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6 0

Consider, please, this solution/explanation:
1. suppose, that number of cars is 'x', numbers of buses is 'y'.
2. according to the condition 'a total of 32', it means that x+y=32. This is the 1st equation.
3. according to the condition in one car are 3x persons, in one bus are 27y persons, and totaly 3x+27y=408 people. This is the 2d equation.
4. if to solve the system of two equations:
$\left \{ {{x+y=32} \atop {3x+27y=408}} \right. \ =\ \textgreater \ \ \left \{ {{x=19 (cars)} \atop {y=13(buses)}} \right.$

Answer: 13 - b., 19 - c.

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<h3>What is a mathematical model?</h3>

A mathematical model is the model which is used to explain the any system, the effect of the components by study and estimate the functions of systems.

Consider a fictitious company, Teslala, that produces a single type of electronic vehicle, Model Z.

The demand for Model Z depends on the gasoline price (q) because customers tend to purchase an electronic vehicle as a substitute for vehicles that run on gasoline when the gasoline price increases.

The demand for Model Z is estimated as

$D(p) = 180 + 10q - 4p$

Here, <em>p</em> is the price of Model Z.

Consider the following two statements:

1. When the average gasoline price q increases by $1, the revenue-maximizing price p" increases by $X.

$D(p) = 180 + 10q - 4p\\D(p) = 180 + 10(q+1) - 4(p+X)$

2. When the average gasoline price q increases by $1, the demand at the revenue-maximizing price (i.e., D(p*)) increases by a factor of Y.

$D(p)+Y = 180 + 10q - 4p\\(D(p)+Y) = 180 + 10(q+1) - 4(p+X)\\Y= 180 + 10(q+1) - 4(p+X)-D(p)\\$

Put the value of demand at the revenue-maximizing price as,

$Y= 180 + 10(q+1) - 4(p+X)-(180 + 10q - 4p)\\Y= 180 + 10q+10 - 4p-pX-180 - 10q + 4p\\Y= 10 -pX\\1=\dfrac{10}{Y}-\dfrac{pX}{Y}\\\dfrac{10}{Y}=\dfrac{pX}{Y}\\\dfrac{10}{pY}=\dfrac{X}{Y}$