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nataly862011 [7]
3 years ago
6

A GMC dealer believes that demand for 2005 Envoys will be normally distributed with a mean of 200 and standard deviation of 30.

His cost of receiving an Envoy is $25,000, and he sells an Envoy for $40,000. Half of all the Envoys not sold at full price can be sold for $30,000. He is considering ordering 200, 220, 240, 260, 280, or 300 Envoys. How many should he order?
Mathematics
1 answer:
irina1246 [14]3 years ago
3 0

Answer:

The correct answer to the following question will be "280 units".

Step-by-step explanation:

The given values are:

Mean = 200

Standard deviation = 30

Let's ask that the supplier needs to reach 99.9% of the standard of operation. To reach this quality of operation, she has to preserve an inventory of

Now,

⇒  Mean + 3\times sigma

On putting the values, we get

⇒  200+3\times 30

⇒  200+90

⇒  290 \ units

290 units would then accomplish a level of service of 99.9%.

Half unpurchased units can indeed be purchased at $30000 i.e. no failure. If he requests 300, there will be losses on 5 products that are unsold.

Hence, ordering 280 units seems to be preferable

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Evaluate a(b - c ^ 2) * if * a = 2/3, b = 3/4, c = 1/2 A: 1/65 B: 1/3 C: 1/4 D: 2/3
DiKsa [7]

Answer:

If a+b+c=1,

a

2

+

b

2

+

c

2

=

2

,

a

3

+

b

3

+

c

3

=

3

then find the value of

a

4

+

b

4

+

c

4

=

?

we know

2

(

a

b

+

b

c

+

c

a

)

=

(

a

+

b

+

c

)

2

−

(

a

2

+

b

2

+

c

2

)

⇒

2

(

a

b

+

b

c

+

c

a

)

=

1

2

−

2

=

−

1

⇒

a

b

+

b

c

+

c

a

=

−

1

2

given

a

3

+

b

3

+

c

3

=

3

⇒

a

3

+

b

3

+

c

3

−

3

a

b

c

+

3

a

b

c

=

3

⇒

(

a

+

b

+

c

)

(

a

2

+

b

2

+

c

2

−

a

b

−

b

c

−

c

a

)

+

3

a

b

c

=

3

⇒

(

a

+

b

+

c

)

(

a

2

+

b

2

+

c

2

−

(

a

b

+

b

c

+

c

a

)

+

3

a

b

c

=

3

⇒

(

1

×

(

2

−

(

−

1

2

)

+

3

a

b

c

)

)

=

3

⇒

(

2

+

1

2

)

+

3

a

b

c

=

3

⇒

3

a

b

c

=

3

−

5

2

=

1

2

⇒

a

b

c

=

1

6

Now

(

a

2

b

2

+

b

2

c

2

+

c

2

a

2

)

=

(

a

b

+

b

c

+

c

a

)

2

−

2

a

b

2

c

−

2

b

c

2

a

−

2

c

a

2

b

=

(

a

b

+

b

c

+

c

a

)

2

−

2

a

b

c

(

b

+

c

+

a

)

=

(

−

1

2

)

2

−

2

×

1

6

×

1

=

1

4

−

1

3

=

−

1

12

Now

a

4

+

b

4

+

c

4

=

(

a

2

+

b

2

+

c

2

)

2

−

2

(

a

2

b

2

+

b

2

c

2

+

c

2

a

2

)

=

2

2

−

2

×

(

−

1

12

)

=

4

+

1

6

=

4

1

6

Extension

a

5

+

b

5

+

c

5

=

(

a

3

+

b

3

+

c

3

)

(

a

2

+

b

2

+

c

2

)

−

[

a

3

(

b

2

+

c

2

)

+

b

3

(

c

2

+

a

2

)

+

c

3

(

a

2

+

c

2

)

]

=

3

⋅

2

−

[

a

3

(

b

2

+

c

2

)

+

b

3

(

c

2

+

a

2

)

+

c

3

(

a

2

+

b

2

)

]

Now

a

3

(

b

2

+

c

2

)

+

b

3

(

c

2

+

a

2

)

+

c

3

(

a

2

+

b

2

)

=

a

2

b

2

(

a

+

b

)

+

b

2

c

2

(

b

+

c

)

+

c

2

a

2

(

a

+

c

)

=

a

2

b

2

(

1

−

c

)

+

b

2

c

2

(

1

−

a

)

+

c

2

a

2

(

1

−

b

)

=

a

2

b

2

+

b

2

c

2

+

c

2

a

2

−

(

a

2

b

2

c

+

b

2

c

2

a

+

c

2

a

2

b

)

=

−

1

12

−

a

b

c

(

a

b

+

b

c

+

c

a

)

=

−

1

12

−

1

6

⋅

(

−

1

2

)

=

0

So

a

5

+

b

5

+

c

5

=

6

−

0

=

6

Step-by-step explanation:

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