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Vilka [71]
3 years ago
5

Clay Corporation manufactures two styles of lampslong dasha Bedford Lamp and a Lowell Lamp. The following per unit data are​ ava

ilable: Bedford Lamp Lowell Lamp Sales price $ 26 $ 38 Variable costs $ 16 $ 23 Machine hours required for one lamp 1 4 Total fixed costs are $ 32 comma 000​, and Clay can sell a maximum of 10 comma 000 units of each style of lamp annually. Machine hour capacity is 25 comma 000 hours per year. What is the contribution margin per machine hour for the Bedford​ Lamp? (Round your answer to the nearest​ cent.)
Business
1 answer:
mote1985 [20]3 years ago
4 0

Answer:

$ 10

Explanation:

Given:

For Bedford lamp

Sales price = $ 26

Variable cost = $ 16

Machine hours required per unit = 1

Now,

the contribution margin per unit = Sales price - Variable cost

= $ 26 - $ 16

= $ 10

therefore,

the contribution margin per machine hour is calculated as:

= contribution margin / machine hours

or

= $ 10 / 1

or

= $ 10

hence,

the contribution margin per machine hour for the bedford lamp is $ 10

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On April 1, the price of gas at Bob’s Corner Station was $4.95 per gallon. On May 1, the price was $5.45 per gallon. On June 1,
AnnyKZ [126]

Answer: Please refer to Explanation

Explanation:

1. a. Between April 1 and May 1, Bob’s price increased by $0.50 , or by ___ %.

To calculate, divide the difference in the amounts by the amount the change occured from.

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= 0.5 / 4.96 * 100%

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b. Between May 1 and June 1, Bob’s price decreased by $ , or ____ %.

The Price by $0.5 from $5.95 to $4.95

= 0.5/5.95

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When Bob's prices are $5.45, there's are,

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= $6.54

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8 0
3 years ago
A bottling operation has a mean fill level of 10.01 ounces with a standard deviation of 0.25 ounces. Random samples of 20 bottle
MAVERICK [17]

Answer:

The upper limit is 10.1

The lower limit is 9.91

Explanation:

Given that:

The mean fill level (μ) = 10.01 ounces,

Standard deviation (σ) = 0.25 ounces

Number of sample bottles (n) = 20

The limits of the sample mean = 92% = 0.92

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\frac{\alpha}{2}=0.04

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z_{\frac{\alpha}{2}}=z_{0.04} = 1.75

The margin of error (e) is given by:

e=z_{0.04}\frac{\sigma}{\sqrt{n} }=1.75*\frac{0.25}{\sqrt{20} }  =0.1

The upper limit = μ + e = 10.01 + 0.1= 10.1

The lower limit = 10.01 - 0.1 = 9.91

6 0
3 years ago
Entonces cualquier pregunta
maxonik [38]
Pero, ¿cuál es tu pregunta?
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Answer:

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