Question

13. The population of Maryland was 5.17 million in 1999, and it grew to 6.05 million in 2019.(a) Assuming that the population is growing exponentially, find the growth rate r for Maryland's population. Give your answer as a percentage, rounded to the nearest hundredth of a percent.r = %(b) Write an exponential model to describe the population of Maryland from 1999 onward (let t=0 in 1999).Pt = (c) What is Maryland's population expected to be in 2030? Round your answer to one decimal place. million people(d) When do you expect that Maryland's population will reach 7.5 million? Give your answer as a calendar year (ex: 1999).During the year

Answer 1

a) r = 0.79%

b)

`P_t=5.17(1.0079)^t`

c) 6.6 million people

d) 2046

Step-by-step explanation:

We'll use the below formula for exponential growth;

`P_t=a(1+r)^t`

where a = initial amount

r = growth rate

t = number of time intervals

a) From the question, we have that

a = 5.17 million

P(t)= 6.05 million

t = 20 years

Let's go ahead and substitute these values into our formula, and solve for r as shown below;

`\begin{gathered} 6.05=5.17(1+r)^(20) \\ (6.05)/(5.17)=(1+r)^(20) \\ (1+r)=\sqrt[20]{(6.05)/(5.17)} \\ r=\sqrt[20]{(6.05)/(5.17)}-1 \\ r=0.00789 \\ r=0.79\text{\%} \end{gathered}`

b) The exponential model can be written as shown below;

`\begin{gathered} P_t=5.17(1+0.0079)^t \\ P_t=5.17(1.0079)^t \end{gathered}`

c) When t = 31 years, let's go ahead and find P as shown below;

`\begin{gathered} P_t=5.17(1.0079)^(31) \\ P_t=6.6\text{ million people} \end{gathered}`

d) When P = 7.5 million, let's go ahead and solve for t as shown below;

`\begin{gathered} 7.5=5.17(1.0079)^t \\ 1.45=(1.0079)^t \\ \log 1.45=\log (1.0079)^t \\ \log 1.45=t*\log (1.0079) \\ t=(\log 1.45)/(\log (1.0079) \\ t=47.2\text{years} \\ \end{gathered}`

So to get the particular year all we need to do is add 47 years to the initial year. That will us 1999 + 47 = 2046