Oriole, Inc., has collected the following information on its cost of electricity:
Machine Hours | Total Electricity Costs | ||||
January | 500 | $240 | |||
February | 540 | $300 | |||
March | 370 | $210 | |||
April | 440 | $220 | |||
May | 680 | $290 | |||
June | 730 | $330 | |||
July | 330 | $180 | |||
August | 530 | $260 | |||
September | 230 | $118 | |||
October | 740 | $340 | |||
November | 860 | $370 | |||
December | 600 | $340 |
Using the high-low method, compute the variable cost of electricity per machine hour. (Round unit cost to 2 decimal places, e.g. 52.75.)
Variable cost | $ per machine hour |
Compute the total fixed cost of electricity. (Round answer to 2 decimal places, e.g. 52.75.)
Fixed cost | $ |
Represent the electricity cost function in equation form. (Round answers to 2 decimal places, e.g. 52.75.)
Total cost = | $ × MH | + | $ |
What is the expected electricity cost when 740 machine hours are used? (Round answer to 2 decimal places, e.g. 52.75.)
Total cost | $ |
Variable cost per machine hour= (Highest activity cost - Lowest activity cost)/(Highest activity - Lowest activity)
= (370-118)/(860-230)
= 252/630
= $0.4 per machine hour
Variable cost | $0.4 per machine hour |
Fixed cost = Highest activity cost - Highest activity x Variable cost per hour
= 370 - 860 x 0.4
= 370-344
= $26
Fixed cost | $26 |
the electricity cost function in equation form:
Total cost = | $0.4 × MH | + | $26 |
Total expected electricity cost when 740 machine hours are used:
= Variable cost per hour x Activity hour + Fixed cost
= 0.4 x 740 + 26
= 296+26
= $322
Total cost | $322 |
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