A retail outlet for calculators sells 900 calculators per year. It costs $2 to store one calculator for a year. To reorder, there is a fixed cost of $4, plus $1.25 for each calculator. How many times per year should the store order calculators, and in what lot size, in order to minimize inventory costs?

In order to minimize cost the outlet must order 60 units 15 times a year.

Explanation:

Theoretically, the EOQ is the optimal order quantity that a firm should purchase in order to minimize its inventory costs (holding costs are included here), and costs of placing an order.

Mathematically:

EOQ=

Where:

D= demand

S= cost of placing an order.

H= holding cost (per unit and per year).

In the statement, we identify each of these values:

D= 900

S= 4

H= $2

EOQ= =√2*4*900/2= 60

Times per year= 900/60= 15

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step-by-step explanation:

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he used the algebra tiles correctly.

step-by-step explanation:

given : 3x(2x-1)

solution :

since he used algebra tiles .

armando's solution is the same in the solution shown in figure

as he distributed 2x into x and x and wrote them vertically

and also write   -1 vertically

and 3x into x and x and x and wrote it horizontally

performing the multiplication in the algebra tiles we get

and solving the given expression :

3x(2x-1)

since solving the expression using tiles   and solving the expression algebraically we get the same answer

thus   we can choose the first statement as the answer to this problem which is "he used the algebra tiles correctly".