Naturalists findthat the populations of some kinds of predatory animals vary periodically. Assume that the population of foxes in a certain forext vareis sinusoidally iwth time. REcord started being kept when time t = 0 years. A minimum number, 200 foxes, occurred when t = 2.9 years. The next maximum, 800 foxes, occured at t = 5.1 years.

Question

come up with equation

Heres my work what did i do wrong i do not know

period = (5.1 = 2.9)2 = 4.4 years
period = (2pi)/w
w = (5pi)/(11 years)
|A| = (800 foxes - 200 foxes)/2 = 300 foxes

average foxes = |A| + Fo = 300 foxes + 200 foxes

Fo is one term represents 200 foxes

((500 foxes)5 pi)/(11 years) = (2500 pi foxes)/(11 years)

y = 500 foxes + 300 foxes sin( (5pi)/(11 years) + (2500 pi foxes)/(11 years) )

Answer 1

First of all, what is your question?
I will assume that you want a sinusoidal equation.

I am with you on the amplitude and period of the function.
so let it be F = 300sin(5pi/11)t

This will give us the general shape, all we have to do is to move it up and either left or right to fit the data

as it stands the max is 300 but we want our max to be 800, so let's move it up 500 to get

F = 300sin(5pi/11)t + 500

As it stands we will have a max at 1.1 (a quarter of our period of 4.4), but we want our max to be at 5.1. So we will have to move our curve from above 4 units to the right, which will give us

F = 300sin(5pi/11)(t-4) + 500

let's test it:
if t = 2.9
F = 300sin(5pi/10)(-1.1) + 500 = 200 ..... Check!
if t=5.1
F = 300sin(5pi/10)(1.1) + 500 = 800 .... Check!!!!

so
F = 300sin(5pi/11)(t-4) + 500

Answer 2

My denominators in the checks should have been 11 not 10, so ...

let's test it:
if t = 2.9
F = 300sin(5pi/11)(-1.1) + 500 = 200 ..... Check!
if t=5.1
F = 300sin(5pi/11)(1.1) + 500 = 800 .... Check!!!!

(I committed a typo, then cut and pasted that typo)

Answer 3

wtfffff did i just read yall are like geniuses