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3 years ago

Beach Travel rents dune buggies for $50 for 4 hours or $75 for 6 hours. What is the hourly rate?

Mathematics

2 answers:

Mars2501 [29]3 years ago

7 0

Answer: $ 12.5

Step-by-step explanation:

To find the hourly rate, divide the total cost by the total hours, that is

$50 / 4 = $12.5

$75/6 = $12.5

PIT_PIT [208]3 years ago

6 0

Answer:

Step-by-step explanation:

Hourly rate = $50/4 or $75/6

= $12.5/hour

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The score on a trivia game is obtained by subtracting the number of incorrect answers from twice the number of correct answers.

labwork [276]

Answer:

Step-by-step explanation:

A player answers x correct questions and y incorrect questions for a total of 40 questions.

x + y = 40

The score is obtained by subtracting y incorrect answers from two times the x of correct answers. The player's score is 50.

2x - y = 50

Use substitution to solve. Rearrange the first equation so that it is equal to y. Then, substitute the y-value into the second equation.

x + y = 40

y = 40 - x

2x - y = 50

2x - (40 - x) = 50

2x - 40 + x = 50

3x - 40 = 50

3x = 90

x = 30

Since x was made to represent correct answers, the player had 30 correct answers.

8 0

3 years ago

Express 48 + 72 using the Distributive Property

tiny-mole [99]

Answer:

12( 6+ 4)

Step-by-step explanation:

7 0

3 years ago

Read 2 more answers

Mis Meyer's paid $3600 all together for the equipment, furniture and decorations for her restaurant. The equipment cost $500 mor

Tju [1.3M]

Answer:

The equipment will cost $1740.

Step-by-step explanation:

Let the Cost of equipment, furniture and decoration be x, y and z.

Now, According to question,

x + y + z = 3600 ...... (1) (cost of all items)

x = 500 + y (∵ equipment cost 500 more than furniture)

and y = 2z ( ∵ furniture twice as much as decoration)

so, z = y/2

Now substituting the value of x and z in eq (1)

$x + y + z = 3600$

$500 + y + y + \frac{y}{2} = 3600$

$2y + \frac{y}{2} = 3600 - 500$

$\frac{5y}{2} = 3100$

$y = \frac{3100\times 2}{5} = 1240$

So, the cost of furniture (y) = 1240

∴ Cost of equipment = y + 500 = 1240 + 500 = 1740

Therefore the cost of equipment was $1740.

3 0

3 years ago

Mai received and spent money in the following ways last month. For each example, write a signed number to represent the change i

atroni [7]

The signed numbers that represent a change in money from her perspective is:

+$25

+$14

-$10

-$3

+$2

<h3>What are the signs to put in front of the amounts?</h3>

When a person receives money, earns interst or his paid wages, the amount of money the person has increases. Thus, the sign to be used is the addition sign(+).

When a person pays for an item, donates to charity or loses money, the amount of money the person has reduces. Thus, the sign to be used is the subtraction sign (-).

To learn more about addition, please check: brainly.com/question/349488

6 0

3 years ago

Consider the following hypothesis test:

postnew [5]

Answer:

a. P-value = 0.039.

The null hypothesis is rejected.

At a significance level of 0.05, there is enough evidence to support the claim that the population mean significantly differs from 100.

b. P-value = 0.013.

The null hypothesis is rejected.

At a significance level of 0.05, there is enough evidence to support the claim that the population mean significantly differs from 100.

c. P-value = 0.130.

The null hypothesis failed to be rejected.

At a significance level of 0.05, there is not enough evidence to support the claim that the population mean significantly differs from 100.

Step-by-step explanation:

This is a hypothesis test for the population mean.

The claim is that the population mean significantly differs from 100.

Then, the null and alternative hypothesis are:

$H_0: \mu=100\\\\H_a:\mu\neq 100$

The significance level is 0.05.

The sample has a size n=65.

The degrees of freedom for this sample size are:

$df=n-1=65-1=64$

a. The sample mean is M=103.

As the standard deviation of the population is not known, we estimate it with the sample standard deviation, that has a value of s=11.5.

The estimated standard error of the mean is computed using the formula:

$s_M=\dfrac{s}{\sqrt{n}}=\dfrac{11.5}{\sqrt{65}}=1.4264$

Then, we can calculate the t-statistic as:

$t=\dfrac{M-\mu}{s/\sqrt{n}}=\dfrac{103-100}{1.4264}=\dfrac{3}{1.4264}=2.103$

This test is a two-tailed test, with 64 degrees of freedom and t=2.103, so the P-value for this test is calculated as (using a t-table):

$\text{P-value}=2\cdot P(t>2.103)=0.039$

As the P-value (0.039) is smaller than the significance level (0.05), the effect is significant.

The null hypothesis is rejected.

At a significance level of 0.05, there is enough evidence to support the claim that the population mean significantly differs from 100.

b. The sample mean is M=96.5.

As the standard deviation of the population is not known, we estimate it with the sample standard deviation, that has a value of s=11.

The estimated standard error of the mean is computed using the formula:

$s_M=\dfrac{s}{\sqrt{n}}=\dfrac{11}{\sqrt{65}}=1.3644$

Then, we can calculate the t-statistic as:

$t=\dfrac{M-\mu}{s/\sqrt{n}}=\dfrac{96.5-100}{1.3644}=\dfrac{-3.5}{1.3644}=-2.565$

This test is a two-tailed test, with 64 degrees of freedom and t=-2.565, so the P-value for this test is calculated as (using a t-table):

$\text{P-value}=2\cdot P(t$

As the P-value (0.013) is smaller than the significance level (0.05), the effect is significant.

The null hypothesis is rejected.

At a significance level of 0.05, there is enough evidence to support the claim that the population mean significantly differs from 100.

c. The sample mean is M=102.

As the standard deviation of the population is not known, we estimate it with the sample standard deviation, that has a value of s=10.5.

The estimated standard error of the mean is computed using the formula:

$s_M=\dfrac{s}{\sqrt{n}}=\dfrac{10.5}{\sqrt{65}}=1.3024$

Then, we can calculate the t-statistic as:

$t=\dfrac{M-\mu}{s/\sqrt{n}}=\dfrac{102-100}{1.3024}=\dfrac{2}{1.3024}=1.536$

This test is a two-tailed test, with 64 degrees of freedom and t=1.536, so the P-value for this test is calculated as (using a t-table):

$\text{P-value}=2\cdot P(t>1.536)=0.130$

As the P-value (0.13) is bigger than the significance level (0.05), the effect is not significant.

The null hypothesis failed to be rejected.

At a significance level of 0.05, there is not enough evidence to support the claim that the population mean significantly differs from 100.

8 0

3 years ago