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3 years ago

Helppp i dont get it

Mathematics

1 answer:

sasho [114]3 years ago

6 0

Answer: Your answer will be 8 s = (8,3) but since there isn’t no 3 for your answers choice then it should be 8! Hope this help!

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Find the set of solutions for the linear system.

Illusion [34]

Answer:

The system has infinitely many solutions.

$\left\begin{array}{ccc}x_1&=&-\frac{1}{3}x_2-\frac{16}{9}x_4-\frac{2}{9} \\x_2&=&s_1\\x_3&=&-\frac{4}{3}x_4+\frac{1}{3} \\x_4&=&s_2\end{array}\right$

Step-by-step explanation:

To find the solution for this system of linear equations $-3x_1-x_2+4x_3=2\\-3x_3 - 4x_4 = -1$ you must:

Step 1: Transform the augmented matrix to the reduced row echelon form.

A matrix is a rectangular arrangement of numbers into rows and columns.

A system of equations can be represented by an augmented matrix.

In an augmented matrix, each row represents one equation in the system and each column represents a variable or the constant terms.

This is matrix that represents the system

$\left[ \begin{array}{ccccc} -3 & -1 & 4 & 0 & 2 \\\\ 0 & 0 & -3 & 4 & -1 \end{array} \right]$

The augmented matrix can be transformed by a sequence of elementary row operations to the matrix.

There are three kinds of elementary matrix operations.

Interchange two rows (or columns).
Multiply each element in a row (or column) by a non-zero number.
Multiply a row (or column) by a non-zero number and add the result to another row (or column).

Using elementary matrix operations, we get that

Row Operation 1: Multiply the 1st row by -1/3

Row Operation 2: Multiply the 2nd row by -1/3

Row Operation 3: Add 4/3 times the 2nd row to the 1st row

$\left[ \begin{array}{ccccc} 1 & \frac{1}{3} & 0 & \frac{16}{9} & - \frac{2}{9} \\\\ 0 & 0 & 1 & \frac{4}{3} & \frac{1}{3} \end{array} \right]$

Step 2: Interpret the reduced row echelon form

The reduced row echelon form of the augmented matrix is

$\left[ \begin{array}{ccccc} 1 & \frac{1}{3} & 0 & \frac{16}{9} & - \frac{2}{9} \\\\ 0 & 0 & 1 & \frac{4}{3} & \frac{1}{3} \end{array} \right]$

which corresponds to the system

$x_1+\frac{1}{3}x_2+ \frac{16}{9}x_4=-\frac{2}{9} \\x_3+ \frac{4}{3}x_4=\frac{1}{3}$

We see that the variables $x_2, x_4$ can take arbitrary numbers; they are called free variables. Let $x_2=s_1$ , $x_4=s_2$ . All solutions of the system are given by

$\left\begin{array}{ccc}x_1&=&-\frac{1}{3}x_2-\frac{16}{9}x_4-\frac{2}{9} \\x_2&=&s_1\\x_3&=&-\frac{4}{3}x_4+\frac{1}{3} \\x_4&=&s_2\end{array}\right$

The system has infinitely many solutions.

8 0

4 years ago

Can someone please explain how to solve for a positive slope? Thank you <3

sukhopar [10]

Rise over run i think lol

4 0

3 years ago

Why is math important in the world todaywhy is math important in the world today

lesya692 [45]

Math helps us have better problem-solving skills.
Math helps us think analytically and have better reasoning abilities. Analytical thinking refers to the ability to think critically about the world around us. ... Analytical and reasoning skills are important because they help us solve problems and look for solutions.

5 0

3 years ago

Read 2 more answers

What is the circumference of a circle with a radius of 6 inches?

kirill [66]

Answer:

37.7in

Step-by-step explanation:

C=2πr=2·π·6≈37.69911in

5 0

3 years ago

Read 2 more answers

Many elementary school students in a school district currently have ear infections. A random sample of children in two different

Marta_Voda [28]

Answer:

Step-by-step explanation:

The summary of the given data includes;

sample size for the first school $n_1$ = 42

sample size for the second school $n_2$ = 34

so 16 out of 42 i.e $x_1$ = 16 and 18 out of 34 i.e $x_2$ = 18 have ear infection.

the proportion of students with ear infection Is as follows:

$\hat p_1 = \dfrac{16}{42}$ = 0.38095

$\hat p_2 = \dfrac{18}{34}$ = 0.5294

Since this is a two tailed test , the null and the alternative hypothesis can be computed as :

$H_0 :p_1 -p_2 = 0 \\ \\ H_1 : p_1 - p_2 \neq 0$

level of significance ∝ = 0.05,

Using the table of standard normal distribution, the value of z that corresponds to the two-tailed probability 0.05 is 1.96. Thus, we will reject the null hypothesis if the value of the test statistics is less than -1.96 or more than 1.96.

The test statistics for the difference in proportion can be achieved by using a pooled sample proportion.

$\bar p = \dfrac{x_1 +x_2}{n_1 +n_2}$

$\bar p = \dfrac{16 +18}{42 +34}$

$\bar p = \dfrac{34}{76}$

$\bar p = 0.447368$

$\bar p + \bar q = 1 \\ \\ \bar q = 1 -\bar p \\ \\\bar q = 1 - 0.447368 \\ \\\bar q = 0.552632$

The pooled standard error can be computed by using the formula:

$S.E = \sqrt{ \dfrac{ \bar p \bar q}{ n_1} + \dfrac{\bar p \bar p}{n_2} }$

$S.E = \sqrt{ \dfrac{ 0.447368 * 0.552632}{ 42} + \dfrac{ 0.447368 * 0.447368}{34} }$

$S.E = \sqrt{ \dfrac{ 0.2472298726}{ 42} + \dfrac{ 0.2001381274}{34} }$

$S.E = \sqrt{ 0.01177284105}$

$S.E = 0.1085$

The test statistics is ;

$z = \dfrac{\hat p_1 - \hat p_2}{S.E}$

$z = \dfrac{0.38095- 0.5294}{0.1085}$

$z = \dfrac{-0.14845}{0.1085}$

z = - 1.368

Decision Rule: Since the test statistics is greater than the rejection region - 1.96 , we fail to reject the null hypothesis.

Conclusion: There is insufficient evidence to support the claim that a difference exists between the proportions of students who have ear infections at the two schools

5 0

4 years ago