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Anna11 [10]
3 years ago
11

Plz help and thank you

Mathematics
2 answers:
Nastasia [14]3 years ago
6 0

Answer:

60%

Also known as 330 students

Step-by-step explanation:

Plz mark brainliest!!!

rjkz [21]3 years ago
3 0

Answer:

Work shown below!

Step-by-step explanation:

If 40% of students take the bus, then 60% of 550 students don't take the bus;

\frac{60}{100}*\frac{550}{1} =6*55=330

330 students do not take the bus

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Write1/64 as a power of 2
Akimi4 [234]

Answer:

2^-6.

Step-by-step explanation:

64 = 2^6

so 1/64 = 1 / 2^6

= 2^-6.

3 0
2 years ago
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A salesperson earns a commission of $252 for selling $2100 in merchandise.
shutvik [7]

Answer:

The commission rate is of 12%.

Step-by-step explanation:

Commission rate:

The commission rate is given by the commission value multiplied by 100% and divided by the value of merchandise sold.

In this question:

$252 in commission for selling $2100 in merchandise. So

252*100%/2100 = 12%

The commission rate is of 12%.

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3 years ago
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29 minutes

Step-by-step explanation:

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Does anyone know this?
lakkis [162]

Answer:

obtuse scalene

Step-by-step explanation:

8 0
3 years ago
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Solve the following systems of equations using the matrix method: a. 3x1 + 2x2 + 4x3 = 5 2x1 + 5x2 + 3x3 = 17 7x1 + 2x2 + 2x3 =
lara [203]

Answer:

a. The solutions are

\left[\begin{array}{c}x_1&x_2&x_3\\\end{array}\right]=\begin{pmatrix}\frac{11}{13}\\ \frac{50}{13}\\ -\frac{17}{13}\end{pmatrix}

b. The solutions are

\left[\begin{array}{c}x&y&z\\\end{array}\right]=\begin{pmatrix}\frac{54}{235}\\ \frac{6}{47}\\ \frac{24}{235}\end{pmatrix}

c. The solutions are

\left[\begin{array}{c}x_1&x_2&x_3&x_4\\\end{array}\right]=\begin{pmatrix}\frac{22}{9}\\ \frac{164}{9}\\ \frac{139}{9}\\ -\frac{37}{3}\end{pmatrix}

Step-by-step explanation:

Solving a system of linear equations using matrix method, we may define a system of equations with the same number of equations as variables as:

A\cdot X=B

where X is the matrix representing the variables of the system,  B is the matrix representing the constants, and A is the coefficient matrix.

Then the solution is this:

X=A^{-1}B

a. Given the system:

3x_1 + 2x_2 + 4x_3 = 5 \\2x_1 + 5x_2 + 3x_3 = 17 \\7x_1 + 2x_2 + 2x_3 = 11

The coefficient matrix is:

A=\left[\begin{array}{ccc}3&2&4\\2&5&3\\7&2&2\end{array}\right]

The variable matrix is:

X=\left[\begin{array}{c}x_1&x_2&x_3\\\end{array}\right]

The constant matrix is:

B=\left[\begin{array}{c}5&17&11\\\end{array}\right]

First, we need to find the inverse of the A matrix. To find the inverse matrix, augment it with the identity matrix and perform row operations trying to make the identity matrix to the left. Then to the right will be inverse matrix.

So, augment the matrix with identity matrix:

\left[ \begin{array}{ccc|ccc}3&2&4&1&0&0 \\\\ 2&5&3&0&1&0 \\\\ 7&2&2&0&0&1\end{array}\right]

This matrix can be transformed by a sequence of elementary row operations to the matrix

\left[ \begin{array}{ccc|ccc}1&0&0&- \frac{2}{39}&- \frac{2}{39}&\frac{7}{39} \\\\ 0&1&0&- \frac{17}{78}&\frac{11}{39}&\frac{1}{78} \\\\ 0&0&1&\frac{31}{78}&- \frac{4}{39}&- \frac{11}{78}\end{array}\right]

And the inverse of the A matrix is

A^{-1}=\left[ \begin{array}{ccc} - \frac{2}{39} & - \frac{2}{39} & \frac{7}{39} \\\\ - \frac{17}{78} & \frac{11}{39} & \frac{1}{78} \\\\ \frac{31}{78} & - \frac{4}{39} & - \frac{11}{78} \end{array} \right]

Next, multiply A^ {-1} by B

X=A^{-1}\cdot B

\left[\begin{array}{c}x_1&x_2&x_3\\\end{array}\right]=\left[ \begin{array}{ccc} - \frac{2}{39} & - \frac{2}{39} & \frac{7}{39} \\\\ - \frac{17}{78} & \frac{11}{39} & \frac{1}{78} \\\\ \frac{31}{78} & - \frac{4}{39} & - \frac{11}{78} \end{array} \right] \cdot \left[\begin{array}{c}5&17&11\end{array}\right]

\left[\begin{array}{c}x_1&x_2&x_3\\\end{array}\right]=\begin{pmatrix}-\frac{2}{39}&-\frac{2}{39}&\frac{7}{39}\\ -\frac{17}{78}&\frac{11}{39}&\frac{1}{78}\\ \frac{31}{78}&-\frac{4}{39}&-\frac{11}{78}\end{pmatrix}\begin{pmatrix}5\\ 17\\ 11\end{pmatrix}=\begin{pmatrix}\frac{11}{13}\\ \frac{50}{13}\\ -\frac{17}{13}\end{pmatrix}

The solutions are

\left[\begin{array}{c}x_1&x_2&x_3\\\end{array}\right]=\begin{pmatrix}\frac{11}{13}\\ \frac{50}{13}\\ -\frac{17}{13}\end{pmatrix}

b. To solve this system of equations

x -y - z = 0 \\30x + 40y = 12 \\30x + 50z = 12

The coefficient matrix is:

A=\left[\begin{array}{ccc}1&-1&-1\\30&40&0\\30&0&50\end{array}\right]

The variable matrix is:

X=\left[\begin{array}{c}x&y&z\\\end{array}\right]

The constant matrix is:

B=\left[\begin{array}{c}0&12&12\\\end{array}\right]

The inverse of the A matrix is

A^{-1}=\left[ \begin{array}{ccc} \frac{20}{47} & \frac{1}{94} & \frac{2}{235} \\\\ - \frac{15}{47} & \frac{4}{235} & - \frac{3}{470} \\\\ - \frac{12}{47} & - \frac{3}{470} & \frac{7}{470} \end{array} \right]

The solutions are

\left[\begin{array}{c}x&y&z\\\end{array}\right]=\begin{pmatrix}\frac{54}{235}\\ \frac{6}{47}\\ \frac{24}{235}\end{pmatrix}

c. To solve this system of equations

4x_1 + 2x_2 + x_3 + 5x_4 = 0 \\3x_1 + x_2 + 4x_3 + 7x_4 = 1\\ 2x_1 + 3x_2 + x_3 + 6x_4 = 1 \\3x_1 + x_2 + x_3 + 3x_4 = 4\\

The coefficient matrix is:

A=\left[\begin{array}{cccc}4&2&1&5\\3&1&4&7\\2&3&1&6\\3&1&1&3\end{array}\right]

The variable matrix is:

X=\left[\begin{array}{c}x_1&x_2&x_3&x_4\\\end{array}\right]

The constant matrix is:

B=\left[\begin{array}{c}0&1&1&4\\\end{array}\right]

The inverse of the A matrix is

A^{-1}=\left[ \begin{array}{cccc} - \frac{1}{9} & - \frac{1}{9} & - \frac{1}{9} & \frac{2}{3} \\\\ - \frac{32}{9} & - \frac{5}{9} & \frac{13}{9} & \frac{13}{3} \\\\ - \frac{28}{9} & - \frac{1}{9} & \frac{8}{9} & \frac{11}{3} \\\\ \frac{7}{3} & \frac{1}{3} & - \frac{2}{3} & -3 \end{array} \right]

The solutions are

\left[\begin{array}{c}x_1&x_2&x_3&x_4\\\end{array}\right]=\begin{pmatrix}\frac{22}{9}\\ \frac{164}{9}\\ \frac{139}{9}\\ -\frac{37}{3}\end{pmatrix}

7 0
3 years ago
Read 2 more answers
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