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SCORPION-xisa [38]
3 years ago
10

Given the following geometric sequence, find the common ratio: {225, 45, 9...}

Mathematics
2 answers:
Tom [10]3 years ago
8 0

Answer:

\frac{1}{5}  is the common ratio

Step-by-step explanation:

Common ratio(r) states that the ratio of each term of a geometric sequence to the term preceding it.

r = \frac{a_2}{a_1} =\frac{a_3}{a_2}........\frac{a_{n+1}}{a_n}

Given the sequence:

225, 45, 9, .........

Here,

a_1 = 225

a_2 = 45

a_3 = 9 and so on...

Using definition to find r.

r = \frac{45}{225}=\frac{9}{45}......

After solving we get;

r = \frac{1}{5}

Therefore, the common ratio is, \frac{1}{5}

Maksim231197 [3]3 years ago
5 0

Answer:

r=\frac{1}{5}

Step-by-step explanation:

We are given that a geometric sequence

225,45,9,...

We have to find the common ratio of the geometric sequence.

a=225,a_2=45,a_3=9

r=\frac{a_2}{a_1}

a_2=ar

Substitute the values then we get

45=225r

r=\frac{45}{225}

r=\frac{1}{5}

Hence, the common ratio of the geometric sequence =\frac{1}{5}

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Answer:

n=(\frac{1.64(14.7)}{3})^2 =64.57 \approx 65

So the answer for this case would be n=65 rounded up to the nearest integer

And the sample size would decrease by 160-65=95 subjects

Step-by-step explanation:

1) Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

\bar X represent the sample mean for the sample  

\mu population mean (variable of interest)

s represent the sample standard deviation

n represent the sample size  

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The confidence interval for the mean is given by the following formula:

\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}   (1)

2) Solution to the problem

Since the new Confidence is 0.90 or 90%, the value of \alpha=0.1 and \alpha/2 =0.05, and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-NORM.INV(0.05,0,1)".And we see that z_{\alpha/2}=1.64

Assuming that the deviation is known we can express the margin of error is given by this formula:

ME=z_{\alpha/2}\frac{\sigma}{\sqrt{n}}    (a)

And on this case we have that ME =\pm 3 and we are interested in order to find the value of n, if we solve n from equation (a) we got:

n=(\frac{z_{\alpha/2} \sigma}{ME})^2   (b)

Replacing into formula (b) we got:

n=(\frac{1.64(14.7)}{3})^2 =64.57 \approx 65

So the answer for this case would be n=65 rounded up to the nearest integer

And the sample size would decrease by 160-65=95 subjects

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Can someone help me find the area of this circle and round to the nearest tenth
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