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BartSMP [9]
3 years ago
5

Please answer correctly !!!! Will mark brainliest !!!!!!!!!!!!!!

Mathematics
2 answers:
alukav5142 [94]3 years ago
6 0

Answer:

A, C

Step-by-step explanation:

10x+ 4y = -2

5x -2y = 2

We can multiply the first equation by 1/2

5x+2y = -1

And then add the equations to eliminate y

5x+2y = -1

5x -2y = 2

--------------------

10x +0y = 1

Adding the equations will not eliminate any variables

15x +2y =0

Multiply the bottom equation by 2 and then subtract

10x -4y = 4

10x+ 4y = -2

-10x +4y = -4

------------------

0x +8y = -6  will eliminate x

ICE Princess25 [194]3 years ago
3 0

Answer:

Answer A, C

Step-by-step explanation:

When multiplying the bottom equation by two, you get 10x-4y=4

After subtracting 10x-4y=4 and the top equation you will be able to eliminate the variable x.

You can also also get the same result and eliminate a variable with answer A.

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A counterexample is something that proves the statement false.

January is a month that starts with J that is not a summer month.

That proves the statement false

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The function f(x)=3x+12x+11 can be written in vertex form as?
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 y = -3(x<span> - 2)^2 + 1 </span>x<span>-coordinate of vertex: </span>x<span> = -b/(2a) = -12/-6 = 2 y-coordintae of vertex: y(2) = -12 + 24 - 11 = 1 </span>Vertex form: y = -3(x<span> - 2)^2 + 1 Check. Develop y to get back to standard form: y = -3(</span>x^2 - 4x + 4) + 1 = -3x<span>^2 + </span>12x<span> - </span>11<span>. </span>
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a car rental company charges $34 per day for a rented car and $0.50 for every mile driven. A second car rental company charges $
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Answer:

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8/10 lies _____ on number line.
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8/10 = 0.8

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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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