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3 years ago

11) If A = {19, 20, 21, 24} and B = {17, 19, 20, 22), list the elements of A U B.

Mathematics

1 answer:

Elodia [21]3 years ago

3 0

Answer:

A U B = {17, 19, 20, 21, 22, 24}

Step-by-step explanation:

The symbol U is the union symbol, which denotes the union of two sets.

This means that given two sets A and B, the expression

A U B

indicates all the elements that belong to either set A or set B.

In this problem, we have the following two sets:

A = {19, 20, 21, 24}

B = {17, 19, 20, 22}

We can therefore list all the elements belonging to either set A or set B, they are:

17, 19, 20, 21, 22, 24

So, the expression A U B indicates the set:

A U B = {17, 19, 20, 21, 22, 24}

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Solve by taking the square root of both sides.

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Answer:

Step-by-step explanation:

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5 × 1/ 2 A) More than 5 because you are multiplying by a whole number

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3 years ago

What is the image of (-8,6) after a translation over the line y=-x

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Answer:

(-6, 8)

Step-by-step explanation:

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3 years ago

Advertising expenses are a significant component of the cost of goods sold. Listed below is a frequency distribution showing the

Alexxandr [17]

Answer:

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

55 up to 65 16

65 up to 75 9

TOTAL 75

Let's find the probabilities first:

$P(25 - 35) = P \Big(\dfrac{25-50.93}{10.80}$

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

35-45 19 0.2218 75*0.2218 = 16.635 5.5932 0.3362

45-55 27 0.3568 75*0.3568 = 26.76 0.0576 0.021

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

65-75 9 0.0839 75*0.0839 = 6.2925 7.331 1.1650

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

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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3 years ago

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Answer:

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Step-by-step explanation:

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