properties of logarithms, we can simplify: -kt < ln(0.005).
Solving for t, we get: t > ln(0.005) / -k.
Therefore, the person died more than ln(0.005) / -k years ago.
10.The yearly rate of appreciation can be found using the formula for compound interest:
A = P(1 + r/n)^(nt)
Where A is the final value, P is the initial value, r is the annual interest rate (expressed as a decimal), n is the number of compounding periods per year, and t is the number of years.
In this case, the initial value is $120,000, the final value is $191,000, and the number of years is 10. We need to find the annual interest rate, r.
Rearranging the formula, we get: r = (A/P)^(1/nt) - 1.
Substituting the given values, we get: r = (191000/120000)^(1/(10*1)) - 1.
Evaluating this expression will give us the annual rate of appreciation.