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marishachu [46]
3 years ago
6

Un empresario textil de Gamarra desea distribuir un bono de productividad entre sus empleados por su buen desempeño en la semana

. Haciendo cálculos, se percata de que si entregara a cada uno 800 soles, le sobrarían 200, y si les diera 900 soles, le faltarían 400. ¿Cuántos empleados hay en su fábrica? ¿Cuánto dinero tiene para repartir? ¿Cómo resolverías el problema sin usar ecuaciones?
Mathematics
1 answer:
spin [16.1K]3 years ago
8 0

Answer:

There are 6 employees in the factory

The amount of money distributed to the employees = 5000 soles

To solve the problem without using equations, We would try different numbers for the employees starting from 2 employees.

Question:

A textile entrepreneur from Gamarra wants to distribute a productivity bonus among his employees for his good

performance in the week. Making calculations, he realizes that if he gave each one 800 soles, he would have

200 leftovers , and if he gave them 900 soles, he would lack 400. How many employees are there in your factory? How much money do you have to distribute? How would you solve the problem without using equations?

Step-by-step explanation:

For the first two questions, we would determine the number of employees that are in the factory and the amount of money that was distributed by solving using equations. Each statement would be written in the form of an equation.

Let the amount of money distributed to the employees = p

The number of employees in the factory = q

First statement showing the relationship of the variables:

800q + 200 = p ...equation 1

2nd statement showing the relationship of the variables:

900q - 400 = p ...equation 2

Equating both equations: p = p

800q + 200 = 900q - 400

900q-800q = 200+400

100q = 600

q = 600/100 = 6

Substitute for q = 6 in any if the equation.

Using equation 1

800(6) + 200 = p

p = 5000

Therefore, the amount of money distributed to the employees = 5000soles

The number of employees in the factory = 6

To solve the problem without using equations, We would try different numbers for the employees starting from 2 employees (as they are more than 1) to arrive at a particular amount distributed:

(800×2) + 200 = 1800

(900×2) - 400 = 1400

By increasing the numbers using consecutive even numbers (2,4,6...) and multiplying, we would arrive at a value that is equal to both.

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worty [1.4K]

Answer:

a) P(X \leq 100) = 1- e^{-0.01342*100} =0.7387

P(X \leq 200) = 1- e^{-0.01342*200} =0.9317

P(100\leq X \leq 200) = [1- e^{-0.01342*200}]-[1- e^{-0.01342*100}] =0.1930

b) P(X>223.547) = 1-P(X\leq 223.547) = 1-[1- e^{-0.01342*223.547}]=0.0498

c) m = \frac{ln(0.5)}{-0.01342}=51.65

d) a = \frac{ln(0.05)}{-0.01342}=223.23

Step-by-step explanation:

Previous  concepts

The exponential distribution is "the probability distribution of the time between events in a Poisson process (a process in which events occur continuously and independently at a constant average rate). It is a particular case of the gamma distribution". The probability density function is given by:

P(X=x)=\lambda e^{-\lambda x}

Solution to the problem

For this case we have that X is represented by the following distribution:

X\sim Exp (\lambda=0.01342)

Is important to remember that th cumulative distribution for X is given by:

F(X) =P(X \leq x) = 1-e^{-\lambda x}

Part a

For this case we want this probability:

P(X \leq 100)

And using the cumulative distribution function we have this:

P(X \leq 100) = 1- e^{-0.01342*100} =0.7387

P(X \leq 200) = 1- e^{-0.01342*200} =0.9317

P(100\leq X \leq 200) = [1- e^{-0.01342*200}]-[1- e^{-0.01342*100}] =0.1930

Part b

Since we want the probability that the man exceeds the mean by more than 2 deviations

For this case the mean is given by:

\mu = \frac{1}{\lambda}=\frac{1}{0.01342}= 74.516

And by properties the deviation is the same value \sigma = 74.516

So then 2 deviations correspond to 2*74.516=149.03

And the want this probability:

P(X > 74.516+149.03) = P(X>223.547)

And we can find this probability using the complement rule:

P(X>223.547) = 1-P(X\leq 223.547) = 1-[1- e^{-0.01342*223.547}]=0.0498

Part c

For the median we need to find a value of m such that:

P(X \leq m) = 0.5

If we use the cumulative distribution function we got:

1-e^{-0.01342 m} =0.5

And if we solve for m we got this:

0.5 = e^{-0.01342 m}

If we apply natural log on both sides we got:

ln(0.5) = -0.01342 m

m = \frac{ln(0.5)}{-0.01342}=51.65

Part d

For this case we have this equation:

P(X\leq a) = 0.95

If we apply the cumulative distribution function we got:

1-e^{-0.01342*a} =0.95

If w solve for a we can do this:

0.05= e^{-0.01342 a}

Using natural log on btoh sides we got:

ln(0.05) = -0.01342 a

a = \frac{ln(0.05)}{-0.01342}=223.23

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