Respuesta :
[tex]\bf \qquad \textit{Amount for Exponential Decay} \\\\ A=P(1 - r)^t\qquad \begin{cases} A=\textit{accumulated amount}\dotfill&50.14\\ P=\textit{initial amount}\dotfill &1200\\ r=rate\to r\%\to \frac{r}{100}\dotfill\\ t=\textit{elapsed time}\dotfill &16\\ \end{cases}[/tex]
[tex]\bf 50.14=1200(1-r)^{16}\implies \cfrac{50.14}{1200}=(1-r)^{16}\implies \sqrt[16]{\cfrac{50.14}{1200}}=1-r \\\\\\ r=1-\sqrt[16]{\cfrac{50.14}{1200}}\implies r\approx 0.18\implies \stackrel{\textit{converting to percent}}{r\approx 0.18\cdot 100}\implies r\approx \stackrel{\%}{18}\quad \leftarrow x[/tex]
The change in the price at the UT Math Department is an exponential
function.
The daily percentage decrease that the UT Math Department set for the
iPhone 3.14 is approximately 18%.
Reasons:
The given original price of the new iPhone 3.14 = $1,200
The daily percentage decrease in price = x
The price of the new iPhone on the 16th day = $50.14
Required:
The daily percentage decrease that the UT Math Department set for the
iPhone.
Solution:
Decreasing (decay) price rate formula is given as follows;
y = a·(1 - x)ⁿ
Where:
y = The price on day 16 = 50.14
a = Original price = 1200
x = Percentage decrease rate
n = The number of days = 16
Which gives;
50.14 = 1200·(1 - x)¹⁶
[tex]ln(1 - x)^{16} = \mathbf{ln \left(\dfrac{50.14}{1200} \right)}[/tex]
[tex]x = 1 - e^{\dfrac{ln \left(\dfrac{50.14}{1200} \right)}{16} } \approx0.18[/tex]
[tex]0.18 = \dfrac{18}{100} = 18 \%[/tex]
Therefore, daily percentage decrease rate, x ≈ 0.18 = 18%
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