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ad-work [718]
3 years ago
13

White shapes and black shapes are used in a game some of the shapes are circles all other shapes are sqaures The ratio of the nu

mber of white shapes to the number of black shapes is 5:11 The ratio of the number of white circles to the number of white squares is 3:7 The ratio of the number of black circles to the number of black squares is 3:8 work out what fraction of all the shapes is circles
Mathematics
1 answer:
melisa1 [442]3 years ago
3 0

Answer:

The fraction of all the shapes are circles is \frac{9}{32}

Step-by-step explanation:

There are two types of the shapes in a game.

Circle

Square

The ratio of number of white shapes to the number of black shapes is 5 : 11.

Consider there are 50 white shapes and 110 black shapes.

The ratio of the number of white circles to the number of white squares is      3 : 7.

Here 3/10 of the white shapes are circle and 7/10 are white squares.

The number of white circles can be obtained as follows,

White circles = 3/10 × 50

= 50

The number of white squares can be obtained as follows,

White squares = 7/10 × 50

= 35

The number of black circles to the number of black squares is 3/8.

The number of black circles can be obtained as follows,

Black circles = 3/11 × 110

= 30

The number of black squares can be obtained as follows,

Black squares = 8/11 × 110

= 80

The total number of circles is 30 + 15 = 45 and the total number of shapes is 50 + 110 = 160.

The fractions of all the shapes are circles are 45/160 which is simplified to 9/32

The fraction of all the shapes are circles is 9/32

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Answer:

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Step-by-step explanation:

<u>Part A </u>

Price Function= 580 - 10x

The revenue function

R(x)=x\cdot (580-10x)\\R(x)=580x-x^2

The marginal revenue function

\dfrac{dR}{dx}= \dfrac{d}{dx}(R(x))=\dfrac{d}{dx}(580x-x^2)=580-2x\\R'(x)=580-2x

<u>Part B </u>

<u>(Fixed Cost)</u>

The total cost function of the company is given by c=(30+5x)^2

We expand the expression

(30+5x)^2=(30+5x)(30+5x)=900+300x+25x^2

Therefore, the fixed cost is 900 .

<u> Marginal Cost Function</u>

If  c=900+300x+25x^2

Marginal Cost Function, \frac{dc}{dx}= (900+300x+25x^2)'=300+50x

<u>Part C </u>

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Profit=Revenue -Total cost

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<u> Part D </u>

To maximize profit, we find the derivative of the profit function, equate it to zero and solve for x.

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