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3 years ago

1. How do the trend lines and how well those represent the data

Mathematics

2 answers:

m_a_m_a [10]3 years ago

5 0

Answer:

every year, there is a steady increase in the frequency the fires occur. it started off with less than 250 fires per year when this graphs info first started, and by the end, it had increased dramatically with more frequent fires

pshichka [43]3 years ago

3 0

Every year and there is an increase

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Please please please help me with this answer

Rasek [7]

x = number of packages he bought

y = number of popsicles in each package

Here are the two equations:

xy = 54

x = 3 + y

Using the substitution method, you can solve them to find x and y.

(3 + y)y = 54,

3y + y^2 = 54

y^2 + 3y - 54 = 0

Solve the quadriatic equation:

(y +9)(y - 6) = 0

y = -9, 6.

Plug y into: xy = 54

x(-9) = 54, x = -6.

x(6) = 54, x = 9.

So he bought 9 packages of popsicles, bc you can't bye -6 packages.

3 0

4 years ago

Matteo is following this recipe to make a cake.

stepan [7]

Answer:

6×3+5×3+7×3+3×3+2×3=69

5 0

3 years ago

Lindsay and Menon have 1240 stickers.Menon has 4 times as many stickers as Lindsay.How many stickers does Menon have.

Nezavi [6.7K]

I do believethe answer would e 4960 stickers is what Menon would have.

8 0

3 years ago

Read 2 more answers

Solve for x x- 2x (12 - 1/2) =2 (4-2x) +20 A.-14/9 B.-9/14 C.14/9 D.9/14

MrMuchimi

Answer:

Step-by-step explanation:

-22x=-4x+28

-18x=28

x=-28/18=-14/9

7 0

3 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