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4 years ago

1. A square ABCD has the vertices A(n, n), B(n, -n), C(-n, -n), and D(-n, n). Which vertex is in Quadrant III?

Mathematics

2 answers:

andrew-mc [135]4 years ago

5 0

The four quadrants and the sign of coordinates is given as follows:

Quadrant I : (n, n ) both x and y positive.

Quadrant II : ( -n,n) = x negative and y positive.

Quadrant III : ( -n,-n) = both x and y negative.

Quadrant IV: (n, -n) = x positive and y negative.

So Based on the above concept , we can say that vertex C (-n,-n) has both x and y negative and so it lies in quadrant III.

Answer is Vertex C (-n,-n)

Elden [556K]4 years ago

5 0

The four quadrants and the sign of coordinates is given as follows:

Quadrant I : (n, n ) both x and y positive.

Quadrant II : ( -n,n) = x negative and y positive.

Quadrant III : ( -n,-n) = both x and y negative.

Quadrant IV: (n, -n) = x positive and y negative.

So Based on the above concept , we can say that vertex C (-n,-n) has both x and y negative and so it lies in quadrant III.

Answer is Vertex C (-n,-n)

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Step-by-step explanation:

The table can be computed as:

Advertising Expenses ($ million) Number of companies

25 up to 35 4

35 up to 45 19

45 up to 55 27

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Let's find the probabilities first:

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For 35 up to 45

$P(35 - 45) = P \Big(\dfrac{35-50.93}{10.80}$

For 45 up to 55

$P(45 - 55) = P \Big(\dfrac{45-50.93}{10.80}$

For 55 up to 65

$P(55 - 65) = P \Big(\dfrac{55-50.93}{10.80}$

For 65 up to 75

$P(65 - 75) = P \Big(\dfrac{65-50.93}{10.80}$

Chi-Square Table can be computed as follows:

Expense No of Probabilities(P) Expe $(O-E)^2$ $\dfrac{(O-E)^2}{E}$

compa cted E (n*p)

nies (O)

25-35 4 0.0612 75*0.0612 = 4.59 0.3481 0.0758

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55-65 16 0.2552 75*0.2552 = 19.14 9.8596 0.5151

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$\sum \dfrac{(O-E)^2}{E}= 2.0492$

Using the Chi-square formula:

$X^2 = \dfrac{(O-E)^2}{E} \\ \\ Chi-square \ X^2 = 2.0942$

Null hypothesis:

$H_o: \text{The population of advertising expenses follows a normal distribution}$

Alternative hypothesis:

$H_a: \text{The population of advertising expenses does not follows a normal distribution}$

Assume that:

$\alpha = 0.02$

degree of freedom:

= n-1

= 5 -1

= 4

Critical value from $X^2 = 11.667$

Decision rule: To reject $H_o \ if \ X^2$ test statistics is greater than $X^2$ tabulated.

Conclusion: Since $X^2 = 2.0942$ is less than critical value 11.667. Then we fail to reject $H_o$

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