_(9)=(2(1-2(2)^(9)))/(1-2(2))PLEASE HELP ME OUT.

Mathematics

1 answer:

USPshnik [31]3 years ago

4 0

Answer:

the answer = um ok

Step-by-step explanation:

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Given a table below which gives the years in A.D. for an archaeological excavation site using the method of tree ring dating:

$\begin{tabular}
{|c|c|c|c|c|c|c|c|c|}
1229&1292&1187&1257&1268&1316&1275& 1317&1275
\end{tabular}$

Part A:

The sample mean is given by:

$\bar{x}= \frac{1229+1292+1187+1257+1268+1316+1275+ 1317+1275}{9} \\ \\ = \frac{11416}{9} =1268$

Therefore, the sample mean is 1268.

Part B:

We calculate the sample standard deviation as follows:

$s= \frac{1}{9-1} \left(\sqrt{(1229-1268)^2+(1292-1268)^2+(1187-1268)^2+(1257-1268)^2+(1268-1268)^2+(1316-1268)^2+(1275-1268)^2+(1317-1268)^2+(1275-1268)^2}\right)\\ \\ = \frac{1}{8}\left(\sqrt{(-39)^2+24^2+(-81)^2+(-11)^2+0^2+48^2+7^2+49^2+7^2}\right)\\ \\ = \frac{1}{8}\left(\sqrt{1521+576+6561+121+2304+49+2401+49}\right)\\ \\ = \frac{1}{8}\left(\sqrt{13,582}\right)\\ \\ = \frac{1}{8}(116.54)=14.57\approx15$

Part C:

The 90% confidence interval is given by:

$90\% \ C. \ I.=1268\pm1.65\left(\frac{15}{\sqrt{9}}\right) \\ \\ =1268\pm\left(\frac{15}{3}\right)=1268\pm5 \\ \\ =(1268-5, \ 1268+5)=(1263, \ 1273)$

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