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Ivan
2 years ago
7

56· 328+162· 44+166· 44−x =0 find x the dot is to mutiply

Mathematics
2 answers:
Slav-nsk [51]2 years ago
8 0

Answer:

x = 32800

Step-by-step explanation:

56 × 328 + 162 × 44 + 166 × 44 - x = 0

18368 + 7128 + 7304 − x = 0

32800 - x = 0

-x = -32800

x = 32800

Eduardwww [97]2 years ago
6 0

Answer:

x = 32,800

Step-by-step explanation:

(56)(328)+(162)(44)+(166)(44)−x=0

18368+7128+7304+−x=0

(−x)+(18368+7128+7304)=0

−x+32800=0

−x+32800=0

−x+32800−32800=0−32800

−x=−32800

−x

−1

=−32800

−1

x=32800

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An aircraft manufacturer wants to determine the best selling price for a new airplane. The company estimates that the initial co
Blizzard [7]

Answer:

(a) C(x)=500+20x-5x^{\frac{3}{4}}+0.01x^2

    p(x)=320-7.7p

    R(x)=(320-7.7p)p=320p-7.7p^2

(b) x=82 \text{planes}

(c) p=\$30.91 M\;\; \text{per plane}

(d) maximum profit =\$ 15.90M

Step-by-step explanation:

Given that,

The company estimates that the initial cost of designing the aeroplane and setting up the factories in which to build it will be 500 million dollars.

The additional cost of manufacturing each plane can be modelled by the function.

m(x)=20x-5x^{\frac{3}{4}}+0.01x^2

(a)  Find the cost, demand (or price), and revenue functions.

   C(x)=500+20x-5x^{\frac{3}{4}}+0.01x^2

   p(x)=320-7.7p

   R(x)=(320-7.7p)p=320p-7.7p^2

(b)  Find the production level that maximizes profit.

    f=R(x)-C(x)

 \Rightarrow f=320p-7.7p^2-(500+20x-5x^{\frac{3}{4}}+0.01x^2)

\Rightarrow df=320dp-15.4pdp-20dx+5(\frac{3}{4} )x^{\frac{-1}{4} }dx-0.02xdx

     x=320-7.7p

     p=\frac{320-x}{7.7}

    \frac{dp}{dx} = \frac{-1}{7.7}

\frac{df}{dx}=\frac{320}{-7.7} -\frac{15.4(320-x) }{7.7(\frac{-1}{7.7} )}-20+5\frac{3}{4} x^{\frac{-1}{4}} -0.02x=0

    \Rightarrow -41.5584+83.1169-0.2597x-20+3.75x^{\frac{-1}{4} }-0.02x=0

   \Rightarrow 21.5585+3.75x^{\frac{-1}{4} }-0.279x=0

   \Rightarrow x=82 \text{planes}

(c)  Find the associated selling price of the aircraft that maximizes profit.

  p=\frac{320-82}{7.7}

\Rightarrow p=\$30.91 M\;\; \text{per plane}

(d)  Find the maximum profit.

Manufacturing cost of one plane is:

m(1)=20-5+0.01

         =\$15.01 M

maximum profit =\$(30.91-15.01)M

                           =\$15.90M

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2 years ago
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Answer:

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3 years ago
A statistics professor plans classes so carefully that the lengths of her classes are uniformly distributed between 46.046.0 and
Gala2k [10]

Answer:

The probability of selecting a class that runs between 50.2550.25 and 51.2551.25 minutes is 0.10

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The formula of probability if given by:

f(x)=

\left \{ {{\frac{1}{b-a}; \ a \leq x \leq b  } \atop {0}; \ x \ otherwise } \right.

In this exercise a= 46.0 and b= 56.0

The probability of selecting a class that runs between 50.2550.25 and 51.2551.25 minutes is:

\int\limits^{51.25}_{50.25} {\frac{1}{56-46} } \, dx = \int\limits^{51.25}_{50.25} {\frac{1}{10} } \, dx = \frac{1}{10} \times (51.25 - 50.25)=\frac{1}{10}=0.1

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Step-by-step explanation:

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