8. Consider the following regressions: + B3X3 + u Y=Bo + B1X₁ + B₂X₂ Y = 10 + 1X₁ + 72(X2X1)+73X3 Y=00 +0₁X₁ +0₂X2 + u +03 (X3 + X₁ - 2X₂) + u. (a) Rewrite the hypothesis 1 = 0 in te

b = 6cos(100°) ≈ -1.94 (rounded to the nearest thousandth), A = 0° and C ≈ 170.697°.

c² = a² + b² - 2ab cos(C)

Substitute the given values:

a = 2c = 6B = 100°

We know that the sum of the angles of a triangle is 180°A + B + C = 180°

Solving for A:180° - 100° - A = 80°A = 180° - 100° - 80°A = 0°

Solving for C:c² = a² + b² - 2ab cos(C)6² = 2² + b² - 2(2)(b) cos(C)36 = 4 + b² - 4b cos(C)32 = b² - 4b cos(C)

Using the given values, we know that:

cos(B) = adjacent / hypotenuse

cos(100°) = b / 6b = 6cos(100°)b ≈ -1.94 (rounded to the nearest thousandth)

32 = b² - 4b cos(C)

32 = (-1.94)² - 4(-1.94) cos(C)32

≈ 4.108 + 7.76 cos(C)27.892

≈ 7.76 cos(C)cos(C)

≈ 3.589 (rounded to the nearest thousandth)A and B are 0° and 100° respectively, while C is approximately 170.697°.

Therefore, b = 6cos(100°) ≈ -1.94 (rounded to the nearest thousandth), A = 0° and C ≈ 170.697°.

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The vectors v1 = (1, c, 0), v2 = (1, 0, 4), and v3 = (0, 1, -c) are linearly independent for all values of c except when c = 0.

To determine the values of c for which the vectors v1 = (1, c, 0), v2 = (1, 0, 4), and v3 = (0, 1, -c) are linearly independent, we need to check if there exists a non-trivial solution to the equation:

a1v1 + a2v2 + a3*v3 = 0

where a1, a2, and a3 are scalars, not all equal to zero.

Substituting the vectors into the equation, we have:

a1*(1, c, 0) + a2*(1, 0, 4) + a3*(0, 1, -c) = (0, 0, 0)

Expanding this equation component-wise, we get:

(a1 + a2, a1, 4a2 - a3c) = (0, 0, 0)

From the first component, we have a1 + a2 = 0, which implies a1 = -a2.

From the second component, we have a1 = 0, which means a1 must be zero.

Combining these two conditions, we find a1 = a2 = 0.

Now, let's look at the third component:

4a2 - a3c = 0

Since a2 = 0, the equation simplifies to -a3*c = 0.

This equation holds for any value of c as long as a3 = 0. Therefore, we conclude that the vectors v1, v2, and v3 are linearly independent for all values of c except when c = 0.

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The estimated regression equation for predicting the closing price based on the value of Period is Price = 82.114 + 0.392 * Period. The Durbin-Watson statistic for testing positive autocorrelation in the data is 1.174, indicating the presence of positive autocorrelation.

a. The estimated regression equation for predicting the closing price based on the value of Period is: Price = 82.114 + 0.392 * Period. This equation suggests that for each increase in Period by 1 unit, the predicted closing price is expected to increase by 0.392 dollars per share.

b. To test for positive autocorrelation in the data, we can use the Durbin-Watson statistic. The Durbin-Watson statistic measures the presence of autocorrelation in the residuals of a regression model. The critical values for the Durbin-Watson statistic at the 0.05 level of significance are typically between 1.5 and 2.5.

We have that the calculated Durbin-Watson statistic is 1.174, which is less than the lower critical value of 1.5, it suggests the presence of positive autocorrelation in the data.

Positive autocorrelation indicates that there is a systematic relationship between the residuals of the regression model, meaning that the residuals are not independently distributed.

This finding of positive autocorrelation in the data implies that the current closing prices are influenced by the past closing prices, and there may be some time-dependent pattern in the data that is not captured by the regression model.

It is important to account for this autocorrelation when interpreting the results and making predictions based on the regression equation.

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Option D) .10, .30, .50 accurately represents Cohen's guidelines for interpreting the strength of a correlation coefficient.

Cohen's guidelines for interpreting the strength of a correlation coefficient, typically denoted as r, categorize small, medium, and large values based on the magnitude of the correlation. These guidelines help assess the strength and practical significance of the relationship between variables. Let's examine the options provided:

A) .46, .67, .84

B) .10, .20, .30

C) .00, .50, 1.00

D) .10, .30, .50

Among these options, the correct choice is D) .10, .30, .50.

According to Cohen's guidelines, small, medium, and large values of r correspond to approximately:

Small: r = 0.10

Medium: r = 0.30

Large: r = 0.50

Therefore, option D) .10, .30, .50 accurately represents Cohen's guidelines for interpreting the strength of a correlation coefficient.

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The correct equation of the circle with center B(4,-6) and radius 7 is:

b. [tex](X-4)^2 + (Y-6)^2 = 49[/tex]

This equation represents a circle centered at (4,-6) with a radius of 7. The general equation for a circle with center (h,k) and radius r is given by [tex](X-h)^2 + (Y-k)^2 = r^2[/tex]. By comparing the given information with the general equation, we can see that the correct answer is option b.

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The chi-square test statistic has a p-value of 0.0007, which is less than the significance level of 0.05, indicating that we have sufficient evidence to reject the null hypothesis and accept the alternative hypothesis that there is a statistically significant relationship between grades and involvement in extracurricular activities and grades.

Chi-square test of independence is used to check whether there is a significant association between two or more categorical variables. This test is carried out using a contingency table which presents data for two or more categorical variables in the form of frequency counts. In our case, we are checking whether there is an association between grades and extracurricular activities.To carry out the chi-square test of independence, we start by stating the null and alternative hypotheses. The null hypothesis states that there is no relationship between the two variables, whereas the alternative hypothesis states that there is a relationship between the two variables.The next step is to determine the expected frequencies for each cell in the contingency table. Expected frequencies are calculated under the assumption that there is no association between the two variables.

Then, we calculate the chi-square test statistic, which measures how much the observed frequencies deviate from the expected frequencies.Finally, we compare the calculated chi-square value with the critical value from the chi-square distribution. If the calculated chi-square value is greater than the critical value, we reject the null hypothesis and accept the alternative hypothesis. On the other hand, if the calculated chi-square value is less than the critical value, we fail to reject the null hypothesis.The results of the chi-square test are presented in the table given in the question.

The p-value of the test is 0.0007, which is less than the significance level of 0.05. Therefore, we have sufficient evidence to reject the null hypothesis and accept the alternative hypothesis that there is a statistically significant relationship between grades and involvement in extracurricular activities. The observed count appears above the expected count in each cell of the table. Thus, the conclusion is that there is a statistically significant association between involvement in extracurricular activities and grades.

Therefore, the chi-square test conducted on the given data concludes that there is a statistically significant association between involvement in extracurricular activities and grades.

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