Aimathic – Percent growth and decay models

Percent growth and decay models

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Difficulty: 1 – 5

5100999

A test tube contains $400$ bacteria. The number of bacteria increases by $2\%$ each hour. Which function correctly models the number of bacteria $N$ after $t$ hours? a) $N=400^{1.02t}$ b) $N=400t^{0.02}$ c) $N=400(1.02)^t$ d) $N=400(0.02)^t$

Hints

- What is the value when $t=0$? - What growth factor represents a $2\%$ increase? - Is a constant percent increase linear or exponential?

Solution

1. The initial value is $N_0=400$. 2. A $2\%$ increase gives a growth factor of $1+\frac{2}{100}=1.02$. 3. Substituting into $N(t)=N_0b^t$ gives $N(t)=400(1.02)^t$, which is choice c).

Answer

c) $N=400(1.02)^t$

5283009

A new electric car loses $18\%$ of its value each year. Its purchase price is $\$45{,}000$. Write a function $V(t)$ for the car's value after $t$ years.

Hints

- Identify the initial value. - What percent of the value remains after an $18\%$ decrease? - Convert that percent to a decimal factor. - Use the general form of an exponential function.

Solution

1. The initial value is $45{,}000$. 2. After an $18\%$ decrease, $82\%$ of the previous value remains, so the yearly decay factor is $0.82$. 3. Therefore, the model is $V(t)=45{,}000(0.82)^t$.

Answer

$V(t)=45{,}000(0.82)^t$

5283039

A forest covers $4500\,\text{ha}$. Through reforestation, its area increases by an average of $2.4\%$ each year. a) Find the annual growth factor $b$. b) Find the forest area after $3$ years and after $10$ years. c) By what total factor has the forest area changed after $5$ years? Round the factor to four decimal places.

Hints

- How is a percent increase converted to a growth factor? - How many times is the annual factor applied over several years? - A total factor compares the future value directly with the initial value.

Solution

1. The annual growth factor is $b=1+0.024=1.024$. 2. After $3$ years, the area is $4500\cdot(1.024)^3\approx4831.84\,\text{ha}$. 3. After $10$ years, the area is $4500\cdot(1.024)^{10}\approx5704.43\,\text{ha}$. 4. The total factor after $5$ years is $(1.024)^5\approx1.1259$.

Answer

a) $b=1.024$ b) After $3$ years: about $4831.84\,\text{ha}$; after $10$ years: about $5704.43\,\text{ha}$ c) $1.1259$

5283069

A bacteria culture grows exponentially according to $N(t)=N_0a^t$, where $t$ is measured in hours. At $t=0$, the culture contains $800$ bacteria. At $t=1$, it contains $1040$ bacteria. a) Find the hourly growth factor $a$. b) Find the hourly percent increase. c) How many bacteria are present after $6$ hours? Round to the nearest whole bacterium.

Hints

- Use the initial value in the given model. - What factor takes the initial population to the population after one hour? - Convert the growth factor to a percent increase.

Solution

1. Since $N(1)=N_0a$, $1040=800a$, so $a=\frac{1040}{800}=1.3$. 2. The factor $1.3=1+0.30$ represents a $30\%$ hourly increase. 3. After $6$ hours, $N(6)=800\cdot(1.3)^6\approx3861.45$, or about $3861$ bacteria.

Answer

a) $a=1.3$ b) $30\%$ c) About $3861$ bacteria

5283109

In 2020, a wildlife preserve contained an estimated $1200$ plants of a rare orchid species. Biologists model the population with a constant annual increase of $5.4\%$. a) Find the annual growth factor $b$. b) Predict the population in 2021, 2022, and 2023. c) Predict the population in 2030. What single growth factor represents the entire $10$-year period? d) Predict the population in 2045. Round population values to the nearest whole plant.

Hints

- A growth factor for a percent increase is greater than $1$. - First find how many years have passed since 2020. - Use an exponent of $10$ for the 2030 prediction. - Round counts of plants only after completing the calculation.

Solution

1. The annual growth factor is $b=1+0.054=1.054$, so the model is $P(t)=1200(1.054)^t$, where $t$ is years after 2020. 2. $P(1)=1264.8\approx1265$, $P(2)\approx1333.10\approx1333$, and $P(3)\approx1405.09\approx1405$. 3. For 2030, $t=10$. Then $P(10)\approx2030.43$, or about $2030$ plants. The $10$-year factor is $(1.054)^{10}\approx1.6920$. 4. For 2045, $t=25$. Then $P(25)\approx4468.86$, or about $4469$ plants.

Answer

a) $b=1.054$ b) 2021: about $1265$; 2022: about $1333$; 2023: about $1405$ c) 2030: about $2030$ plants; $10$-year factor: about $1.6920$ d) About $4469$ plants

5283479

A forest contains $8400\,\text{m}^3$ of timber. Because of an insect infestation, the timber volume decreases by $6.2\%$ each year. a) Write a function $V(t)$ for the timber volume after $t$ years. b) What percent of the original timber volume remains after $8$ years?

Hints

- What factor represents a decrease of $6.2\%$? - Identify the value at $t=0$. - The power of the decay factor gives the fraction remaining.

Solution

1. The annual decay factor is $1-0.062=0.938$. 2. With initial volume $8400\,\text{m}^3$, the model is $V(t)=8400(0.938)^t$. 3. The fraction remaining after $8$ years is $(0.938)^8\approx0.5993$, so about $59.93\%$ remains.

Answer

a) $V(t)=8400(0.938)^t$ b) About $59.93\%$

5283489

Algae cover $0.5\,\text{m}^2$ of a lake's surface at the beginning of an observation. The covered area increases by $22\%$ each week. a) Write a function $A(t)$ for the covered area after $t$ weeks. b) Find the covered area after $5$ weeks. Round to the nearest hundredth of a square meter.

Hints

- Convert the percent increase to a growth factor. - Use the general form of an exponential function. - Substitute the given time into your function.

Solution

1. A $22\%$ increase gives a weekly growth factor of $1.22$. 2. The model is $A(t)=0.5(1.22)^t$. 3. After $5$ weeks, $A(5)=0.5\cdot(1.22)^5\approx1.3514$, so the covered area is about $1.35\,\text{m}^2$.

Answer

a) $A(t)=0.5(1.22)^t$ b) About $1.35\,\text{m}^2$

5283529

The number of people reached by a social media post is recorded each hour: <table> <tr> <td>Time in hours ($t$)</td> <td>$0$</td> <td>$1$</td> <td>$2$</td> <td>$3$</td> </tr> <tr> <td>People reached ($N$)</td> <td>$400$</td> <td>$600$</td> <td>$900$</td> <td>$1350$</td> </tr> </table> a) Verify numerically that the reach grows exponentially. b) State the hourly growth factor $r$ and the hourly percent increase. c) How many people will the post reach after $6$ hours if the growth rate remains constant? Round to the nearest whole person.

Hints

- What must be true of consecutive ratios for an exponential model? - How is a growth factor related to a percent increase? - Use the exponential model to predict a value beyond the table.

Solution

1. The consecutive ratios are $\frac{600}{400}=1.5$, $\frac{900}{600}=1.5$, and $\frac{1350}{900}=1.5$. Since the ratio is constant, the data follow an exponential model. 2. The hourly growth factor is $r=1.5$. Because $1.5-1=0.5$, the hourly percent increase is $50\%$. 3. The model is $N(t)=400(1.5)^t$. Thus $N(6)=400\cdot(1.5)^6=4556.25$, which rounds to about $4556$ people.

Answer

a) Yes. Each consecutive ratio is $1.5$. b) $r=1.5$; $50\%$ increase per hour c) About $4556$ people

5283989

A protected plant population grows by a factor of $1.4$ each year. At the beginning of the study, there are $250$ plants. Write a function $B(t)$ for the population after $t$ years. Then find the predicted population after $5$ years and after $10$ years. Round to the nearest whole plant.

Hints

- Identify the initial value and the yearly multiplication factor. - Substitute each requested time into the model. - Round population predictions to whole plants.

Solution

1. The initial value is $250$, and the yearly growth factor is $1.4$, so $B(t)=250(1.4)^t$. 2. After $5$ years, $B(5)=250\cdot(1.4)^5=1344.56$, or about $1345$ plants. 3. After $10$ years, $B(10)=250\cdot(1.4)^{10}\approx7231.37$, or about $7231$ plants.

Answer

$B(t)=250(1.4)^t$ After $5$ years: about $1345$ plants After $10$ years: about $7231$ plants

5283999

Complete the table for exponential models of the form $f(x)=Ca^x$. <table> <thead> <tr> <th>Percent change per step</th> <th>Factor $a$</th> <th>Initial value $C$</th> <th>Function $f(x)$</th> </tr> </thead> <tbody> <tr> <td>$+35\%$</td> <td></td> <td>$120$</td> <td></td> </tr> <tr> <td></td> <td>$0.82$</td> <td>$50$</td> <td></td> </tr> <tr> <td></td> <td></td> <td></td> <td>$8(1.045)^x$</td> </tr> </tbody> </table>

Hints

- A factor above $1$ represents growth; a factor below $1$ represents decay. - The initial value is the coefficient outside the exponential power. - Use $a=1\pm\frac{p}{100}$.

Solution

1. In row 1, a $35\%$ increase gives $a=1.35$, so $f(x)=120(1.35)^x$. 2. In row 2, $a=0.82$ means $82\%$ remains, so the change is an $18\%$ decrease. The function is $f(x)=50(0.82)^x$. 3. In row 3, the function shows $C=8$ and $a=1.045$. This factor represents a $4.5\%$ increase.

Answer

Row 1: $a=1.35$; $f(x)=120(1.35)^x$ Row 2: $18\%$ decrease; $f(x)=50(0.82)^x$ Row 3: $4.5\%$ increase; $a=1.045$; $C=8$

5284009

Consider the three functions $f(x)=400(1.015)^x$ $g(x)=25(0.6)^x$ $h(x)=2(3)^x$ a) For each function, identify the initial value and the percent increase or decrease per unit. b) Find $f(2)$.

Hints

- The initial value is the function value at $x=0$. - Compare each exponential factor with $1$. - Evaluate the exponent before multiplying.

Solution

1. For $f$, the initial value is $400$. The factor $1.015$ represents a $1.5\%$ increase. 2. For $g$, the initial value is $25$. The factor $0.6$ represents a $40\%$ decrease. 3. For $h$, the initial value is $2$. The factor $3$ represents a $200\%$ increase. 4. $f(2)=400\cdot(1.015)^2=412.09$.

Answer

a) $f$: initial value $400$, $1.5\%$ increase; $g$: initial value $25$, $40\%$ decrease; $h$: initial value $2$, $200\%$ increase b) $f(2)=412.09$

5284039

A rare orchid population in a wildlife preserve decreases by $6\%$ each year. At the beginning of the study, $450$ plants are counted. a) Write a function $N(t)$ for the population after $t$ years. b) Find the predicted population after $5$ years and after $10$ years. Round to the nearest whole plant.

Hints

- Identify the initial population. - Convert the percent decrease to a decay factor. - Raise the factor to the number of elapsed years.

Solution

1. The yearly decay factor is $1-0.06=0.94$, so $N(t)=450(0.94)^t$. 2. After $5$ years, $N(5)=450\cdot(0.94)^5\approx330.26$, or about $330$ plants. 3. After $10$ years, $N(10)=450\cdot(0.94)^{10}\approx242.38$, or about $242$ plants.

Answer

a) $N(t)=450(0.94)^t$ b) After $5$ years: about $330$ plants; after $10$ years: about $242$ plants

5344939

A medication is modeled as decreasing in the body after it is taken. The graph shows the concentration $C$, in $\text{mg/L}$, after $t$ hours. a) Find the factor $a$ in $C(t)=ca^t$. b) By what percent does the concentration decrease each hour?

Hints

- The factor compares the value after one time unit with the initial value. - Read the graph at $t=0$ and $t=1$. - Find the percent that is lost when $80\%$ remains.

Solution

1. The graph shows an initial concentration of $c=20$. 2. After $1$ hour, the concentration is $16\,\text{mg/L}$. Thus $a=\frac{16}{20}=0.8$. 3. A factor of $0.8$ means $80\%$ remains each hour, so the concentration decreases by $20\%$ per hour.

Answer

a) $a=0.8$ b) $20\%$ decrease per hour

5115969

Lucas wants to invest $\$250.00$ for two years. Interest is added to the balance at the end of each year, so it earns interest during the following year. He compares two options: Option A: $3\%$ interest each year. Option B: $2\%$ interest in the first year and $4\%$ interest in the second year. Create a table showing the balance for each year under both options. What is the balance after two years for each option, and which option earns more? Round each year’s interest to the nearest cent.

Hints

- Find the first year’s interest and add it to the starting balance. - Use the ending balance from one year as the beginning balance for the next year. - Organize the two options in separate tables. - Pay attention to the different second-year interest rates.

Solution

For Option A: | Year | Beginning balance | Interest rate | Interest | Ending balance | |---|---:|---:|---:|---:| | 1 | $\$250.00$ | $3\%$ | $\$7.50$ | $\$257.50$ | | 2 | $\$257.50$ | $3\%$ | $\$7.73$ | $\$265.23$ | For Option B: | Year | Beginning balance | Interest rate | Interest | Ending balance | |---|---:|---:|---:|---:| | 1 | $\$250.00$ | $2\%$ | $\$5.00$ | $\$255.00$ | | 2 | $\$255.00$ | $4\%$ | $\$10.20$ | $\$265.20$ | Option A ends with $\$265.23$, while Option B ends with $\$265.20$. Therefore, Option A earns $\$0.03$ more.

Answer

Option A: $\$265.23$ Option B: $\$265.20$ Option A earns $\$0.03$ more.

5115979

A three-year savings account starts with $\$600.00$ and has a different annual interest rate each year: - Year 1: $1.5\%$ - Year 2: $2.5\%$ - Year 3: $3.5\%$ Interest is added to the balance at the end of each year and earns interest in later years. Create a table showing the beginning balance, interest earned, and ending balance for each year. How much total interest is earned after three years? Round each year’s interest to the nearest cent.

Hints

- Use columns for year, beginning balance, interest rate, interest, and ending balance. - The ending balance for one year becomes the beginning balance for the next year. - You can find the total interest by adding the yearly interest amounts or subtracting the original deposit from the final balance.

Solution

| Year | Beginning balance | Interest rate | Interest | Ending balance | |---|---:|---:|---:|---:| | 1 | $\$600.00$ | $1.5\%$ | $\$9.00$ | $\$609.00$ | | 2 | $\$609.00$ | $2.5\%$ | $\$15.23$ | $\$624.23$ | | 3 | $\$624.23$ | $3.5\%$ | $\$21.85$ | $\$646.08$ | The total interest is $\$9.00 + \$15.23 + \$21.85 = \$46.08$. This agrees with $\$646.08 - \$600.00 = \$46.08$.

Answer

The ending balance is $\$646.08$, and the total interest earned is $\$46.08$.

5127849

Ms. Weber deposits $\$4500$ in an account that pays $2\%$ interest per year. At the end of each year, the interest is added to the balance and earns interest in later years. Find the interest earned and the ending balance for each of the first three years. Round to the nearest cent at the end of each year.

Hints

- Find the first year’s interest from the original deposit. - Use each year’s ending balance as the next year’s beginning balance. - Interest added to the account increases the amount that earns interest later. - Round money to the nearest cent at the end of each year.

Solution

1. Year 1 interest: $\$4500.00 \cdot 0.02 = \$90.00$. The ending balance is $\$4500.00 + \$90.00 = \$4590.00$. 2. Year 2 interest: $\$4590.00 \cdot 0.02 = \$91.80$. The ending balance is $\$4590.00 + \$91.80 = \$4681.80$. 3. Year 3 interest: $\$4681.80 \cdot 0.02 = \$93.636 \approx \$93.64$. The ending balance is $\$4681.80 + \$93.64 = \$4775.44$.

Answer

Year 1: Interest $\$90.00$; balance $\$4590.00$ Year 2: Interest $\$91.80$; balance $\$4681.80$ Year 3: Interest $\$93.64$; balance $\$4775.44$

5127859

A savings account begins the year with $\$1200$. At the end of the year, $\$18$ in interest is added to the balance. a) Find the annual interest rate. b) If the new balance earns interest at the same rate during the second year, how much interest will be earned in the second year?

Hints

- Compare the first-year interest with the original balance. - Add the first-year interest to find the second year’s beginning balance. - Apply the same rate to the new balance.

Solution

1. For part a, the rate is $\frac{18}{1200} = 0.015 = 1.5\%$. 2. The balance at the beginning of the second year is $\$1200 + \$18 = \$1218$. 3. For part b, the second-year interest is $\$1218 \cdot 0.015 = \$18.27$.

Answer

a) The annual interest rate is $1.5\%$. b) The second-year interest is $\$18.27$.

5142709

Mr. Smith invests $\$3000$ and earns exactly $\$75$ in interest after one year. a) Find the annual interest rate. b) How much interest would $\$4500$ earn in one year at the same rate? c) Return to the original $\$3000$ investment. Suppose the first-year interest is added to the account, but the rate drops to $2\%$ for the second year. How much interest is earned during the second year?

Hints

- Use the first-year interest and principal to find the rate. - For part b, the rate stays the same while the principal changes. - For part c, find the balance at the beginning of the second year.

Solution

1. For part a, the rate is $\frac{75}{3000} = 0.025 = 2.5\%$. 2. For part b, the interest is $\$4500 \cdot 0.025 = \$112.50$. 3. For part c, the balance at the beginning of the second year is $\$3000 + \$75 = \$3075$. 4. The second-year interest is $\$3075 \cdot 0.02 = \$61.50$.

Answer

a) The annual interest rate is $2.5\%$. b) The interest would be $\$112.50$. c) The second-year interest would be $\$61.50$.

5149319

A bacteria culture is modeled by $N(t)=N_0b^t$, where $N_0$ is the initial population, $t$ is time in hours, and $b$ is the hourly growth factor. In an experiment, the population increases from $1200$ to $10{,}000$ in $5$ hours. a) Find the growth factor $b$, rounded to three decimal places. b) Find the hourly percent increase.

Hints

- What number raised to the fifth power equals the ratio of the final population to the initial population? - How is a growth factor converted to a percent increase?

Solution

1. Substitute the given values: $10{,}000=1200b^5$. 2. Divide by $1200$: $b^5=\frac{10{,}000}{1200}=\frac{25}{3}$. 3. Take the fifth root: $b=\left(\frac{25}{3}\right)^{1/5}\approx1.528$. 4. The percent increase is $(b-1)\cdot100\%$. Using the unrounded factor gives approximately $52.814\%$, or $52.8\%$ to the nearest tenth of a percent.

Answer

a) $b\approx1.528$ b) About $52.8\%$ per hour

5251989

Two friends each begin with $\$1000$ in a savings account. - Account A grows by $4\%$ in each of two consecutive years. - Account B grows by $2\%$ in the first year and $6\%$ in the second year. Determine which account has the greater balance after two years. Briefly explain why the balances differ even though the listed rates add to $8\%$ for each account.

Hints

- Find each account’s balance after each year. - Use the first-year ending balance as the second-year beginning balance. - Compare the products of the growth factors.

Solution

1. Account A has a balance of $\$1000(1.04)^2 = \$1081.60$. 2. Account B has a balance of $\$1000 \cdot 1.02 \cdot 1.06 = \$1081.20$. 3. Since $\$1081.60 > \$1081.20$, Account A has the greater balance. 4. Successive percent changes are combined by multiplying growth factors, not by adding the rates. Here, $1.04 \cdot 1.04 = 1.0816$, while $1.02 \cdot 1.06 = 1.0812$.

Answer

Account A has the greater balance: $\$1081.60$, compared with $\$1081.20$ in Account B. The balances differ because successive growth factors are multiplied.

5281459

A manufacturing system is purchased for $\$54{,}000$. Its value decreases by $12\%$ each year. a) Find its value after $5$ years. Round to the nearest dollar. b) Determine the first full year when its value is less than $\$20{,}000$.

Hints

- What decay factor represents a $12\%$ decrease? - What type of model represents a constant percent change? - Test consecutive full years near the threshold. - Identify the first year for which the value is below the limit.

Solution

1. The yearly decay factor is $1-0.12=0.88$, so $V(t)=54{,}000(0.88)^t$. 2. After $5$ years, $V(5)=54{,}000\cdot(0.88)^5\approx28{,}497.52$, which rounds to $\$28{,}498$. 3. $V(7)\approx22{,}068.48$, which is still above $\$20{,}000$. $V(8)\approx19{,}420.26$, which is below $\$20{,}000$. Therefore, the value first falls below the threshold after $8$ full years.

Answer

a) $\$28{,}498$ b) $8$ years

5282959

A bacteria culture is modeled by $N(t)=200(1.5)^t$, where $t$ is measured in hours. a) By what factor does the population change over $2$ hours, $3$ hours, and $5$ hours? b) Find the population after half an hour and after $15$ minutes. Round to the nearest whole bacterium.

Hints

- How does the output change when the exponent increases by a given amount? - Keep the time unit in the exponent consistent. - Convert minutes to hours. - Use fractional exponents for fractional time intervals.

Solution

1. Over a time interval of $\Delta t$ hours, the growth factor is $(1.5)^{\Delta t}$. 2. The factors are $(1.5)^2=2.25$, $(1.5)^3=3.375$, and $(1.5)^5=7.59375$. 3. Half an hour is $0.5$ hour, so $N(0.5)=200\cdot(1.5)^{0.5}\approx244.95$, or about $245$ bacteria. 4. Fifteen minutes is $0.25$ hour, so $N(0.25)=200\cdot(1.5)^{0.25}\approx221.34$, or about $221$ bacteria.

Answer

a) $2$ hours: $2.25$; $3$ hours: $3.375$; $5$ hours: $7.59375$ b) Half an hour: about $245$ bacteria; $15$ minutes: about $221$ bacteria

5282969

The value of a collectible increases by $6\%$ each year. Its initial value is $\$500$, and the model has the form $W(t)=W_0b^t$, where $t$ is measured in years. a) Write the function. Find the total growth factor after $2$, $5$, and $10$ years. b) Find the value after $3$ months and after $1.5$ years. Round to the nearest cent.

Hints

- Convert the percent increase to a growth factor. - Express a number of months as a fraction of a year. - The total growth factor depends on the elapsed time, not the initial value.

Solution

1. A $6\%$ increase gives a yearly growth factor of $1.06$, so $W(t)=500(1.06)^t$. 2. The total growth factors are $(1.06)^2=1.1236$, $(1.06)^5\approx1.3382$, and $(1.06)^{10}\approx1.7908$. 3. Three months is $\frac{3}{12}=0.25$ year, so $W(0.25)=500\cdot(1.06)^{0.25}\approx507.34$. 4. After $1.5$ years, $W(1.5)=500\cdot(1.06)^{1.5}\approx545.67$.

Answer

a) $W(t)=500(1.06)^t$. Factors: $1.1236$ after $2$ years, about $1.3382$ after $5$ years, and about $1.7908$ after $10$ years. b) After $3$ months: $\$507.34$; after $1.5$ years: $\$545.67$

5283049

A manufacturing machine is purchased for $\$12{,}500$. Its value decreases by $15\%$ each year. a) What annual decay factor $b$ models the depreciation? b) Find the machine's value after $2$ years and after $5$ years. c) By what percent has the machine's value decreased after $3$ years compared with its purchase price?

Hints

- A decay factor is less than $1$. How do you find it from the percent decrease? - Raise the annual factor to the number of years. - First find the fraction of the original value that remains after $3$ years.

Solution

1. The annual decay factor is $b=1-0.15=0.85$. 2. After $2$ years, the value is $12{,}500\cdot(0.85)^2=9031.25$. 3. After $5$ years, the value is $12{,}500\cdot(0.85)^5\approx5546.32$. 4. After $3$ years, the remaining-value factor is $(0.85)^3=0.614125$. Therefore, the total percent decrease is $(1-0.614125)\cdot100\%=38.5875\%\approx38.59\%$.

Answer

a) $b=0.85$ b) After $2$ years: $\$9031.25$; after $5$ years: about $\$5546.32$ c) About $38.59\%$

5283059

The balance of a savings account over three years is shown in the table. <table> <tr><td>Year $n$</td><td>$0$</td><td>$1$</td><td>$2$</td><td>$3$</td></tr> <tr><td>Balance $K_n$ in dollars</td><td>$5000.00$</td><td>$5125.00$</td><td>$5253.13$</td><td>$5384.46$</td></tr> </table> a) Divide consecutive table values to confirm that the balance grows by approximately the same factor $b$ each year. b) Find the annual interest rate as a percent. c) Find the balance after $10$ years if the rate remains constant. Round to the nearest cent.

Hints

- Divide each year's balance by the previous year's balance. - How does a growth factor greater than $1$ relate to a percent increase? - Use compound growth to calculate many years at once.

Solution

1. The consecutive ratios are $\frac{5125.00}{5000.00}=1.025$, $\frac{5253.13}{5125.00}\approx1.025$, and $\frac{5384.46}{5253.13}\approx1.025$. Thus the annual growth factor is $b=1.025$. 2. Since $1.025=1+0.025$, the annual interest rate is $2.5\%$. 3. After $10$ years, $K_{10}=5000\cdot(1.025)^{10}\approx6400.42$.

Answer

a) Each ratio is approximately $1.025$. b) $2.5\%$ c) $\$6400.42$

5283099

A small town currently uses $850\,\text{GWh}$ of energy per year. Efficiency measures are expected to reduce this annual use by $2.1\%$ each year compared with the previous year. a) Find the annual decay factor $b$. b) Find the expected annual energy use after $4$ years. c) Find the factor that multiplies the current use to give the use after $10$ years. Then find the annual energy use after $10$ years. d) By what total percent has annual energy use decreased after $10$ years?

Hints

- Decide whether the situation describes growth or decay. - Convert the percent decrease to a decimal factor. - Use a power for repeated yearly changes. - The multi-year factor is the annual factor raised to the number of years.

Solution

1. The annual decay factor is $b=1-0.021=0.979$. 2. After $4$ years, the annual use is $850\cdot(0.979)^4\approx780.82\,\text{GWh}$. 3. The $10$-year factor is $(0.979)^{10}\approx0.8088$. Therefore, the annual use after $10$ years is $850\cdot(0.979)^{10}\approx687.46\,\text{GWh}$. 4. The total percent decrease is $(1-(0.979)^{10})\cdot100\%\approx19.12\%$.

Answer

a) $b=0.979$ b) About $780.82\,\text{GWh}$ c) Factor: about $0.8088$; energy use: about $687.46\,\text{GWh}$ d) About $19.12\%$

5283159

Use $K_n=K_0\left(1+\frac{p}{100}\right)^n$ for each compound-interest case. Round dollar amounts to the nearest cent and the unknown interest rate to the nearest tenth of a percent. a) $K_0=\$4000.00$, $p=3.5\%$, $n=6$. Find $K_n$. b) $p=2.25\%$, $n=10$, $K_n=\$8000.00$. Find $K_0$. c) $K_0=\$2500.00$, $n=8$, $K_n=\$4000.00$. Find $p$.

Hints

- Identify the unknown quantity in each part. - Rearrange the compound-interest formula for the unknown. - Distinguish between the growth factor and the interest rate. - Solving for the interest rate in part c) requires an eighth root.

Solution

1. For a), $K_6=4000\cdot(1.035)^6\approx4917.02$. 2. For b), solve for the initial principal: $K_0=\frac{8000}{(1.0225)^{10}}\approx6404.08$. 3. For c), the growth factor satisfies $b^8=\frac{4000}{2500}=1.6$. Thus $b=(1.6)^{1/8}\approx1.0605106$, so $p=(b-1)\cdot100\%\approx6.1\%$.

Answer

a) $K_n\approx\$4917.02$ b) $K_0\approx\$6404.08$ c) $p\approx6.1\%$

5283179

An initial deposit of $\$2500$ earns a fixed annual interest rate of $1.75\%$, compounded annually. a) Find the account balance after $5$ years and after $10$ years. b) Find the interest earned during the $10$th year alone, from the end of year $9$ to the end of year $10$. c) After how many full years has the account gained more than $\$500$ in total?

Hints

- Convert the annual interest rate to a growth factor. - Separate the total balance from the increase during one specific year. - What total balance corresponds to a gain of more than $\$500$? - Check consecutive full years near that target.

Solution

1. The growth factor is $1.0175$, so $K_n=2500(1.0175)^n$. 2. $K_5\approx2726.54$, and $K_{10}\approx2973.61$. 3. The interest earned during year $10$ is $K_{10}-K_9\approx2973.61-2922.47=51.14$. 4. A gain greater than $500$ requires $K_n>3000$. Since $K_{10}\approx2973.61<3000$ and $K_{11}\approx3025.65>3000$, the account first exceeds the target after $11$ full years.

Answer

a) After $5$ years: $\$2726.54$; after $10$ years: $\$2973.61$ b) $\$51.14$ c) $11$ years

5283189

Compare two savings accounts. Both compound interest annually. Account A starts with $\$4000$ and earns $3.0\%$ per year. Account B starts with $\$4500$ and earns $2.0\%$ per year. a) Find both balances after $5$ years. b) Find both balances after $15$ years. Which account has the greater balance then? c) Determine the first full year when Account A has a greater balance than Account B.

Hints

- Write a compound-growth function for each account. - Compare the balances after $5$ and $15$ years. - A lower initial balance can eventually become greater if its growth factor is larger. - Check consecutive whole years near where the order changes.

Solution

1. The models are $A(n)=4000(1.03)^n$ and $B(n)=4500(1.02)^n$. 2. After $5$ years, $A(5)\approx4637.10$, and $B(5)\approx4968.36$. 3. After $15$ years, $A(15)\approx6231.87$, and $B(15)\approx6056.41$. Account A is greater. 4. Since $\frac{A(n)}{B(n)}=\frac{4000}{4500}\left(\frac{1.03}{1.02}\right)^n$ increases with $n$, it is enough to check consecutive years near the crossover. At $n=12$, $A(12)\approx5703.04<B(12)\approx5707.09$. At $n=13$, $A(13)\approx5874.13>B(13)\approx5821.23$. Therefore, Account A first becomes greater after $13$ years.

Answer

a) Account A: about $\$4637.10$; Account B: about $\$4968.36$ b) Account A: about $\$6231.87$; Account B: about $\$6056.41$. Account A is greater. c) $13$ years

5283279

A threatened bird population in a wildlife preserve is currently estimated at $1200$. a) Under favorable conditions, the population is modeled with annual growth of $6\%$. Find the predicted population after $5$, $10$, and $15$ years. b) Under a second scenario, annual growth slows to $3.5\%$. Find the predicted population after $5$, $10$, and $15$ years. c) Find the difference between the two predicted populations after $15$ years. Round population values to the nearest whole bird.

Hints

- Convert each percent increase to a growth factor. - Use the exponential model to calculate each future value directly. - Round counts only after completing each calculation. - Subtract the two unrounded $15$-year predictions before rounding.

Solution

1. The growth factors are $1.06$ and $1.035$. 2. In the $6\%$ scenario, the predictions are $1200\cdot(1.06)^5\approx1606$, $1200\cdot(1.06)^{10}\approx2149$, and $1200\cdot(1.06)^{15}\approx2876$. 3. In the $3.5\%$ scenario, the predictions are $1200\cdot(1.035)^5\approx1425$, $1200\cdot(1.035)^{10}\approx1693$, and $1200\cdot(1.035)^{15}\approx2010$. 4. Using unrounded values, the difference after $15$ years is $1200\cdot(1.06)^{15}-1200\cdot(1.035)^{15}\approx865.45$, or about $865$ birds.

Answer

a) About $1606$, $2149$, and $2876$ birds b) About $1425$, $1693$, and $2010$ birds c) About $865$ birds

5283289

An initial $\$30{,}000$ is invested for retirement. a) Bank A pays $3.2\%$ annual interest, compounded annually. Find the balance after $10$ years and after $20$ years. b) Bank B pays $2.1\%$ annual interest, compounded annually. Find the balance after $10$ years and after $20$ years. c) How much greater is the Bank A balance than the Bank B balance after $20$ years?

Hints

- Use the compound-interest model. - Consider how a small difference in annual rates accumulates over a long period. - Round money to the nearest cent. - Subtract the two $20$-year balances.

Solution

1. Bank A has growth factor $1.032$. Its balances are $30{,}000\cdot(1.032)^{10}\approx41{,}107.23$ and $30{,}000\cdot(1.032)^{20}\approx56{,}326.82$. 2. Bank B has growth factor $1.021$. Its balances are $30{,}000\cdot(1.021)^{10}\approx36{,}929.95$ and $30{,}000\cdot(1.021)^{20}\approx45{,}460.70$. 3. After $20$ years, the difference is $56{,}326.82-45{,}460.70=10{,}866.12$.

Answer

a) After $10$ years: $\$41{,}107.23$; after $20$ years: $\$56{,}326.82$ b) After $10$ years: $\$36{,}929.95$; after $20$ years: $\$45{,}460.70$ c) $\$10{,}866.12$

5283429

In a lake, light intensity decreases by $12\%$ for each meter of water depth. a) Explain the relationship between the percent decrease and the decay factor $b$. Find $b$ for this situation. b) At the surface, the light intensity is $100\%$. Find the remaining intensity at depths of $2\,\text{m}$, $5\,\text{m}$, and $10\,\text{m}$. c) A diver claims, “If the intensity decreases by $12\%$ per meter, then at $5\,\text{m}$ it has decreased by $60\%$ in all.” Evaluate the claim mathematically.

Hints

- What percent remains after removing $12\%$? - Each decrease is based on the newly reduced value. - Compare repeated exponential decay with simply multiplying $12\%$ by the number of meters.

Solution

1. After a $12\%$ decrease, $88\%$ remains, so the decay factor is $b=1-0.12=0.88$. 2. The percent intensity at depth $d$ is $I(d)=100(0.88)^d$. 3. $I(2)=77.44\%$, $I(5)\approx52.77\%$, and $I(10)\approx27.85\%$. 4. The diver's claim is false because the $12\%$ decrease is applied to the remaining intensity at each meter, not repeatedly to the original amount. At $5\,\text{m}$, about $52.77\%$ remains, so the total decrease is about $47.23\%$, not $60\%$.

Answer

a) $b=0.88$; the factor is the fraction that remains after each $12\%$ decrease. b) $2\,\text{m}$: $77.44\%$; $5\,\text{m}$: about $52.77\%$; $10\,\text{m}$: about $27.85\%$ c) The claim is false. The total decrease after $5\,\text{m}$ is about $47.23\%$.

5283499

A filter system reduces a pollutant's concentration by $18\%$ in each filter layer. a) After $4$ filter layers, what percent of the original concentration remains? b) Find the decay factor for $1$ layer, $2$ layers, and $5$ layers. c) What is the minimum number of filter layers needed for the concentration to be less than $10\%$ of its original value?

Hints

- Convert the percent decrease to a decay factor. - Repeated filtering raises the one-layer factor to a power. - For part c, test whole-number exponents until the remaining fraction is below $0.10$.

Solution

1. The decay factor for one layer is $b=1-0.18=0.82$. 2. After $4$ layers, the remaining fraction is $(0.82)^4\approx0.4521$, so about $45.2\%$ remains. 3. The factors are $0.82$ for $1$ layer, $(0.82)^2=0.6724$ for $2$ layers, and $(0.82)^5\approx0.3707$ for $5$ layers. 4. Test consecutive whole numbers near the threshold: $(0.82)^{11}\approx0.1127>0.10$, while $(0.82)^{12}\approx0.0924<0.10$. Therefore, at least $12$ layers are required.

Answer

a) About $45.2\%$ b) $1$ layer: $0.82$; $2$ layers: $0.6724$; $5$ layers: about $0.3707$ c) $12$ layers

5283599

A contrast agent is modeled as decreasing in a patient's bloodstream by $20\%$ each hour. Immediately after injection, $120\,\text{mg}$ is present. a) How much remains after $3$ hours? b) Find the decay factor for a $4$-hour interval. c) After how many full hours is the amount first less than one-fourth of the initial amount?

Hints

- Find the factor that remains after a $20\%$ decrease. - Raise that factor to a power for a multi-hour interval. - One-fourth of the starting amount corresponds to a remaining fraction of $0.25$. - Test consecutive whole hours near the threshold.

Solution

1. The hourly decay factor is $0.8$, so $A(t)=120(0.8)^t$. 2. After $3$ hours, $A(3)=120\cdot(0.8)^3=61.44\,\text{mg}$. 3. The $4$-hour decay factor is $(0.8)^4=0.4096$. 4. One-fourth of the initial amount is $30\,\text{mg}$. Since $A(6)\approx31.46\,\text{mg}>30\,\text{mg}$ and $A(7)\approx25.17\,\text{mg}<30\,\text{mg}$, the amount is first below one-fourth after $7$ full hours.

Answer

a) $61.44\,\text{mg}$ b) $0.4096$ c) $7$ hours

5283609

A filter system reduces the pollutant concentration in an artificial lake by $12\%$ each day. The initial concentration is $250\,\text{mg/L}$. a) Find the concentration after $5$ days. b) State the daily decay factor and the decay factor for one week. c) A technician claims, “After two days, the concentration has decreased by exactly $24\%$.” Show that the claim is false.

Hints

- Convert the daily percent decrease to a factor less than $1$. - Raise the daily factor to a power for longer intervals. - On day two, the $12\%$ decrease is applied to the already reduced concentration.

Solution

1. The daily decay factor is $b=1-0.12=0.88$. 2. After $5$ days, the concentration is $250\cdot(0.88)^5\approx131.93\,\text{mg/L}$. 3. The one-week decay factor is $(0.88)^7\approx0.4087$. 4. After two days, the remaining fraction is $(0.88)^2=0.7744$. Therefore, the total decrease is $1-0.7744=0.2256$, or $22.56\%$, not $24\%$.

Answer

a) About $131.93\,\text{mg/L}$ b) Daily factor: $0.88$; one-week factor: about $0.4087$ c) The actual two-day decrease is $22.56\%$, so the claim is false.

5283629

A ball is dropped from a height of $2.50\,\text{m}$. After each bounce, it reaches $80\%$ of its previous height. a) Find the height after the 3rd, 4th, and 6th bounces. Give each answer in meters and centimeters. b) Estimate the height after the 10th bounce using $(0.8)^{10}\approx0.1$.

Hints

- What factor represents retaining $80\%$ of a height? - How many times is the factor applied after a given number of bounces? - How many centimeters are in one meter? - Substitute the given power estimate for part b.

Solution

1. The height factor is $0.8$, and the initial height is $2.50\,\text{m}=250\,\text{cm}$. 2. After the 3rd bounce, the height is $2.50\cdot(0.8)^3=1.28\,\text{m}=128\,\text{cm}$. 3. After the 4th bounce, the height is $2.50\cdot(0.8)^4=1.024\,\text{m}=102.4\,\text{cm}$. 4. After the 6th bounce, the height is $2.50\cdot(0.8)^6=0.65536\,\text{m}=65.536\,\text{cm}$. 5. Using the given estimate, the height after the 10th bounce is approximately $2.50\cdot0.1=0.25\,\text{m}=25\,\text{cm}$.

Answer

a) 3rd bounce: $1.28\,\text{m}$ or $128\,\text{cm}$; 4th bounce: $1.024\,\text{m}$ or $102.4\,\text{cm}$; 6th bounce: $0.65536\,\text{m}$ or $65.536\,\text{cm}$ b) About $0.25\,\text{m}$, or $25\,\text{cm}$

5283689

An algae culture is modeled by $A(t)=A_0(1.15)^t$, where $t$ is measured in days. Find the percent change in the population when a) the time increases by $1$ day; b) the time increases by $2$ days; c) the time increases by $0.5$ day. Round to the nearest tenth of a percent when needed.

Hints

- Relate the daily growth factor to a percent increase. - Over two days, apply the daily factor twice. - A half-day interval corresponds to an exponent of $0.5$.

Solution

1. Over a time increase of $h$ days, the growth factor is $(1.15)^h$, so the percent change is $((1.15)^h-1)\cdot100\%$. 2. For $h=1$, the factor is $1.15$, giving a $15\%$ increase. 3. For $h=2$, the factor is $(1.15)^2=1.3225$, giving a $32.25\%\approx32.3\%$ increase. 4. For $h=0.5$, the factor is $\sqrt{1.15}\approx1.07238$, giving an increase of about $7.2\%$.

Answer

a) $15\%$ increase b) About $32.3\%$ increase c) About $7.2\%$ increase

5283699

The table shows the population of a bacteria culture under ideal conditions. <table> <tr> <td>Time $t$ (hours)</td> <td>$0$</td> <td>$1$</td> <td>$2$</td> <td>$3$</td> </tr> <tr> <td>Population $N(t)$</td> <td>$400$</td> <td>$600$</td> <td>$900$</td> <td>$1350$</td> </tr> </table> a) Show that the population grows exponentially. Find the growth factor $b$ and initial value $a$. b) Write a function in the form $N(t)=ab^t$. c) Predict the population after $5$ hours and after $8$ hours. Round to the nearest whole bacterium. d) By what percent does the population increase each hour?

Hints

- Compare consecutive values using differences and ratios. - The value at time zero is the initial value. - Relate the growth factor to the percent increase. - Substitute the requested times into your function.

Solution

1. The ratios of consecutive values are $\frac{600}{400}=1.5$, $\frac{900}{600}=1.5$, and $\frac{1350}{900}=1.5$. The constant ratio shows exponential growth with $b=1.5$. 2. The value at $t=0$ is $a=400$, so $N(t)=400(1.5)^t$. 3. $N(5)=400\cdot(1.5)^5=3037.5\approx3038$, and $N(8)=400\cdot(1.5)^8=10{,}251.5625\approx10{,}252$. 4. The percent increase is $(1.5-1)\cdot100\%=50\%$.

Answer

a) $a=400$, $b=1.5$ b) $N(t)=400(1.5)^t$ c) After $5$ hours: about $3038$; after $8$ hours: about $10{,}252$ d) $50\%$ per hour

5283729

Let $g(x)=100(0.8)^x$. Find the percent change in the function value when $x$ (1) increases by $1$; (2) increases by $2$; (3) decreases by $1$; (4) increases by $0.5$. Round to the nearest tenth of a percent when needed.

Hints

- Relate the multiplicative factor to the percent change. - A factor less than $1$ represents a decrease. - Use negative and fractional exponents for the last two changes. - Start from $g(x+h)=g(x)(0.8)^h$.

Solution

1. For a change of $h$ in the input, the change factor is $(0.8)^h$, and the percent change is $((0.8)^h-1)\cdot100\%$. 2. For $h=1$, the factor is $0.8$, so the value decreases by $20\%$. 3. For $h=2$, the factor is $0.64$, so the value decreases by $36\%$. 4. For $h=-1$, the factor is $(0.8)^{-1}=1.25$, so the value increases by $25\%$. 5. For $h=0.5$, the factor is $\sqrt{0.8}\approx0.8944$, so the value decreases by about $10.6\%$.

Answer

(1) $20\%$ decrease (2) $36\%$ decrease (3) $25\%$ increase (4) About $10.6\%$ decrease

5283769

The number of fir trees in a forest decreases by $10\%$ each year because of environmental stress. 1. State the one-year decay factor $b$. 2. Find the percent of the original population remaining after $5$ years. 3. Find the factor that represents the total change after $10$ years. 4. An observer claims, “If the population has decreased by a certain percent after $5$ years, then the percent loss after $10$ years must be exactly twice as large.” Test the claim by comparing the two percent losses.

Hints

- Convert the annual percent decrease to a multiplication factor. - Use a power for repeated yearly decay. - Distinguish between the percent remaining and the percent lost. - Each year's decrease is based on the new, smaller population.

Solution

1. A $10\%$ decrease gives a decay factor of $b=0.9$. 2. After $5$ years, the remaining factor is $(0.9)^5=0.59049$, so $59.049\%\approx59.05\%$ remains. 3. After $10$ years, the factor is $(0.9)^{10}\approx0.348678$. 4. The loss after $5$ years is $100\%-59.049\%=40.951\%$. The loss after $10$ years is about $100\%-34.8678\%=65.1322\%$. Since $65.1322\%\neq2\cdot40.951\%$, the claim is false.

Answer

1. $b=0.9$ 2. About $59.05\%$ 3. About $0.3487$ 4. The claim is false. The losses are about $40.95\%$ after $5$ years and $65.13\%$ after $10$ years.

5283839

Let $f(x)=(1.44)^x$. Find the percent change in $f(x)$ when $x$ a) increases by $0.5$; b) increases by $2$; c) decreases by $1$. Round part c to the nearest tenth of a percent.

Hints

- Use the exponent rule for adding numbers in an exponent. - Convert a multiplication factor to a percent change. - An exponent of $0.5$ represents a square root. - Decreasing $x$ corresponds to a negative input change.

Solution

1. If the input changes by $k$, then $f(x+k)=(1.44)^kf(x)$. The change factor is $(1.44)^k$. 2. For $k=0.5$, the factor is $\sqrt{1.44}=1.2$, so the value increases by $20\%$. 3. For $k=2$, the factor is $(1.44)^2=2.0736$, so the value increases by $107.36\%$. 4. For $k=-1$, the factor is $(1.44)^{-1}\approx0.6944$, so the value decreases by about $30.6\%$.

Answer

a) $20\%$ increase b) $107.36\%$ increase c) About $30.6\%$ decrease

5300239

A reforestation project increased the area of a mixed forest. At the beginning of the study, the forest covered $150\,\text{ha}$. After $12$ years, it covered $215\,\text{ha}$. Assume the area increased by the same percent each year. Find the average annual percent increase.

Hints

- A constant percent change suggests an exponential model. - Solve for the yearly growth factor. - Convert the growth factor to a percent rate.

Solution

1. Use $A(n)=150q^n$. The data give $215=150q^{12}$. 2. Thus, $q=\left(\frac{215}{150}\right)^{1/12}\approx1.0305$. 3. The annual percent increase is $(q-1)\cdot100\%\approx3.05\%$.

Answer

About $3.05\%$ per year

5300249

A company buys an industrial robot for $\$85{,}000$. After $7$ years, its value is $\$18{,}400$. Assuming exponential depreciation, find the average annual percent decrease in value.

Hints

- Model the value with a constant yearly factor. - Solve for the factor using the initial and final values. - A decay factor is less than $1$; subtract it from $1$ to find the percent decrease.

Solution

1. Use $V(n)=85{,}000q^n$. Then $18{,}400=85{,}000q^7$. 2. Thus, $q=\left(\frac{18{,}400}{85{,}000}\right)^{1/7}\approx0.8036$. 3. The annual percent decrease is $(1-q)\cdot100\%\approx19.64\%$.

Answer

About $19.64\%$ per year

5335139

A cup contains freshly brewed tea at $90\,\text{°C}$. The room temperature is constant at $18\,\text{°C}$. Each minute, the difference between the tea's temperature and the room temperature decreases by $12\%$. a) Find the tea's temperature after $5$ minutes and after $10$ minutes. b) Write a function $T(x)$ for the tea's temperature after $x$ minutes. c) Examine the graph. What value will the temperature approach after a very long time? Explain.

Hints

- First find the initial difference between the tea temperature and room temperature. - What fraction remains after a $12\%$ decrease? - Add the remaining temperature difference to the fixed room temperature. - Look for the horizontal value that the graph approaches.

Solution

1. The initial temperature difference is $90-18=72\,\text{°C}$. 2. A $12\%$ decrease leaves $88\%$, so the decay factor for the difference is $0.88$. 3. The temperature model is $T(x)=18+72(0.88)^x$. 4. After $5$ minutes, $T(5)=18+72\cdot(0.88)^5\approx56.00\,\text{°C}$. 5. After $10$ minutes, $T(10)=18+72\cdot(0.88)^{10}\approx38.05\,\text{°C}$. 6. As $x$ increases, $72(0.88)^x$ approaches $0$, so $T(x)$ approaches the room temperature of $18\,\text{°C}$.

Answer

a) After $5$ minutes: about $56.00\,\text{°C}$; after $10$ minutes: about $38.05\,\text{°C}$ b) $T(x)=18+72(0.88)^x$ c) The temperature approaches $18\,\text{°C}$, because the remaining temperature difference approaches $0$.

5115989

Mia has saved $\$800.00$ and wants at least $\$860.00$ in her account after three years. Interest is added to the balance at the end of each year and earns interest in later years. a) Determine whether a fixed annual interest rate of $2.5\%$ is enough to reach her goal. b) Another account pays $1\%$ in the first year and $2\%$ in the second year. What is the minimum dollar amount of interest Mia must earn in the third year to reach $\$860.00$? c) If the third-year rate must be stated in increments of $0.1\%$, what is the minimum annual interest rate that will ensure Mia reaches her goal?

Hints

- For part a), apply the same growth factor once for each year. - For part b), find the balance after the first two years, then compare it with the goal. - For part c), first find the exact required rate, then check the two nearest tenth-percent rates to determine which one guarantees the goal.

Solution

1. For part a), the balance after three years is $\$800.00(1.025)^3 = \$861.5125 \approx \$861.51$. Because $\$861.51 > \$860.00$, the fixed rate is sufficient. 2. For part b), the balance after two years under the second account is $\$800.00 \cdot 1.01 \cdot 1.02 = \$824.16$. The minimum third-year interest needed is $\$860.00 - \$824.16 = \$35.84$. 3. For part c), the exact required rate is $\frac{35.84}{824.16} \cdot 100\% \approx 4.3487\%$. A rate of $4.3\%$ would give $\$824.16 \cdot 1.043 \approx \$859.60$, which is below the goal. A rate of $4.4\%$ gives $\$824.16 \cdot 1.044 \approx \$860.42$, so the minimum rate in increments of $0.1\%$ is $4.4\%$.

Answer

a) Yes. The balance would be approximately $\$861.51$. b) Mia must earn at least $\$35.84$ in the third year. c) The minimum rate stated in increments of $0.1\%$ is $4.4\%$.

5127959

An investment of $\$4000$ earns interest for three years. In both options, the interest is added to the balance each year. - Option 1: A constant annual rate of $2\%$ for all three years. - Option 2: An annual rate of $1\%$ in Year 1, $2\%$ in Year 2, and $3\%$ in Year 3. a) Find the ending balance for each option after three years. b) Without recalculating, explain whether Option 2 would have a different ending balance if the rates occurred in the order $3\%$, $2\%$, $1\%$.

Hints

- Use each year’s ending balance as the next year’s beginning balance. - Write each annual rate as a growth factor. - For part b, consider whether changing the order of factors changes a product.

Solution

1. For Option 1, the ending balance is $\$4000(1.02)^3 = \$4244.832 \approx \$4244.83$. 2. For Option 2, the ending balance is $\$4000 \cdot 1.01 \cdot 1.02 \cdot 1.03 = \$4244.424 \approx \$4244.42$. 3. Option 1 is greater by approximately $\$4244.83 - \$4244.42 = \$0.41$. 4. Reversing the order of the rates in Option 2 would not change the product of the growth factors because multiplication is commutative. Therefore, the ending balance would be unchanged.

Answer

a) Option 1: approximately $\$4244.83$ Option 2: approximately $\$4244.42$ b) The ending balance would be the same because changing the order of the growth factors does not change their product.

5283709

The value of an electric bike is modeled by exponential depreciation. <table> <tr> <td>Time $t$ (years)</td> <td>$0$</td> <td>$2$</td> <td>$4$</td> </tr> <tr> <td>Value $V(t)$</td> <td>$\$1200$</td> <td>$\$768$</td> <td>$\$491.52$</td> </tr> </table> a) Find the annual decay factor $b$ and write a function $V(t)=ab^t$. b) Predict the bike's value after $6$ years. c) According to the same model, what was the bike worth one year before $t=0$? d) Find the value after $10$ years and the percent of the original value that remains.

Hints

- The table uses two-year intervals, so convert the two-year factor to a one-year factor. - A negative time represents extrapolation into the past. - The percent remaining depends on the decay factor and elapsed time, not the initial value. - A depreciation model should give a value below the initial value for positive times.

Solution

1. Over two years, the value factor is $\frac{768}{1200}=0.64$. Thus $b^2=0.64$, so the annual factor is $b=0.8$. 2. The initial value is $1200$, so $V(t)=1200(0.8)^t$. 3. $V(6)=1200\cdot(0.8)^6\approx314.57$, so the value is about $\$314.57$. 4. $V(-1)=1200\cdot(0.8)^{-1}=1500$, so the modeled value one year earlier was $\$1500$. 5. $V(10)=1200\cdot(0.8)^{10}\approx128.85$. The remaining fraction is $(0.8)^{10}\approx0.1074$, or about $10.74\%$.

Answer

a) $b=0.8$; $V(t)=1200(0.8)^t$ b) About $\$314.57$ c) $\$1500$ d) About $\$128.85$, which is about $10.74\%$ of the original value

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