1. the profits of an airline company can be modeled as ( p(t)=2(1.1)^t ) where ( p(t) ) represents annual profit, in millions, t years since 1997.
a. use this function to approximate the profit in 2003. state your answer in dollars.
( 2(1.1)^6 = 3.5 ) million
b. what is the y - intercept as an ordered pair? what is the practical meaning of the intercept in this situation?
c. use your calculator (graph or table) to estimate the first year that the companys profit will be at least double its 1997 profit. show your method.
2. a chemical process is used to purify waste water. each treatment removes 11% of the pollutants. the waste water initially has 16 micrograms of pollutant per gallon.
a. let ( p(n) ) represent the level of pollutants after n treatments. write the exponential function rule for ( p(n) ).
( 16(.89)^n )
b. use the function rule to find the level of pollutants after 3 treatments algebraically.
( 16(.89)^3 = 11.27 )
c. use the graph intersection method to solve for the number of treatments needed to bring the level of pollutants to half the initial amount. show the supporting work. write your answer in a complete sentence.

Question

Accepted Answer

# Explanation:
## Step1: Determine the value of $t$ for 2003
The number of years $t$ since 1997 for 2003 is $t = 2003 - 1997=6$.
## Step2: Substitute $t = 6$ into the profit function $P(t)=2(1.1)^{t}$
$$
\begin{align*}
P(6)&=2\times(1.1)^{6}\
&=2\times1.771561\
& = 3.543122
\end{align*}
$$
Since $P(t)$ is in millions, the profit in dollars is $P(6)=3543122$ dollars.

## Step3: Find the $y -$intercept
For the $y -$intercept, set $t = 0$ in $P(t)=2(1.1)^{t}$. Then $P(0)=2(1.1)^{0}=2\times1 = 2$. The ordered pair is $(0,2)$.
The practical meaning: In 1997 ($t = 0$ years since 1997), the annual profit of the airline company was 2 million dollars.

## Step4: Find when $P(t)\geq4$ (double of 1997 profit $P(0) = 2$)
Set $P(t)=2(1.1)^{t}\geq4$, then $(1.1)^{t}\geq2$.
Take the natural logarithm of both sides: $\ln(1.1^{t})\geq\ln(2)$, using the property $\ln(a^{b})=b\ln(a)$, we get $t\ln(1.1)\geq\ln(2)$. So $t\geq\frac{\ln(2)}{\ln(1.1)}\approx7.27$. Since $t$ represents the number of years since 1997, and $t$ must be an integer, $t = 8$. The year is $1997 + 8=2005$.

For problem 2:
## Step1: Write the exponential function rule
If each treatment removes $11\%$ of the pollutants, then the remaining fraction is $1-0.11 = 0.89$. The initial amount $P(0)=16$. So the function $P(n)=16(0.89)^{n}$

## Step2: Find the level of pollutants after $n = 3$ treatments
Substitute $n = 3$ into $P(n)=16(0.89)^{n}$
$$
\begin{align*}
P(3)&=16\times(0.89)^{3}\
&=16\times0.704969\
&=11.279504\approx11.28
\end{align*}
$$

## Step3: Find $n$ when $P(n)=8$ (half of the initial amount $16$)
Set $P(n)=16(0.89)^{n}=8$, then $(0.89)^{n}=\frac{8}{16}=0.5$.
Take the natural logarithm of both sides: $\ln(0.89^{n})=\ln(0.5)$, using $\ln(a^{b})=b\ln(a)$, we have $n\ln(0.89)=\ln(0.5)$. So $n=\frac{\ln(0.5)}{\ln(0.89)}\approx5.91$. Using the graph - intersection method: graph $y = 16(0.89)^{x}$ and $y = 8$ on a calculator. The $x -$coordinate of the intersection point (where $x=n$) is approximately $n = 6$.

# Answer:
1.
a. The profit in 2003 is $3543122$ dollars.
b. The $y -$intercept is $(0,2)$. It means in 1997, the annual profit was 2 million dollars.
c. The first year is 2005.
2.
a. $P(n)=16(0.89)^{n}$
b. The level of pollutants after 3 treatments is approximately $11.28$ micrograms per gallon.
c. Approximately 6 treatments are needed to bring the level of pollutants to half the initial amount.

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QUESTION IMAGE

Question

Explanation:

Step1: Determine the value of \(t\) for 2003

Step2: Substitute \(t = 6\) into the profit function \(P(t)=2(1.1)^{t}\)

Step3: Find the \(y -\)intercept

Step4: Find when \(P(t)\geq4\) (double of 1997 profit \(P(0) = 2\))

Step1: Write the exponential function rule

Step2: Find the level of pollutants after \(n = 3\) treatments

Step3: Find \(n\) when \(P(n)=8\) (half of the initial amount \(16\))

Answer: