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Viefleur [7K]
2 years ago
4

Lester Co. produces toy kites. It has a fixed cost of $62,150. If the selling price per unit is $9.50 and the variable cost per

unit is $6.25, the breakeven point is:
Mathematics
2 answers:
Vera_Pavlovna [14]2 years ago
6 0
The selling price per unit is a variable of 33.89$ cost per the breakeven point.
Allisa [31]2 years ago
6 0

Answer:

Lester Co. Produces 19124 units to make breakeven point.

Step-by-step explanation:

Given:

Fixed cost = $62,150

Selling price per unit = $9.50

Variable cost per unit = $6.25

Production cost = 6.25x + 62,150

Revenue = 9.50 x, where "x" is the number of units.

To find the break even point, we need to equivate revenue and production cost.

9.50x = 6.25x + 62,150

Now we have to find the value of x.

9.50x - 6.25x = 62150

3.25x = 62150

Dividing both sides by 3.25, we get

x = 62150 ÷ 3.25

x = 19123.08

The number of toy kits cannot be in decimal.

So, x = 19124

Therefore, Lester Co. Produces 19124 units to make breakeven point.

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Now, \cos\theta could be positive or negative, which means \sqrt{\cos^2\theta}=|\cos\theta|. If we specifically knew the sign of \cos\theta was positive, then we can simplify and write

\dfrac{\sin\theta}{\sqrt{1-\sin^2\theta}}=\dfrac{\sin\theta}{\cos\theta}=\tan\theta

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Jordan used the distributive property to write an expression that is equivalent to 6c – 48. 6c - 48 is equivalent to 6(c - 48) I
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A machine produces parts that are either defect free (90%), slightly defective (3%), or obviously defective (7%). Prior to shipm
AURORKA [14]

Answer:

(a) 0.0686

(b) 0.9984

(c) 0.0016

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Given that a machine produces parts that are either defect free (90%), slightly defective (3%), or obviously defective (7%).

Let A, B, and C be the events of defect-free, slightly defective, and the defective parts produced by the machine.

So, from the given data:

P(A)=0.90, P(B)=0.03, and P(C)=0.07.

Let E be the event that the part is disregarded by the inspection machine.

As a part is incorrectly identified as defective and discarded 2% of the time that a defect free part is input.

So, P\left(\frac{E}{A}\right)=0.02

Now, from the conditional probability,

P\left(\frac{E}{A}\right)=\frac{P(E\cap A)}{P(A)}

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P\left(\frac{E}{A}\right)=0.40

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This is the probability of disregarding the partially defective parts by inspection machine.

P\left(\frac{E}{A}\right)=0.98

and \Rightarrow P(E\cap C)=0.98\times 0.07=0.0686\cdots(iii)

This is the probability of disregarding the defective parts by inspection machine.

(a) The total probability that a part is marked as defective and discarded by the automatic inspection machine

=P(E\cap C)

= 0.0686 [from equation (iii)]

(b) The total probability that the parts produced get disregarded by the inspection machine,

P(E)=P(E\cap A)+P(E\cap B)+P(E\cap C)

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So, the total probability that the part produced get shipped

=1-P(E)=1-0.0986=0.9014

The probability that the part is good (either defect free or slightly defective)

=\left(P(A)-P(E\cap A)\right)+\left(P(B)-P(E\cap B)\right)

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So, the probability that a part is 'good' (either defect free or slightly defective) given that it makes it through the inspection machine and gets shipped

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=\frac{0.9}{0.9014}

=0.9984

(c) The probability that the 'bad' (defective} parts get shipped

=1- the probability that the 'good' parts get shipped

=1-0.9984

=0.0016

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2 years ago
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Check the picture below.

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\bf tan(16.38^o)=\cfrac{200}{a+b}\implies a+b=\cfrac{200}{tan(16.38^o)}
\\\\\\
\boxed{b=\cfrac{200}{tan(16.38^o)}-a}
\\\\\\
tan(49.48^o)=\cfrac{200}{b}\implies \boxed{b=\cfrac{200}{tan(49.48^o)}}\\\\
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make sure your calculator is in Degree mode.

4 0
2 years ago
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