How do i graph that equation

What is the square root of y^3

tia_tia [17]

Answer:

$y^{\frac{3}{2}$

Step-by-step explanation:

Let $m=\sqrt{y^3}$

$m=(y^3)^{\frac{1}{2}}\\\\m=y^{3\times \frac{1}{2}}\ \ \ \ \ \ \ \ \ [as\ (x^a)^b=x^{ab}]\\\\m=y^{\frac{3}{2}$

7 0

3 years ago

The point slope equation of a line is

EastWind [94]

Step-by-step explanation:

Point-slope is the general form y-y₁=m(x-x₁) for linear equations. It emphasizes the slope of the line and a point on the line (that is not the y-intercept).

3 0

3 years ago

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Help please! the question is in the picture below for reference when answering! thank you so much!

stepan [7]

Answer:

x = -1

Step-by-step explanation:

AB + BC = AC = 28

the way from A to C goes through B.

so, the total way of A to C is the sum of the way from A to B and then from B to C.

=>

(5x + 10) + (2x + 25) = 28

7x + 35 = 28

7x = -7

x = -1

4 0

3 years ago

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5270÷312 with remainder

aleksley [76]

This is the answer for the calculation

6 0

3 years ago

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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