Proof that the identities are equal to eachother

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 $H_o$

6 0

3 years ago

4+2(8²×4²)+6+4+6+6+6+2+2

mash [69]

Answer: =2084

4+2(82)(42)+6+4+6+6+6+2+2

=4+264(42)+6+4+6+6+6+2+2

=4+2(64)(16)+6+4+6+6+6+2+2

=4+(2)(1024)+6+4+6+6+6+2+2

=4+2048+6+4+6+6+6+2+2

=2052+6+4+6+6+6+2+2

=2058+4+6+6+6+2+2

=2062+6+6+6+2+2

=2068+6+6+2+2

=2074+6+2+2

=2080+2+2

=2082+2

=2084

4 0

4 years ago

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Simplify the expression given below.<br> (732 - 83 + 21) + (-1363 - 11.2 + 133 + 19)

Vera_Pavlovna [14]

Answer:

732−83+21−1363−11.2+133+19= −552.2

8 0

3 years ago

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112 is 70% of _______

aliina [53]

Answer:

160

Step-by-step explanation:

70% of 160=112.

A handy trick is to divide the number by the percentage.

Brainliest please I need those points.

8 0

3 years ago

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A spinner is divided into eight equal-sized sections, numbered 1 to 8, inclusive. What is true about spinning the spinner one at

zvonat [6]

Answer with explanation:

The number of section into which Spinner is divided =Number from 1 to 8={1,2,3,4,5,6,7,8}

Now, Spinner is Spun once .

The Correct statement among the given statements are:

→1. S={1,2,3,4,5,6,7,8}

→2. If A is a subset of S, A could be {1,2,3} , because , {1,2,3}⊂{1,2,3,4,5,6,7,8}.

→5.If a subset A represents the complement of spinning an odd number, it’s sample space is A={2,,4,6,8}

A=odd number={1,3,5,7}

A's Complement=Even Number={2,4,6,8}

Why, not 3 and 4, because, The number of elements in the set is from 1 to 8, but the subset has three elements, 7,8, and 9.Element, 7 ∈ S, 8∈S, but , 9∈S.

Also, number less than 4, is equal to ={1,2,3}, but in the statement, it is given that,number less than 4, it’s sample space is A={1,2,3,4},which is Incorrect.

7 0

3 years ago

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Proof that the identities are equal to eachother​

Proof that the identities are equal to eachother