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2 years ago

Find the missing value. Hint: Use the number line to find the missing value. Helppp

Mathematics

2 answers:

REY [17]2 years ago

7 0

Answer:

-2

Step-by-step explanation:

the multiplication of -×- will be + so if-×-will be plus then -7 after multiplying will be + so

_2+7=5

stira [4]2 years ago

7 0

Answer is 12! Gl on what ur doing!

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frozen [14]

Answer:

A y=2x + 3

Step-by-step explanation:

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3 years ago

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What are the different types of triangles ?

sp2606 [1]

Right-angle triangle

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3 years ago

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Solve pls brainlies

MrRa [10]

Answer:

2) 11

3) 21

4) 7

Step-by-step explanation:

2) (-6) -(-17) = 17 - 6 = 11

3) 19 - (-2) = 19 + 2 = 21

4) (-13) + 20 = 20 - 13 = 7

6 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

I need help so much please help the attachment below is the question i need help on

leonid [27]

Answer:

$\dfrac{\sqrt[12]{55296}}{2}$

Step-by-step explanation:

Rationalize the denominator, then use a common root for the numerator.

$\dfrac{\sqrt[4]{6}}{\sqrt[3]{2}}=\dfrac{(2\cdot 3)^{\frac{1}{4}}}{2^{\frac{1}{3}}}\\\\=\dfrac{(2\cdot 3)^{\frac{1}{4}}}{2^{\frac{1}{3}}}\cdot\dfrac{2^{\frac{2}{3}}}{2^{\frac{2}{3}}}=\dfrac{2^{\frac{1}{4}+\frac{2}{3}}3^{\frac{1}{4}}}{2}\\\\=\dfrac{2^{\frac{11}{12}}3^{\frac{3}{12}}}{2}=\dfrac{\sqrt[12]{2^{11}3^{3}}}{2}\\\\=\dfrac{\sqrt[12]{55296}}{2}$

3 0

3 years ago