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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Suppose that we randomly sample 122 high school students to investigate an association between involvement in extracurricular activities and grades. We use the data to conduct a chi-square test of independence at the 5% level. In this results table, the observed count appears above the expected count in each cell. The expected counts are in parenthesis. What can we conclude? Good grades Poor grades Total Extracurricular activity What can we conclude? 16 11 (11.016) 5 (4.9836) Low a) There is a statistically significant association between involvement in extracurricular activities and grades. 93 70 (64.033) 23 (28.967) Moderate 13 10 (4,0492) b) There is not a statistically significant association between involvement in extracurricular activities and grades. 3 (8.9508) 84 High 38 122 Total c) Involvement in extracurricular activities helps students to perform better academically. d) Nothing, because the conditions for use of the chi-square test are not met. Statistic DF Value P-value Chi-square 2 14.487 0.0007
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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find all values of c for which the vectors v1 = 1, c, 0 , v2 = 1, 0, 4 , and v3 = 0, 1, −c are linearly independent
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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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
If all three models produce straight lines when graphed, they are linear models
Rewrite hypothesis 1 = 0 in terms of the other two models.
Hypothesis 1 in the first model is:B3 = 0.
Hypothesis 1 in the second model is: B2 = 0.
Hypothesis 1 in the third model is: B1 + 2B2 + 3B3 = 0.
All three models are linear in the parameters because each explanatory variable has a coefficient attached to it. Linear regression is the process of fitting a linear equation to a given set of data points.
This equation can be used to predict the value of the dependent variable (Y) based on the value of the independent variable (X).
A linear equation is an equation that produces a straight line when graphed.
Therefore, if all three models produce straight lines when graphed, they are linear models
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Identify the equation of the circle B with center B(4,−6) and radius 7. a. (X – 4)^2 + (Y-6)^2 = 49
b. (X-4)^2 + (Y-6)^2 = 49
c. (X-4)^2 + (Y-6)^2 = 7
d. (X-4)^2 + (Y-6)^2 = 7
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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