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3 years ago

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a matinee ticket costs $6 per adult and $1 per child. On a certain day, the total number of adults (a) and children (c) who saw

a movie was 35, and the total money collected was $70. which of the following options represents the number of adults who saw a movie that day, and the pair of equations that can be solved to find the numbers?

1 answer:

vovikov84 [41]3 years ago

5 0

Answer:

The number of adults who saw a movie that day is 7

and

the pair of equations that can be solved to find the numbers is

6x+y = 70

x + y = 35

Step-by-step explanation:

Let the total no. of adults who saw the move be x

Let the total no. of child who saw the move be y

Given that the total number of adults (a) and children (c) who saw a movie was 35

thus. x + y = 35

cost of ticket for 1 adult = $6

cost of ticket for 1*x adult = $6*x = $6x

cost of ticket for x adult = $6x

cost of ticket for 1 child = $1

cost of ticket for 1*y child = $1*y = $y

cost of ticket for y child = $y

Total cost for x adult and y child = $(6x+y) _______(2)

Given that he total money collected was $70

Thus, we have two equations

6x+y = 70 -------(1)

x + y = 35

y = 35 - x (using this value of y in equation 1 we have)

6x + 35 - x = 70

=> 5x = 70 -35 = 35

=> x = 35/5 = 7

The number of adults who saw a movie that day is 7

and

the pair of equations that can be solved to find the numbers is

6x+y = 70

x + y = 35

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Distance between any two points, say, $(x_1, y_1)$ and $(x_2, y_2)$ is:

$\sqrt{\bigg ( \textbf{x}_{\textbf{2}} \hspace{1mm} \textbf{- x}_{\textbf{1}} \bigg )^{\textbf{2}} \textbf{+} \bigg( \textbf{y}_{\textbf{2}} \hspace{1mm} \textbf{- y}_{\textbf{1}} \bigg)^ {\textbf{2}}$

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Length of Side 1 = $\sqrt{\textbf{53}}$ units.

Distance between $P_1$ and $P_2$ , (Side 2):

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Length of Side 3 = $\sqrt{\textbf{53}}$ units.

Note that the length of Side 1 = Length of Side 3.

That means the triangle is Isosceles.

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(Side 1)² + (Side 3)² = (Side 2)²

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Hence, the triangle is a right - angled triangle as well.

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Given

A survey found that 12 out of 30 people preferred bacon over sausage.

Find out the what percent of people prefer bacon over sausage.

To proof

As given in the question

survey found that 12 out of 30 people preferred bacon over sausage.

Total number of people = 30

people preferred bacon over sausage = 12

Formula

$Percentage =\frac{Part}{Total} \times 100$

put the value in the formula

$Percentage\ of\ prefer\ bacon\ over\ sausage = \frac{12}{30} \times 100$

solve the above

percent of people prefer bacon over sausage = 40 %

Hence proved

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