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irina [24]
3 years ago
14

28)Hace un año compré una finca que se ha revalorizado el 20% (es decir que el precio de la finca hoy es el 20% más caro que su

precio de compra), razón por la cual he decidido venderla. Si me han pagado por ella 150 000 €, ¿a qué precio la compré y cuánto he ganado vendiéndola?
Mathematics
1 answer:
Tcecarenko [31]3 years ago
7 0

Answer:

Precio de compra=$125.000

Ganancia= $25.000

Step-by-step explanation:

Dada la siguiente información:

Revalorización= 20%

Precio de venta= $150.000

<u>Para calcular el precio de compra, tenemos que usar la siguiente formula:</u>

Precio de compra= precio de venta / (1 + revalorización)

Precio de compra= 150.000 / (1,2)

Precio de compra=$125.000

Ganancia= 150.000 - 125.000

Ganancia= $25.000

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The process standard deviation is 0.27, and the process control is set at plus or minus one standard deviation. Units with weigh
mr_godi [17]

Answer:

a) P(X

And for the other case:

tex] P(X>10.15)[/tex]

P(X>10.15)= P(Z > \frac{10.15-10}{0.15}) = P(Z>1)=1-P(Z

So then the probability of being defective P(D) is given by:

P(D) = 0.159+0.159 = 0.318

And the expected number of defective in a sample of 1000 units are:

X= 0.318*1000= 318

b) P(X

And for the other case:

tex] P(X>10.15)[/tex]

P(X>10.15)= P(Z > \frac{10.15-10}{0.05}) = P(Z>3)=1-P(Z

So then the probability of being defective P(D) is given by:

P(D) = 0.00135+0.00135 = 0.0027

And the expected number of defective in a sample of 1000 units are:

X= 0.0027*1000= 2.7

c) For this case the advantage is that we have less items that will be classified as defective

Step-by-step explanation:

Assuming this complete question: "Motorola used the normal distribution to determine the probability of defects and the number  of defects expected in a production process. Assume a production process produces  items with a mean weight of 10 ounces. Calculate the probability of a defect and the expected  number of defects for a 1000-unit production run in the following situation.

Part a

The process standard deviation is .15, and the process control is set at plus or minus  one standard deviation. Units with weights less than 9.85 or greater than 10.15 ounces  will be classified as defects."

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".  

Solution to the problem

Let X the random variable that represent the weights of a population, and for this case we know the distribution for X is given by:

X \sim N(10,0.15)  

Where \mu=10 and \sigma=0.15

We can calculate the probability of being defective like this:

P(X

And we can use the z score formula given by:

z=\frac{x-\mu}{\sigma}

And if we replace we got:

P(X

And for the other case:

tex] P(X>10.15)[/tex]

P(X>10.15)= P(Z > \frac{10.15-10}{0.15}) = P(Z>1)=1-P(Z

So then the probability of being defective P(D) is given by:

P(D) = 0.159+0.159 = 0.318

And the expected number of defective in a sample of 1000 units are:

X= 0.318*1000= 318

Part b

Through process design improvements, the process standard deviation can be reduced to .05. Assume the process control remains the same, with weights less than 9.85 or  greater than 10.15 ounces being classified as defects.

P(X

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tex] P(X>10.15)[/tex]

P(X>10.15)= P(Z > \frac{10.15-10}{0.05}) = P(Z>3)=1-P(Z

So then the probability of being defective P(D) is given by:

P(D) = 0.00135+0.00135 = 0.0027

And the expected number of defective in a sample of 1000 units are:

X= 0.0027*1000= 2.7

Part c What is the advantage of reducing process variation, thereby causing process control  limits to be at a greater number of standard deviations from the mean?

For this case the advantage is that we have less items that will be classified as defective

5 0
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egoroff_w [7]

Answer:

(1, 3)

Step-by-step explanation:

You are given the h coordinate of the vertex as 1, but in order to find the k coordinate, you have to complete the square on the parabola.  The first few steps are as follows.  Set the parabola equal to 0 so you can solve for the vertex.  Separate the x terms from the constant by moving the constant to the other side of the equals sign.  The coefficient HAS to be a +1 (ours is a -2 so we have to factor it out).  Let's start there.  The first 2 steps result in this polynomial:

-2x^2+4x=-1.  Now we factor out the -2:

-2(x^2-2x)=-1.  Now we complete the square.  This process is to take half the linear term, square it, and add it to both sides.  Our linear term is 2x.  Half of 2 is 1, and 1 squared is 1.  We add 1 into the set of parenthesis.  But we actually added into the parenthesis is +1(-2).  The -2 out front is a multiplier and we cannot ignore it.  Adding in to both sides looks like this:

-2(x^2-2x+1)=-1-2.  Simplifying gives us this:

-2(x^2-2x+1)=-3

On the left we have created a perfect square binomial which reflects the h coordinate of the vertex.  Stating this binomial and moving the -3 over by addition and setting the polynomial equal to y:

-2(x-1)^2+3=y

From this form,

y=-a(x-h)^2+k

you can determine the coordinates of the vertex to be (1, 3)

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