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luda_lava [24]
3 years ago
8

Find the sum of the first 9 terms in the following geometric series 7+21+63+

Mathematics
1 answer:
Zielflug [23.3K]3 years ago
4 0

Answer: 68887

Step-by-step explanation: the geometics series is (the previous number in the series) x 3, so 63 x 3 =189 x 3 =567 x 3 = 1701 x 3 = 5103 x 3 =15309 x 3 = 45927 and you add all those numbers up to get 68887

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Assume that population mean is to be estimated from the sample described. Use the sample results to approximate the margin of er
Vikentia [17]

Answer:

ME=1.998 \frac{4}{\sqrt{64}}=0.999 \approx 1

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".

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\bar X =69 represent the sample mean for the sample  

\mu population mean

s=4.0 represent the sample standard deviation

n=64 represent the sample size  

Assuming the X follows a normal distribution

X \sim N(\mu, \sigma)

The sample mean \bar X have the following distribution:

\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})

The margin of error for the sample mean is given by this formula:

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

Since we have the sample standard deviation we can estimate the margin of error like this:

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

The next step would be find the value of \z_{\alpha/2}, \alpha=1-0.95=0.05 and \alpha/2=0.025

We can find the degrees of freedom like this:

df=n-1=64-1=63

And we can find the critical value with the following code in excel for example: "=T.INV(0.025,63)" and we got:

t_{\alpha/2}=\pm 1.998

And the margin of error would be :

ME=1.998 \frac{4}{\sqrt{64}}=0.999 \approx 1

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