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lutik1710 [3]
3 years ago
8

0.5s +1=7 + 4.5s S =

Mathematics
1 answer:
kvv77 [185]3 years ago
3 0

Answer:

s = -\frac{3}{2}

Step-by-step explanation:

Solve for the value of s :

0.5s + 1 = 7 + 4.5s

-Combine both 0.5s and 4.5s by subtracting 0.5s by 4.5s :

0.5s + 1 - 4.5s = 7 + 4.5s - 4.5s

-4s + 1 = 7

-Subtract 1 to both sides:

-4s + 1 - 1 = 7 - 1

-4s = 6

-Divide both sides by -4 :

\frac{-4s}{-4} = \frac{6}{-4}

s = -\frac{3}{2}

So, the value of s is -\frac{3}{2}.

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Step-by-step explanation:

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2. Make a ratio of that number and the number of all the students in Music/Drama

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\frac{25}{100}

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3 years ago
Suppose that a company needs 1, 200,000 items during a year and that preparation for each production run costs $500. Suppose als
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Answer:

The number of items in each production run so that the total costs of production and storage are minimized is 8165 items/run

Step-by-step explanation:

We will use the following variables:

Q = Quantity being ordered

Q* = the optimal order Quantity: the result being sought

D = annual Demand for the item, over the year

P = unit Production cost

S = cost of setting up a production run, regardless of the number of units in the production run (fixed cost per production run)

H = annual cost to Hold one unit

It is important to note which variables are annualized, which are per-order and which are per-unit.

Using the variables, here are the components of the first equation

Total Cost, TC = PC + SC + HC

PC = P x D :  Production Cost = unit Production cost times the annual Demand

SC = (D x S)/Q : Setting up Cost = annual Demand times cost per production setup, divided by the order Quantity (number of units)

HC = (H x Q)/2: Holding Cost = annual unit Holding cost times order Quantity (number of units), divided by 2 (because throughout the year, on average the warehouse is half full).

So TC = PC + SC + HC =  (P x D) + ((D x S)/Q) + ((H x Q)/2) = PD + (DS/Q) + HQ/2

To obtain the optimal order quantity, Q* that minimizes TC, at the minimum TC, dTC/dQ = 0

dTC/dQ = (H/2) – (D x S)/(Q²) = 0

(H/2) – (D x S)/(Q²) = 0

Solving for Q, which is Q* at this point.

(Q*)² = 2DS/H

Q* = √(2DS/H)

D = annual demand for the item = 200000

S = cost of setting up a production run, regardless of the number of units in the production run (fixed cost per production run) = $500

H = annual cost to Hold one unit = $3

Q* = √(2×200000×500/3) = 8164.97 = 8165 items.

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From my research the answer for this question is 5
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