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puteri [66]
3 years ago
12

Solve the system of equations by finding the reduced row-echelon form of the augmented matrix for the system of equations.

Mathematics
1 answer:
Anuta_ua [19.1K]3 years ago
3 0

Answer:

c

Step-by-step explanation:

First, we can transform this into a matrix. The x coefficients will be the first ones for each row, the y coefficients the second column, etc.

\left[\begin{array}{cccc}1&-2&3&-2\\6&2&2&-48\\1&4&3&-38\end{array}\right]

Next, we can define a reduced row echelon form matrix as follows:

With the leading entry being the first non zero number in the first row, the leading entry in each row must be 1. Next, there must only be 0s above and below the leading entry. After that, the leading entry of a row must be to the left of the leading entry of the next row. Finally, rows with all zeros should be at the bottom of the matrix.

Because there are 3 rows and we want to solve for 3 variables, making the desired matrix of form

\left[\begin{array}{ccc}1&0&0\\0&1&0\\0&0&1\end{array}\right] for the first three rows and columns. This would make the equation translate to

x= something

y= something

z = something, making it easy to solve for x, y, and z.

Going back to our matrix,

\left[\begin{array}{cccc}1&-2&3&-2\\6&2&2&-48\\1&4&3&-38\end{array}\right] ,

we can start by removing the nonzero values from the first column for rows 2 and 3 to reach the first column of the desired matrix. We can do this by multiplying the first row by -6 and adding it to the second row, as well as multiplying the first row by -1 and adding it to the third row. This results in

\left[\begin{array}{cccc}1&-2&3&-2\\0&14&-16&-36\\0&6&0&-36\end{array}\right]

as our matrix. * Next, we can reach the second column of our desired matrix by first multiplying the second row by (2/14) and adding it to the first row as well as multiplying the second row by (-6/14) and adding it to the third row. This eliminates the nonzero values from all rows in the second column except for the second row. This results in

\left[\begin{array}{cccc}1&0&10/14&-100/14\\0&14&-16&-36\\0&0&96/14&-288/14\end{array}\right]

After that, to reach the desired second column, we can divide the second row by 14, resulting in

\left[\begin{array}{cccc}1&0&10/14&-100/14\\0&1&-16/14&-36/14\\0&0&96/14&-288/14\end{array}\right]

Finally, to remove the zeros from all rows in the third column outside of the third row, we can multiply the third row by (16/96) and adding it to the second row as well as multiplying the third row by (-10/96) and adding it to the first row. This results in

\left[\begin{array}{cccc}1&0&0&-5\\0&1&0&-6\\0&0&96/14&-288/14\end{array}\right]

We can then divide the third row by -96/14 to reach the desired third column, making the reduced row echelon form of the matrix

\left[\begin{array}{cccc}1&0&0&-5\\0&1&0&-6\\0&0&1&-3\end{array}\right]

Therefore,

x=-5

y=-6

z=-3

* we could also switch the second and third rows here to make the process a little simpler

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We reject H₀ at 95 % of CI the group of Gen Xers who do not pay their credit card each month, is greater than this proportion for Millennians

Step-by-step explanation:

Sample proportion 1: Born between 1980 and 1996

sample size    n₁  = 450

p₁  =  0,48        p₁ = 48 %

x₁ = 0,48 * 450        x₁ = 216

Sample proportion 2:  Born between 1965 and 1971

sample size    n₂  = 300

p₂ =  0,60         p₂ = 60%

x₂  = 0,6* 300     x₂  = 180

Test Hypothesis:

Null Hypothesis                          H₀               p₂  =   p₁

Alternative Hypothesis              H                p₂  >  p₁

Samples both big enough to use the normal distribution as an approximation to the binomial distribution.

We assume  CI = 95 %   then significance level α   = 5 %  the alternative hypothesis tells us that we need to develop a one-tail test to the right.

z-score  is z(c)  for  α = 0,05       from z-table   z(c) = 1,64

To calculate z(s)

z(s)  =  ( p₂  -  p₁ ) / √ p*q ( 1 /n₁ + 1 / n₂)

p₂ - p₁  =  0,60  - 0,48

p₂ - p₁  =  0,12

p = ( x₁ + x₂ ) / n₁ + n₂

p = ( 216  + 300) / 450 + 300

p = 516 / 750     p = 0,688      then   q = 1 - p     q = 0,312

z(s) = 0,12 / √ 0,688*0,312* ( 1/450) + 1/ 300)

z(s) = 0,12 / √ 0,2146* ( 0,0022 + 0,0033)

z(s) = 0,12 / √ 0,001192

z(s) = 0,12/ 0,03452

z(s) = 3,47

Comparing |z(c)|   and |z(s)|

z(s) > z(c)

Then z(s) is in the rejection region. We reject H₀.

We can support the  claim that the proportion of the group Gen Xers who do not pay their credit card is greater than Millennians

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