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Reptile [31]
3 years ago
14

The regular price of a child's entry ticket to a water park is $8 less than that for an adult's. The park offers half off of all

entry tickets during the off-peak season. The Sandlers paid a total of $194 for 1 adult ticket and 4 child's tickets to the water park during the off-peak season. The following equation represents this situation, where x represents the regular price of an adult ticket.
194 = 1/2x + 2(x - 8)

What is the regular price of a child's ticket?

$68

$76

$84

$89
Mathematics
1 answer:
valkas [14]3 years ago
3 0
<span>194 = 1/2x + 2(x - 8)
194 = 1/2x + 2x - 16 = 5/2x - 16
5/2x = 194 + 16 = 210
5x = 210 x 2 = 420
x = 420/5 = 84

Therefore, the regular price of a child's ticket is $84 - $8 = $76.
</span>
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Answer: a. 1.981 < μ < 2.18

              b. Yes.

Step-by-step explanation:

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First, we calculate mean of the sample:

\overline{x}=\frac{\Sigma x}{n}

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\overline{x}= 2.08

Now, we estimate standard deviation:

s=\sqrt{\frac{\Sigma (x-\overline{x})^{2}}{n-1} }

s=\sqrt{\frac{(2.31-2.08)^{2}+...+(2.02-2.08)^{2}}{11} }

s = 0.1564

For t-score, we need to determine degree of freedom and \frac{\alpha}{2}:

df = 12 - 1

df = 11

\alpha = 1 - 0.95

α = 0.05

\frac{\alpha}{2}= 0.025

Then, t-score is

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The interval will be

\overline{x} ± t.\frac{s}{\sqrt{n} }

2.08 ± 2.201\frac{0.1564}{\sqrt{12} }

2.08 ± 0.099

The 95% two-sided CI on the mean is 1.981 < μ < 2.18.

B. We are 95% confident that the true population mean for this clinic is between 1.981 and 2.18. Since the mean number performed by all clinics has been 1.95, and this mean is less than the interval, there is evidence that this particular clinic performs more scans than the overall system average.

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Answer:

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Step-by-step explanation:

To find the option that is equivalent to this equation, we can begin by narrowing down our options. The slope contains a negative number, meaning the second point must have a lower y value than the first.

Using this criteria, we can eliminate the first and the third answer. Now, we can test the second and fourth answer.

We can test using the slope formula and plugging in values. When the slope formula: m= \frac{y_{2}-y_{1}  }{x_{2} -x_{1} }

We get a slope of -5 for both of these equations. Let's plug in the point values to check whether they work:

Choice 2:

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