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3 years ago

8

Anna is a teacher at an elementary school. She purchased 73 tickets to take the first-grade children and some parents on a field

trip to the zoo. She purchased
children's tickets for $9 each and adult tickets for $15 each. She spent a total of $771. How many of each ticket did she buy?
Adults
Children
Total

1 answer:

Svetllana [295]3 years ago

4 0

She bought 54 child tickets and 19 adult tickets.

Step-by-step explanation:

No. of tickets purchased = 73

Cost of tickets = $771

Cost of one child ticket = $9

Cost of one adult ticket = $15

Let,

Child ticket = x

Adult ticket = y

According to given statement;

x+y=73 Eqn 1

9x+15y=771 Eqn 2

Multiplying Eqn 1 by 9;

$9(x+y=73)\\9x+9y=657\ \ \ Eqn\ 3\\$

Subtracting Eqn 3 from Eqn 2;

$(9x+15y)-(9x+9y)=771-657\\9x+15y-9x-9y=114\\6y=114\\$

Dividing both sides by 6;

$\frac{6y}{6}=\frac{114}{6}\\y=19$

Putting y=19 in Eqn 1

$x+19=73\\x=73-19\\x=54\\$

She bought 54 child tickets and 19 adult tickets.

Keywords: linear equations, subtraction

Learn more about linear equations at:

#LearnwithBrainly

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Classwork 12.1 surface area

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Answer:

6) LA = 90 units² and SA = 202 units²

7) LA = 377 units² and SA = 477.5 units²

8) LA = 208 units² and SA = 232 units²

9) LA = 455 units² and SA = 591 units²

Step-by-step explanation:

* Lets revise the rules of lateral area and surface area

- Lateral area of the solid = perimeter of its base × its height

- Surface area = lateral area + 2 × area of its base

* Now lets solve the problems

6) The solid is a rectangular prism

- Its base is a rectangle with dimensions 8 units and 7 units

- Its height is 3 units

∵ Perimeter of the rectangle = 2(L + W)

∴ Perimeter of the base = 2(8 + 7) = 2(15) = 30 units

∴ LA = 30 × 3 = 90 units²

∵ Area of the rectangle = L × W

∴ Area of the base = 8 × 7 = 56 units²

∴ SA = 90 + 2 × 56 = 90 + 112 = 202 units²

7) The solid is a cylinder

- Its base is a circle with diameter 8 units

∴ Its radius = 8 ÷ 2 = 4 units

- Its height is 15 units

∵ The perimeter of the circle is 2πr

∴ The perimeter of the base = 2π(4) = 8π

∴ LA = 8π(15) = 120π = 376.99 ≅ 377 units²

∵ The area of the circle = πr²

∴ The area of the base = π(4)² = 16π

∴ SA = 120π + 2 × 16π = 120π + 32π = 152π = 477.5 units²

6) The solid is a triangular prism

- Its base is a triangle with sides 5 , 5 , 6 units and height 4 units

- Its height is 13 units

∵ Perimeter of the triangle is the sum of the 3 sides

∴ Perimeter of the base = 5 + 5 + 6 = 16 units

∴ LA = 16 × 13 = 208 units²

∵ Area of the triangle = 1/2 × base × height

∴ Area of the base = 1/2 × 6 × 4 = 12 units²

∴ SA = 208 + 2 × 12 = 208 + 24 = 232 units²

9) The solid is a prism

- Its base is an isosceles trapezium with parallel bases 7 units and 10

units, 2 non-parallel bases 9 units and height 8 units

- Its height is 13 units

∵ Perimeter of the trapezium is the sum of its sides

∴ Perimeter of the base = 7 + 10 + 9 + 9 = 35 units

∴ LA = 35 × 13 = 455 units²

∵ Area of the trapezium = 1/2(b1 + b2) × h

∴ Area of the base = 1/2(7 + 10) × 8 = 68 units²

∴ SA = 455 + 2 × 68 = 455 + 136 = 591 units²

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4 years ago

Barbara is 142 cm tall. This is 2 cm less than 3 times

krok68 [10]

The answer for your question is 45 cm long

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If 5 over 6 gallon covers 1 over 4 of the house then how much paint is needed for the entire house

MissTica

If 5/6 gallons covers a quarter of the house, then you would need 4 times as much paint:
$\frac{5}{6} \times 4$
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Suppose that W1, W2, and W3 are independent uniform random variables with the following distributions: Wi ~ Uni(0,10*i). What is

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I'll leave the computation via R to you. The $W_i$ are distributed uniformly on the intervals $[0,10i]$ , so that

$f_{W_i}(w)=\begin{cases}\dfrac1{10i}&\text{for }0\le w\le10i\\\\0&\text{otherwise}\end{cases}$

each with mean/expectation

$E[W_i]=\displaystyle\int_{-\infty}^\infty wf_{W_i}(w)\,\mathrm dw=\int_0^{10i}\frac w{10i}\,\mathrm dw=5i$

and variance

$\mathrm{Var}[W_i]=E[(W_i-E[W_i])^2]=E[{W_i}^2]-E[W_i]^2$

We have

$E[{W_i}^2]=\displaystyle\int_{-\infty}^\infty w^2f_{W_i}(w)\,\mathrm dw=\int_0^{10i}\frac{w^2}{10i}\,\mathrm dw=\frac{100i^2}3$

so that

$\mathrm{Var}[W_i]=\dfrac{25i^2}3$

Now,

$E[W_1+W_2+W_3]=E[W_1]+E[W_2]+E[W_3]=5+10+15=30$

and

$\mathrm{Var}[W_1+W_2+W_3]=E\left[\big((W_1+W_2+W_3)-E[W_1+W_2+W_3]\big)^2\right]$

$\mathrm{Var}[W_1+W_2+W_3]=E[(W_1+W_2+W_3)^2]-E[W_1+W_2+W_3]^2$

We have

$(W_1+W_2+W_3)^2={W_1}^2+{W_2}^2+{W_3}^2+2(W_1W_2+W_1W_3+W_2W_3)$

$E[(W_1+W_2+W_3)^2]$

$=E[{W_1}^2]+E[{W_2}^2]+E[{W_3}^2]+2(E[W_1]E[W_2]+E[W_1]E[W_3]+E[W_2]E[W_3])$

because $W_i$ and $W_j$ are independent when $i\neq j$ , and so

$E[(W_1+W_2+W_3)^2]=\dfrac{100}3+\dfrac{400}3+300+2(50+75+150)=\dfrac{3050}3$

giving a variance of

$\mathrm{Var}[W_1+W_2+W_3]=\dfrac{3050}3-30^2=\dfrac{350}3$

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# # #

A faster way, assuming you know the variance of a linear combination of independent random variables, is to compute

$\mathrm{Var}[W_1+W_2+W_3]$

$=\mathrm{Var}[W_1]+\mathrm{Var}[W_2]+\mathrm{Var}[W_3]+2(\mathrm{Cov}[W_1,W_2]+\mathrm{Cov}[W_1,W_3]+\mathrm{Cov}[W_2,W_3])$

and since the $W_i$ are independent, each covariance is 0. Then

$\mathrm{Var}[W_1+W_2+W_3]=\mathrm{Var}[W_1]+\mathrm{Var}[W_2]+\mathrm{Var}[W_3]$

$\mathrm{Var}[W_1+W_2+W_3]=\dfrac{25}3+\dfrac{100}3+75=\dfrac{350}3$

and take the square root to get the standard deviation.

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