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3 years ago

(5xy)^5 fully simplify

Mathematics

2 answers:

AfilCa [17]3 years ago

5 0

Answer:

Ok so i'm pretty sure its 3125x^5y^5

Step-by-step explanation:

BaLLatris [955]3 years ago

3 0

Answer:

5^3 xy

Step-by-step explanation:

i looked it up hehe

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If a cone is 6 inches tall and has a radius of 4 inches. What is its volume?

Doss [256]

Answer:

100.53

Step-by-step explanation:

Following the formula, V=πr2h 3, you would get 100.53 with 6 as the height and 4 as the radius.

5 0

3 years ago

A line has a slope of 3/5. It passes through (3,5) and (x,9). What is the value of x?

blagie [28]

This is in 2 point form
(y2-y1)=m(x2-x1)
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4 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

I need help on this question

rewona [7]

x = -7

hope this helps!

8 0

3 years ago

an amusement park charge 35.50 for admission.on saturday,6,789 people visited the park.about how much money did the park earn fr

sp2606 [1]

6,789
-
35.50
______
32.39

3 0

3 years ago

Read 2 more answers