A company is planning to manufacture snowboards. The fixed costs are $136 per day and the total costs are $7,427 per daily output of 25 boards. What is the average costs per board tend to as production increases?

Algebra and Trigonometry (6th Edition)
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ChapterP: Prerequisites: Fundamental Concepts Of Algebra
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Problem 1MCCP: In Exercises 1-25, simplify the given expression or perform the indicated operation (and simplify,...
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**Transcription for Educational Website**

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**Cost Analysis for Snowboard Manufacturing**

A company is planning to manufacture snowboards. The fixed costs are $136 per day, and the total costs amount to $7,427 for a daily output of 25 boards. What do the average costs per board tend to as production increases?

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**Explanation**

In this scenario, the company aims to determine the behavior of average costs per snowboard as production ramps up. Initially, fixed costs remain constant regardless of the number of units produced. The key is to evaluate how average costs change with increasing output, leveraging economies of scale to potentially reduce costs per unit.

No graphs or diagrams are included in this problem statement. The task involves calculating and understanding cost trends in a production setting.
Transcribed Image Text:**Transcription for Educational Website** --- **Cost Analysis for Snowboard Manufacturing** A company is planning to manufacture snowboards. The fixed costs are $136 per day, and the total costs amount to $7,427 for a daily output of 25 boards. What do the average costs per board tend to as production increases? --- **Explanation** In this scenario, the company aims to determine the behavior of average costs per snowboard as production ramps up. Initially, fixed costs remain constant regardless of the number of units produced. The key is to evaluate how average costs change with increasing output, leveraging economies of scale to potentially reduce costs per unit. No graphs or diagrams are included in this problem statement. The task involves calculating and understanding cost trends in a production setting.
**Problem Statement:**

Karen wishes to have $19,390 cash for a new car 7 years from now. How much should be placed in an account now, if the account pays a 5.3% annual interest rate, compounded weekly?

**Solution:**

To find out how much should be placed in the account now, we'll use the formula for compound interest:

\[ 
A = P \left(1 + \frac{r}{n}\right)^{nt} 
\]

Where:
- \( A \) is the amount of money accumulated after n years, including interest.
- \( P \) is the principal amount (the initial amount of money).
- \( r \) is the annual interest rate (decimal).
- \( n \) is the number of times that interest is compounded per year.
- \( t \) is the time in years.

Given:
- \( A = 19,390 \)
- \( r = 5.3\% = 0.053 \)
- \( n = 52 \) (since the interest is compounded weekly)
- \( t = 7 \)

Rearrange the formula to solve for \( P \):

\[ 
P = \frac{A}{\left(1 + \frac{r}{n}\right)^{nt}} 
\]

Plug in the values:

\[ 
P = \frac{19,390}{\left(1 + \frac{0.053}{52}\right)^{52 \times 7}} 
\]

Calculate:

1. Compute \(\frac{0.053}{52}\).
2. Add 1 to the result.
3. Raise the result to the power of \(52 \times 7\).
4. Divide 19,390 by the result obtained in step 3 to find \(P\).

By solving the calculation, you would find the amount that needs to be placed in the account today.
Transcribed Image Text:**Problem Statement:** Karen wishes to have $19,390 cash for a new car 7 years from now. How much should be placed in an account now, if the account pays a 5.3% annual interest rate, compounded weekly? **Solution:** To find out how much should be placed in the account now, we'll use the formula for compound interest: \[ A = P \left(1 + \frac{r}{n}\right)^{nt} \] Where: - \( A \) is the amount of money accumulated after n years, including interest. - \( P \) is the principal amount (the initial amount of money). - \( r \) is the annual interest rate (decimal). - \( n \) is the number of times that interest is compounded per year. - \( t \) is the time in years. Given: - \( A = 19,390 \) - \( r = 5.3\% = 0.053 \) - \( n = 52 \) (since the interest is compounded weekly) - \( t = 7 \) Rearrange the formula to solve for \( P \): \[ P = \frac{A}{\left(1 + \frac{r}{n}\right)^{nt}} \] Plug in the values: \[ P = \frac{19,390}{\left(1 + \frac{0.053}{52}\right)^{52 \times 7}} \] Calculate: 1. Compute \(\frac{0.053}{52}\). 2. Add 1 to the result. 3. Raise the result to the power of \(52 \times 7\). 4. Divide 19,390 by the result obtained in step 3 to find \(P\). By solving the calculation, you would find the amount that needs to be placed in the account today.
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x=number of boards

C=total cost

the fixed cost is b=136 dollars

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