Because the population of the city is decreasing every year at 10.5% we can find the population after each year by using

$\displaystyle \small 13,529\cdot(1-0.105) = 13,529\cdot(0.895)$

Because this decrease will continue every year for the 6 years, we can continue to multiply the population by the decay for every year.

$\displaystyle \small \small \small 13,529\cdot0.895\cdot0.895\cdot0.895\cdot0.895\cdot0.895\cdot0.895 \Rightarrow$

$\displaystyle \small \small \Rightarrow 13,529\cdot0.895^6 \approx 6953$

Because the water is leaking at a continuous rate, we can use the exponential decay equation.

$\displaystyle \small W_{new} = W\cdot(1-D)^t$

$\displaystyle \small D$ is the decay of the problem, 12% or 0.12. $\displaystyle \small t$ is equal to how many times the water will have a 12% decay. This can be calculated as

$\displaystyle \small t = \frac{4min 12secs}{1min}$

To calculate this we must first convert both time to seconds

$\displaystyle \small t = \frac{4\cdot60+12}{1\cdot60} = \frac{240+12}{60} = \frac{252}{60}$

Our equation is then

$\displaystyle \small \large \large W_{new} = 8\cdot(1-0.12)^{\frac{252}{60}} \approx 4.68$

$\displaystyle \small W_{new} = 4.68$

This is an exponential decay problem. Therefore, we can use this equation.

$\displaystyle \small P_{7} = P_{0}\cdot(1-d)^t$

$\displaystyle \small P_{7}$ is the animal population after the 7 years. $\displaystyle \small P_{0}$ is the animal population right now. $\displaystyle \small d$ is the decay of the animal population every year. $\displaystyle \small t$ is the time period of the animal populations decay. 

From the problem we know after the 7 years the animal population will be 80, so

$\displaystyle \small P_{7} = 80$

The population is decreasing by 8% every year, therefore

$\displaystyle \small d$ = 8% = 0.08

$\displaystyle \small t$ is equal to the number of times a decay of 8% has occurred. Since an 8% decay happens every year, and the population is 7 years from now, the population has decayed 8%, 7 times. In other words,

$\displaystyle \small t = \frac{7 \text{ years}}{1 \text{ year}} = 7$

These values give us,

$\displaystyle \small 80 = P_{0}\cdot(1-0.08)^7$

Rearranging this equation we can solve for $\displaystyle \small P_{0}$

$\displaystyle \small P_{0} = \frac{80}{(1-0.08)^7} = \frac{80}{0.92^7} = 143$

This is an exponential decay problem, meaning that the decay of the school's students can be found using

$\displaystyle \small S_{now} = S_{past}\cdot(1-d)^t$

$\displaystyle \small S_{now}$ is the number of students currently at the school, which is 242

$\displaystyle \small S_{past}$ is the number of students that were at the school 4 years ago, which is 591

$\displaystyle \small t$ is the number of times the decay has occurred. Since, we are trying to find the yearly decay, the decay that happened to the school from one year to the next, and we have the number of students from 4 years ago, $\displaystyle \small t =4$. 

$\displaystyle \small d$ is the decay rate of the school that we are trying to find. Because we have every number except for $\displaystyle \small d$, we can plug the values into the equation to solve for $\displaystyle \small d$.

$\displaystyle \small 242 = 591\cdot(1-d)^4$

$\displaystyle \small (1-d)^4 = \frac{242}{591} \approx 0.41$

$\displaystyle \small 1-d = \sqrt[4]{0.41} \approx 0.80$

$\displaystyle \small d = 0.20$ = 20%

This is an exponential decay problem. That means that after 20 minutes 3% of the cells decay and 97% of the cells in the dish are left. To find the new amount of cells in the dish, we multiple the original number by the 97%.

$\displaystyle \small 1000 \cdot(0.97)$

The number of cells after every 20 minute interval can be calculated this way. Therefore, to find how long the cell were decaying we use,   

$\displaystyle \small \small \small 1000 \cdot(0.97) \cdot(0.97)\cdot... = 92$

Which can be rewritten as,

$\displaystyle \small \small 1000(0.97)^t = 92$

Now we can solve for $\displaystyle \small t$, which is the amount of time that you were gone

$\displaystyle \small (1-0.03)^t = \frac{92}{1000}$

To solve for $\displaystyle \small t$, we must take the log of both sides to base 10. This will give us,

$\displaystyle \small t\cdotlog(0.97) = log\bigg(\frac{92}{1000}\bigg)$

$\displaystyle \small \small t = \frac{log(\frac{92}{1000})}{log(0.97)} \approx 78.333$

Remember! $\displaystyle \small t$ is the number of decays that the cells went through. Each decay took 20 minutes to get through, however.

Therefore, the time that you were gone is

$\displaystyle \small \small t = 78.333\cdot\frac{20}{60} \approx 26 \text{ hours and} \ 6 \text{ minutes}$

Write the formula for radioactive decay.

$\displaystyle X=X_0e^{-\lambda T }$

$\displaystyle X= \textup{final amount} = 3.2$

$\displaystyle X_0=\textup{ initial amount }= 5$

$\displaystyle T=\textup{time in days} = \frac{48 \textup{ hours}}{ \frac{24\textup{ hours}}{1\textup{ day}}} = 2$

Substitute the values in the equation and solve for lambda.

$\displaystyle 5=3.2e^{-2\lambda }$

Divide by 3.2 on both sides.

$\displaystyle 1.5625=e^{-2\lambda }$

Take the natural log of both sides to eliminate the exponential.

$\displaystyle ln(1.5625)=ln(e^{-2\lambda })$

$\displaystyle ln(1.5625)=-2\lambda$

Divide by negative two on both sides.

$\displaystyle \frac{ln(1.5625)}{-2}=\frac{-2\lambda}{-2}$

$\displaystyle \lambda=\frac{ln(1.5625)}{-2}=-0.223144$

Convert this to a percentage.

This element's decay rate is approximately:  $\displaystyle 22\%$

Write the formula for half life.

$\displaystyle y=a(\frac{1}{2})^x$

$\displaystyle a=\textup{ sample size} = 50$

$\displaystyle x=\textup{ time in days}$

Since the time requested is five out of the six day of the half life, the value of $\displaystyle x$ is:

$\displaystyle x=\frac{5}{6}$

Substitute all the known values into the equation.

$\displaystyle y=50(\frac{1}{2})^{\frac{5}{6}} = 28.06155$

The answer is:  $\displaystyle 28.06155$

Because the butterflies are decreasing exponentially, we can use this equation

$\displaystyle \small F = O\cdot(1-\text{decay})^{\text{time}}$

$\displaystyle \small F$ is the final value

$\displaystyle \small O$ is the original value

The decay for this problem is 5% or 0.05

The period of time is 7 years

Using this equation we can solve for $\displaystyle \small F$

$\displaystyle \small F = 500(1-0.05)^7 = 500\cdot0.95^7 \approx 349$

$\displaystyle \small F = 349$

Set $\displaystyle P = 5,000$ and solve for $\displaystyle t$ :

$\displaystyle 5,000 = 253 \cdot 1.04^{t}$

$\displaystyle 1.04 ^{t} = \frac{5,000 }{253 }$

$\displaystyle 1.04^{t} \approx 19.7628$

$\displaystyle \ln 1.04^{t} \approx \ln 19.7628$

$\displaystyle t \ln 1.04 \approx \ln 19.7628$

$\displaystyle t \approx \frac{\ln 19.7628}{\ln 1.04 } \approx \frac{2.9838}{0.0392} \approx 76.1$

January and February have 59 days total; add March, and this is 90 days. The 76th day is in March.

Set $\displaystyle P = 100$ and solve for $\displaystyle t$:

$\displaystyle 100 = 318\cdot 1.07 ^{t}$

$\displaystyle \frac{100}{318} = 1.07 ^{t}$

$\displaystyle 1.07 ^{t} \approx 0.3145$

$\displaystyle \ln 1.07 ^{t} \approx \ln 0.3145$

$\displaystyle t \ln 1.07 \approx \ln 0.3145$

$\displaystyle t \approx \frac{\ln 0.3145}{ \ln 1.07} \approx \frac{-1.1568}{0.0677} \approx -17.1$

17 days before January 1 was in December of 2014.