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8. the value of a laptop after it is purchased depreciates according to…

Question

  1. the value of a laptop after it is purchased depreciates according to the formula: (f(x) = 1090(0.35)^x), where (x) is the time after purchase in years and (f(x)) is the value of the laptop.

a) what is the purchase price of the laptop? a1

b) what is the annual depreciation rate? explain your thought process. a2

c) what is the laptops value at the end of 2 years? show your work. a2

  1. a culture of bacteria doubles every 20 minutes. a culture originally consisting of 200 bacteria is left to grow in a petri dish for 2 hours. what is the final number of bacteria in the petri dish after 2 hours? explain your thought process. a3

Explanation:

Response

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"quality": "clear",
"question_count": 4,
"skills_matched": [
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],
"needs_drawing": false,
"concepts_used": [
"Exponential Functions",
"Exponential Growth and Decay",
"Exponential Depreciation"
],
"new_concepts": [],
"current_concepts": [
"Exponential Functions",
"Exponential Growth and Decay",
"Exponential Depreciation"
]
}
</pre_analysis>

<reasoning>

Evaluate the initial value for the laptop

\[

$$\begin{aligned} &f(x) = 1090(0.35)^x \\ &f(0) = 1090(0.35)^0 = 1090 \end{aligned}$$

\]

Determine the annual depreciation rate

\[

$$\begin{aligned} &b = 1 - r = 0.35 \\ &r = 1 - 0.35 = 0.65 = 65\% \end{aligned}$$

\]

Calculate the laptop value at 2 years

\[

$$\begin{aligned} &f(2) = 1090(0.35)^2 \\ &f(2) = 1090(0.1225) = 133.525 \approx 133.53 \end{aligned}$$

\]

Calculate the final bacteria population

\[

$$\begin{aligned} &t = 2 \text{ hours} = 120 \text{ minutes} \\ &n = \frac{120}{20} = 6 \text{ doubling periods} \\ &N = 200 \times 2^6 = 200 \times 64 = 12800 \end{aligned}$$

\]
</reasoning>

<answer>

Question 8a

The purchase price of the laptop is \$1,090.

Question 8b

The annual depreciation rate is 65%.

Question 8c

The laptop's value at the end of 2 years is \$133.53.

Question 9

The final number of bacteria after 2 hours is 12,800.
</answer>

<post_analysis>
{
"subject": "Mathematics",
"question_type": "Multi-part",
"knowledge_point": [
"Mathematics",
"Algebra",
"Exponential Growth and Decay"
]
}
</post_analysis>

Answer:

<pre_analysis>
{
"quality": "clear",
"question_count": 4,
"skills_matched": [
"step_cot",
"explain_then_answer",
"quick"
],
"needs_drawing": false,
"concepts_used": [
"Exponential Functions",
"Exponential Growth and Decay",
"Exponential Depreciation"
],
"new_concepts": [],
"current_concepts": [
"Exponential Functions",
"Exponential Growth and Decay",
"Exponential Depreciation"
]
}
</pre_analysis>

<reasoning>

Evaluate the initial value for the laptop

\[

$$\begin{aligned} &f(x) = 1090(0.35)^x \\ &f(0) = 1090(0.35)^0 = 1090 \end{aligned}$$

\]

Determine the annual depreciation rate

\[

$$\begin{aligned} &b = 1 - r = 0.35 \\ &r = 1 - 0.35 = 0.65 = 65\% \end{aligned}$$

\]

Calculate the laptop value at 2 years

\[

$$\begin{aligned} &f(2) = 1090(0.35)^2 \\ &f(2) = 1090(0.1225) = 133.525 \approx 133.53 \end{aligned}$$

\]

Calculate the final bacteria population

\[

$$\begin{aligned} &t = 2 \text{ hours} = 120 \text{ minutes} \\ &n = \frac{120}{20} = 6 \text{ doubling periods} \\ &N = 200 \times 2^6 = 200 \times 64 = 12800 \end{aligned}$$

\]
</reasoning>

<answer>

Question 8a

The purchase price of the laptop is \$1,090.

Question 8b

The annual depreciation rate is 65%.

Question 8c

The laptop's value at the end of 2 years is \$133.53.

Question 9

The final number of bacteria after 2 hours is 12,800.
</answer>

<post_analysis>
{
"subject": "Mathematics",
"question_type": "Multi-part",
"knowledge_point": [
"Mathematics",
"Algebra",
"Exponential Growth and Decay"
]
}
</post_analysis>