节: 指数增长和衰减模型 | 11.12 指数增长和衰减模型

章节大纲

In 1885, a psychologist named Hermann Ebbinghaus described that forgetting is an exponential function of the form $\begin{aligned} R = e^{- \frac{t}{S}}, \end{aligned}$ where $\begin{aligned} R \end{aligned}$ is the proportion of the memory retained , $\begin{aligned} S \end{aligned}$ is the strength of the memory, and $\begin{aligned} t \end{aligned}$ is time ¹ . If the strength of a memory is a 30, how much time will pass until someone retains 67% of the memory?
::1885年,一位名叫Hermann Ebinghaus的心理学家描述说,遗忘是R=e-tS形式的指数函数,R是保留记忆的比例,S是记忆的强度,t是时间1。如果记忆的强度是30,那么要多久才能有人保留67%的记忆?

This is an example of an exponential model, and we discuss how to work with these in this section.
::这是一个指数模型的例子,我们在本节讨论如何与这些模型合作。

Exponential Growth and Decay Models
::指数增长和衰减模型

There are many real-world situations that can be modeled exponentially. We will consider two approaches in this section described below.
::许多现实世界局势可以成倍地模拟,我们将在本节中考虑以下两种办法。

Approach 1: When a quantity increases by a percentage over a period of time, the final amount can be modeled by the equation $\begin{aligned} A = A_{0} (1 + r)^{t} \end{aligned}$ , where $\begin{aligned} A \end{aligned}$ is the final amount, $\begin{aligned} A_{0} \end{aligned}$ is the initial amount, $\begin{aligned} r \end{aligned}$ is the rate (or percentage), and $\begin{aligned} t \end{aligned}$ is the time. $\begin{aligned} 1 + r \end{aligned}$ is known as the growth factor . A quantity can also decrease by a percentage over a period of time. The final amount can be modeled by the equation $\begin{aligned} A = A_{0} (1 - r)^{t} \end{aligned}$ , where $\begin{aligned} 1 - r \end{aligned}$ is the decay factor .
::办法1:当数量在一段时间内增加一个百分比时,最终数额可以用公式A=A0(1+r)t作为模型,其中A0是最终数额,A0是初始数额,r是比率(或百分比),t是时间。1+r称为增长系数。1+r也可以在一段时间内减少一个百分比。最后数额可以用公式A=A0(1+r)t作为模型,1-r是衰变系数。

Approach 2: and decay can be modeled by $\begin{aligned} A = A_{0} e^{k t}, \end{aligned}$ where $\begin{aligned} k \end{aligned}$ is the growth or decay rate. If $\begin{aligned} k > 0 \end{aligned}$ , this is a growth model. If $\begin{aligned} k < 0 \end{aligned}$ , this is a decay model.
::方法2: 和衰变可以由 A=A0ekt 模拟, k 是生长或衰变率。如果 k>0, 这是增长模式。如果 k<0, 这是衰变模式。 k<0 则是一个衰变模式。

We begin by looking at exponential growth models.
::我们首先研究指数增长模式。

Exponential Growth Models
::指数增长模型

Example 1
::例1

The population of Big Lake , Texas, grows at an annual rate of 2% ² . If the population in 2000 was 2,885, what was the population in 2010? Round up to the nearest person.
::得克萨斯州大湖的人口年增长率为2%,如果2000年的人口为2 885人,那么2010年的人口是多少?

Solution: Since we know the growth rate, let's use the 1st approach. Set up a model using the growth factor, $\begin{aligned} r = 0.02 \end{aligned}$ and $\begin{aligned} A_{0} = 2885 \end{aligned}$ . Next, substitute $\begin{aligned} t = 10 \end{aligned}$ .
::解决方案: 由于我们知道增长率, 让我们使用第一种方法。使用增长系数 r= 0.02 和 A0= 2885 建立一个模型。接下来, 替代 t= 10 。

$\begin{aligned} A & = 2, 885 (1 + 0.02)^{t} \\ A & = 2, 885 (1.02)^{10} \\ = 2, 885 (1.02)^{10} \\ = 3, 517 people \end{aligned}$

::A=2,885(1+0.02)t,A=2,885(1.02)10=2,885(1.02)10=2,885(1.02)10=3,517人

The population of Big Lake was expected to increase by about 600 people in the decade between 2000 and 2010.
::2000年至2010年这十年间,大湖人口预计将增加约600人。

Example 2
::例2

In a lab, a scientist is testing the growth of bacteria. If she has a sample that initially weighs 10 grams, goes to lunch, returns two hours later, and finds that the sample now weighs 18 grams, how much should she expect the sample to weigh when she leaves work in five more hours?
::在实验室里,科学家正在测试细菌的生长。如果她有样本,最初体重为10克,去吃午饭,两小时后返回,发现样本现在的重量为18克,那么在5小时后离开工作时,她应该期望样本能称多少?

Solution: Since we do not know the growth rate, let's use the 2nd approach. We have two points of data: at time 0, the sample weighed 10 grams, and at time 2, the sample weighed 18 grams. $\begin{aligned} A_{0} = 10 \end{aligned}$ because $\begin{aligned} t = 0 \end{aligned}$ is our initial or starting time. Now we need to find $\begin{aligned} k \end{aligned}$ . To do that, we use the other data point.
::解答: 由于我们不知道增长率, 让我们使用第2种方法。我们有两个数据点: 时间为0, 样本重量为10克, 时间为2, 样本重量为18克。 A0=10, 因为 t=0 是我们的初始或起始时间。现在我们需要找到 k。要做到这一点, 我们使用另一个数据点。

$\begin{aligned} A & = 10 e^{k t} \\ 18 & = 10 e^{k \cdot 2} \\ 1.8 & = e^{k \cdot 2} \\ \ln 1.8 & = \ln e^{2 k} \\ \ln 1.8 & = 2 k \\ \frac{\ln 1.8}{2} & = k \\ 0.29 & = k \end{aligned}$

::A= 10ekt18= 10ekt1821.8= 10ek_21.8=ek_2ln_1.8= e2kln_1.8=2kn_1.8=2kn_1.22=k0.29=k

With $\begin{aligned} k = 0.29 \end{aligned}$ , we can set up the model and address the question that asks for the weight after five additional hours or at seven hours.
::k=0.29, 我们可以设置模型, 并解决在额外5小时或7小时后要求加权的问题。

$\begin{aligned} A & = 10 e^{0.29 t} \\ A & = 10 e^{0.29 \cdot 7} \\ A & = 76.14 \end{aligned}$

::A=10e0.29.A=10e0.29A=10e0.29_7A=76.14

The sample will weigh about 76.14 grams at the end of the workday.
::在工作日结束时,样品将重约76.14克。

by CK-12 shows how to set up a growth model when the rate is known.
::使用 CK-12 显示当速率为已知时如何建立增长模式。

Example 3
::例3

In September 2014, the Ebola epidemic was continuing to spread in West Africa. If the number of people reported to be infected on September 1 was 4,000 people, and the number of people with reported cases was doubling every 24 days, how many people would be infected at the end of December without medical intervention ³ ?
::2014年9月,埃博拉疫情在西非继续蔓延。如果9月1日报告的感染人数为4 000人,且每24天报告病例人数翻一番,那么到12月底,如果没有医疗干预3 将有多少人感染?

Solution: Here, we have two data points, so we can use the 2nd approach. At time 0, there were 4,000 people infected, and at time 24 days, 8,000 people were infected.
::解决方案:在这里,我们有两个数据点, 所以我们可以使用第2种方法。在0个时,有4,000人感染, 在24天时,有8,000人感染。

$\begin{aligned} A & = 4, 000 e^{k t} \\ 8, 000 & = 4, 000 e^{k \cdot 24} \\ 2 & = e^{24 k} \\ \ln 2 & = \ln e^{24 k} \\ \ln 2 & = 24 k \\ \frac{\ln 2}{24} & = k \\ 0.029 & = k \end{aligned}$

::A=4 000ekt8 000=4 000eknort8 000=4 000ek_242=e24kln @2=024kln @2=2=24kln @2=224kln @224=k0.029=k

December 31 is 121 days after September 1. Using our model, we have
::12月31日是9月1日之后121天的12月31日。

$\begin{aligned} A & = 4, 000 e^{0.029 t} \\ A & = 4, 000 e^{0.029 \cdot 121} \\ A & = 133, 659 \end{aligned}$

::A=4,000e0.029tA=4,000e0.029-121A=133,659

If the virus had continued to spread unabated, the number of people predicted to be infected after four months would have been 133,659 people ³ . The rate of growth of the Ebola virus began to decline after September 2014.
::如果病毒继续有增无减地传播,预计4个月后感染的人数将达到133 659人。埃博拉病毒的增长率在2014年9月后开始下降。

by HCCMathHelp demonstrates how to solve exponential growth and decay problems.
::HCCMathHelp展示了如何解决指数增长和衰变问题。

Exponential Decay Models
::指数衰减模型

Example 5
::例5

Japan's population declined from 2010 to 2015 with a decay rate of 0.74%. If there are about 127,110,000 people in Japan according to the 2015 census, how many people will there be in 2050 if the decay rate stays the same ⁴ ?
::日本人口从2010年到2015年有所下降,衰落率为0.74%。如果根据2015年人口普查,日本人口约为127,11万人,那么如果衰变率保持不变,那么2050年日本人口会有多少? 4

Solution: Since we know the decay rate, we can use the 1st approach.
::解决方案:既然我们知道衰变率, 我们可以使用第一种方法。

$\begin{aligned} A & = 127, 110, 000 (1 - 0.0074)^{t} \\ A & = 127, 110, 000 (0.9926)^{t} \\ = 127, 110, 000 (0.9926)^{35} \\ = 98, 011, 988 people \end{aligned}$

::A=127,110,000(1-0.0074)tA=127,110,000(0.9926)t=127,110,000(0.9926)t=127,110,000(0.9926)35=98,011,988人

If the rate between 2010 and 2015 continues, there will be less than 100 million people in Japan in 2050.
::如果2010年至2015年之间的比率继续下去,日本2050年的人口将少于1亿。

Example 6
::例6

You buy a new car for $35,000. If the value of the car decreases by 12% each year, what will be the value of the car in five years?
::你买一辆新车要三万五千美元,如果汽车每年减少12%,五年内汽车的价值会是多少?

Solution: We use approach 1.
::解决办法:我们采用办法1。

$\begin{aligned} A & = 35, 000 (1 - 0.12)^{5} \\ = 35, 000 (0.88)^{5} \\ = 18, 470.62 \end{aligned}$

::A=35 000(1-0.12.5)5=35 000(0.885=18 470.62)

The car will be worth $18,470.62 after five years.
::五年后车价将达18 470.62美元

by Mathispower4u shows how to find the growth or decay rate and the initial amount from a function .
::通过 Mathispower4u 显示如何从函数中找到增减率和初始值。

Example 7
::例7

As the altitude increases, the atmospheric pressure (the pressure of the air around you) decreases. For every 1,000 feet up, the atmospheric pressure decreases about 4%. The atmospheric pressure at sea level is 101.3. If you are on top of Heavenly Mountain at Lake Tahoe (elevation about 10,000 feet), what is the atmospheric pressure ^5,6 ?
::随着高度升高,大气压力(你周围的空气压力)下降。每1000英尺,大气压力下降约4%。海平面的大气压力为101.3。如果你在塔霍湖的天堂山顶(高度约10,000英尺),那么大气压力是多少?

Solution: The equation will be $\begin{aligned} A = 101, 325 (1 - 0.04)^{100} = 1, 709.39. \end{aligned}$ The decay factor is only raised to the power of 100 because for every 1,000 feet, the pressure decreased. Therefore , $\begin{aligned} 10, 000 \div 1, 000 = 100 \end{aligned}$ .
::解答: 方程式为 A= 101 325 (1- -0.04) 100 = 1 709. 39。衰变系数仅提高到100 的功率, 因为每1000英尺, 压力就会降低。因此, 10,000 1 000 = 100 。

Atmospheric pressure is what you feel when you are at a higher altitude—it can make you feel light-headed. The picture below demonstrates the atmospheric pressure on a plastic bottle when it was sealed at 14,000 feet elevation (1), and when the resulting pressure was 9,000 feet (2) and 1,000 feet (3). The lower the elevation, the higher the atmospheric pressure, thus the bottle was crushed at 1,000 feet.
::大气压力是高海拔时你感受到的大气压力——它可以使你感到轻头,下面的图象显示了塑料瓶在14,000英尺高的高度(1)封住时的大气压力,所产生的压力是9,000英尺高(2)和1,000英尺高(3)。高海拔越低,大气压力越高,瓶子被压在1,000英尺高。

Example 8
::例8

Marie Curie was the 1st woman to win the Nobel prize, and the only woman to win the prize twice. She was a pioneer in the study of radioactivity. Among her accomplishments was the discovery of radium ⁷ . Radium can be a radioactive isotope, which means that it decays exponentially. We can measure how much a sample of radium decays by knowing the half-life , or time until one-half of the original amount remains.
::Marie Curie是获得诺贝尔奖的第一位女性,也是获得该奖两次的唯一一位女性。她是辐射学研究的先锋。她的成就之一是发现了放射性同位素。放射性同位素可以是放射性同位素,这意味着放射性同位素会成倍衰减。我们可以通过了解半衰期或直至半衰期的剩余时间来测量辐射衰减的样本。

The half-life of radium is 1,620 years ⁸ . How long will it take for 10% of a 100-gram sample to remain?
::的半衰期为1,620年。 100克样本的10%要保留多久?

Solution: We have two data points, so we will use the 2nd approach. The data point from the half-life is that there is 50 grams of the sample at time 1,620 years. We can use this to find $\begin{aligned} k \end{aligned}$ .
::解答: 我们有两个数据点, 所以我们将使用第2种方法。从半衰期得出的数据点是, 1,620 年时样本中有50克。我们可以用这个来找到 k 。

$\begin{aligned} A & = 100 e^{k t} \\ 50 & = 100 e^{k \cdot 1620} \\ 0.5 & = e^{1620 k} \\ \ln 0.5 & = \ln e^{1620 k} \\ \ln 0.5 & = 1620 k \\ \frac{\ln 0.5}{1620} & = k \\ - 0.00043 & = k \end{aligned}$

::A=100ekt50 = 100ekt50 = 100ekä16200.5 = e1620kln 0.5 = 0.620kln 0.5 = 1620kln 0.5 = 1620kln 0.51620 = k-0.00043=k

Now that we have $\begin{aligned} k \end{aligned}$ , we can find the time it takes to have 10 grams of the sample remaining.
::现在我们有了K, 就能找到剩下的10克样本所需的时间。

$\begin{aligned} A & = 100 e^{- 0.00043 t} \\ 10 & = 100 e^{- 0.00043 t} \\ 0.1 & = e^{- 0.00043 t} \\ \ln 0.1 & = \ln e^{- 0.00043 t} \\ \ln 0.1 & = - 0.00043 t \\ \frac{\ln 0.1}{- 0.00043} & = t \\ 5354.85 & = k \end{aligned}$

::A=100e-000043t10 =100e-000043t0.1 = e-000043tln @0.1 = ln -000043tln @%0.1 =-000043tln_0.1 =-0.0043tn_0.1-0.0043=t5354.85=k

It will take about 5,355 years.
::大约需要5 355年的时间。

by Professor Elvis Zap demonstrates how to solve a problem with carbon dating.
::Elvis Zap教授展示了如何解决碳约会问题。

Example 9
::例9

In 1885, a psychologist named Hermann Ebbinghaus described that forgetting is an exponential function of the form $\begin{aligned} R = e^{- \frac{t}{S}}, \end{aligned}$ where $\begin{aligned} R \end{aligned}$ is the proportion of the memory retained, $\begin{aligned} S \end{aligned}$ is the strength of the memory, and $\begin{aligned} t \end{aligned}$ is time in days ¹ . If the strength of a memory is a 30, how much time will pass until someone retains 67% of the memory?
::1885年,一位名叫Hermann Ebinghaus的心理学家描述说,遗忘是R=e-tS形式的指数函数,R是保留记忆的比例,S是记忆的强度,t是天数的时间。1 如果记忆的强度是30,那么在有人保留67%的记忆之前,会持续多久?

Solution: We are already given the model. We can substitute the known values and solve.
::解答:我们已经得到了模型。我们可以替代已知的值和解答。

$\begin{aligned} R & = e^{- \frac{t}{S}} \\ 0.67 & = e^{- \frac{t}{30}} \\ \ln 0.67 & = \ln e^{- \frac{t}{30}} \\ \ln 0.67 & = - \frac{t}{30} \\ - 30 \ln 0.67 & = t \\ 12.014 & = t \end{aligned}$

::R= et-tS0.67= e-t30ln0.67= e-t30ln0.67= T30-t30ln0.67= T30-30ln0.67= t12.014= t12.014=t

It will take about 12 days for a person to retain about two-thirds of the memory.
::一个人大约需要12天才能保留约三分之二的记忆。

Feature: Melting Down
::特色: 向下熔化

by Deirdre Mundy
::由Deirdre Mundy 编辑

The worst nuclear disaster in history occurred on April 26, 1986, in present-day Ukraine. After a series of accidents and explosions, the reactor of the Chernobyl Nuclear Power Plant melted down. It took two days for the Soviet authorities to evacuate the surrounding towns. A cloud of radioactive debris spread across Europe. Today the plant is surrounded by a large area of land that is still too radioactive for human habitation.
::历史上最严重的核灾难发生在1986年4月26日,即今天的乌克兰。在一系列事故和爆炸之后,切尔诺贝利核电厂的反应堆融化了。苏联当局花了两天时间才撤离周围城镇。放射性碎片的云散布在欧洲各地。今天,该工厂被大片仍然对人类居住来说太放射性的土地包围。

Half Lives, Isotopes, and Toxic Fungi
::半活半活,伊索托普和毒菌菌

Radioactive materials are unstable. As they emit radiation, they decay into other materials made of smaller atoms. The half-life of a radioactive isotope is the amount of time that passes before half of it has decayed. F or instance, if an isotope has a half-life of 8 days, after 8 days there would be only half of the original quantity left. After 16 days there would be $\begin{aligned} \frac{1}{4} \end{aligned}$ left, after 24 days there would be $\begin{aligned} \frac{1}{8} \end{aligned}$ left, and so on. One of the isotopes that currently contaminates the land around Chernobyl has a half-life of 2 million years. People will not be able to live safely near the plant anytime soon.
::放射性物质是不稳定的,随着辐射的释放,它们会腐蚀成由较小原子制成的其他材料。放射性同位素的半衰期是半数同位素衰减前的半衰期。例如,如果同位素的半衰期为8天,则8天后将只剩下原来数量的一半。16天后将剩下14天,24天后将剩下18天,等等。目前污染切尔诺贝利周围土地的同位素之一的半衰期为200万年。很快,人们将无法安全地生活在工厂附近。

The radioactive waste from the accident spread across Europe. Fungi grew in contaminated soil and became tainted with the radiation. Sheep and pigs ate the fungi, which made the livestock unsafe for human consumption. Farms as far away the United Kingdom and Norway suffered severe effects from Chernobyl's radiation. In 2012, both countries finally declared that the radiation crisis had passed. It had taken 26 years for their farms to be safe again.
::事故产生的放射性废物散布于欧洲各地。真菌在受污染的土壤中生长,并受到辐射的污染。羊和猪吃了真菌,这使得牲畜不安全供人类食用。远在英国和挪威的农场受到切尔诺贝利辐射的严重影响。 2012年,两国终于宣布辐射危机已经过去。它们的农场需要26年才能恢复安全。

Today, some elderly people have returned to the towns around Chernobyl. They are willing to risk the health problems that deter younger people. Meanwhile, scientists are trying to learn how Chernobyl's radiation has been affecting the plants and animals living near the abandoned plant.
::今天,一些老年人已经回到切尔诺贝利周围的城镇,他们愿意冒着阻吓年轻人的健康问题的风险。与此同时,科学家们正在试图了解切尔诺贝利辐射如何影响了生活在废弃工厂附近的动植物。

by Radio Free Europe/Radio Liberty discusses the Chernobyl nuclear disaster.
::欧洲自由电台/自由电台讨论切尔诺贝利核灾难。

Summary
::摘要

To find exponential growth or decay models, identify the initial amount and the growth or decay rate.
::为了找到指数增长或衰变模型,确定初始数量和增长率或衰变率。

You can use either $\begin{aligned} A (t) = A_{0} (1 + r)^{t} \end{aligned}$ (growth), $\begin{aligned} A (t) = A_{0} (1 - r)^{t} \end{aligned}$ (decay), or $\begin{aligned} A (t) = A_{0} e^{k t} \end{aligned}$ (growth or decay depending on $\begin{aligned} k \end{aligned}$ ) for exponential models.
::对于指数模型,您可以使用 A(t) = A0(1+r)t(生长)、 A(t) = A0(1-r)t(衰减), 或 A(t) = A0ekt(生长或衰减取决于 k) 。

If you need to find the amount at a certain time, evaluate the function for that particular time.
::如果您在某个时间需要找到金额,请对特定时间的函数进行评价。

If you need to find the time until there is a certain amount, solve the exponential equation after substituting the amount into the function.
::如果您需要找到时间直到有一定数量,请在将数量替换为函数后解析指数方程式。

Review
::回顾

Use an exponential growth or exponential decay function to model the following scenarios and answer the questions:
::使用指数增长或指数衰减函数模拟下列假设情况并回答问题:

1. Sonya's salary increases at a rate of 4% per year. Her starting salary was $45,000. What is her annual salary, to the nearest $100, after 8 years of service?
::1. Sonya的年薪增加率为4%,其起始工资为45,000美元,服务8年后,其年薪是多少,最接近的100美元?

2. The value of Sam's car depreciates at a rate of 8% per year. The initial value was $22,000. What will his car be worth after 12 years, to the nearest dollar?
::2. Sam的汽车年折旧率为8%,最初价值为22,000美元,12年后他的汽车价值是多少,最接近的美元?

3. Rebecca is training for a marathon. Her weekly run is currently 5 miles. If she increases her mileage each week by 10%, will she complete a 20-mile training run within 15 weeks?
::3. Rebecca正在训练马拉松,她每周的跑程目前为5英里,如果她每星期的里程增加10%,她能否在15周内完成20英里的训练?

4. An investment grows at a rate of 6% per year. How much, to the nearest $100, should Noel invest if he wants to have $100,000 at the end of 20 years?
::4. 投资以每年6%的速度增长,如果Noel想要在20年结束时获得10万美元,他应该投资多少,再到最接近的100美元?

5. Charlie purchases a used, 7-year-old RV for $54,000. If the rate of depreciation was 13% per year during those 7 years, how much was the RV worth when it was new? Give your answer to the nearest $1,000.
::5. Charlie以54 000美元购买一辆7岁的旧房车,如果这7年的折旧率为每年13%,新房车的价值是多少?回答最近的1 000美元。

6. The value of homes in a neighborhood increases an average of 3% per year. What will a home purchased for $180,000 be worth in 25 years to the nearest $1,000?
::6. 邻里住宅的价值平均每年增加3%,以180 000美元购买的房屋在25年内价值多少,再到最接近的1 000美元?

7. The population of a community is decreasing at a rate of 2% per year. The current population is 152,000. How many people lived in the town 5 years ago?
::7. 社区人口每年以2%的速度下降,目前人口为152,000人,5年前有多少人居住在该镇?

8. The value of a particular piece of land worth $40,000 is increasing at a rate of 1.5% per year. Assuming the rate of appreciation continues, how long will the owner need to wait to sell the land if he hopes to get $50,000 for it? Give your answer to the nearest year.
::8. 价值40 000美元的一块土地的价值以每年1.5%的速度增长,假设升值率继续,如果业主希望得到50 000美元的地皮,需要等待多久才能出售土地?回答最近的一年。

9. The half-life of an isotope of barium is about 10 years ⁹ . If a nuclear scientist starts with 200 grams of barium, how many grams will remain after 100 years?
::9. 同位素的半衰期约为10年,如果核科学家从200克开始,在100年后还剩下多少克?

10. In an effort to control the overgrowth of vegetation in an isolated forest (free of predators), 100 rabbits are released into the forest. After one year, it is estimated that the rabbit population has increased to 500 rabbits. Assuming exponential population growth, what will the population be after another 3 months? Round to the nearest rabbit.
::10. 为了控制一个孤立的森林(无食肉动物)植被的过度增长,100只兔子被放入森林,一年之后,估计兔子人口已增加到500只兔子,假设人口指数增长,再过3个月人口会有多少?回合到最近的兔子。

11.

a. The half-life of gold-194 is approximately 1.6 days ¹⁰ . Find a model that describes the amount of gold-194 left after $\begin{aligned} t \end{aligned}$ days. Round to six decimal places.
::a. 金-194的半衰期约为1.6天10。找到一种模型,说明天后剩下的黄金-194的数量。小数点后四舍五入至六位。

b. How much of a 15-gram sample of gold-194 will remain after 5 days? Round to six decimal places.
::b. 5天后,15克黄金-194抽样将保留多少?小数点至小数点后六位。

c. How much of a 25 gram sample of gold-194 will remain after 3 days? Round to six decimal places.
::c. 3天后,25克黄金-194样本中有多少将保留在小数点后几到6位。

12 . The concentration $\begin{aligned} C (t) \end{aligned}$ of a certain drug in the bloodstream after $\begin{aligned} t \end{aligned}$ minutes is given by the formula $\begin{aligned} C (t) = 0.08 (1 - e^{- 0.2 t}) \end{aligned}$ . What is the concentration after 15 minutes? Round to six decimal places.
::12. 以公式C(t)=0.08(1-e-0.2t)表示的2分钟后血液中某种药物的浓度C(t)值为0.9分钟。15分钟后浓度是多少?小数点到小数点后6个位数。

13. Seven years ago Tommy bought a truck that is now worth $12,348. If the value of his truck decreased 14% each year, how much did he buy it for? Round to the nearest dollar.
::13. 七年前,汤米购买了一辆现在价值12 348美元的卡车,如果他的卡车价值每年下降14%,他买多少?

Answers for Review Problems
::回顾问题的答复

Please see the Appendix.
::请参看附录。

References
::参考参考资料

1. "Forgetting Curve," last edited February 18, 2017,
::1. 2017年2月18日编辑的《放弃曲线》

2. "Big Lake, Texas," last edited April 21, 2017,
::2. “得克萨斯州大湖”, 2017年4月21日编辑,

3. "West African Ebola Virus Epidemic," last edited May 25, 2017,
::3. “西非埃博拉病毒流行病”, 2017年5月25日最后一次编辑,

4. "Demography of Japan," last edited June 3, 2017,
::4. "日本人口统计", 上次编辑于2017年6月3日, 2017年6月3日,

5. "Atmospheric Pressure," last edited May 22, 2017,
::5. “大气压力”, 2017年5月22日最后一次编辑,

6. "Heavenly Mountain Resort," last edited June 3, 2017,
::2017年6月3日编辑

7. "Marie Curie," last edited June 4, 2017,
::2017年6月4日编辑的《Marie Curie》

8. "Radium," last edited May 13, 2017,
::8. 2017年5月13日最后一次编辑的""

9. "Barium," last edited May 11, 2017,
::9. 2017年5月11日上一次编辑的“Bior”

10. "Isotopes of Gold, last edited April 15, 2017,
::10. 2017年4月15日编辑的《黄金问题》

章节大纲

Exponential Growth and Decay Models ::指数增长和衰减模型

Exponential Growth Models ::指数增长模型

Example 1 ::例1

Example 2 ::例2

Example 3 ::例3

Exponential Decay Models ::指数衰减模型

Example 5 ::例5

Example 6 ::例6

Example 7 ::例7

Example 8 ::例8

Example 9 ::例9

Feature: Melting Down ::特色: 向下熔化

Half Lives, Isotopes, and Toxic Fungi ::半活半活,伊索托普和毒菌菌

Summary ::摘要

Review ::回顾

Answers for Review Problems ::回顾问题的答复

References ::参考参考资料

Exponential Growth and Decay Models
::指数增长和衰减模型

Exponential Growth Models
::指数增长模型

Example 1
::例1

Example 2
::例2

Example 3
::例3

Exponential Decay Models
::指数衰减模型

Example 5
::例5

Example 6
::例6

Example 7
::例7

Example 8
::例8

Example 9
::例9

Feature: Melting Down
::特色: 向下熔化

Half Lives, Isotopes, and Toxic Fungi
::半活半活,伊索托普和毒菌菌

Summary
::摘要

Review
::回顾

Answers for Review Problems
::回顾问题的答复

References
::参考参考资料