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patriot [66]
3 years ago
10

Which equation describes the relationship between the variables in the table below? x y -1 1 9 2 81 4 6,561 5 59,049 ; each y-va

lue is the previous y-value multiplied by 9. ; each y-value is the previous y-value plus 9 more than was added previously. ; each y-value is the previous y-value plus 9. ; each y-value is the previous y-value multiplied by another 9.
Mathematics
2 answers:
Minchanka [31]3 years ago
5 0
<span>x y
  1
1 9
2 81
4 6,561
5 59,049 

Based on the variables on the table, I can say that </span><span>each y-value is the previous y-value multiplied by 9.

x y
0 1 
1 9 </span>→ 1 * 9 = 9
2 81 → 9 * 9 = 81
3 729 → 81 * 9 = 729  NOT WRITTEN ON THE TABLE
4 6,561 → 729 * 9 = 6,561
5 59,049 → 6,561 * 9 = 59,049

y = 9^x
Mrac [35]3 years ago
4 0

Answer:

9^x

Step-by-step explanation:

Got it right on the test

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a) The 95% confidence interval for the population proportion of college students who work to pay for tuition andliving expenses is (0.4239, 0.5161).

b) The 99% confidence interval for the population proportion of college students who work to pay for tuition and living expenses is (0.4094, 0.5306).

c)The margin of error increases as the confidence level increases.

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of \pi, and a confidence interval 1-\alpha, we have the following confidence interval of proportions.

\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}

In which

z is the zscore that has a pvalue of 1 - \frac{\alpha}{2}.

For this problem, we have that:

n = 450, p = 0.47

a) Provide a 95% confidence interval for the population proportion of college students who work to pay for tuition andliving expenses.

95% confidence interval

So \alpha = 0.05, z is the value of Z that has a pvalue of 1 - \frac{0.05}{2} = 0.975, so Z = 1.96.

The lower limit of this interval is:

\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.47 - 1.96\sqrt{\frac{0.47*0.53}{450}} = 0.4239

The upper limit of this interval is:

\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.47 + 1.96\sqrt{\frac{0.47*0.53}{450}}{119}} = 0.5161

The 95% confidence interval for the population proportion of college students who work to pay for tuition andliving expenses is (0.4239, 0.5161).

b. Provide a 99% confidence interval for the population proportion of college students who work to pay for tuition andliving expenses.

95% confidence interval

So \alpha = 0.01, z is the value of Z that has a pvalue of 1 - \frac{0.01}{2} = 0.995, so Z = 2.575.

The lower limit of this interval is:

\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.47 - 2.575\sqrt{\frac{0.47*0.53}{450}} = 0.4094

The upper limit of this interval is:

\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.47 + 2.575\sqrt{\frac{0.47*0.53}{450}}{119}} = 0.5306

The 99% confidence interval for the population proportion of college students who work to pay for tuition and living expenses is (0.4094, 0.5306).

c What happens to the margin of error as the confidence is increased from 95% to 99%?

The margin of error is the subtraction of the upper end by the lower end of the interval, divided by 2. So

95% confidence interval

M = \frac{(0.5161 - 0.4239)}{2} = 0.0461

99% confidence interval

M = \frac{(0.5306 - 0.4094)}{2} = 0.0606

The margin of error increases as the confidence level increases.

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