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Leni [432]
3 years ago
12

A jeweler wants to minimize business costs and has found that the average cost in dollars per necklace is given by C(x)=0.5x^2-7

x+65.5, where x is the number of necklaces that are created.
Mathematics
1 answer:
Vika [28.1K]3 years ago
8 0

Answer:

The maximum cost is 41 dollars per necklace.

Step-by-step explanation:

The average cost in dollars/necklace is given by :

C(x)=0.5x^2-7x+65.5  .....(1)

Where

x is the number of necklaces that are created

To maximize the cost, find the value of x by putting dC/dx = 0

\dfrac{d}{dx}(0.5x^2-7x+65.5)=0\\\\x-7=0\\\\x=7

Now put the value of x = 7 in equation (1).

C(x)=0.5(7)^2-7(7)+65.5\\\\=41\ \text{dollars per necklace}

Hence, the maximum cost is 41 dollars per necklace.

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Answer

P(A) = 0.30

P(B) = 0.77

P(A\ n\ B) = 0.22

P(A\ u\ B) = 0.85

Explanation:

Given

See attachment for proper data presentation

n = 100 --- Sample

A = Supplier 1

B = Conforms to specification

Solving (a): P(A)

Here, we only consider data in sample 1 row.

Here:

Yes = 22 and No = 8

n(A) = Yes + No

n(A) = 22 + 8

n(A) = 30

P(A) is then calculated as:

P(A) = \frac{n(A)}{Sample}

P(A) = \frac{30}{100}

P(A) = 0.30

Solving (b): P(B)

We only consider data in the Yes column.

Here:

(1) = 22    (2) = 25 and (3) = 30

n(B) = (1) + (2) + (3)

n(B) = 22 + 25 + 30

n(B) = 77

P(B) is then calculated as:

P(B) = \frac{n(B)}{Sample}

P(B) = \frac{77}{100}

P(B) = 0.77

Solving (c): P(A n B)

Here, we only consider the similar cell in the yes column and sample 1 row.

i.e. [Supplier 1][Yes]

This is represented as: n(A n B)

n(A\ n\ B) = 22

The probability is then calculated as:

P(A\ n\ B) = \frac{n(A\ n\ B)}{Sample}

P(A\ n\ B) = \frac{22}{100}

P(A\ n\ B) = 0.22

Solving (d): P(A u B)

This is calculated as:

P(A\ u\ B) = P(A) + P(B) - P(A\ n\ B)

This gives:

P(A\ u\ B) = \frac{30}{100} + \frac{77}{100} - \frac{22}{100}

Take LCM

P(A\ u\ B) = \frac{30+77-22}{100}

P(A\ u\ B) = \frac{85}{100}

P(A\ u\ B) = 0.85

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Answer:

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