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Liono4ka [1.6K]
3 years ago
5

Change 0.307 to a percent

Mathematics
1 answer:
Nookie1986 [14]3 years ago
4 0
It would be 30.7% because when you change a decimal to a percent, the decimal point moves 2 places to the right. For a percent to a decimal, move the decimal point 2 places to the left
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Fill in the missing values in the ratio table:
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A firm produces two types of earphones per year: x thousand of type A and y thousand of type B. The cost function (in thousands
tino4ka555 [31]

Answer:

  • 9000 A and 7000 B should be produced
  • p = 100, q = 170
  • $350,000 is the maximum profit

Step-by-step explanation:

We assume that "p" and "q" are given in dollars, so that the profit function (in thousands of dollars) can be written:

  P = xp +yq -C(x, y)

Profit will be maximized when the partial derivatives of P with respect to x and y are both zero:

  ∂P/∂x = p +x(∂p/∂x) +y(∂q/∂x) -∂C/∂x = 0

  (120 -3x +y) +x(-3) +y(1) -80 = 0

  -6x +2y +40 = 0

  3x -y = 20 . . . . put in standard form

and ...

  ∂P/∂y = x(∂p/∂y) +q +y(∂q/∂y) -∂C/∂y = 0

  x(1) +(210 +x -7y) +y(-7) -(130) = 0

  2x -14y +80 = 0

  x -7y = -40 . . . . put in standard form

__

The solution to these simultaneous equations can be found a variety of ways. Using Cramer's Rule, we have ...

  x = ((-1)(-40) -(-7)(20))/(-1·1-(-7)(3)) = 180/20 = 9

  y = (20(1) -(-40)(3))/20 = 140/20 = 7

9000 type A and 7000 type B earphones should be produced.

__

The corresponding selling prices are ...

  p = 120 -3(9) +7 = 100

  q = 210 +9 -7(7) = 170

The selling prices should be ...

  p = $100, q = $170

__

The maximum profit is ...

  P = xp +yq -C(x, y) = (9)(100) +(7)(170) -(110 +80(9) +130(7))

  P = 9(100 -180) +7(170 -130) -110 = 180 +280 -110

  P = 350

The maximum profit is $350,000.

4 0
3 years ago
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