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Maurinko [17]
4 years ago
14

Gina Chuez has considered starting her own custom greeting card business. With an initial start up cost of $1500, she figures it

will cost $0.45 to produce each card. In order to remain competitive with the larger greeting card companies, Gina must sell her cards for no more than $1.70 each. To make a profit, her income must exceed her costs. Determine the number of cards she must sell before making a profit.
Mathematics
2 answers:
nordsb [41]4 years ago
8 0
1500 + 0.45x < 1.70x
1500 < 1.70x - 0.45x
1500 < 1.25x
1500/1.25 < x
1200 < x.....so she would have to sell 1201 cards to make a profit

ioda4 years ago
4 0
The more she charges for her cards, the fewer cards she'll have to sell 
<span>to recover her start-up costs.  So naturally, she wants to charge as much </span>
<span>for each card as she can get away with. </span>

<span>Let's start out assuming she charges the maximum of $1.70 for each one </span>
<span>(and that there are customers willing to pay $1.70 to buy one.) </span>

It costs Gina $0.45 to produce a card, and she sells it for $1.70.
<span>Her profit from selling each card is    ($1.70 - $0.45)  =  </span>$1.25 profit.

<span>How many times does she need $1.25 in profit to get back the $1500 </span>
that she sank into the business to get it started ?

<span>                 ($1500) / ($1.25 per card)  =  </span>1,200 cards .

<span>That's the </span>minimum<span> number she must sell, and it only works if she </span>
<span>charges the full $1.70 for each card.  If she charges a lower price </span>
<span>for them, then she'll need to sell </span>more<span> cards to make up the $1500 . </span>
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Answer:

Question 1) 17.4

Question 2) 1.7

Question 3) 2.8

Question 4) 8.6

Step-by-step explanation:

Mean absolute deviation is the average distance between the data points from the set to the mean point of the data set. It shows the variability in data or how much the data points are spread.

method to calculate Mean absolute deviation:

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mean absolute deviation = ∑lx_{i}-xl / n

The given problem has four sub-parts

Solution 1:

Mean= (78+99+90+80+55+56+102+88+60+42)/10

            =  750/10

            =  75

Mean absolute deviation= ( I 78-75 I + I 99-75 I+I 90-75 I + I 80-75 I +

                                             I 55-75 I + I 56-75 I + I 102-75 I + I 88- 75 I +

                                             I 60-75 I + I 42-75 I) /10

         =  (174)/10

        =  17.4

Solution 2:

Mean = (10+13+7+12+9+8+12+10+11+13)/10

              =  (105)/10

              = 10.5

Mean absolute deviation = ( I 10-10.5 I + I 13-10.5 I + I 7-10.5 I + I 12-10.5 I +

                                              I 9-10.5 I + I 8-10.5 I + I 12-10.5 I + I 10-10.5 I +

                                               I 11-10.5 I + I 3-10.5 I) /10

       =   17/10

       = 1.7

Solution 3:

Mean= (1+7+10+5+3+3+6+12+9+4)/10

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            =  6

Mean absolute deviation  = ( I 1-6 I + I 7-6 I + I 10-6 I + I 5-6 I + I 3-6 I +

                                               I 3-6 I + I 6-6 I + I 12- 6 I + I 9-6 I + I 4-6 I) /10

         = (28)/10

        =  2.8

Solution 4:

Mean= (30+46+25+45+18+25+15+32+40+24)/10

            =  300/10

            =  30

Mean absolute deviation  = ( I 30-30 I + I 46-30 I + I 25-30 I + I 45-30 I +

                                               I 18-30 I + I 25-30 I + I 15-30 I + I 32-30 I +

                                               I 40-30 I + I 24-30 I) /10

         = (86)/10

        =  8.6

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