Question

The mean average temperature for Fairbanks, Alaska, is 27° F. The monthly average temperatures vary between 36.5° above and below this value. If t = 0 represents February, the phase shift of the sine function is 2.
a. Write a model for the average monthly temperature in Fairbanks, Alaska.
b. According to your model, what is the average temperature in November?

Answer 1

so you're iven the amplitude as 36.5

Answer 2

the average is say... your starting point

Answer 3

equation of sine in terms of a,b,c

\[sin(ax+b)+c\]

Answer 4

actually
\[asin(bx+c)+d\]

Answer 5

a is your amplitude, d is your shift upwards and downwards,, so now putting those in you get...

Answer 6

now with phase shifts , which is c

so now youhae

\[36.5=a\]
\[27=d\]
\[2=c\]

Answer 7

36.5sin(bx+2)+27?

Answer 8

i believe so...

Answer 9

Sorry for being so like dumb right now, it;s 4am :l

and thank you n.n

Answer 10

it's 4 here also

Answer 11

ugh I just looked at the answers and t has to fit in there :l want the answers?


And why are you awake? Just wanting to help people? cx

Answer 12

Isn't x = t here? I think you're just using x in your formula instead of t...

Answer 13

YES

Answer 14

sorry caps lol

Answer 15

but the period part is what i'm wondering upon

Answer 16

Oh that would make more sense.

Answer 17

Period is one month... i think. Or two months maybe.

"average temperature for Fairbanks, Alaska, is 27° F. The monthly average temperatures vary between 36.5° above and below this value."

I think this is saying that one month is 36.5 above, the next is 36.5 below... making the period 2 months.

Answer 18

yeah worded very vague

Answer 19

but in that case that wouldn't make it 2 months a period... it'd make it 4

Answer 20

possible answers: y=18.3sin[t/6pi + pi/3] +27
y=27sin[t/6pi + 2pi/3] +36.5
y=18.3sin[t/12pi - pi/6] +27
y=36.5sin[t/6pi - pi/3] +27

Answer 21

Feb-27, March-27+36 april-27, may-27+36 june 27

Answer 22

Yeah i'm really not sure what the period is meant to be... but yeah four months might actually be right

Answer 23

I'm just confused at this point honestly.

Answer 24

was the phase shift suposed to be in pi and it didn't show up?

Answer 25

Looks like it has to be last option. That's the only one with amplitude 36.5.

Answer 26

nope, all of it is there :l and yeah I was thinking the same thing @agent0smith

Answer 27

y=36.5sin[t/6pi - pi/3] +27

from that one, phase shift = (pi/3)/(pi/6) = 2

d = 27

A = 36.5

It all fits with the question.

Answer 28

oh I see c: