Ernstfalld

2020-11-02

A car has a purchase price of $24,800. The value declines continuously at an exponential rate of %23 annually. A. Wat is an equation modeling the value of this car after t years? B. What is its value after 6 years? c. How long will it take for its value to be$2000? 2. A bacteria colony grows continuously at an exponential rate. There are initially 1.1 million sent. After 5 days, there are 6.8 million present . a. What is the an equation modeling the number of bacteria present after d days? b. How many bacteria will be present after 7 days? c. How long will it take the number of bacteria to reach 92 million?

Malena

Step 1: Consider the provided information, A car has purchase on price $24,800 and the value of car declines continuously at an exponential rate of 23% annually. Step 2: (A) Consider the function for exponential decay is, $⇑$ $P={P}_{0}{\left(1-r\mathrm{%}\right)}^{t}$ $=24800{\left(1-\frac{23}{100}\right)}^{t}$ $=24800{\left(\frac{77}{100}\right)}^{t}$ $=24800{\left(0.77\right)}^{t}$ Therefore, the exponential funktion is $P=24800\left(0.77{\right)}^{t}$. Step 3: (B) Substitute t=6 in above function, $P=24800{\left(0.77\right)}^{6}$ $\approx \mathrm{}5168.88$ Therefore, the prise after 6 years is$5168.88 Step 4: (c) Substitute P=2000 in the above function.
$2000=24800{\left(0.77\right)}^{t}$
${\left(0.77\right)}^{t}=\frac{2000}{24800}$
${\left(0.77\right)}^{t}=\frac{5}{62}$
$t\mathrm{ln}\left(0.77\right)=\mathrm{ln}\left(\frac{5}{65}\right)$
$t=\frac{In\left(\frac{5}{62}\right)}{In\left(\frac{0}{77}\right)}$
$\approx 9.6$ years Therefore, the time requarite for value of car become \$2000 is 9.6 years.

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