attendance at a state park throughout the year is found to be periodic and can be modeled by a sine…

attendance at a state park throughout the year is found to be periodic and can be modeled by a sine function. the attendance ranges from a low of approximately 1,000,000 visitors in september to a high of approximately 2,000,000 visitors in march. if t is the month number, where t = 1 is january, and n(t) is the attendance, in millions, of visitors, which of the functions can be used to model this behavior? n(t)=1.5 sin(π/6 t)+0.5 n(t)=2 sin(π/6 t)-1 n(t)=0.5 sin(2πt)+1.5 n(t)=0.5 sin(π/6 t)+1.5
Answer
Explanation:
Step1: Determine the amplitude
The amplitude $A$ of a sinusoidal function $y = A\sin(Bx - C)+D$ is given by $A=\frac{\text{max}-\text{min}}{2}$. Here, $\text{max} = 2$ (2 million visitors) and $\text{min}=1$ (1 million visitors), so $A=\frac{2 - 1}{2}=0.5$.
Step2: Determine the vertical - shift
The vertical - shift $D$ is given by $D=\frac{\text{max}+\text{min}}{2}$. So $D=\frac{2 + 1}{2}=1.5$.
Step3: Determine the period
The period of a sinusoidal function $y = A\sin(Bx - C)+D$ is $T=\frac{2\pi}{B}$. The attendance has a period of 12 months. So $T = 12=\frac{2\pi}{B}$, which gives $B=\frac{\pi}{6}$.
Step4: Check the functions
The general form of the sinusoidal function for this problem is $N(t)=A\sin(Bt)+D$. Substituting $A = 0.5$, $B=\frac{\pi}{6}$, and $D = 1.5$, we get $N(t)=0.5\sin(\frac{\pi}{6}t)+1.5$.
Answer:
$N(t)=0.5\sin(\frac{\pi}{6}t)+1.5$