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3 years ago

Zahra wants the equation below to have an infinite number of solutions when the missing number is placed in the box.

Mathematics

2 answers:

zhannawk [14.2K]3 years ago

7 0

The answer is -3.

To find this, we simply have to solve the equation and leave only the blank and the parenthesis that it is with on the right side. Then we can divide by the parenthesis. The work for this is done for you below.

____(x - 3) + 2x = -(x - 5) + 4 ----> Distribute the -1 to the parenthesis.

____(x + 3) + 2x = -x + 5 + 4 ----> Simplify

____(x + 3) + 2x = -x + 9 ----> Subtract 2x from both sides

____(x + 3) = -3x + 9 ------> Divide both sides by x + 3

____ = -3.

erastovalidia [21]3 years ago

5 0

Let the missing number be 'y'

$= > y(x - 3) + 2x = - (x - 5) + 4 \\ \\ = > y(x - 3) + 2x= - x + 5 + 4 \\ \\ = > y(x - 3) = - x - 2x + 9 \\ \\ = > y(x - 3) = - 3x + 9 \\ \\ = > y = \frac{ - 3(x - 3)}{(x - 3)} \\ \\ = > y = - 3$

$\boxed{Missing \: number = -3}$

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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 $H_o$

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3 years ago

The figure shows three quadrilaterals on a coordinate grid:

makkiz [27]

Answer:

E and G are similar but not congruent.

Step-by-step explanation: I don't really know how to explain this but I am positive this is the answer.

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2 years ago

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erastova [34]

3y - 12 = -6
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y = 2 | Divide by 3 on both sides

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Answer:

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2 years ago

Which function pass through the points (1, 4), (2, 9), and (3, 16)?

swat32

ANSWER

$y= {(x+ 1)}^{2}$

EXPLANATION

The given points are:

(1, 4), (2, 9), and (3, 16)

It is obvious that the function is not linear because there is no constant difference among the y-values.

We can however manipulate the y-values to quickly identify the function.

$4 = {2}^{2} = {(1 + 1)}^{2}$

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We can infer from the pattern that, the function is;

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