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3 years ago

What is 1/6 x 4? Please help

Mathematics

2 answers:

Tom [10]3 years ago

8 0

Answer:

2/3

Step-by-step explanation:

kobusy [5.1K]3 years ago

4 0

Answer: 2/3

Step-by-step explanation:

Hope this helps

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Elodia [21]

Answer:

It would be the 2nd one

Step-by-step explanation:

4 0

3 years ago

At a football championship game,the home team gave a football to each of the first 100 fans who arrived at the stadium. How much

Ilia_Sergeevich [38]

What grade level is this

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3 years ago

Read 2 more answers

Pablo mixes three units white paint and one unit of black paint to make gray paint. What two equations show the relationship bet

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2 years ago

Solve the system by graphing.

igor_vitrenko [27]

Answer:

Hi There the correct answer is {x,y} = {-1,-10}

System of Linear Equations entered :

[1] 3x - y = 7

[2] 4x - 2y = 16

Graphic Representation of the Equations :

y + 3x = 7 -2y + 4x = 16

Solve by Substitution :

// Solve equation [1] for the variable y

[1] y = 3x - 7

// Plug this in for variable y in equation [2]

[2] 4x - 2•(3x-7) = 16

[2] -2x = 2

// Solve equation [2] for the variable x

[2] 2x = - 2

[2] x = - 1

// By now we know this much :

x = -1

y = 3x-7

// Use the x value to solve for y

y = 3(-1)-7 = -10

Solution :

{x,y} = {-1,-10}

Hope it helps!

8 0

2 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