Chapter 7.7, Problem 39PFA

a.

To determine

To state: whether the given statement is true or false.

Answer to Problem 39PFA

Option A is true..

Explanation of Solution

Given information: The population of a small town, in thousands, represented by $y=2.6{(1.02)}^t$ , where t is the number of years since 2010. Based on above given equation we have to choose correct option.

This option is true because for year 2010 the value of t=0, putting this value in equation $y=2.6{(1.02)}^t$ , we get,

  $\begin{array}{l}$ y=2.6{(1.02)}^o$\text{ =2}\text{.6}\times \text{1=2}\text{.6}\end{array}$

This is in thousand, so the population is $\text{2}\text{.6}\times \text{1000=2600}$

Hence option A is true.

b.

To determine

To state: whether the given statement is true or false.

Answer to Problem 39PFA

Option B is true.

Explanation of Solution

Given information: The population of a small town, in thousands, represented by $y=2.6{(1.02)}^t$ , where t is the number of years since 2010. Based on above given equation we have to choose correct option.

The equation $y=2.6{(1.02)}^t$ can be written as $y=2.6{\left(1+\frac{2}{100}\right)}^t$ ,this equation is like compound interest formula and hence the population of the town increases every year.

Hence option B is correct.

c.

To determine

To state: whether the given statement is true or false.

Answer to Problem 39PFA

Option C is false.

Explanation of Solution

Given information: The population of a small town, in thousands, represented by $y=2.6{(1.02)}^t$ , where t is the number of years since 2010. Based on above given equation we have to choose correct option.

The equation $y=2.6{(1.02)}^t$ can be written as

  $y=2.6{\left(1+\frac{2}{100}\right)}^t$ Here the value 2 indicates that population of town increases every year by 2% and not by 20%.

Hence option C is false.

d.

To determine

To state: whether the given statement is true or false.

Answer to Problem 39PFA

Option D is true.

Explanation of Solution

Given information: The population of a small town, in thousands, represented by $y=2.6{(1.02)}^t$ , where t is the number of years since 2010. Based on above given equation we have to choose correct option.

For the year 2020 the value of t=10, putting this value in $y=2.6{(1.02)}^t$ , we get

  $y=2.6{(1.02)}^{10}\approx 3.169$

Thus the population is nearly 3.169 in thousands

  $\Rightarrow$ The population is nearly 3.169×1000=3169

Which is clearly greater than 3100

e.

To determine

To state: whether the given statement is true or false.

Answer to Problem 39PFA

Option E is false.

Explanation of Solution

Given information: The population of a small town, in thousands, represented by $y=2.6{(1.02)}^t$ , where t is the number of years since 2010. Based on above given equation we have to choose correct option.

For the year 2015 the value of t=5, putting this value in $y=2.6{(1.02)}^t$ , we get

  $y=2.6{(1.02)}^5\approx 2.870$

Thus the population is nearly 2.870 in thousands

  $\Rightarrow$ The population is nearly 2.870×1000=2870

Hence option E is false.

f.

To determine

To state: whether the given statement is true or false.

Answer to Problem 39PFA

Option F is false.

Explanation of Solution

Given information: The population of a small town, in thousands, represented by $y=2.6{(1.02)}^t$ , where t is the number of years since 2010. Based on above given equation we have to choose correct option.

Since the population increases by 2% every year then,

For the year 2010 to 2011 the population is increases by 2%