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3 years ago

Ruth found 37 tickets for $7. Please help Ruth by figuring out the ratio. (round to 2 decimal places)

Mathematics

1 answer:

Amanda [17]3 years ago

3 0

Answer:

Ratio of ticket to its price is 1 : 0.189.

Step-by-step explanation:

Here, the cost of 37 tickets = $7

Let us find the unit cost of the ticket.

$\textrm{Unit Price } = \frac{\textrm{Total Price of n units}}{\textrm{n}}$

$\implies U = \frac{7}{37}=0.189$

So, the unit price of each ticket = $0.189

Now, ratio of Each Ticket: Unit Price = 1 : 0.189

Hence, the ratio of ticket to its price is 1 : 0.189.

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A group of college students are going to a lake house for the weekend and plan on renting small cars and large cars to make the

Alexus [3.1K]

4x+6y=56 and y=4x can be used to determine the number of small cars rented and the number of large cars rented, where x represents number of small cars and y represents number of large cars.

Step-by-step explanation:

Number of people hold by small cars = 4

Number of people hold by large cars = 6

Let,

Number of small cars = x

Number of large cars = y

The students rented 4 times as many large cars as small cars,

y = 4x Eqn 1

which altogether can hold 56 people.

4x+6y=56 Eqn 2

4x+6y=56 and y=4x can be used to determine the number of small cars rented and the number of large cars rented, where x represents number of small cars and y represents number of large cars.

Keywords: linear equations, addition

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3 years ago

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4 years ago

Len's recipe for blueberry muffins calls for 110.25 g of brown sugar for the muffin batter and an additional 55.45 g of brown su

nasty-shy [4]

This may be wrong but this is how id do it.

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4 years ago

Which of the follow is is trinomial

sasho [114]

Answer:

A. 3x^2 + 4x + 1

Step-by-step explanation:

A is the correct answer because there are three terms within the trinomial

6 0

3 years ago

Read 2 more answers

Given the function f(x) = x^4 + 3x^3 - 2x^2 - 6x - 1, use intermediate theorem to decide which of the following intervals contai

marta [7]

$f(x) = x^4 + 3x^3 - 2x^2 - 6x - 1$

Lets check with every option

(a) [-4,-3]

We plug in -4 for x and -3 for x

$f(-4) = (-4)^4 + 3(-4)^3 - 2(-4)^2 - 6(-4) - 1= 55$

$f(-3) = (-3)^4 + 3(-3)^3 - 2(-3)^2 - 6(-3) - 1= -1$

f(-4) is positive and f(-3) is negative. there is some value at x=c on the interval [-4,-3] where f(c)=0. so there exists atleast one zero on this interval.

(b) [-3,-2]

We plug in -3 for x and -2 for x

$f(-3) = (-3)^4 + 3(-3)^3 - 2(-3)^2 - 6(-3) - 1= -1$

$f(-2) = (-2)^4 + 3(-2)^3 - 2(-2)^2 - 6(-2) - 1= -5$

f(-2) is negative and f(-3) is negative. there is no value at x=c on the interval [-3,-2] where f(c)=0.

(c) [-2,-1]

We plug in -2 for x and -1 for x

$f(-2) = (-2)^4 + 3(-2)^3 - 2(-2)^2 - 6(-2) - 1= -5$

$f(-1) = (-1)^4 + 3(-1)^3 - 2(-1)^2 - 6(-1) - 1= 1$

f(-2) is negative and f(-1) is positive. there is some value at x=c on the interval [-2,-1] where f(c)=0. so there exists atleast one zero on this interval.

(d) [-1,0]

We plug in -1 for x and 0 for x

$f(-1) = (-1)^4 + 3(-1)^3 - 2(-1)^2 - 6(-1) - 1= 1$

$f(0) = (0)^4 + 3(0)^3 - 2(0)^2 - 6(0) - 1= -1$

f(-1) is positive and f(0) is negative. there is some value at x=c on the interval [-1,0] where f(c)=0. so there exists atleast one zero on this interval.

(e) [0,1]

We plug in 0 for x and 1 for x

$f(0) = (0)^4 + 3(0)^3 - 2(0)^2 - 6(0) - 1= -1$

$f(1) = (1)^4 + 3(1)^3 - 2(1)^2 - 6(1) - 1= -5$

f(0) is negative and f(1) is negative. there is no value at x=c on the interval [0,1] where f(c)=0.

(f) [1,2]

We plug in 1 for x and 2 for x

$f(1) = (1)^4 + 3(1)^3 - 2(1)^2 - 6(1) - 1= -5$

$f(2) = (2)^4 + 3(2)^3 - 2(2)^2 - 6(2) - 1= 19$

f(-4) is positive and f(-3) is negative. there is some value at x=c on the interval [-4,-3] where f(c)=0. so there exists atleast one zero on this interval.

so answers are (a) [-4,-3], (c) [-2,-1], (d) [-1,0], (f) [1,2]

8 0

3 years ago

Read 2 more answers