Question

1.) The deer population increases at a rate of 3% per year. There are 1253 deer this year. Write a function that models the deer population. About how many deer will there be in 6 years? A. 9134. . B. 1044. . C. 1496. . D. 1777. . . . . 2.) An investment of $200 increases at a rate of 9.5% per year. What is the growth factor, b. . A. 0.095. . B. 1.095. . C. 1.95. . D. 9.5.

Answer 1

Answer: 1.)  C. 1496

2.) B. 1.095

Step-by-step explanation:

The exponential growth function is modeled by :-

$f(x)=A(1+r)^x$, where A is the initial value , r is the rate of growth and  x is the time period .

1.) Given : The deer population increases at a rate of 3% per year. There are 1253 deer this year.

Rate = 3% = 0.03.

A = 1253

Now, the function that models the deer population :-

$f(x)=1253(1+0.03)^x\\\\\Rightarrow\ f(x)=1253(1.03)^x$

The population of deer in 6 years :

$f(6)=1253(1.03)^6=1496.1475\approx1496$

2.) Given : An investment of $200 increases at a rate of 9.5% per year.

Rate = 9.5%=0.095

A = $200

Growth factor = $(1+r)=(1+0.095)=(1.095)$

Answer 2

Exponential growth/decay rule with general function is:y=y(0)ekt with k is growth/decay constant rate.
(1) Deer population function y(0) = 1253 k
= 3% t
 = 6 (years)
  y(6) = 1253 . e^(0.03*6) ~ 1500 (deers)
: 1496 deers
 (2) Growth factor b is rate 9.5% = 0.095