A 1000 gallon tank, initially full of water, develops a leak at the bottom. Given that 300 gallons of water leak out in the first 30 minutes, find the amount of water, A(t), left in the tank t minutes after the leak develops if the water drains off at a rate proportional to the amount of water present.

a. A(t)= 1000(1/3)^1/10t
b. A(t)= 1000(1/2)^-1/10t
c. A(t)= 1000(1001/2)^t
d. A(t)= 1000e ^-1/10t
e. A(t)= 1000(1/2)^1/10t
f. None of these

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wagonbelleville wagonbelleville · Answer 1 · 2019-12-16T18:03:15+01:00

Answer:

option F

Step-by-step explanation:

given,

capacity of tank = 1000 gallon

volume of water leak = 300 gallon

time = 30 min

water is leaking at the rate

formula used

[tex]A = B e^{-kt}[/tex]

At t = 0 A = 1000 gallons

[tex]1000= B e^{-k\times 0}[/tex]

B = 1000

[tex]A = 1000 e^{-kt}[/tex]

at t = 30 A = 700

[tex]700= 1000 e^{-k\times 30}[/tex]

[tex]e^{-k\times 30}=0.7[/tex]

taking ln both side

[tex]-k\times 30=ln(0.7)[/tex]

[tex]k=-\dfrac{ln(0.7)}{30}[/tex]

now,

[tex]A = 1000 e^{\dfrac{ln(0.7)}{30}t}[/tex]

[tex]A = 1000\times 0.7 e^{\dfrac{t}{30}}[/tex]

[tex]A = 700 e^{\dfrac{t}{30}}[/tex]

hence, the correct answer is option F