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3 years ago

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I conclude from the information given that he is just very lucky. I believe he is lucky, because it is very hard to perform tric

ks or deceive people with a dice. He may be cheating, but a dice can be very unpredictable, and he may just be skilled at the game.

1 answer:

VikaD [51]3 years ago

7 0

Answer:

I believe that the owner of the dice is actually not lucky. There are people who practice throwing dice in a controlled way in order to get a specific side of the dice. If he practices his dice throwing a lot, then he would have a much higher chance of winning at a game of dice.

Step-by-step explanation:

Since you said he was lucky, I decided to make an answer saying that he wasn't lucky so you can choose between both of them. Your answer was good as well.

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For the following linear equations, determine which inverse operation allows you to determine he solution of he equation. In you

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Answer:

N = -2 1/6

Step-by-step explanation:

1. n/5 = -0.3

5* 5*

n = -1.5

2. -2n= 4 1/3

-2n = 13/3

-2n/ -3 = (13/3)/-2

n = -13/6

n = -2 1/6

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3 years ago

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Quadrilateral ABCD is inscribed in this circle.

Svetllana [295]

Answer:

132°

Step-by-step explanation:

Quadrilateral ABCD is inscribed in a circle,

So, it is a cyclic quadrilateral.

Opposite angles of a cyclic quadrilateral are supplementary.

$\implies \: x \degree + (3x - 12) \degree = 180 \degree \\ \\ \implies \: (4x - 12) \degree = 180 \degree \\ \\ \implies \: 4x - 12 = 180 \\ \\ \implies \: 4x = 180 + 12 \\ \\ \implies \: 4x = 192 \\ \\ \implies \: x = \frac{192}{4} \\ \\ \implies \: x = 48\\\\$

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3 years ago

Let production be given by P = bLαK1−α where b and α are positive and α < 1. If the cost of a unit of labor is m and the cost

Nana76 [90]

Answer:

The proof is completed below

Step-by-step explanation:

1) Definition of info given

We have the function that we want to maximize given by (1)

$P(L,K)=bL^{\alpha}K^{1-\alpha}$ (1)

And the constraint is given by $mL+nK=p$

2) Methodology to solve the problem

On this case in order to maximize the function on equation (1) we need to calculate the partial derivates respect to L and K, since we have two variables.

Then we can use the method of Lagrange multipliers and solve a system of equations. Since that is the appropiate method when we want to maximize a function with more than 1 variable.

The final step will be obtain the values K and L that maximizes the function

3) Calculate the partial derivates

Computing the derivates respect to L and K produce this:

$\frac{dP}{dL}=b\alphaL^{\alpha-1}K^{1-\alpha}$

$\frac{dP}{dK}=b(1-\alpha)L^{\alpha}K^{-\alpha}$

4) Apply the method of lagrange multipliers

Using this method we have this system of equations:

$\frac{dP}{dL}=\lambda m$

$\frac{dP}{dK}=\lambda n$

$mL+nK=p$

And replacing what we got for the partial derivates we got:

$b\alphaL^{\alpha-1}K^{1-\alpha}=\lambda m$ (2)

$b(1-\alpha)L^{\alpha}K^{-\alpha}=\lambda n$ (3)

$mL+nK=p$ (4)

Now we can cancel the Lagrange multiplier $\lambda$ with equations (2) and (3), dividing these equations:

$\frac{\lambda m}{\lambda n}=\frac{b\alphaL^{\alpha-1}K^{1-\alpha}}{b(1-\alpha)L^{\alpha}K^{-\alpha}}$ (4)

And simplyfing equation (4) we got:

$\frac{m}{n}=\frac{\alpha K}{(1-\alpha)L}$ (5)

4) Solve for L and K

We can cross multiply equation (5) and we got

$\alpha Kn=m(1-\alpha)L$

And we can set up this last equation equal to 0

$m(1-\alpha)L-\alpha Kn=0$ (6)

Now we can set up the following system of equations:

$mL+nK=p$ (a)

$m(1-\alpha)L-\alpha Kn=0$ (b)

We can mutltiply the equation (a) by $\alpha$ on both sides and add the result to equation (b) and we got:

$Lm=\alpha p$

And we can solve for L on this case:

$L=\frac{\alpha p}{m}$

And now in order to obtain K we can replace the result obtained for L into equations (a) or (b), replacing into equation (a)

$m(\frac{\alpha P}{m})+nK=p$

$\alpha P +nK=P$

$nK=P(1-\alpha)$

$K=\frac{P(1-\alpha)}{n}$

With this we have completed the proof.

5 0

3 years ago