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A hot-air balloon was decommissioned after a successful flight by letting the 3000 m^3 of air in it leak out. The loss in the volume of air in the balloon was exponential, decreasing at x percent per minute. If there was 1626 m^3 of air in the balloon 15 minutes after it began to be decommissioned, at what percent per minute to the nearest percent did the balloon lose air?
A) 1%
B) 2%
C) 3%
D) 4%

Respuesta :

dragonbro1015
for the first minute, the new volume of air is 3000 x [tex] \frac{X}{100} [/tex]because X% volume of air is lost every minute.What would've happened is a total of 3000 x [tex] (\frac{X}{100})^{15}[/tex] would've been lost.

thus, what we have so far is:
3000 x [tex](\frac{100 - X}{100})^{15}[/tex] = 1626
[tex](\frac{100 - X}{100})^{15}[/tex] = [tex] \frac{1626}{3000} [/tex]
[tex]\frac{100 - X}{100} = \sqrt[15]{\frac{1626}{3000} }[/tex]


which is same as:
[tex]\frac{100 - X}{100} = 0.95999[/tex] (rounded)
[tex]100 - X = 0.95999[/tex] x 100
100 - X = 95.999
-X = 95.999 - 100
X ≈ 4

Therefore, 4% of air was lost from the balloon every minute.

BTW 
The reason I did [tex] (\frac{X}{100})^{15}[/tex] was because in multiplications, the order doesn't matter, so instead of doing
[tex]\frac{X}{100}(\frac{X}{100}(\frac{X}{100}3000))....[/tex]
15 times, I just merged them into that equation

UPDATED sorry forgot some steps...

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