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8090 [49]
3 years ago
8

Which point would not be a solution to the system of linear inequalities shown below?

Mathematics
1 answer:
Wittaler [7]3 years ago
6 0

Answer:

(5, 8)

Step-by-step explanation:

A solution to a system of linear inequalities is a point that, when its x and y values are substituted into both inequalities, makes <u>both</u> of them true. (Knowing this, if a point makes one inequality true but not the other, it is not a solution to that specific system of linear inequalities.)

So, let's plug in the x and y values of (5, 8) into the inequality y > 3x - 7 and solve like so:

(8) > 3(5)  - 7 \\8 > 15 - 7 \\8 > 8

However, 8 is equal to 8, not greater than it. So, we know that this point is not a solution to this inequality since it does not make it true, therefore (5, 8) is not a solution to the whole system.

Hope this helps!

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Se x^2+y^2=12 e xy=9, qual é o valor de (x+y)^2?
Dmitry_Shevchenko [17]
I hope this helps you

3 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
Can someone help me pls
Mandarinka [93]

Answer:

It would be 4/a

Step-by-step explanation:

3 0
3 years ago
I need help with these two questions
Kitty [74]
Question 1
The answer is last one. Y would equal 7+24 not 7-24.
3 0
3 years ago
Just give the answer plz 0_o
GalinKa [24]

Answer:

44.5

Step-by-step explanation:

5 0
3 years ago
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