Question

The following data represents the normal monthly precipitation for a certain city.( I will post the table)

Answer 1

Answer:

$y=2.14sin(\frac{ \pi}{6}(x-4))+6.05$

Step-by-step explanation:

Use a sinusoidal function of the form

$y=Asin(B(x-C))+D$

A is the amplitude, obtain the difference between the highest and the lowest value, and then divide by two:

$A=\frac{ 8.19-3.91}{2}=\frac{4.28}{2}=2.14$

D is the vertical shift, is equal to the lowest value plus the amplitude:

$D=3.91+2.14=6.05$

B is the frequency, being 12 months for each cycle (period T=12) and the relation between the period and the frequency is:

$B=\frac{2\pi}{T}$

$B=\frac{2\pi}{12}=\frac{\pi}{6}$

Finally, for C the phase shift, the highest value in a  sine function the highest point is in the first quartes of the period if the period is 12, then this maximum is in x=3. In the data the highest is in x=7, so the phase shift is to the right and equal in value to 4 (C=4).

So:

$y=Asin(B(x-C))+D$

$y=2.14sin(\frac{\pi}{6}(x-4))+6.05$

For the form $y=Asin(Bx-E)+D$

multiply C by B to find E ($4*\frac{\pi}{6}=\frac{4\pi}{6} =\frac{2\pi}{3}$):

$y=2.14sin(\frac{\pi}{6}x-\frac{2\pi}{3})+6.05$

Answer 2

In your question where as the table represent the following data of the normal monthly precipitation for a certain city. To draw its diagram you must first analyze the table and pick a variable that represent the trend that could show the raising and failing of data. i hope this helps