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3 years ago

What is the common difference between ratio between successive terms in the sequence 1.5, 1.2, 0.96, 0.768

Mathematics

2 answers:

nikklg [1K]3 years ago

6 0

First, we should see if there are any common differences. We do this by subtracting each term from its previous:

1.5 - 1.2 = 0.3

1.2 - 0.96 = 0.24

Already, we can tell that there is no common difference.

Let's look at common ratio:

1.2/1.5 = 0.8

0.96/1.2 = 0.8

0.768/0.96 = 0.8

Therefore, there is a common ratio and it is 0.8.

sladkih [1.3K]3 years ago

4 0

Answer:

Answer is D: 0.8

Step-by-step explanation:

r = 1.2 / 1.5 = 0.8

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a) A sample size of 5615 is needed.

b) 0.012

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 level of $1-\alpha$ , we have the following confidence interval of proportions.

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

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So $\alpha = 0.005$ , z is the value of Z that has a pvalue of $1 - \frac{0.005}{2} = 0.9975$ , so $Z = 2.81$ .

(a) Past studies suggest that this proportion will be about 0.2. Find the sample size needed if the margin of the error of the confidence interval is to be about 0.015.

This is n for which M = 0.015.

We have that $\pi = 0.2$

$M = z\sqrt{\frac{\pi(1-\pi)}{n}}$

$0.015 = 2.81\sqrt{\frac{0.2*0.8}{n}}$

$0.015\sqrt{n} = 2.81\sqrt{0.2*0.8}$

$\sqrt{n} = \frac{2.81\sqrt{0.2*0.8}}{0.015}$

$(\sqrt{n})^{2} = (\frac{2.81\sqrt{0.2*0.8}}{0.015})^{2}$

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A sample size of 5615 is needed.

(b) Using the sample size above, when the sample is actually contacted, 12% of the sample say they are not satisfied. What is the margin of the error of the confidence interval?

Now $\pi = 0.12, n = 5615$ .

We have to find M.

$M = z\sqrt{\frac{\pi(1-\pi)}{n}}$

$M = 2.81\sqrt{\frac{0.12*0.88}{5615}}$

$M = 0.012$

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