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Mashutka [201]
3 years ago
13

A retired auto mechanic hopes to open a rustproofing shop. Customers would be local new-car dealers. Two locations are being con

sidered, one in the center of the city and one on the outskirts. The central city location would involve fixed monthly costs of $6,980 and labor, materials, and transportation costs of $30 per car. The outside location would have fixed monthly costs of $4,200 and labor, materials, and transportation costs of $40 per car. Dealer price at either location will be $90 per car.
a.

Which location will yield the greatest profit if monthly demand is (1) 200 cars? (2) 300 cars?
200 cars: (City or Outside?) yields the greatest profit.
300 cars: (City or Outside?) yields the greatest profit.

b.

At what volume of output will the two sites yield the same monthly profit?

Volume of output = _______cars

Mathematics
1 answer:
const2013 [10]3 years ago
3 0

It is convenient to let a spreadsheet or graphing calculator do the math for this. Functions can be defined for cost and profit, and evaluated at each of the volumes of interest.

a1) For 200 cars, the Outside location yields the greatest profit

a2) For 300 cars, the City location yields the greatest profit

b) The sites yield the same profit for a volume of 278 cars.

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The initial outlay or cost is $1,000,000 for a four-year project. The respective future cash inflows for years 1, 2, 3 and 4 are
trasher [3.6K]

Answer:

The Payback Period from non discounted cash flows  is 2 Years and 8 months

Step-by-step explanation:

With an initial outlay of $1,000,000

and Cash inflows for 4 years consecutively of $500,000 + $300,000 + $300,000 + $300,000

Pay back Period = the Period where Initial Outlay minus Cash Inflows equal Zero

= 1,000,000 - 500,000 (yr 1) - 300,000 (yr 2) - 200,000 (2/3 of Year 3) = 0

Pay back Period therefore is equal to 2 years & 8 months.

3 0
4 years ago
Are lines e and f parallel? Show your work.
SashulF [63]

Answer:

last two both should be = 180

Step-by-step explanation:

5 0
3 years ago
When Helen Mirren won the Oscar for Best Actress, she was 61 years old. The Oscar-winning Best Actresses had a mean age of 35.8
saveliy_v [14]

Answer:

a) 61-35.8=25.3

b) \frac{25.2}{11.3}=2.23 deviations

c) z = \frac{61- 35.8}{11.3}= 2.23

d) For this case since we have that z>2 we can consider this value as unusual, since is outside of the interval considered usual.

Step-by-step explanation:

Assuming this complete question : "Helen Mirren was 61 when she earned her Oscar-winning Best Actress award. The Oscar-winning Best Actresses have a mean age of 35.8 years and a standard deviation of 11.3 years"

a) What is the difference between Helen Mirren’s age and the mean age?

For this case we can do this:

61-35.8=25.2

b) How many standard deviations is that?

We just need to take the difference and divide by the deviation and we got:

\frac{25.2}{11.3}=2.23 deviations

c) Convert Helen Mirren’s age to a z score.

The z score is defined as:

z = \frac{x- \mu}{\sigma}

And if we replace the values given we got:

z = \frac{61- 35.8}{11.3}= 2.23

d) If we consider “usual” ages to be those that convert to z scores between –2 and 2, is Helen Mirren’s age usual or unusual?

For this case since we have that z>2 we can consider this value as unusual, since is outside of the interval considered usual.

6 0
4 years ago
PLZ HELP!!
Marysya12 [62]

3*10 = 30 - 48*2 = 96

30 - 90

= -60

8 0
3 years ago
Read 2 more answers
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