the period is 12 (months)
sin(kx) has period 2π/k, so k = π/6
max is 15
min is 9.25
so, amplitude = (max-min)/2 = 2.875
center line is (max+min)/2 = 12.125
max occurs when x=6, so we can model with a cosine curve shifted right by 6
y = 12.125 + 2.875 cos(π/6 (x-6))
check:
https://www.wolframalpha.com/input/?i=12.125+%2B+2.875cos%28%CF%80%2F6+%28x-6%29%29+for+0+%3C%3D+x+%3C%3D+12
Needing a little help with this one:
The data in the table represents the average number of daylight hours each month in Springfield in 2015, rounded to the nearest quarter-hour.
January - 9.5
February - 10.5
March - 12
April - 13.25
May - 14.5
June - 15
July - 14.75
August - 13.75
September - 12.5
October - 11.0
November - 9.75
December - 9.25
Write an equation that best models the data.
What is the expected number of daylight hours in March 2020? Explain.
Please advise (and please show work so I know how to do this)!
4 answers
well the period seems to be 12 months :) It better repeat the next year more or less. so make it a sine wave with period of 12 moths december = 0 and next december = one period = 12 months
in other words redraw starting at month = 0
0 -- December last year = 9.25
1 --- January - 9.5
2 ----February - 10.5
March - 12
April - 13.25
May - 14.5
June - 15
July - 14.75
August - 13.75
September - 12.5
October - 11.0
November - 9.75
12 ---- December - 9.25
so it will be 9.25 + sin (2 pi t/T) where t is the time after december in months and T is 12
in other words redraw starting at month = 0
0 -- December last year = 9.25
1 --- January - 9.5
2 ----February - 10.5
March - 12
April - 13.25
May - 14.5
June - 15
July - 14.75
August - 13.75
September - 12.5
October - 11.0
November - 9.75
12 ---- December - 9.25
so it will be 9.25 + sin (2 pi t/T) where t is the time after december in months and T is 12
ignore what I did --- I left amplitude out and should be cosine about june use oobleck solution.
thanks you guys! oobleck is right
1 answer