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3 years ago

12

3. A game console originally is priced at $120. Demand increases the cost by 75%. What is the new price? Show your work

1 answer:

morpeh [17]3 years ago

8 0

Answer:

New price = $210

Step-by-step explanation:

Given:

Original price = $120

Increase in cost = 75%

Find:

New price

Computation:

New price = Original price + Increase in cost

New price = 120 + 120(75%)

New price = 120 + 90

New price = $210

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For some transformation having kinetics that obey the Avrami equation (Equation 10.17), the parameter n is known to have a value

Nikolay [14]

Given Information:

constant = n = 1.7

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Required Information:

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Answer:

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Step-by-step explanation:

The Avrami equation is used to model the transformation of solids that is from one phase to another provided that temperature is constant.

The equation is given by

$y=1 - e^{{-kt}^{n}}$

Where t is the transformation time in seconds and n, k are constants.

Let us first find the constant k, since after 100 s transformation is 50% complete,

$0.50=1 - e^{{-k*100}^{1.7}}$

$0.50 - 1= - e^{{-k*100}^{1.7}}$

$-0.50= - e^{{-k*100}^{1.7}}$

$0.50 = e^{{-k*100}^{1.7}}$

Take ln on both sides,

$ln(0.50) = ln(e^{{-k*100}^{1.7}})$

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$0.693 = k*100}^{1.7}$

$k = 0.693/100}^{1.7}$

$k = 2.759*10^{-4}$

Now we can find out the time when the transformation is 99% complete.

$0.90=1 - e^{{-kt}^{n}}$

$0.90 - 1= - e^{{-k*t}^{n}}$

$-0.10= - e^{{-kt}^{n}}$

$0.10 = e^{{-kt}^{n}}$

Take ln on both sides,

$ln(0.10) = ln(e^{{-k*t}^{n}})$

$-2.303 = -kt}^{n}$

$\frac{2.303}{k} = t^{n}$

$\frac{2.303}{2.759*10^{-4} } = t^{n}$

Again take ln on both sides

$ln(\frac{2.303}{2.759*10^{-4} }) =ln( t^{n})$

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$\frac{9.03}{n} = ln(t)$

$\frac{9.03}{1.7} = ln(t)$

$5.312 = ln(t)$

Take exponential on both sides

$e^{5.312} = e^{ln(t)}$

$202.75 = t$

$t = 202.75$ $seconds$

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