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3 years ago

What is 5m-7=5(m-2)+3

Mathematics

1 answer:

Ymorist [56]3 years ago

4 0

Answer: follow this!

Step-by-step explanation:

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4(n+2+2n+5+n+6)=-32 How do you show the problem?

yarga [219]

Answer:

n = -21/4 <em>or </em>-5.25

Step-by-step explanation:

4 0

3 years ago

Read 2 more answers

Nico wants to use a horizontal number line to compare 1 3/4

e-lub [12.9K]

Answer:

1 3/4 is greater than 1.5.

Step-by-step explanation:

1. To make a line that measures 1 3/4, make 2 regular lines.

2. Divide one of the lines into 4 equal sections and erase the one section and leaving you with 1 3/4.

To make the 1.5:

1. Make a line and make another line but that is half of the length of the previous line.

6 0

4 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

18+6/-4-16 HELP only d part

Ivahew [28]

Answer:

$\frac{1}{2}$ is the answer

Step-by-step explanation:

First we need to substitute X and Y with their respective numbers

$3\cdot \:6+\frac{6}{-4}-\left(-4\right)^2$

Then, following PEMDAS, We just need to simplify

$18+\frac{6}{-4} -16$

$16.5-16=0.5=\frac{1}{2}$

3 0

3 years ago

Read 2 more answers

Maisie bought a house.

sergeinik [125]

Answer:

55.52+10=65.52%

65.52 divide 3=21.84%

Ans:21.84%

7 0

3 years ago