A certain radioactive substance decays according to the function $N(t) = N_0 e^{-kt}$, where $N_0$ is the initial quantity and $k$ is the decay constant. If the initial quantity is 500 grams and the quantity reduces to 250 grams in 3 hours, find the decay constant $k$. - 2.5 Exponential Function Context and Data Modeling - Review - Use this assignment in your class!

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Given the data points

(1, 3)

(2, 9)

, and

(3, 27)

, fit an exponential function of the form

y = a b^{x}

to the data. Show your work and explain how you determined the values of

a

and

b

A certain radioactive substance decays according to the function

N (t) = N_{0} e^{- k t}

, where

N_{0}

is the initial quantity and

k

is the decay constant. If the initial quantity is 500 grams and the quantity reduces to 250 grams in 3 hours, find the decay constant

k

A car depreciates in value according to the function

V (t) = V_{0} e^{- k t}

, where

V_{0}

is the initial value and

k

is the depreciation constant. If a car is worth $20,000 initially and $15,000 after 2 years, determine the depreciation constant

k

A population of fish in a lake grows according to the function

P (t) = P_{0} e^{r t}

, where

P_{0}

is the initial population and

r

is the growth rate. If the initial population is 200 fish and the population grows to 400 fish in 5 years, find the growth rate

r

A certain radioactive substance decays according to the function

N (t) = N_{0} e^{- k t}

, where

N_{0}

is the initial quantity and

k

is the decay constant. If the initial quantity is 500 grams and the quantity reduces to 250 grams in 3 hours, find the decay constant

k

Given

N (3) = 250

and

N_{0} = 500

, we substitute into the decay function:

250 = 500 e^{- 3 k}

. Dividing both sides by 500, we get

0.5 = e^{- 3 k}

. Taking the natural logarithm of both sides, we get

\ln (0.5) = - 3 k

. Solving for

k

, we get

k = - \frac{\ln (0.5)}{3} \approx 0.231

If the decay constant

k

is approximately 0.231, what is the quantity of the substance after 6 hours?

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