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Arturiano [62]
3 years ago
11

the average u.s. citizen throws away about 1,606 lb of trash each year. find this rate in pounds per month, to the nearest tenth

Mathematics
2 answers:
erik [133]3 years ago
8 0

Answer:

The average U.S. citizen throws away about 133.8 pounds of garbage a month.

Step-by-step explanation:

First take the total amount of trash thrown away in a year (1,606), and divide it by the number of months in a year (12).

1,606 / 12 = 133.833333

You would get 133.833333 and would round it to the nearest tenth and get 133.8.

atroni [7]3 years ago
8 0

Answer:

133.83 lb/month, or  133.8 lb/month to the nearest tenth.

Step-by-step explanation:

Here we need to find the unit rate of throwing stuff away.

To do this, divide 1,606 lb by 12 months, since we want the answer in pounds per month:

1,606 lb

------------- = 133.83 lb/month, or 133 5/6 lb/month

12 mos

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A 75-gallon tank is filled with brine (water nearly saturated with salt; used as a preservative) holding 11 pounds of salt in so
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Let A(t) = amount of salt (in pounds) in the tank at time t (in minutes). Then A(0) = 11.

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\left(0.6\dfrac{\rm lb}{\rm gal}\right) \left(3\dfrac{\rm gal}{\rm min}\right) = \dfrac95 \dfrac{\rm lb}{\rm min}

and flows out at a rate

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Then the net rate of salt flow is given by the differential equation

\dfrac{dA}{dt} = \dfrac95 - \dfrac{13A}{300-t}

which I'll solve with the integrating factor method.

\dfrac{dA}{dt} + \dfrac{13}{300-t} A = \dfrac95

-\dfrac1{(300-t)^{13}} \dfrac{dA}{dt} - \dfrac{13}{(300-t)^{14}} A = -\dfrac9{5(300-t)^{13}}

\dfrac d{dt} \left(-\dfrac1{(300-t)^{13}} A\right) = -\dfrac9{5(300-t)^{13}}

Integrate both sides. By the fundamental theorem of calculus,

\displaystyle -\dfrac1{(300-t)^{13}} A = -\dfrac1{(300-t)^{13}} A\bigg|_{t=0} - \frac95 \int_0^t \frac{du}{(300-u)^{13}}

\displaystyle -\dfrac1{(300-t)^{13}} A = -\dfrac{11}{300^{13}} - \frac95 \times \dfrac1{12} \left(\frac1{(300-t)^{12}} - \frac1{300^{12}}\right)

\displaystyle -\dfrac1{(300-t)^{13}} A = \dfrac{34}{300^{13}} - \frac3{20}\frac1{(300-t)^{12}}

\displaystyle A = \frac3{20} (300-t) - \dfrac{34}{300^{13}}(300-t)^{13}

\displaystyle A = 45 \left(1 - \frac t{300}\right) - 34 \left(1 - \frac t{300}\right)^{13}

After 1 hour = 60 minutes, the tank will contain

A(60) = 45 \left(1 - \dfrac {60}{300}\right) - 34 \left(1 - \dfrac {60}{300}\right)^{13} = 45\left(\dfrac45\right) - 34 \left(\dfrac45\right)^{13} \approx 34.131

pounds of salt.

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