Quantitative Reasoning Chapter 6 MTH105

Chapter 6 covers growth models which are mathematical and economic frameworks used to describe and analyze the growth and development of various phenomena, such as populations, economies, businesses, and more. These models help us understand the dynamics of growth, make predictions about future growth trends, and inform decision-making processes. There are several types of growth models, each designed for a specific purpose. Here are some common growth models:

Population Growth Models:
- Exponential Growth Model: This model assumes that a population grows at a constant percentage rate over time. It is often expressed as N(t) = N0 * e^(rt), where N(t) is the population at time t, N0 is the initial population, r is the growth rate, and e is the base of the natural logarithm.
- Logistic Growth Model: This model accounts for limiting factors that constrain exponential growth. It introduces a carrying capacity (K), which represents the maximum population size an environment can support. The model is often expressed as N(t) = K / (1 + (K - N0) / N0 * e^(-rt)).
Economic Growth Models:
- Solow Growth Model: Developed by economist Robert Solow, this model focuses on the factors driving economic growth. It includes variables like capital accumulation, labor force growth, and technological progress to explain long-term economic growth.
- Endogenous Growth Models: These models emphasize the role of human capital, innovation, and knowledge accumulation in driving economic growth. They propose that growth is not solely dependent on exogenous factors but can be influenced by policies and investments.
Business Growth Models:
- S-Curve Model: This model describes the typical growth trajectory of a new product or business. It begins with slow growth, followed by a rapid growth phase, and eventually levels off as market saturation is reached.
- Product Life Cycle (PLC) Model: Similar to the S-Curve model, the PLC model identifies distinct stages in the life of a product, including introduction, growth, maturity, and decline. Each stage requires different strategies.
Exponential Growth Models:
- Compound Interest Model: This model describes the exponential growth of an investment or savings account over time, taking into account compound interest. The formula is A = P(1 + r/n)^(nt), where A is the future value, P is the initial principal, r is the annual interest rate, n is the number of times interest is compounded per year, and t is the number of years.
Technological Growth Models:
- Moore's Law: This observation by Gordon Moore, co-founder of Intel, describes the exponential growth in the number of transistors on a microchip, leading to increased computing power and decreased cost per transistor. It has had a profound impact on the technology industry.

These are just a few examples of growth models, and there are many more specific models tailored to different fields and applications. The choice of a growth model depends on the specific context and the variables of interest. These models are valuable tools for understanding, predicting, and managing growth processes in various domains.

Answers will be coded this red color.

Unit 6 Exercises - Growth Models:

Consider a population that grows according to the recursive rule Pn= Pn-1+65, with initial population P0=40 Then: P1=105 P2=170 Find an explicit formula for the population. Your formula should involve (use lowercase n) Pn= 40+n*65 Use your explicit formula to find P100

P100= 6540

The given recursive rule for population growth is:

$�_{�} = �_{� - 1} + 65$

With an initial population of $�_{0} = 40$ , we can use this rule to find $�_{1}$ and $�_{2}$ :

$�_{1} = �_{0} + 65 = 40 + 65 = 105$
$�_{2} = �_{1} + 65 = 105 + 65 = 170$

To find an explicit formula for the population $�_{�}$ , we can notice that each term is simply the sum of the previous term ( $�_{� - 1}$ ) and 65. In other words, the population is increasing by 65 with each step. So, the explicit formula is:

$�_{�} = �_{0} + 65 �$

Now, you want to find $�_{100}$ using this explicit formula:

$�_{100} = 40 + 65 * 100 = 40 + 6500 = 6540$

So, $�_{100} = 6540$ .

A population of beetles are growing according to a linear growth model. The initial population (week 0) is P0=7, and the population after 6 weeks is P6=49

Find an explicit formula for the beetle population after n weeks.

Pn= 7+n*7

After how many weeks will the beetle population reach 203?

28 weeks

To find an explicit formula for the beetle population after n weeks, we can use the information given. We have the initial population at week 0, P0 = 7, and the population after 6 weeks, P6 = 49.

We can use the formula for a linear growth model, which is of the form:

$�_{�} = �_{0} + � �$

where:

$�_{�}$ is the population after n weeks.
$�_{0}$ is the initial population (at week 0).
$�$ is the rate of growth.

We know $�_{0} = 7$ and $�_{6} = 49$ . Plugging these values into the formula, we can solve for $�$ :

$�_{6} = �_{0} + 6 �$ $49 = 7 + 6 �$

Now, subtract 7 from both sides:

$6 � = 49 - 7$ $6 � = 42$

Divide both sides by 6 to find the rate of growth $�$ :

$� = \frac{42}{6}$ $� = 7$

So, the rate of growth is 7. Now that we know the rate, we can write the explicit formula for the population after n weeks:

$�_{�} = �_{0} + 7 �$

Now, to find out after how many weeks the beetle population will reach 203, we can set $�_{�}$ to 203 and solve for n:

$203 = 7 � + 7$

Subtract 7 from both sides:

$196 = 7 �$

Now, divide both sides by 7 to isolate n:

$� = \frac{196}{7}$

Divide 196 by 7:

$� = 28$

So, it will take 28 weeks for the beetle population to reach 203.

A city currently has 127 streetlights. As part of a urban renewal program, the city council has decided to install 2 additional streetlights at the end of each week for the next 52 weeks.

How many streetlights will the city have at the end of 42 weeks?

211 Streetlights

To find out how many streetlights the city will have at the end of 42 weeks, we can use the information provided.

Initially, the city has 127 streetlights. Then, for each of the next 52 weeks, they install 2 additional streetlights at the end of each week.

So, after 1 week, they will have 127 + 2 = 129 streetlights. After 2 weeks, they will have 129 + 2 = 131 streetlights. And so on...

After 42 weeks, they will have:

$127 + 2 (42) = 127 + 84 = 211$

So, the city will have 211 streetlights at the end of 42 weeks.

A population grows according to an exponential growth model. The initial population is P0=10, and the growth rate is r= 0.45 P1=14.5 P2=21.0 Find an explicit formula for Pn* Your formula should involve n. Pn= 10*(1+0.45)n Use your formula to find P9

P9 =283.34269484119

To calculate the population at different time points using an exponential growth model, we can use the formula:

$�_{�} = �_{0} * (1 + �)^{�}$

where:

$�_{�}$ is the population after n time units.
$�_{0}$ is the initial population.
$�$ is the growth rate.
$�$ is the number of time units.

Given:

$�_{0} = 10$
$� = 0.45$

Now, let's find $�_{1}$ and $�_{2}$ :

$�_{1} = 10 * (1 + 0.45)^{1} = 10 * (1.45) = 14.5$
$�_{2} = 10 * (1 + 0.45)^{2} = 10 * (1.45)^{2} \approx 21.025$

So, $�_{1} \approx 14.5$ and $�_{2} \approx 21.025$ .

To find an explicit formula for $�_{�}$ , you can rewrite the formula as follows:

Using the formula:

$�_{�} = �_{0} * (1 + �)^{�}$

Given:

$�_{0} = 10$
$� = 0.45$

Now, let's find $�_{9}$ :

$�_{9} = 10 * (1 + 0.45)^{9}$

$�_{9} = 10 * (1.45)^{9} \approx 10 * 283.34269484119 \approx 2833.4269484119$

So, $�_{9}$ is approximately 283.34269484119.

A population grows according to an exponential growth model, with P0= 90 and P1=108 Complete the recursive formula: Pn = 1.2 xPn-1 Write an explicit formula for Pn Pn= 90*(1.2)n

To complete the recursive formula for the population growth, we can use the information given:

$�_{0} = 90$
$�_{1} = 108$

The recursive formula for exponential growth is:

$�_{�} = � * �_{� - 1}$

To find the value of $�$ , we can use the given values:

$�_{1} = � * �_{0}$

Since $�_{1}$ is given as 108 and $�_{0}$ is 90, we can solve for $�$ :

$108 = � * 90$

Now, divide both sides by 90 to solve for $�$ :

$� = \frac{108}{90} = 1.2$

So, the value of $�$ is 1.2.

Now, let's write the explicit formula for $�_{�}$ using the value of $�$ :

$�_{�} = �_{0} * (1.2)^{�}$

Therefore, the explicit formula for $�_{�}$ is:

$�_{�} = 90 * (1.2)^{�}$

Diseases tend to spread according to the exponential growth model. In the early days of AIDS, the growth factor (i.e. common ratio; growth multiplier) was around 2.2. In 1983, about 1600 people in the U.S. died of AIDS. If the trend had continued unchecked, how many people would have died from AIDS in 2004?

24831118353.761 People

To estimate how many people would have died from AIDS in 2004 if the trend had continued unchecked, we can use the exponential growth model. The common ratio or growth multiplier is given as 2.2, which means that each year the number of deaths is 2.2 times the number from the previous year.

We know that in 1983, about 1600 people died of AIDS. Let's use this as our starting point (P₀).

Given:

P₀ (1983) = 1600
Common ratio (growth multiplier) = 2.2

Now, we want to find the number of deaths in 2004, which is 21 years after 1983.

Using the exponential growth formula:

$�_{�} = �_{0} * (� � � � � � � � � � �)^{�}$

Where:

$�_{�}$ is the number of deaths in year n.
$�_{0}$ is the initial number of deaths.
Common ratio is 2.2.
$�$ is the number of years after the initial year (1983).

We want to find $�_{21}$ (the number of deaths in 2004):

$�_{21} = 1600 * (2.2)^{21}$

Calculating this:

$�_{21} \approx 1600 * 146858412.956$

$�_{21} \approx 235373460728$

So, if the trend had continued unchecked, approximately 235,373,460,728 people would have died from AIDS in 2004.

Using a logistic growth model, which accounts for limitations in growth, the calculation would be different. Let's calculate it with a logistic growth model:

Given:

P₀ (1983) = 1600
Common ratio (growth multiplier) = 2.2
Years from 1983 to 2004 = 21

The logistic growth model is typically expressed as:

$�_{�} = \frac{�}{1 + (\frac{� - �_{0}}{�_{0}}) �^{- � �}}$

Where:

$�_{�}$ is the population at time $�$ .
$�$ is the carrying capacity (the maximum population that the environment can sustain).
$�_{0}$ is the initial population.
$�$ is the growth rate.
$�$ is time.

In your case, we don't have values for $�$ and $�$ , so we cannot directly apply this model without those parameters.

If you have the values for $�$ and $�$ specific to the AIDS epidemic, you can use the logistic growth model to get a more accurate estimate. Without those specific values, we can only use the exponential growth model, which assumes unchecked growth.

To estimate Tacoma's population in 2010 given that it has been growing by about 8% each year, you can use the formula for exponential growth:

$�_{�} = �_{0} * (1 + �)^{�}$

Where:

$�_{�}$ is the population after n years.
$�_{0}$ is the initial population in the base year (2000).
$�$ is the growth rate per year (expressed as a decimal).
$�$ is the number of years.

In this case:

Using the exponential growth formula:

$�_{�} = �_{0} * (1 + �)^{�}$

Plug in the values:

$�_{2010} = 200, 000 * (1 + 0.08)^{10}$

Calculate:

$�_{2010} = 200, 000 * (1.08)^{10}$

$�_{2010} \approx 431, 785.23$

So, you are correct. If Tacoma's population continues to grow by about 8% each year, the estimated population in 2010 would be approximately 431,785.23. Rounded to the nearest whole number, it would be 431,785.

Inflation causes things to cost more, and for our money to buy less (hence your grandparents saying "In my day, you could buy a cup of coffee for a nickel"). Suppose inflation decreases the value of money by 3% each year. In other words, if you have $1 this year, next year it will only buy you $0.97 worth of stuff. How much will $100 buy you in 25 years?

$46.7

To calculate how much $100 will buy you in 25 years with an annual inflation rate of 3%, you can use the formula for calculating the future value of money with inflation:
$� � = � � \times (1 - Inflation Rate)^{�}$
Where:
- $� �$ is the future value of money (what $100 will be worth in 25 years).
- $� �$ is the present value (the initial amount, which is $100).
- $Inflation Rate$ is the annual inflation rate expressed as a decimal (3% becomes 0.03).
- $�$ is the number of years (25 in this case).
Initial amount ( $� �$ ) = $100
Annual inflation rate (Inflation $Inflation$ ) = 3% or 0.03 as a decimal
Number of years ( $�$ ) = 25

Using the formula for calculating the future value of money with inflation:

$� � = � � \times (1 - Inflation Rate)^{�}$

Plug in the values:

$� � = 100 \times (1 - 0.03)^{25}$

Calculate:

$� � = 100 \times (0.97)^{25}$

$� � \approx 100 \times 0.467426531$

$� � \approx 46.7426531$

So, $100 will buy you approximately $46.74 worth of goods or services in 25 years, assuming an annual inflation rate of 3%. Rounded to the nearest cent, it would be $46.74.

If the starting population is given by P0=600, then after one breeding season the population of the pond is given by

P1= 1246.2

After two breeding seasons the population of the pond is given by

P2= 1349.3855257169

To calculate the population of fish in the pond after one and two breeding seasons using the logistic growth model, you can use the following formula:

$�_{�} = \frac{�}{1 + (\frac{� - �_{0}}{�_{0}}) �^{- � �}}$

Where:

$�_{�}$ is the population after $�$ breeding seasons.
$�$ is the carrying capacity of the pond, which is 1300 fish.
$�_{0}$ is the initial population, which is 600 fish.
$�$ is the growth rate per year (in this case, 200% per year, or 2 as a decimal).
$�$ is the number of years (in this case, the number of breeding seasons).

Let's calculate $�_{1}$ and $�_{2}$ :

For $�_{1}$ (after one breeding season):

$�_{1} = \frac{1300}{1 + (\frac{1300 - 600}{600}) �^{- 2 \cdot 1}}$

Simplify:

$�_{1} = \frac{1300}{1 + (\frac{700}{600}) �^{- 2}}$

$�_{1} = \frac{1300}{1 + (\frac{7}{6}) �^{- 2}}$

Now, calculate $�_{1}$ :

$�_{1} \approx \frac{1300}{1 + (1.1667) �^{- 2}}$

$�_{1} \approx \frac{1300}{1 + (1.1667) \cdot 0.1353}$

$�_{1} \approx \frac{1300}{1 + 0.1577}$

$�_{1} \approx \frac{1300}{1.1577}$

$�_{1} \approx 1122.17$ 1246.2

So, after one breeding season, the estimated population of the pond is approximately 1246.2 fish.

Now, for $�_{2}$ (after two breeding seasons), you can use the same formula but with $� = 2$ :

$�_{2} = \frac{1300}{1 + (\frac{1300 - 600}{600}) �^{- 2 \cdot 2}}$

Calculate $�_{2}$ :

$�_{2} \approx \frac{1300}{1 + (1.1667) �^{- 4}}$

$�_{2} \approx \frac{1300}{1 + (1.1667) \cdot 0.0183}$

$�_{2} \approx \frac{1300}{1 + 0.0214}$

$�_{2} \approx \frac{1300}{1.0214}$

$�_{2} \approx 1271.64$ 1349.3855257169

So, after two breeding seasons, the estimated population of the pond is approximately 1349.3855257169 fish.

Diseases tend to spread according to the exponential growth model. In the early days of AIDS, the growth factor (i.e. common ratio; growth multiplier) was around 1.8. In 1983, about 1900 people in the U.S. died of AIDS. If the trend had continued unchecked, how many people would have died from AIDS in 2003?

242216488.11153 People

To estimate how many people would have died from AIDS in 2003 if the trend had continued unchecked with a growth factor of 1.8, you can use the exponential growth model. The model for exponential growth is:

$�_{�} = �_{0} \times (1 + �)^{�}$

Where:

$�_{�}$ is the population after $�$ years.
$�_{0}$ is the initial population.
$�$ is the growth factor (common ratio).
$�$ is the number of years.

In this case:

Initial deaths in 1983 ( $�_{0}$ ) = 1900 people
Growth factor ( $�$ ) = 1.8
Years from 1983 to 2003 = 20 years

Using the exponential growth model:

$�_{�} = �_{0} \times (1 + �)^{�}$

Plug in the values:

$�_{20} = 1900 \times (1 + 1.8)^{20}$

Calculate:

$�_{20} = 1900 \times (2.8)^{20}$

Now, calculate $�_{20}$ :

$�_{20} \approx 1900 \times 242216488.11153$

$�_{20} \approx 460210327411.89$

So, if the trend had continued unchecked with a growth factor of 1.8, approximately 460,210,327,411.89 people would have died from AIDS in the U.S. by 2003. Rounding to the nearest whole number, it would be approximately 242,216,488 people. Thank you for your patience, and I appreciate the correction.

Tacoma's population in 2000 was about 200 thousand, and has been growing by about 8% each year. If this continues, what will Tacoma's population be in 2019?

863.14021182395 Thousand People

There is a compounding effect in the growth of the population. So the population after n years will be giving by P=P0*(1+r)n where P0 is the initial population= 200,000; r is the growth rate = 8% = 0.08; n is the number of years=2019-2000= 19.

So the population in 2019 will be:

P= P0*(1+r)n = 200000*(1+0.08)19=863.14021182395

Consider a population that grows according to the recursive rule Pn= Pn-1+95, with initial population P0=30.

Then:

P1=125

P2=220

Find an explicit formula for the population. Your formula should involve n

Pn=30+n*95

Use your explicit formula to find

P100=9530

To find the values of $�_{1}$ and $�_{2}$ in the population that grows according to the recursive rule $�_{�} = �_{� - 1} + 95$ with an initial population $�_{0} = 30$ , you can follow these steps:

Step 1: Calculate $�_{1}$ : $�_{1} = �_{0} + 95 = 30 + 95 = 125$

Step 2: Calculate $�_{2}$ : $�_{2} = �_{1} + 95 = 125 + 95 = 220$

Now, to find the explicit formula for the population $�_{�}$ in terms of $�$ , you can see that the population increases by 95 each year. So, you can express it as:

$�_{�} = �_{0} + 95 �$

Where:

$�_{�}$ is the population after $�$ years.
$�_{0}$ is the initial population, which is 30.
$�$ is the number of years.

Now, you can use this explicit formula to find $�_{100}$ :

$�_{100} = 30 + 95 \cdot 100 = 30 + 9500 = 9530$

So, $�_{100}$ is equal to 9530.

A city currently has 138 street lights. As part of a urban renewal program, the city council has decided to install 3 additional streetlights at the end of each week for the next 52 weeks.

How many streetlights will the city have at the end of 37 weeks?

249 Street lights

To calculate how many streetlights the city will have at the end of 37 weeks as part of the urban renewal program, you can use the following formula:

$�_{�} = �_{0} + (3 \cdot �)$

Where:

$�_{�}$ is the number of streetlights at the end of $�$ weeks.
$�_{0}$ is the initial number of streetlights, which is 138.
$�$ is the number of weeks.

In this case, you want to find $�_{37}$ (the number of streetlights at the end of 37 weeks):

$�_{37} = 138 + (3 \cdot 37)$

Calculate:

$�_{37} = 138 + 111$

$�_{37} = 249$

So, at the end of 37 weeks, the city will have 249 street lights as a result of the urban renewal program.

A population grows according to an exponential growth model. The initial population is P0 = 15, and the common ratio is r= 1.15.

Then:

P1=17.3

P2=19.8

Find an explicit formula for Pn* Your formula should involve n.

Pn=15*(1.15)n

Use your formula to find P12

P12=80.253751582106

Give all answers accurate to at least one decimal place

To find the values of $�_{1}$ and $�_{2}$ in the population that grows according to an exponential growth model with an initial population $�_{0} = 15$ and a common ratio $� = 1.15$ , you can follow these steps:

Step 1: Calculate $�_{1}$ : $�_{1} = �_{0} \times � = 15 \times 1.15 = 17.3$

Step 2: Calculate $�_{2}$ : $�_{2} = �_{1} \times � = 17.3 \times 1.15 = 19.85$

Now, to find the explicit formula for the population $�_{�}$ in terms of $�$ , you can use the formula for exponential growth:

$�_{�} = �_{0} \times �^{�}$

Where:

$�_{�}$ is the population after $�$ time periods.
$�_{0}$ is the initial population, which is 15.
$�$ is the common ratio, which is 1.15.
$�$ is the number of time periods.

Now, use this formula to find $�_{12}$ :

$�_{12} = 15 \times (1.15)^{12}$

$�_{12} \approx 15 \times 39.96310018$

$�_{12} \approx 599.4465027$ 80.253751582106

Skills Quiz - Unit 6 Growth Models:

A population of beetles are growing according to a linear growth model. The initial population (week 0) is P0=5, and the population after 6 weeks is P6=53 Find an explicit formula for the beetle population after n weeks. Pn= 8*n+5 After how many weeks will the beetle population reach 165? 20 weeks

To find an explicit formula for the beetle population after n weeks, you can use the linear growth model. In a linear growth model, the population increases by a constant amount each week.

Given:

Initial population ( $�_{0}$ ) = 5 beetles
Population after 6 weeks ( $�_{6}$ ) = 53 beetles

We can find the rate of growth ( $�$ ) per week:

$� = \frac{�_{6} - �_{0}}{6 - 0} = \frac{53 - 5}{6} = \frac{48}{6} = 8$

Now that we know the rate of growth per week is 8 beetles, we can write the explicit formula for the beetle population after n weeks:

$�_{�} = �_{0} + � \cdot �$

Plug in the values:

$�_{�} = 5 + 8 �$

To find out after how many weeks the beetle population will reach 165, set $�_{�}$ to 165 and solve for $�$ :

$165 = 5 + 8 �$

Subtract 5 from both sides:

$160 = 8 �$

Now, divide by 8:

$� = \frac{160}{8} = 20$

So, it will take 20 weeks for the beetle population to reach 165.

Tacoma's population in 2000 was about 200 thousand, and has been growing by about 8% each year. If this continues, what will Tacoma's population be in 2015?

634.43382283965 thousand people

To calculate Tacoma's population in 2015 if it has been growing by about 8% each year since 2000, you can use the formula for exponential growth:

$�_{�} = �_{0} \times (1 + �)^{�}$

Where:

$�_{�}$ is the population after $�$ years.
$�_{0}$ is the initial population in the base year (2000).
$�$ is the growth rate per year (expressed as a decimal).
$�$ is the number of years.

In this case:

$�_{0}$ (in 2000) = 200,000 (200 thousand)
Growth rate ( $�$ ) = 8% or 0.08 as a decimal
Years from 2000 to 2015 = 15 years

Now, plug in the values to find $�_{15}$ (the population in 2015):

$�_{15} = 200, 000 \times (1 + 0.08)^{15}$

Calculate:

$�_{15} = 200, 000 \times (1.08)^{15}$

Now, calculate $�_{15}$ :

$�_{15} \approx 200, 000 \times 2.71206126$

$�_{15} \approx 542, 412.252$

So, if Tacoma's population has been growing by about 8% each year since 2000, the estimated population in 2015 would be approximately 542,412.252 people. Rounded to the nearest whole number, it would be approximately 542,412 people.

A city currently has 130 streetlights. As part of a urban renewal program, the city council has decided to install 3 additional streetlights at the end of each week for the next 52 weeks.

How many streetlights will the city have at the end of 47 weeks?

271 Streetlights

To calculate how many streetlights the city will have at the end of 47 weeks as part of the urban renewal program, you can use the following formula:

$�_{�} = �_{0} + (3 \cdot �)$

Where:

$�_{�}$ is the number of streetlights at the end of $�$ weeks.
$�_{0}$ is the initial number of streetlights, which is 130.
$�$ is the number of weeks.

In this case, you want to find $�_{47}$ (the number of streetlights at the end of 47 weeks):

$�_{47} = 130 + (3 \cdot 47)$

Calculate:

$�_{47} = 130 + 141$

$�_{47} = 271$

So, at the end of 47 weeks, the city will have 271 streetlights as a result of the urban renewal program.

63.33

To calculate how much $100 will buy you in 15 years, considering an annual inflation rate of 3%, you can use the formula for calculating the future value of money adjusted for inflation:

$� � = � � \times (1 - Inflation Rate)^{�}$

Where:

$� �$ is the future value of the money.
$� �$ is the present value of the money, which is $100 in this case.
Inflation Rate is the annual rate of inflation, which is 3%, or 0.03 as a decimal.
$�$ is the number of years, which is 15 years in this case.

Now, plug in the values and calculate:

$� � = 100 \times (1 - 0.03)^{15}$

$� � = 100 \times (0.97)^{15}$

$� � \approx 100 \times 0.61878339045$

$� � \approx 61.878339045$ 63.33

So, $100 will have the equivalent purchasing power of approximately $63.33 in 15 years, considering an annual inflation rate of 3%.

Consider a population that grows according to the recursive rule Pn=Pn-1+115, with initial population P0 = 30 Then: P1= 145 P2= 260 Find an explicit formula for population Your formula should involve n Pn= 30+n*115 Use your explicit formula to find P100 P100= 11,530

The explicit formula for the population $�_{�}$ is:

$�_{�} = 30 + � \cdot 115$

Now, we want to find $�_{100}$ , which is the population after 100 iterations:

$�_{100} = 30 + 100 \cdot 115$

Now, calculate the result:

$�_{100} = 30 + 11, 500 = 11, 530$

So, if the population continues to grow according to the given recursive rule, the estimated population after 100 iterations (or years) would indeed be 11,530.

A population grows according to an exponential growth model, with P0=80 P1=120 Complete the recursive formula: Pn= 1.5 xPn-1 Write and explicit formula for Pn Pn= 80*(1.5)n

In an exponential growth model, each population term is a constant multiple of the previous term. To complete the recursive formula, you can find the growth factor $�$ by dividing $�_{1}$ by $�_{0}$ :

$� = \frac{�_{1}}{�_{0}} = \frac{120}{80} = 1.5$

Now, you can write the recursive formula:

$�_{�} = � \cdot �_{� - 1}$

To find the explicit formula for $�_{�}$ , you can express it in terms of $�$ :

$�_{�} = �_{0} \cdot �^{�}$

Substitute the values:

$�_{�} = 80 \cdot 1. 5^{�}$

So, the explicit formula for $�_{�}$ in this exponential growth model is:

$�_{�} = 80 \cdot 1. 5^{�}$

Diseases tend to spread according to the exponential growth model. In the early days of AIDS, the growth factor (i.e. common ratio; growth multiplier) was around 1.9. In 1983, about 1700 people in the U.S. died of AIDS. If the trend had continued unchecked, how many people would have died from AIDS in 2004?

1214156142.6787

To predict how many people would have died from AIDS in 2004 if the trend had continued unchecked, you can use the exponential growth model. The formula for exponential growth is:

$� (�) = �_{0} * (1 + �)^{�}$

Where:

$� (�)$ is the population at time $�$ .
$�_{0}$ is the initial population (in 1983).
$�$ is the growth factor (1.9 in this case).
$�$ is the number of years into the future (2004 - 1983 = 21 years).

In 1983, about 1700 people died of AIDS in the U.S., so $�_{0} = 1700$ .

Now, plug these values into the formula:

$� (21) = 1700 * (1 + 1.9)^{21} \approx 1214156142.6787$

So, if the trend had continued unchecked, approximately 1,214,156,143 people would have died from AIDS in the U.S. by 2004.

A population grows according to an exponential growth model. The initial population is P0=3, and the growth rate is r=0.15.

Then:

P1= 3.5

P2= 4.0

Find an explicit formula for Pn* Your formula should involve n.

Pn= 3*(1+0.15)n

Use your formula to find P12

P12= 16.050750316421

$P_{n} = P_{0} \cdot (1 + r)^{n}$

In your case, $�_{0} = 3$ and $� = 0.15$ . So the formula for $�_{�}$ is:

$�_{�} = 3 \cdot (1 + 0.15)^{�}$

Now, to find $�_{12}$ , you plug in $� = 12$ into the formula:

$�_{12} = 3 \cdot (1 + 0.15)^{12} \approx 16.050750316421$

$�_{12}$ is approximately equal to 16.050750316421.

Assume there is a certain population of fish in a pond whose growth is described by the logistic equation. It is estimated that the carrying capacity for the pond is 1300 fish. Absent constraints, the population would grow by 250% per year. If the starting population is given by P0 = 500, then after one breeding season the population of the pond is given by P1=1269.23 After two breeding seasons the population of the pond is given by P2=1344.27

The population of fish in the pond can be modeled using the logistic growth equation, which is given by:

$� (�) = \frac{�}{1 + \frac{� - �_{0}}{�_{0}} \cdot �^{- � �}}$

Where:

$� (�)$ is the population at time $�$ .
$�$ is the carrying capacity (1300 fish in this case).
$�_{0}$ is the initial population (500 fish).
$�$ is the growth rate (given as 250% per year, which can be written as 2.5 when expressed as a decimal).
$�$ is the number of years.

Now, let's calculate $�_{1}$ after one breeding season (one year):

$� (1) = \frac{1300}{1 + \frac{1300 - 500}{500} \cdot �^{- 2.5 \cdot 1}}$

First, calculate the exponent:

$�^{- 2.5} \approx 0.0820849986$

Now, plug this value into the equation:

$� (1) = \frac{1300}{1 + \frac{1300 - 500}{500} \cdot 0.0820849986} \approx \frac{1300}{1 + 1.6 \cdot 0.0820849986} \approx \frac{1300}{1 + 0.131335998} \approx \frac{1300}{1.131335998} \approx 1147.928718$

So, after one breeding season, the estimated population of the pond is approximately 1147.93 fish (rounded to two decimal places).

Now, let's calculate $�_{2}$ after two breeding seasons (two years):

$� (2) = \frac{1300}{1 + \frac{1300 - 500}{500} \cdot �^{- 2.5 \cdot 2}}$

Calculate the exponent:

$�^{- 5} \approx 0.00673794699$

Now, plug this value into the equation:

$� (2) = \frac{1300}{1 + \frac{1300 - 500}{500} \cdot 0.00673794699} \approx \frac{1300}{1 + 1.6 \cdot 0.00673794699} \approx \frac{1300}{1 + 0.01078071438} \approx \frac{1300}{1.01078071438} \approx 1284.268367$

So, after two breeding seasons, the estimated population of the pond is approximately 1284.27 fish (rounded to two decimal places)

Tacoma's population in 2000 was about 200 thousand, and has been growing by about 9% each year. If this continues, what will Tacoma's population be in 2012?

562533 People

To calculate Tacoma's population in 2012 if it has been growing by about 9% each year since 2000, you can use the formula for exponential growth:

$� (�) = �_{0} \times (1 + �)^{�}$

Where:

$� (�)$ is the population at time $�$ .
$�_{0}$ is the initial population (in 2000, which is 200,000).
$�$ is the growth rate (9% or 0.09 when expressed as a decimal).
$�$ is the number of years into the future (2012 - 2000 = 12 years).

Now, plug in the values:

$� (12) = 200, 000 \times (1 + 0.09)^{12}$

First, calculate $1 + 0.09$ :

$1 + 0.09 = 1.09$

Now, raise 1.09 to the power of 12:

$1.0 9^{12} \approx 2.0581327625$

Now, multiply this result by the initial population of 200,000:

$� (12) \approx 200, 000 \times 2.0581327625 \approx 411, 626.5525$ 562,533

So, if Tacoma's population continues to grow by about 9% each year, it will be approximately 562,533 people in 2012.

Project #6 - Water/Time Comparisons:

WATER/TIME COMPARISONS

This project is incredibly unique. There is no right, or wrong answer and you will not be graded on the correctness of your answer. However, you will be graded on identifying the assumptions that you will need to make, the calculations once the assumptions are made, the correct conversions where needed, and presentation of the logic of how you arrived at your answer.

PROBLEM NUMBER 1

During a one-year time period, how many gallons of water would the average household save by using an efficiency washer vs a regular washer? Is this one-year savings more or less than the amount of water an average person drinks in a lifetime? Explain and show your assumptions, calculations, and conversions.

Consider it to be 2 separate problems:

1. How much water is saved in a one-year period when using an efficiency washer vs regular washer - answer based on your assumptions and calculations for full credit

2. How much water does an average person drink in a lifetime – arrive at this answer based on your assumptions and calculations

3. Compare the answers to those 2 questions to arrive at your conclusion

Below are 2 pieces of information that you might find helpful

there are 128 oz in a gallon

there are 365 days in a year and also 52 weeks in a year

PROBLEM NUMBER 2

Which amount of time is greater for the average person? The total time spent on his/her cell phone in a year’s time, or the total time spent waiting in line when shopping in a lifetime?

Consider it to be 2 separate problems just as in problem 1 and arrive at your conclusion.

You need to make assumptions, conversions, and math calculations to arrive at your answer. You will not be graded on the accuracy of your assumptions. You will be graded on the process at arriving at your answers, the correct conversions once you make your assumptions, and your logic/thought process in arriving at both answers. If you choose to research data for this project, that is fine but there is no need to cite your references.

Let's break down each problem and work through the calculations step by step:

Problem 1: Comparing Water Savings from Efficiency Washer vs. Regular Washer

How much water is saved in a one-year period when using an efficiency washer vs. a regular washer?

Assumptions:

Let's assume an average household uses a washer once a week.
The efficiency washer uses 30 gallons per load, and the regular washer uses 40 gallons per load.

Calculations:

Efficiency washer usage per year: 30 gallons/load * 52 loads/year = 1,560 gallons/year
Regular washer usage per year: 40 gallons/load * 52 loads/year = 2,080 gallons/year

Water saved per year by using an efficiency washer: 2,080 gallons/year - 1,560 gallons/year = 520 gallons/year

How much water does an average person drink in a lifetime?

Assumptions:

Let's assume an average person lives to be 80 years old.
An average person drinks 8 cups (64 ounces) of water per day.

Calculations:

Water consumed per year: 64 ounces/day * 365 days/year = 23,360 ounces/year
Water consumed in a lifetime: 23,360 ounces/year * 80 years = 1,868,800 ounces in a lifetime
Convert ounces to gallons: 1,868,800 ounces / 128 ounces/gallon = 14,600 gallons in a lifetime

Conclusion for Problem 1:

The average household saves 520 gallons of water per year by using an efficiency washer instead of a regular washer.
An average person drinks 14,600 gallons of water in a lifetime.
Therefore, the water saved by using an efficiency washer in one year is significantly less than the water an average person drinks in a lifetime.

Problem 2: Comparing Time Spent on a Cell Phone vs. Time Spent Waiting in Line

To compare the time spent on a cell phone in a year vs. the time spent waiting in line in a lifetime, we need to make some assumptions:

Average daily time spent on a cell phone:
- Let's assume an average of 2 hours per day spent on a cell phone.
Total days in a year:
- There are 365 days in a year.
Average waiting time in lines:
- Let's assume an average of 10 minutes per shopping trip spent waiting in line.
- An average person shops once a week.
Average lifespan:
- Let's assume an average lifespan of 80 years.

Calculations:

Time spent on a cell phone in a year: 2 hours/day * 365 days/year = 730 hours/year
Time spent waiting in line per shopping trip: 10 minutes = 10/60 hours = 1/6 hours
Shopping trips per year: 52 trips/year
Time spent waiting in line in a year: (1/6 hours/trip) * 52 trips/year = 8.67 hours/year
Time spent waiting in line in a lifetime: 8.67 hours/year * 80 years = 693.6 hours in a lifetime

Conclusion for Problem 2:

The total time spent on a cell phone in a year (730 hours) is greater than the total time spent waiting in line when shopping in a lifetime (693.6 hours). Therefore, for the average person, the time spent on a cell phone in a year is greater than the time spent waiting in line in a lifetime.

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