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Murljashka [212]
3 years ago
14

PLEASE HELPSHSHSJAJSNDMFMMF

Mathematics
1 answer:
Lapatulllka [165]3 years ago
8 0
23) 460,000
24) 0.0092
25) 1.6*10^-2; 16% ; 6^-1 ; .2 ; 3/5; square root of 6
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Answer: A on Edge :)

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an ordered pair where the x and y coordinate are the same lies in the 1st or 3rd quadrant. true, sometimes true or not true?. ex
Zarrin [17]

The quadrants are as follows:

The 1st quadrant has the points which have both x and y positive and the 3rd quadrant has the points which have both x and y negative. If the ordered pair and the same x and y value, if ons is positive, the other also is, and the same for negative.

So, at first we see that there are point where the x and y are the same and that are in the 1st or 3rd quadrant.

However, there is one special case:

When x and y are 0, that is, the ordered pair is (0, 0).

Since this point is the origin, it doesn't lie on any of the quadrants.

Thus, this affirmative is sometimes true. Every point but (0, 0) that have same x and y values are in the 1st or 3rd quadrant except for (0, 0).

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2 years ago
Solve for x<br><br> x/4= 25<br><br> quick
storchak [24]

Answer:

x=100

Step-by-step explanation:

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3 years ago
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How many years is 694.44 days?
antoniya [11.8K]

Answer:

1.9 years

Step-by-step explanation:

There are 365 days in a year

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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
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