Asked by jas
How do I predict an item's price at a future date (what formula do I use?)? given info. is cost of a Fast food Hamburger $0.15 in 1960,$0.30 in 1970, $0.50 in 1980, $0.75 in 1990, $0.89 in 2000.
question#1 find average rate of increase. then predict the price of a fast food hamburger in the future...say 2020.
My work so far is:
a. 1960-1970: (0.30-0.15)/0.15 = 0.15/0.15 = 1 = 100%
b. 1970-1980: (0.50-0.30)/0.30 = 0.20/0.30 = 0.6repeating = 67%
c. 1980-1990: (0.75-0.50)/0.50 = 0.25/0.50 = 0.50 = 50%
d. 1990-2000: (0.89-0.75)/0.75 = 0.14/0.75 = 0.186(with the 6 repeating) = 19%
e. 100% + 67% + 50% +19% = 236/4 = 59% average rate of change from 1960-2000.
Is my work right so far and what formula do I use to predict a future price?
question#1 find average rate of increase. then predict the price of a fast food hamburger in the future...say 2020.
My work so far is:
a. 1960-1970: (0.30-0.15)/0.15 = 0.15/0.15 = 1 = 100%
b. 1970-1980: (0.50-0.30)/0.30 = 0.20/0.30 = 0.6repeating = 67%
c. 1980-1990: (0.75-0.50)/0.50 = 0.25/0.50 = 0.50 = 50%
d. 1990-2000: (0.89-0.75)/0.75 = 0.14/0.75 = 0.186(with the 6 repeating) = 19%
e. 100% + 67% + 50% +19% = 236/4 = 59% average rate of change from 1960-2000.
Is my work right so far and what formula do I use to predict a future price?
Answers
Answered by
Reiny
you could say that the increase is somewhat linear, (straight-lined), and let the equation be
cost = .59n + .15, where n is a multiple of 10 with 1960 represented by n=0
so 2020 would be n = 6
(2020 = 1960 + 60 and 60 = 6(10) )
cost = .59(6) + .15 = 3.69
cost = .59n + .15, where n is a multiple of 10 with 1960 represented by n=0
so 2020 would be n = 6
(2020 = 1960 + 60 and 60 = 6(10) )
cost = .59(6) + .15 = 3.69
Answered by
Ali
I'm confused? Can you explain how you did this?
Answered by
Kit-Kat
Yeah I agree. Can you explain? @Reiny
Answered by
sweaterweather and girl in red:)
It's been five years and I still need an explanation for that lmao
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