Can you help me with this thank you

Sorry for botheribg u guys again but please help

Nostrana [21]

Answer:

With what though?

Step-by-step explanation:

its np

6 0

3 years ago

If you know this please tell me all the answers

sp2606 [1]

1. 3n
2. 9x
3. 9wx
4. 5ac^2
5. 8wx^2
6. 4a^2 c

7 0

3 years ago

A triangular prism container is full of water of 428 cubic cm. The water is 10 cm deep, what is the base area of the container?

Hatshy [7]

<h3>The base area of triangular prism container is 42.8 cubic centimeter</h3>

Solution:

The volume of triangular prism is given as:

$v = base\ area \times height$

Given that,

A triangular prism container is full of water of 428 cubic cm

The water is 10 cm deep

Therefore,

v = 428 cubic cm

h = 10 cm

Substituting the values we get,

$428 = base\ area \times 10\\\\base\ area = \frac{428}{10}\\\\base\ area = 42.8\ cm^2$

Thus the base area of triangular prism container is 42.8 cubic centimeter

3 0

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 $H_o$

6 0

3 years ago

The graph shown is the graph of which function?

alukav5142 [94]

Where the pictures of the graph

5 0

3 years ago