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3 years ago

Question 2

Mathematics

1 answer:

Anni [7]3 years ago

6 0

Answer:

$M = (9,3)$

Step-by-step explanation:

Given

$(x_1,y_1) = (4,8)$

$(x_2,y_2) = (14,-2)$

Required

Find the midpoint (M)

The midpoint is calculated as follows:

$M = (\frac{1}{2}(x_1+x_2),\frac{1}{2}(y_1+y_2))$

Substitute values for x's and y's

$M = (\frac{1}{2}(4+14),\frac{1}{2}(8-2))$

$M = (\frac{1}{2}(18),\frac{1}{2}(6))$

$M = (\frac{1}{2}*18,\frac{1}{2}*6)$

$M = (9,3)$

Hence, the midpoint is (9,3)

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Amount of members = 9

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Step-by-step explanation:

So there are 36 different ways

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Answer:

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3 years ago

Suppose the method of tree ring dating gave the following dates A.D. for an archaeological excavation site. Please show your wor

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Answer:

a) $\bar X=\frac{1245+1245+1321+1191+1295+1330+1239+1250+1228}{9}= 1260.444$

b) $s =\sqrt{\frac{\sum_{i=1}^n (X_i -\bar X)^2}{n-1}}$

And if we replace we got:

$s = 45.580$

c) For this case since the sampel size is n=9 <30 we can use the t distribution in order to find the confidence interval.

d) $1260.444-2.306 \frac{45.580}{\sqrt{9}}= 1225.408$

$1260.444+2.306 \frac{45.580}{\sqrt{9}}= 1295.480$

Step-by-step explanation:

For this case we ave the following data:

1245 1245 1321 1191 1295 1330 1239 1250 1228

Part a

We can calculate the sample mean with the following formula:

$\bar X = \frac{\sum_{i=1}^n X_i}{n}$

And if we replace we got:

$\bar X= \frac{1245+1245+1321+1191+1295+1330+1239+1250+1228}{9}= 1260.444$

Part b

For this case we can use the following formula in order to find the sample standard deviation:

$s =\sqrt{\frac{\sum_{i=1}^n (X_i -\bar X)^2}{n-1}}$

And if we replace we got:

$s = 45.580$

Part c

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".

For this case since the sampel size is n=9 <30 we can use the t distribution in order to find the confidence interval.

Part d

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

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

We need to find the degrees of freedom given by:

$df = n-1 = 9-1=8$

Since the Confidence is 0.95 or 95%, the value of $\alpha=1-0.95=0.05$ and $\alpha/2 =0.025$ , and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.025,8)".And we see that $t_{\alpha/2}=2.306$

Now we have everything in order to replace into formula (1):

$1260.444-2.306 \frac{45.580}{\sqrt{9}}= 1225.408$

$1260.444+2.306 \frac{45.580}{\sqrt{9}}= 1295.480$

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