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schepotkina [342]
3 years ago
7

A company needs to make at least A company needs to make at least 75,000 bottles 20-oz bottles can made from 1 pound if plastic

new plastic cost $.83/pound recycled plastic cost $.55/pound determine the economic Impact of using more materials and or recycled materials
Mathematics
1 answer:
strojnjashka [21]3 years ago
3 0

Answer:

By using the recycled plastic the company will save $1050.

Step-by-step explanation:

A company needs to make at least 75,000 bottles.

Now, 20-oz bottles can be made from 1 pound.

Then, 75000-oz bottles can be made from \frac{75000}{20} = 3750 pounds of plastic.

Now, the cost of new plastic is $0.83 per pound.

So, cost of new plastic to make 75000 bottles will be $(3750 × 0.83) = $3112.5.

Again,  the cost of recycled plastic is $0.55 per pound.

So, cost of recycled plastic to make 75000 bottles will be $(3750 × 0.55) = $2062.5.

Therefore, by using the recycled plastic the company will save $(3112.5 - 2062.5) = $1050. (Answer)

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Which expression correctly represents “six more than the product of five and a number, decreased by one”?
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Step-by-step explanation:

Product of 5 and a number:  5n

Six more than that would be 5n + 6

Finally, "six more than the product of 5 and a number, decreased by one" would be

5n + 6 - 1, or 5n + 5

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0.3

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2 years ago
A swimming pool with a volume of 30,000 liters originally contains water that is 0.01% chlorine (i.e. it contains 0.1 mL of chlo
SpyIntel [72]

Answer:

R_{in}=0.2\dfrac{mL}{min}

C(t)=\dfrac{A(t)}{30000}

R_{out}= \dfrac{A(t)}{1500} \dfrac{mL}{min}

A(t)=300+2700e^{-\dfrac{t}{1500}},$  A(0)=3000

Step-by-step explanation:

The volume of the swimming pool = 30,000 liters

(a) Amount of chlorine initially in the tank.

It originally contains water that is 0.01% chlorine.

0.01% of 30000=3000 mL of chlorine per liter

A(0)= 3000 mL of chlorine per liter

(b) Rate at which the chlorine is entering the pool.

City water containing 0.001%(0.01 mL of chlorine per liter) chlorine is pumped into the pool at a rate of 20 liters/min.

R_{in}=(concentration of chlorine in inflow)(input rate of the water)

=(0.01\dfrac{mL}{liter}) (20\dfrac{liter}{min})\\R_{in}=0.2\dfrac{mL}{min}

(c) Concentration of chlorine in the pool at time t

Volume of the pool =30,000 Liter

Concentration, C(t)= \dfrac{Amount}{Volume}\\C(t)=\dfrac{A(t)}{30000}

(d) Rate at which the chlorine is leaving the pool

R_{out}=(concentration of chlorine in outflow)(output rate of the water)

= (\dfrac{A(t)}{30000})(20\dfrac{liter}{min})\\R_{out}= \dfrac{A(t)}{1500} \dfrac{mL}{min}

(e) Differential equation representing the rate at which the amount of sugar in the tank is changing at time t.

\dfrac{dA}{dt}=R_{in}-R_{out}\\\dfrac{dA}{dt}=0.2- \dfrac{A(t)}{1500}

We then solve the resulting differential equation by separation of variables.

\dfrac{dA}{dt}+\dfrac{A}{1500}=0.2\\$The integrating factor: e^{\int \frac{1}{1500}dt} =e^{\frac{t}{1500}}\\$Multiplying by the integrating factor all through\\\dfrac{dA}{dt}e^{\frac{t}{1500}}+\dfrac{A}{1500}e^{\frac{t}{1500}}=0.2e^{\frac{t}{1500}}\\(Ae^{\frac{t}{1500}})'=0.2e^{\frac{t}{1500}}

Taking the integral of both sides

\int(Ae^{\frac{t}{1500}})'=\int 0.2e^{\frac{t}{1500}} dt\\Ae^{\frac{t}{1500}}=0.2*1500e^{\frac{t}{1500}}+C, $(C a constant of integration)\\Ae^{\frac{t}{1500}}=300e^{\frac{t}{1500}}+C\\$Divide all through by e^{\frac{t}{1500}}\\A(t)=300+Ce^{-\frac{t}{1500}}

Recall that when t=0, A(t)=3000 (our initial condition)

3000=300+Ce^{0}\\C=2700\\$Therefore:\\A(t)=300+2700e^{-\dfrac{t}{1500}}

3 0
3 years ago
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