Question

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

Draw a scatter diagram of the data for one period. Find the sinusoidal function of the form (I will post) that fits the data.

Answer 1

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

Answer 2

`y=2.14sin(( \pi)/(6)(x-4))+6.05`

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=( 8.19-3.91)/(2)=(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=(2\pi)/(T)`

`B=(2\pi)/(12)=(\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((\pi)/(6)(x-4))+6.05`

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

multiply C by B to find E (
`4*(\pi)/(6)=(4\pi)/(6) =(2\pi)/(3)`):

`y=2.14sin((\pi)/(6)x-(2\pi)/(3))+6.05`