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Nitella [24]
3 years ago
8

Jeremy opens a chocolate shop in the city. He pays $1,500 a month for rent and maintenance of the shop. The price of raw materia

ls and manufacturing the chocolates is $6,000 a month. He sells the chocolates individually and in boxes of a dozen. Jeremy understands that his business needs a little time to become a success and decides that he wants to build a customer base initially. He is happy to break even for the first year. After a year, Jeremy prices his chocolates at $3 apiece. He offers a discount of 10% on boxes of chocolate to promote the sale of boxes. How many boxes of chocolate would he have to sell to recover the cost of running the business this year per month? (Assume that he sells no individual chocolates.)
Mathematics
1 answer:
alukav5142 [94]3 years ago
8 0

To solve this problem you must apply the proccedure shown below:

1. Let's call x to the number of boxes of chocolate.

2. You have that he spends for month:

1500dollars+6000dollars=7500dollars

3. The problem says that Jeremy prices his chocolates at 3dollars the piece and offers a discount of 10% on boxes of chocolate. Therefore, the price of a box is:

36dollars-(36dollars)(0.1)=32.4dollars

4. Therefore, you can write the following equation and solve for x:

32.4x=7500\\ x=232

The answer is: 232 boxes of chocolate.

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2 years ago
Write 3.63 x 10^-5 in standard notation.
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Answer:

3.63E-5

Step-by-step explanation:

Convert 3.63 x 10^-5 from a scientific calculator

We do this by raising 10^n power * whole number entered where n=-5

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Answer:

ME= \frac{69.397-60.128}{2}= 4.6345 \approx 4.635

And the best answer on this case would be:

b) m = 4.635

Step-by-step explanation:

Let X the random variable of interest and we know that the confidence interval for the population mean \mu is given by this formula:

\bar X \pm t_{\alpha/2} \frac{s}{\sqrt{n}}

The confidence level on this case is 0.9 and the significance \alpha=1-0.9=0.1

The confidence interval calculated on this case is 60.128 \leq \mu \leq 69.397

The margin of error for this confidence interval is given by:

ME =t_{\alpha/2} \frac{s}{\sqrt{n}}

Since the confidence interval is symmetrical we can estimate the margin of error with the following formula:

ME = \frac{Upper -Lower}{2}

Where Upper and Lower represent the bounds for the confidence interval calculated and replacing we got:

ME= \frac{69.397-60.128}{2}= 4.6345 \approx 4.635

And the best answer on this case would be:

b) m = 4.635

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