The fox population in a certain region has a continuous growth rate of 5% per year. It is estimated that the population in the year 2000 was 10,100 foxes. a) Find a function that models the population, P(t) , after (t) years since year 2000 (i. e. t= 0 for the year 2000).
b) Use your function from part (a) to estimate the fox population in the year 2008.
c) Use your function to estimate the year when the fox population will reach over 18,400 foxes. Round t to the nearest whole year, then state the year.

P(t) = A * (1 + r)^t ;

14,922 ;

Year 2013

Step-by-step explanation:

Given the following :

Continuous growth rate(r) = 5% = 0.05

Population in year 2000 = Initial population (A) = 10,100

Time(t) = period (years since year 2000)

A)

Find a function that models the population,P(t) , after (t) years since year 2000 (i.e. t= 0 for the year 2000).

P(t) = A * (1 + r)^t

Trying out our function for t = year 2000, t =0

P(0) = 10,100 * (1 + 0.05)^0

P(0) = 10,100 * 1.05^0 = 10,100

B.)

Use your function from part (a) to estimate the fox population in the year 2008.

Year 2008, t = 8

P(8) = 10,100 * (1 + 0.05)^8

P(8) = 10,100 * 1. 05^8

P(8) = 10,100 * 1.4774554437890625

= 14922.29

= 14,922

c) Use your function to estimate the year when the fox population will reach over 18,400 foxes. Round t to the nearest whole year, then state the year.

P(t) = A * (1 + r)^t

18400 = 10,100 * (1.05)^t

18400/10100 = 1.05^t

1.8217821 = 1.05^t

1.05^t = 1.8217821

In(1.05^t) = ln(1.8217821)

0.0487901 * t = 0.5998151

t = 0.5998151 / 0.0487901

t = 12.293787

Therefore eit will take 13 years

2000 + 13 = 2013

10/60

step-by-step explanation:

if you divide 10 by 60 the answer is 0.166666 while if you divide or solve the other problems it equals 0.6

can you put the picture up?

step-by-step explanation: