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8_murik_8 [283]
3 years ago
9

Convert 150$ US dollars into Canadian dollars

Mathematics
2 answers:
Grace [21]3 years ago
8 0

Answer:

199 Canadian dollars

Step-by-step explanation:

1 US dollar equals 1.33 Canadian dollar

FinnZ [79.3K]3 years ago
8 0

Answer:

204.60 CAD

Step-by-step explanation:

For this we need to determine the current exchange rate.  Go to your favorite browser and type in "US$150 to Canadian dollar."

According to a site named xrates.com:

Quick Conversions from United States Dollar to Canadian Dollar : 1 USD = 1.36393 CAD

                                                                                  1.364 CAD

Then multiply US$150 by the conversion factor  ---------------------  

                                                                                   1 USD

This comes out to ($150)(1.364 CAD / 1 USD:

204.60 CAD

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A manufacturer has been selling 1000 flat-screen TVs a week at $500 each. A market survey indicates that for A manufacturer has
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Answer:

a) Demand function: q=6000-10\cdot p

b) The rebate should be of $200, so the sale price becomes $300 per unit.

c) The rebate should be of $150, so the sale price becomes $350 per unit.

Step-by-step explanation:

a) In this case, we have a known point of the demand function (1000 units sold at $500), and the slope of a linear function (increase by 100 units fora decrease in $10).

We can express the demand function (linear) as:

q=b+m\cdot p

To calculate the slope m, we use:

m=\Delta q/\Delta p=(+100)/(-10)=-10

To calculate b, we use the known point and the calculated slope:

q=b+m\cdot p\\\\b=q-m\cdot p=1000-(-10)\cdot (500)=1000+5000=6000

Then the demand function is:

q=6000-10\cdot p

b) The revenue can be expressed as:

R=q\cdot p = (6000-10p)\cdot p=6000p-10p^2

To maximize, we can derive and equal to zero

dR/dp=6000-2*10p=0\\\\20p=6000\\\\p=300

The rebate should be of $200, so the sale price becomes $300 per unit.

c) If we take into account the cost, we have that

R=q\cdot p-C=(6000p-10p^2)-(68000+100q)\\\\R=(6000p-10p^2)-(68000+100(6000-10p))\\\\R=6000p-10p^2-(68000+600000-1000p)\\\\R=-10p^2+7000p-668000

To maximize, we can derive and equal to zero

dR/dp=-20p+7000=0\\\\p=7000/20=350

The rebate should be of $150, so the sale price becomes $350 per unit.

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3 years ago
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