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muminat
3 years ago
7

Rotate 90 clockwise around the origin (-6,-2) (-3,1) (-1,1) (2,-4) please help me!

Mathematics
1 answer:
N76 [4]3 years ago
5 0
The rule for a 90 clockwise rotation is basically an 180 degree rotation counterclockwise so (x,y) to (y,-x) so 
(-6,-2) to (-2,6)
(-3,1) to (1,3)
(-1,1) to (1,1)
(2,-4) to (-4,-2)
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Which equation represents a proportional relationship that has a constant of proportionality equal to 2?
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The table showing the stock price changes for a sample of 12 companies on a day is contained in the Excel file below.
AfilCa [17]

Answer:

(a) The sample variance for the daily price change is 0.2501.

(b) The sample standard deviation for the daily price change is 0.5001.

(c) The 95% confidence interval estimates of the population variance is (0.1255, 0.7210).

Step-by-step explanation:

Let the random variable <em>X</em>  denote the stock price changes for a sample of 12 companies on a day.

The data provided is:

<em>X</em> = {0.82 , 1.44 , -0.07 , 0.41 , 0.21 , 1.33 , 0.97 , 0.30 , 0.14 , 0.12 , 0.42 , 0.15}

(a)

The formula to compute the sample variance for the daily price change is:

s^{2}=\frac{1}{n-1}\sum\limits^{12}_{i=1}{(X_{i}-\bar X)^{2}}

The sample mean is computed using the formula:

\bar X=\frac{1}{n}\sum\limits^{12}_{i=1}{X_{i}}

Consider the Excel output attached below.

In Excel the formula to compute the sample mean and sample variance are:

\bar X =AVERAGE(A2:A13)

s^{2} =VAR.S(A2:A13)

Thus, the sample variance for the daily price change is 0.2501.

(b)

The formula to compute the sample standard deviation for the daily price change is:

s=\sqrt{\frac{1}{n-1}\sum\limits^{12}_{i=1}{(X_{i}-\bar X)^{2}}}

Consider the Excel output attached below.

In Excel the formula to compute the sample standard deviation is:

s =STDEV.S(A2:A13)

Thus, the sample standard deviation for the daily price change is 0.5001.

(c)

The (1 - <em>α</em>)% confidence interval for population variance is:

CI=[\frac{(n-1)s^{2}}{\chi^{2}_{\alpha/2} } \leq \sigma^{2}\leq \frac{(n-1)s^{2}}{\chi^{2}_{1-\alpha/2} } ]

Compute the critical value of Chi-square for <em>α</em> = 0.05 and (n - 1) = (12 - 1) = 11 degrees of freedom as follows:

\chi^{2}_{\alpha/2, (n-1)}=\chi^{2}_{0.05/2,11}=21.920

\chi^{2}_{1-\alpha/2, (n-1)}=\chi^{2}_{(1-0.05/2),11}=\chi^{2}_{0.975,11}=3.816

*Use a Chi-square table.

Compute the 95% confidence interval estimates of the population variance as follows:

CI=[\frac{(n-1)s^{2}}{\chi^{2}_{\alpha/2} } \leq \sigma^{2}\leq \frac{(n-1)s^{2}}{\chi^{2}_{1-\alpha/2} } ]

     =[\frac{(12-1)\times 0.2501}{21.920 } \leq \sigma^{2}\leq \frac{(12-1)\times 0.2501}{3.816} ]

     =[0.125506\leq \sigma^{2}\leq 0.720938]\\\approx [0.1255, 0.7210]

Thus, the 95% confidence interval estimates of the population variance is (0.1255, 0.7210).

7 0
4 years ago
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