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olasank [31]
3 years ago
15

Which statements are true the ordered pair (1, 2) and the system of equations? y=−2x+47x−2y=3 HELP ASAP! WILL GIVE BRAINLIEST

Mathematics
2 answers:
const2013 [10]3 years ago
6 0
When (1,2) is substituted into the first equation, the equation is true.
The ordered pair (1,2) is a solution to the system of linear equations.
When (1,2) is substituted into the second equation, the equation is true.

Mkey [24]3 years ago
6 0
To solve, plug the value of the variables in each equation in the system

(1,2) means x = 1 and y = 2

y = -2x + 4

2 = -2(1) + 4

2 = -2 + 4

2 = 2

the ordered pair is a solution to the first equation

7x - 2y = 3

7(1) - 2(2) = 3

7 - 4 = 3

3 = 3

the ordered pair is a solution for the second equation

Because the ordered pair is a solution for BOTH equations in the system, it is a solution to the system. Therefore, the correct answers are:

"When (1,2) is substituted into the first equation, the equation is true."

"The ordered pair (1,2) is a solution to the system of linear equations."

"When (1,2) is substituted into the second equation, the equation is true."

Hope this helps!
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A randomized controlled study was designed to test whether regular drinking of cranberry juice can prevent the recurrence of uri
Kamila [148]

Answer:

1. The chi-squared statistic = 10.36

The degrees of freedom = 17

The p-value for the test = 0.89

2. The range of the p-value from the Chi squared table = 0.75 < p-value < 0.90

Step-by-step explanation:

1. The Chi squared test is given as follows;

\chi ^{2} = \sum \dfrac{\left (Observed - Expected  \right )^{2}}{Expected  }

Therefore,

                              UTI   No UTI    %     Total

Cranberry juice       8           42      84     50

Lactobacillus          19          30       61     49

Control                    18          30      60    50

The chi-squared statistic is given as follows;

\chi ^{2} = \dfrac{\left (8- 18\right )^{2}}{18} +  \dfrac{\left (42 - 30\right )^{2}}{30} = 10.36

The chi-squared statistic = 10.36

The degrees of freedom, df = 18 - 1 = 17 since the all of the expected count have a minimum value of 18

With the aid of the calculator we find the p value as p as follows;

p = 0.9 - \dfrac{10.36 - 10.085}{12.972 - 10.085} \times (0.9 - 0.75)

The p-value for the test = 0.89  

2. The range of the p-value from the Chi squared table is given as follows;

0.75 < p-value < 0.90.

5 0
3 years ago
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