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Solnce55 [7]
2 years ago
8

A una persona le ofrecieron dos trabajos de vendedor. En la empresa A, le ofrecieron al mes un sueldo base de $6 000 más una com

isión de $20 por cada pieza vendida. En la empresa B, le ofrecieron el mismo sueldo base más una comisión igual al cuadrado de la cantidad de piezas vendidas. ¿A partir de qué cantidad de piezas vendidas en un mes le pagan más en la empresa B que en la A?A.
15

B.
19

C.
20

D.
21
Mathematics
1 answer:
Dominik [7]2 years ago
4 0

Concluimos que a partir de 21 piezas se gana más en la empresa B.

<h3>¿Cuanto debe vender para que le paguen más en la empresa B?</h3>

Si definimos el número de piezas vendidas como x, la cantidad que le pagan en la empresa A es:

A(x) = $6000 + $20*x

Y lo que le pagan en la empresa B es:

B(x) = $6000 + x^2

Queremos ver para que valor de x, se da que B(x) > A(x).

Entoces resolvamos esa inequación.

6000 + 20*x < 6000 + x^2

20x < x^2

20 < x

x debe ser más grande que 20, entonces concluimos que a partir de 21 piezas se gana más en la empresa B.

Sí quieres aprender más sobre inecuaciones, puedes leer:

brainly.com/question/23023694

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7. What's the perimeter of a rectangle with length 12 m and width 5 m?
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Option(c) 34m

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Read 2 more answers
In an experiment, college students were given either four quarters or a $1 bill and they could either keep the money or spend it
gavmur [86]

Answer:

a) P(A|B) = \frac{15/83}{44/83} =\frac{15}{44}=0.341

b) P(B|A) = \frac{29/83}{44/83} =\frac{29}{44}=0.659

c)  A. A student given a $1 bill is more likely to have kept the money.

Because the probability 0.659 is atmoslt two times greater than 0.341

Step-by-step explanation:

Assuming the following table:

                                                     Purchased Gum      Kept the Money   Total

Students Given 4 Quarters              25                              14                      39

Students Given $1 Bill                       15                               29                    44

Total                                                   40                              43                     83

a. find the probability of randomly selecting a student who spent the money, given that the student was given a $1 bill.

For this case let's define the following events

B= "student was given $1 Bill"

A="The student spent the money"

For this case we want this conditional probability:

P(A|B) =\frac{P(A and B)}{P(B)}

We have that P(A)= \frac{40}{83} , P(B)= \frac{44}{83}, P(A and B)= \frac{15}{83}

And if we replace we got:

P(A|B) = \frac{15/83}{44/83} =\frac{15}{44}=0.341

b. find the probability of randomly selecting a student who kept the money, given that the student was given a $1 bill.

For this case let's define the following events

B= "student was given $1 Bill"

A="The student kept the money"

For this case we want this conditional probability:

P(A|B) =\frac{P(A and B)}{P(B)}

We have that P(A)= \frac{43}{83} , P(B)= \frac{44}{83}, P(A and B)= \frac{29}{83}

And if we replace we got:

P(B|A) = \frac{29/83}{44/83} =\frac{29}{44}=0.659

c. what do the preceding results suggest?

For this case the best solution is:

A. A student given a $1 bill is more likely to have kept the money.

Because the probability 0.659 is atmoslt two times greater than 0.341

3 0
4 years ago
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