Question

Trying to help my son with his practice activity but we need help

Answer 1

Step-by-step explanation:

Given;

We are given a table of values showing the growth of frog population on a weekly basis over the period of 4 weeks.

Required;

We are required to use the information provided to determine which of the listed functions models the frog population.

Step-by-step solution;

The first thing to bear in mind is the general formula for an exponential growth. This is indicated below;

`f(x)=a(1+r)^x`

Note that the variables are;

`\begin{gathered} f(x)=population\text{ }after\text{ }x\text{ }number\text{ }of\text{ }weeks \\ a=initial\text{ }population\text{ }at\text{ }0\text{ }week \\ r=percentage\text{ }rate\text{ }of\text{ }growth \\ x=time\text{ }period(weeks\text{ }in\text{ }this\text{ }case) \end{gathered}`

The important factor here is the rate of growth which is not expressly stated but, we can determine by dividing the current population by the previous. Alternatively, we can take the difference between two consecutive populations and express the difference as a percentage.

For example,

`\begin{gathered} 110-100=10 \\ \\ (10)/(100)=0.1\text{ } \end{gathered}`

Similarly;

`\begin{gathered} 121-110=11 \\ \\ (11)/(110)=0.1 \end{gathered}`

The same result will be derived for other weeks' population too. That means the frogs grew at the rate of 10% (that is 0.1 expressed as a percentage).

`0.1*100=10\%`

We now have the value of r as 0.1

The function that models the growth of the frogs therefore would be;

`\begin{gathered} f(x)=a(1+r)^x \\ \\ Where: \\ a=100\text{ }(initial\text{ }population) \\ r=0.1\text{ }(rate\text{ }of\text{ }growth) \\ \\ f(x)=100(1+0.1)^x \\ \\ f(x)=100(1.1)^x \end{gathered}`

The correct model therefore is the last option given

ANSWER:

`f(x)=100(1.1)^x`

The graph of this function is now displayed below;