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3 years ago

The ordered pair below is a solution to which equation? (2, 12) A. y = 6x B. y=1/2x C. y = 2x D. y=1/6x

2 answers:

7 0

We can substitute the values of x and y to see which equation is satisfied.

A. 12 = 6(2)
12 = 12, true

B. 12 = 1/2 (2)
12 ≠ 1, false

C. 12 = 2(2)
12 ≠ 4, false

D. 12 = 1/6 (2)
12 ≠ 1/3, false

The answer is A.

Illusion [34]3 years ago

3 0

Substitute the x for 2 and the y for 12 in each equation. The first choice works out. (12)=6(2), 12=12. This means the answer is true. A is your answer.

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For the following linear system, put the augmented coefficient matrix into reduced row-echelon form.

Anni [7]

Answer:

The reduced row-echelon form of the linear system is $\left[\begin{array}{cccc}1&0&-5&0\\0&1&3&0\\0&0&0&1\end{array}\right]$

Step-by-step explanation:

We will solve the original system of linear equations by performing a sequence of the following elementary row operations on the augmented matrix:

Interchange two rows
Multiply one row by a nonzero number
Add a multiple of one row to a different row

To find the reduced row-echelon form of this augmented matrix

$\left[\begin{array}{cccc}2&3&-1&14\\1&2&1&4\\5&9&2&7\end{array}\right]$

You need to follow these steps:

Divide row 1 by 2 $\left(R_1=\frac{R_1}{2}\right)$

$\left[\begin{array}{cccc}1&3/2&-1/2&7\\1&2&1&4\\5&9&2&7\end{array}\right]$

Subtract row 1 from row 2 $\left(R_2=R_2-R_1\right)$

$\left[\begin{array}{cccc}1&3/2&-1/2&7\\0&1/2&3/2&-3\\5&9&2&7\end{array}\right]$

Subtract row 1 multiplied by 5 from row 3 $\left(R_3=R_3-\left(5\right)R_1\right)$

$\left[\begin{array}{cccc}1&3/2&-1/2&7\\0&1/2&3/2&-3\\0&3/9&9/2&-28\end{array}\right]$

Subtract row 2 multiplied by 3 from row 1 $\left(R_1=R_1-\left(3\right)R_2\right)$

$\left[\begin{array}{cccc}1&0&-5&16\\0&1/2&3/2&-3\\0&3/9&9/2&-28\end{array}\right]$

Subtract row 2 multiplied by 3 from row 3 $\left(R_3=R_3-\left(3\right)R_2\right)$

$\left[\begin{array}{cccc}1&0&-5&16\\0&1/2&3/2&-3\\0&0&0&-19\end{array}\right]$

Multiply row 2 by 2 $\left(R_2=\left(2\right)R_2\right)$

$\left[\begin{array}{cccc}1&0&-5&16\\0&2&3&-6\\0&0&0&-19\end{array}\right]$

Divide row 3 by −19 $\left(R_3=\frac{R_3}{-19}\right)$

$\left[\begin{array}{cccc}1&0&-5&16\\0&2&3&-6\\0&0&0&1\end{array}\right]$

Subtract row 3 multiplied by 16 from row 1 $\left(R_1=R_1-\left(16\right)R_3\right)$

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

Add row 3 multiplied by 6 to row 2 $\left(R_2=R_2+\left(6\right)R_3\right)$

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

8 0

3 years ago

What is 6 1/2 + 10? please tell me and explain for me to understand

ipn [44]

Answer:

33/2

Step-by-step explanation:

6 1/2 ---> 13/2

10 ---> 20/2

13/2 + 20/2 =

33/2

5 0

3 years ago

A researcher wants to determine whether the number of minutes adults spend online per day is related to gender. A random sample

suter [353]

Answer:

The expected frequency for the cell E2,2 is = 33

Step-by-step explanation:

The given data is

Gender | 0-30| 30-60| 60-90| 90+ Totals

Male | 23 | 35 | 76 | 46 180

Female | 31 | 42 | 46 | 16 135

Totals 54 77 122 62 315

The expected frequency for the cell E2,2 is :

Expected for the (30-60 box for females) = Total of (30-6)/ (total )* females

= ( 77/315)135= 33

Here p= 77/315 and n= 135

therefor X= pn = 33

χ²= (33-42)²/33= 2.455 ( for the single value of E2,2=33)

Expected for the (30-60 box for males) = Total of (30-6)/ (total )* males

= ( 77/315)180= 44

χ²= (44-42)²/44= 0.09

The critical region is χ² (0.05) 3 ≥ χ²= 11.34

Let the null and alternate hypothesis be

H0:the number of minutes spent online per day is not related to gender

against

Ha: the number of minutes spent online per day is related to gender

The single value of χ² for E2,2 = 2.45 is less than the critical value of 11.34.

6 0

3 years ago

Is the value of 8 in 345,802 ten times greater or smaller than the value of 8 in 758,910

german

Answer:

Yes

Step-by-step explanation:

One is in the hundreds and one is in the thousands. 100x10=1000

7 0

2 years ago

Please help me. I am stuck.

dalvyx [7]

C is greater than or equal to $40 (which means put the > sign with a line under it).

4 0

2 years ago