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Rudiy27
3 years ago
9

A tank contains 30 lb of salt dissolved in 300 gallons of water. a brine solution is pumped into the tank at a rate of 3 gal/min

; it mixes with the solution there, and then the mixture is pumped out at a rate of 3 gal/min. determine a(t), the amount of salt in the tank at time t, if the concentration of salt in the inflow is variable and given by cin(t) = 2 + sin(t/4) lb/gal.
Mathematics
1 answer:
sesenic [268]3 years ago
3 0
A'(t)=(\text{flow rate in})(\text{inflow concentration})-(\text{flow rate out})(\text{outflow concentration})
\implies A'(t)=\dfrac{3\text{ gal}}{1\text{ min}}\cdot\left(2+\sin\dfrac t4\right)\dfrac{\text{lb}}{\text{gal}}-\dfrac{3\text{ gal}}{1\text{ min}}\cdot\dfrac{A(t)\text{ lb}}{300+(3-3)t\text{ gal}}
A'(t)+\dfrac1{100}A(t)=6+3\sin\dfrac t4

We're given that A(0)=30. Multiply both sides by the integrating factor e^{t/100}, then

e^{t/100}A'(t)+\dfrac1{100}e^{t/100}A(t)=6e^{t/100}+3e^{t/100}\sin\dfrac t4
\left(e^{t/100}A(t)\right)'=6e^{t/100}+3e^{t/100}\sin\dfrac t4
e^{t/100}A(t)=600e^{t/100}-\dfrac{150}{313}e^{t/100}\left(25\cos\dfrac t4-\sin\dfrac t4\right)+C
A(t)=600-\dfrac{150}{313}\left(25\cos\dfrac t4-\sin\dfrac t4\right)+Ce^{-t/100}

Given that A(0)=30, we have

30=600-\dfrac{150}{313}\cdot25+C\implies C=-\dfrac{174660}{313}\approx-558.02

so the amount of salt in the tank at time t is

A(t)\approx600-\dfrac{150}{313}\left(25\cos\dfrac t4-\sin\dfrac t4\right)-558.02e^{-t/100}
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Answer:

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Step-by-step explanation:

Given data:

mean number is given as 6.2

correct option is

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we know that variance is given as\sigma^2 = mean = 6.2

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correct option is

Poisson distribution is 6.2 and standard deviation is 2.49

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P(X \leq 2) = P(X =0) + P(X =1) + P(X =2)

                 = \frac{e^{-6.2} 6.2^0}{0!} +\frac{e^{-6.2} 6.2^1}{1!} +\frac{e^{-6.2} 6.2^2}{2!}

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                = 0.0536

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2 years ago
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Answer:

x = -8

Step-by-step explanation:

Given the equation:

27^x=9^{x-4}

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