(b) A somewhat smaller room is available for $200 for 3 hours. The company is co

Question

(b) A somewhat smaller room is available for $200 for 3 hours. The company is considering this possibility. How would this affect the break-even point? 1-25 Zoe Garcia is the manager of a small office-support business that supplies copying, binding, and other services for local companies. Zoe must replace a worn-out copy machine that is used for black-and- white copying. Two machines are being considered, and each of these has a monthly lease cost plus a cost for each page that is copied. Machine I has a monthly lease cost of $600, and there is a cost of $0.010 per page copied. Machine 2 has a monthly lease cost of $400, and there is a cost of $0.015 per page copied. Customers are charged $0.05 per page for copies. (a) What is the break-even point for each machine? (b) If Zoe expects to make 10,000 copies per month. what would be the cost for each machine? (c) If Zoe expects to make 30,000 copies per month. what would be the cost for each machine? (d) At what volume (the number of copies) would the two machines have the same monthly cost? What would the total revenue be for this number of copies? 1-26 Bismarck Manufacturing intends to increase capac- ity through the addition of new equipment. Two vendors have presented proposals. The fixed cost for proposal A is $65,000 and for proposal B, $34,000. for A ?. S 10 and for B, $14. The

Explanation / Answer

Answers to Part 1:

(a)Machine 1 , Monthly lease =600 $

Per page cost =0.01$

Revenue per page =0.05$

Let A be the breakeven num ber of pages

A=(Fixed cost ) /( Sales -variable cost)

= 600 /(0.05 -0.01)

=15000

Machine 2 , Monthly lease =400 $

Per page cost =0.015$

Revenue per page =0.05$

Let B be the breakeven num ber of pages

A=(Fixed cost ) /( Sales -variable cost)

= 400 /(0.05 -0.015)

=11428.57

(b) For 10000 copies

Cost for machine 1 = (600 + 10000*0.01) $ =700 $

Cost for machine 2 = (400 + 10000*0.015) $ =550 $

(c) For 30000 copies

Cost for machine 1 = (600 + 30000*0.01) $ =900 $

Cost for machine 2 = (400 + 30000*0.015) $ =850 $

(d)Let the volume be N

Therefore

600+ N*0.01 = 400 +N*0.015

N= 200/0.005

=40000

Cost of either of the machines at ths volume =600+ 40000*0.01 =1000 $

400+40000*0.015 =1000$

Answer to part 2: In the uploaded image the car sales value for car A is not given. Let it be X

(a)As per question

sales of A /(total car sales) =0.1

X/(80+75+90+X+25) =0.1

Solving the equation

X/ (270+X)=1/10

X=30

Therefore probability of selling car Z = Sales of Z / ( total car sales)

90/300 =0.3

ie 30%

(b)This is equivalent to the conditional probability = P(Y I X)

Ie probability of selling Y given X has been sold

This can be simplified as

P(Y and X) /P(X)

=P(Y )*P(X) / P(X)

Considerin P(Y) and P(X) as independent.

Therefore P(Y I X) =P(Y) = 75/300 =0.25 ie 25 %

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