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Chapter 5: Problem 4

What is the growth or decay factor for each given time period? a. Weight increases by \(0.2 \%\) every 5 days. b. Mass decreases by \(6.3 \%\) every year. c. Population increases \(23 \%\) per decade. d. Profit increases \(300 \%\) per year. e. Blood alcohol level decreases \(35 \%\) per hour.

Short Answer

Expert verified
(a) 1.002, (b) 0.937, (c) 1.23, (d) 4.0, (e) 0.65

Step by step solution

01

- Understanding Growth and Decay Factors

Growth or decay factors are calculated using the formula for exponential growth or decay: \[ \text{Factor} = 1 \text{ (+ or -) Rate} \text{ [expressed as a decimal]} \]For growth, use '+'. For decay, use '-'.
02

- Calculate Growth Factor for (a)

The weight increases by 0.2% every 5 days. Convert the percentage into a decimal:\[ \text{Rate} = 0.2 / 100 = 0.002 \]Then apply the formula for growth:\[ \text{Growth Factor} = 1 + 0.002 = 1.002 \]
03

- Calculate Decay Factor for (b)

The mass decreases by 6.3% every year. Convert the percentage into a decimal:\[ \text{Rate} = 6.3 / 100 = 0.063 \]Then apply the formula for decay:\[ \text{Decay Factor} = 1 - 0.063 = 0.937 \]
04

- Calculate Growth Factor for (c)

The population increases by 23% per decade. Convert the percentage into a decimal:\[ \text{Rate} = 23 / 100 = 0.23 \]Then apply the formula for growth:\[ \text{Growth Factor} = 1 + 0.23 = 1.23 \]
05

- Calculate Growth Factor for (d)

The profit increases by 300% per year. Convert the percentage into a decimal:\[ \text{Rate} = 300 / 100 = 3.0 \]Then apply the formula for growth:\[ \text{Growth Factor} = 1 + 3.0 = 4.0 \]
06

- Calculate Decay Factor for (e)

The blood alcohol level decreases by 35% per hour. Convert the percentage into a decimal:\[ \text{Rate} = 35 / 100 = 0.35 \]Then apply the formula for decay:\[ \text{Decay Factor} = 1 - 0.35 = 0.65 \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

growth factor
Growth factors are essential when dealing with exponential growth problems. They determine how much a quantity increases over a given period. The basic formula for finding a growth factor is: Growth Factor = 1 + Rate where the rate is expressed as a decimal. For instance, if a weight increases by 0.2% every 5 days, you first convert 0.2% to a decimal: Rate = 0.2 / 100 = 0.002 Then, apply the formula: Growth Factor = 1 + 0.002 = 1.002 This means the weight grows by a factor of 1.002 every 5 days. To make the conversion, always divide the percentage by 100 to turn it into a decimal.
decay factor
Decay factors are crucial when addressing exponential decay situations. They tell you how much a quantity decreases over time. The basic formula for finding a decay factor is: Decay Factor = 1 - Rate, where the rate is expressed as a decimal. For example, if a mass decreases by 6.3% every year, you convert 6.3% to a decimal: Rate = 6.3 / 100 = 0.063 Then, apply the formula: Decay Factor = 1 - 0.063 = 0.937 The mass decreases by a factor of 0.937 annually. Always remember to convert the percentage to a decimal by dividing by 100.
percentage conversion
Percentage conversion is the first step to solving growth and decay problems. This process involves turning a percentage into a decimal. To do this, simply divide the percentage by 100. For example: If you have a percentage of 23%, you'd convert it by calculating: 23 / 100 = 0.23 Similarly, for a percentage of 300%, you'd convert it by: 300 / 100 = 3.0 Once you have the decimal, you can use it in your growth or decay formulas. This conversion is a vital step for correct calculations.
exponential functions
Exponential functions describe growth or decay processes, where quantities change at a consistent rate. The general form of an exponential function is: y = a(1 + r)^t for growth or y = a(1 - r)^t for decay. where - y is the final amount - a is the initial amount - r is the growth (or decay) rate as a decimal - t represents time For example, if you start with a population of 1000 that grows by 23% per decade, the function is: y = 1000(1 + 0.23)^t Exponential functions help model real-world phenomena such as population growth, radioactive decay, and financial investments.

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Most popular questions from this chapter

[Part (e) requires use of the Internet and technology to find a best-fit function.] A "rule of thumb" used by car dealers is that the trade-in value of a car decreases by \(30 \%\) each year. a. Is this decline linear or exponential? b. Construct a function that would express the value of the car as a function of years owned. c. Suppose you purchase a car for \(\$ 15,000 .\) What would its value be after 2 years? d. Explain how many years it would take for the car in part (c) to be worth less than \(\$ 1000\). Explain how you arrived at your answer. e. Internet search: Go to the Internet site for the Kelley Blue Book (www.kbb.com). i. Enter the information about your current car or a car you would like to own. Specify the actual age and mileage of the car. What is the Blue Book value? ii. Keeping everything else the same, assume the car is I year older and increase the mileage by 10,000 . What is the new value? iii. Find a best-fit exponential function to model the value of your car as a function of years owned. What is the annual decay rate? iv. According to this function, what will the value of your car be 5 years from now?

Which of the following functions have a fixed doubling time A fixed half-life? a. \(y=6(2)^{x}\) b. \(y=5+2 x\) c. \(Q=300\left(\frac{1}{2}\right)^{T}\) d. \(A=10(2)^{t / 5}\) e. \(P=500-\frac{1}{2} T\) f. \(N=50\left(\frac{1}{2}\right)^{t / 20}\)

(Graphing program recommended.) Below is a table of values for \(y=500(3)^{x}\) and for \(\log y\) $$ \begin{array}{rrr} \hline x & y & \log y \\ \hline 0 & 500 & 2.699 \\ 5 & 121,500 & 5.085 \\ 10 & 29,524,500 & 7.470 \\ 15 & 7.17 \cdot 10^{9} & 9.856 \\ 20 & 1.74 \cdot 10^{12} & 12.241 \\ 25 & 4.24 \cdot 10^{14} & 14.627 \\ 30 & 1.03 \cdot 10^{17} & 17.013 \\ \hline \end{array} $$ a. Plot \(y\) vs. \(x\) on a linear scale. Remember to identify the largest number you will need to plot before setting up axis scales. b. Plot \(\log y\) vs. \(x\) on a semi-log plot with a log scale on the vertical axis and a linear scale on the horizontal axis. c. Rewrite the \(y\) -values as powers of \(10 .\) How do these values relate to \(\log y ?\)

The price of a home in Medford was \(\$ 100,000\) in 1985 and rose to \(\$ 200,000\) in 2005 . a. Create two models, \(f(t)\) assuming linear growth and \(g(t)\) assuming exponential growth, where \(t=\) number of years after \(1985 .\) b. Fill in the following table representing linear growth and exponential growth for \(t\) years after \(1985 .\) c. Which model do you think is more realistic?

According to the Arkansas Democrat Gazette (February \(27,\) 1994): Jonathan Holdeen thought up a way to end taxes forever. It was disarmingly simple. He would merely set aside some money in trust for the government and leave it there for 500 or 1000 years. Just a penny, Holdeen calculated, could grow to trillions of dollars in that time. But the stash he had in mind would grow much bigger-to quadrillions or quintillions-so big that the government, one day, could pay for all its operations simply from the income. Then taxes could be abolished. And everyone would be better off. a. Holdeen died in 1967 , leaving a trust of \(\$ 2.8\) million that is being managed by his daughter, Janet Adams. In 1994 , the trust was worth \(\$ 21.6\) million. The trust was debated in Philadelphia Orphans' Court. Some lawyers who were trying to break the trust said that it is dangerous to let it go on, because "it would sponge up all the money in the world." Is this possible? b. After 500 years, how much would the trust be worth? Would this be enough to pay off the current national debt (over \(\$ 7\) trillion in 2004\() ?\) What about after 1000 years? Describe the model you used to make your predictions.
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