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Serhud [2]
1 year ago
9

The price of a train ticket consists of an initial fee plus a constant fee per stop. The table compares the number of stops and

the price of a ticket (in dollars). Stops Price (dollars) 333 6.506.506, point, 50 777 12.5012.5012, point, 50 111111 18.5018.5018, point, 50 What is the fee per stop?
Mathematics
1 answer:
12345 [234]1 year ago
4 0

Based on the task content given; the initial fee and the fee per stop is $2 and $1.5 respectively.

<h3>Equation</h3>

let

  • Initial fee = x
  • Fee per stop = y

x + 3y = 6.50

x + 7y = 12.50

  • Subtract both equation to eliminate x

7y - 3y = 12.50 - 6.50

4y = 6

y = 6/4

y = 1.5

  • Substitute y = 1.5 into

x + 3y = 6.50

x + 3(1.5) = 6.50

x + 4.5 = 6.50

x = 6.50 - 4.5

x = 2

Therefore, the initial fee and the fee per stop is $2 and $1.5 respectively.

Learn more about equation:

brainly.com/question/11418438

#SPJ1

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7(5x - 13) = 14
vfiekz [6]

Answer:

Part A, one solution

Part B, x=3

Step-by-step explanation:

divide both sides of the equation by 7 (5x-13=2)

move the constant to the right hand side and change its sign (5x=2+13)

add numbers (5x=15)

divide both sides of the equation by 5  (x=3)

hope this helps, have a good day

5 0
2 years ago
10. A manufacturer wanted to know if more coupons would be redeemed if they were mailed to the
Mandarinka [93]

Using the t-distribution, it is found that the data does not provide convincing evidence that the mean number is greater when the coupons were addressed to a female.

<h3>What are the hypothesis tested?</h3>

At the null hypothesis, it is tested if there is no difference, that is, the mean is of 0, hence:

H_0: \mu = 0

At the alternative hypothesis, it is tested if the mean number is greater for females, that is, the mean is greater than 0, hence:

H_1: \mu > 0

<h3>What is the test statistic?</h3>

The test statistic is given by:

t = \frac{\overline{x} - \mu}{\frac{s}{\sqrt{n}}}

The parameters are:

  • \overline{x} is the sample mean.
  • \mu is the value tested at the null hypothesis.
  • s is the standard deviation of the sample.
  • n is the sample size.

In this problem, the parameters are given as follows:

\overline{x} = 1.5, \mu = 0, s = 4.75, n = 50.

Hence, the test statistic is given by:

t = \frac{\overline{x} - \mu}{\frac{s}{\sqrt{n}}}

t = \frac{1.5 - 0}{\frac{4.75}{\sqrt{50}}}

t = 2.23

<h3>What is the conclusion?</h3>

Considering a <em>right-tailed test</em>, as we are testing if the mean is greater than a value, with a <em>significance level of 0.01 and 50 - 1 = 49 df</em>, the critical value is given by t^{\ast} = 2.4.

Since the test statistic is less than the critical value for the right-tailed test, the data does not provide convincing evidence that the mean number is greater when the coupons were addressed to a female.

More can be learned about the t-distribution at brainly.com/question/26454209

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