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saul85 [17]
2 years ago
15

Find the 8th term of the geometric sequence show below

Mathematics
2 answers:
vivado [14]2 years ago
6 0

Answer:

Alli khanaw .................

Fed [463]2 years ago
5 0

Answer:

8x^36

Step-by-step explanation:

first term (a) = 8x^{8}

Second term (t2)= 8x^{12}

common ratio (r)

= \frac{8x^{12} }{8x^{8} }

= x^{4}

now the 8th term of the G.S is

t8 = ar^{n-1}

  = 8x^8 * (x^4)^8^-^1

  = 8x^8 * (x^4)^7\\

  = 8x^8 *x^2^8\\= 8x^3^6

hope it will help

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sqrt{4 • 12}

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4•sqrt{3}.

The geometric mean of 4 and 12 is

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3 years ago
a car dealer sells suvs and sedans at a ratio of 3:2 each month. if the dealer reports selling 95 cars in a month, how many of e
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Step-by-step explanation:

20) 3125.76

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3 years ago
A journal article reports that a sample of size 5 was used as a basis for calculating a 95% CI for the true average natural freq
oksano4ka [1.4K]

Answer:

Lower = 231.134- 3.098=228.036

Upper = 231.134+ 3.098=234.232

Step-by-step explanation:

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=5 represent the sample size  

Solution to the problem

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

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

And for this case we know that the 95% confidence interval is given by:

\bar X=\frac{233.002 +229.266}{2}= 231.134

And the margin of error is given by:

ME = \frac{233.002 -229.266}{2}= 1.868

And the margin of error is given by:

ME= t_{\alpha/2} \frac{s}{\sqrt{n}}

The degrees of freedom are given by:

df = n-1 = 5-1=4

And the critical value for 95% of confidence is t_{\alpha/2}= 2.776

So then we can find the deviation like this:

s = \frac{ME \sqrt{n}}{t_{\alpha/2}}

s = \frac{1.868* \sqrt{5}}{2.776}= 1.506

And for the 99% confidence the critical value is: t_{\alpha/2}= 4.604

And the margin of error would be:

ME = 4.604 *\frac{1.506}{\sqrt{5}}= 3.098

And the interval is given by:

Lower = 231.134- 3.098=228.036

Upper = 231.134+ 3.098=234.232

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She gave 4 because 9+4=13

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