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3 years ago

Solve the following System of Three Equations: x−3y+z=−15 2x+y−z=−2 x+y+2z=1

Mathematics

1 answer:

jeyben [28]3 years ago

6 0

Answer:

x = -3 , y = 4 , z = 0

Step-by-step explanation:

Solve the following system:

{x - 3 y + z = -15

2 x + y - z = -2

x + y + 2 z = 1

Hint: | Choose an equation and a variable to solve for.

In the first equation, look to solve for z:

{x - 3 y + z = -15

2 x + y - z = -2

x + y + 2 z = 1

Hint: | Solve for z.

Subtract x - 3 y from both sides:

{z = 3 y + (-x - 15)

2 x + y - z = -2

x + y + 2 z = 1

Hint: | Perform a substitution.

Substitute z = -15 - x + 3 y into the second and third equations:

{z = -15 - x + 3 y

15 + 3 x - 2 y = -2

x + y + 2 (-15 - x + 3 y) = 1

Hint: | Expand the left hand side of the equation x + y + 2 (-15 - x + 3 y) = 1.

x + y + 2 (-15 - x + 3 y) = x + y + (-30 - 2 x + 6 y) = -30 - x + 7 y:

{z = -15 - x + 3 y

15 + 3 x - 2 y = -2

-30 - x + 7 y = 1

Hint: | Choose an equation and a variable to solve for.

In the second equation, look to solve for x:

{z = -15 - x + 3 y

15 + 3 x - 2 y = -2

-30 - x + 7 y = 1

Hint: | Isolate terms with x to the left hand side.

Subtract 15 - 2 y from both sides:

{z = -15 - x + 3 y

3 x = 2 y - 17

-30 - x + 7 y = 1

Hint: | Solve for x.

Divide both sides by 3:

{z = -15 - x + 3 y

x = (2 y)/3 - 17/3

-30 - x + 7 y = 1

Hint: | Perform a substitution.

Substitute x = (2 y)/3 - 17/3 into the third equation:

{z = -15 - x + 3 y

x = (2 y)/3 - 17/3

(19 y)/3 - 73/3 = 1

Hint: | Choose an equation and a variable to solve for.

In the third equation, look to solve for y:

{z = -15 - x + 3 y

x = (2 y)/3 - 17/3

(19 y)/3 - 73/3 = 1

Hint: | Isolate terms with y to the left hand side.

Add 73/3 to both sides:

{z = -15 - x + 3 y

x = (2 y)/3 - 17/3

(19 y)/3 = 76/3

Hint: | Solve for y.

Multiply both sides by 3/19:

{z = -15 - x + 3 y

x = (2 y)/3 - 17/3

y = 4

Hint: | Perform a back substitution.

Substitute y = 4 into the first and second equations:

{z = -x - 3

x = -3

y = 4

Hint: | Perform a back substitution.

Substitute x = -3 into the first equation:

{z = 0

x = -3

y = 4

Hint: | Sort results.

Collect results in alphabetical order:

Answer: {x = -3 , y = 4 , z = 0

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Answer:

1) H0: There is independence between the marital status and the diagnostic of alcoholic

H1: There is association between the marital status and the diagnostic of alcoholic

2) The statistic to check the hypothesis is given by:

$\sum_{i=1}^n \frac{(O_i -E_i)^2}{E_i}$

3) $\chi^2 = \frac{(21-33.143)^2}{33.143}+\frac{(37-41.429)^2}{41.429}+\frac{(58-41.429)^2}{41.429}+\frac{(59-46.857)^2}{46.857}+\frac{(63-58.571)^2}{58.571}+\frac{(42-58.571)^2}{58.571} =19.72$

4) $df=(rows-1)(cols-1)=(3-1)(2-1)=2$

And we can calculate the p value given by:

$p_v = P(\chi^2_{2} >19.72)=5.22x10^{-5}$

And we can find the p value using the following excel code:

"=1-CHISQ.DIST(19.72,2,TRUE)"

Since the p value is lower than the significance level so then we can reject the null hypothesis at 5% of significance, and we can conclude that we have association between the two variables analyzed.

Step-by-step explanation:

A chi-square goodness of fit test "determines if a sample data matches a population".

A chi-square test for independence "compares two variables in a contingency table to see if they are related. In a more general sense, it tests to see whether distributions of categorical variables differ from each another".

Assume the following dataset:

Diag. Alcoholic Undiagnosed Alcoholic Not alcoholic Total

Married 21 37 58 116

Not Married 59 63 42 164

Total 80 100 100 280

Part 1

We need to conduct a chi square test in order to check the following hypothesis:

H0: There is independence between the marital status and the diagnostic of alcoholic

H1: There is association between the marital status and the diagnostic of alcoholic

The level os significance assumed for this case is $\alpha=0.05$

Part 2

The statistic to check the hypothesis is given by:

$\sum_{i=1}^n \frac{(O_i -E_i)^2}{E_i}$

Part 3

The table given represent the observed values, we just need to calculate the expected values with the following formula $E_i = \frac{total col * total row}{grand total}$

And the calculations are given by:

$E_{1} =\frac{80*116}{280}=33.143$

$E_{2} =\frac{100*116}{280}=41.429$

$E_{3} =\frac{100*116}{280}=41.429$

$E_{4} =\frac{80*164}{280}=46.857$

$E_{5} =\frac{100*164}{280}=58.571$

$E_{6} =\frac{100*164}{280}=58.571$

And the expected values are given by:

Diag. Alcoholic Undiagnosed Alcoholic Not alcoholic Total

Married 33.143 41.429 41.429 116

Not Married 46.857 58.571 58.571 164

Total 80 100 100 280

And now we can calculate the statistic:

$\chi^2 = \frac{(21-33.143)^2}{33.143}+\frac{(37-41.429)^2}{41.429}+\frac{(58-41.429)^2}{41.429}+\frac{(59-46.857)^2}{46.857}+\frac{(63-58.571)^2}{58.571}+\frac{(42-58.571)^2}{58.571} =19.72$

Part 4

Now we can calculate the degrees of freedom for the statistic given by:

$df=(rows-1)(cols-1)=(3-1)(2-1)=2$

And we can calculate the p value given by:

$p_v = P(\chi^2_{2} >19.72)=5.22x10^{-5}$

And we can find the p value using the following excel code:

"=1-CHISQ.DIST(19.72,2,TRUE)"

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<h3>What exactly is an equation?</h3>

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brainly.com/question/2972832

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