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myrzilka [38]
3 years ago
12

Determine whether the statement is true or false. If it is false, rewrite it as a true statement. Data at the ratio level cannot

be put in order.
Mathematics
1 answer:
ziro4ka [17]3 years ago
5 0

Answer:

False

Step-by-step explanation:

The statement is a false statement. We're asked to rewrite it as a true statement if it were false. Since it's false, we then try to rewrite it as a true statement, in so doing we are going to have it like this

"Data at the ratio level can be placed in a meaningful​ order."

Now we've successfully written it as a true statement

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Solve the following matrix equations: (matrices)
Masja [62]

Step-by-step explanation:

a)

3X + \begin{pmatrix} 2 & 3 \\ 4 & 5 \end{pmatrix} = \begin{pmatrix}  - 1 & 6 \\ 10 & 14 \end{pmatrix} \\  \\  3X  = \begin{pmatrix}  - 1 & 6 \\ 10 & 14 \end{pmatrix} -  \begin{pmatrix} 2 & 3 \\ 4 & 5 \end{pmatrix}  \\  \\ 3X  = \begin{pmatrix}  - 1 - 2 & 6 - 3 \\ 10 - 4 & 14 - 5 \end{pmatrix}\\  \\ 3X  = \begin{pmatrix}   - 3 & 3 \\ 6 & 9\end{pmatrix}\\  \\ X  =  \frac{1}{3} \begin{pmatrix}   - 3 & 3 \\ 6 & 9\end{pmatrix}\\  \\ X  =  \begin{pmatrix}  \frac{ - 3}{3}  & \frac{3}{3}  \\  \\ \frac{6}{3}  & \frac{9}{3} \end{pmatrix}\\  \\ \huge \red{ X}  =  \purple{ \begin{pmatrix}  - 1  &1  \\ 2 & 3 \end{pmatrix}}

b)

3X + 2I_3=\begin{pmatrix} 5 & 0 & -3 \\6 & 5 & 0\\ 9 & 6 & 5\end{pmatrix} \\\\3X + 2\begin{pmatrix} 1 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & 1\end{pmatrix} =\begin{pmatrix} 5 & 0 & - 3\\6 & 5 & 0\\ 9 & 6 & 5 \end{pmatrix} \\\\3X + \begin{pmatrix} 2 & 0 & 0\\ 0 & 2 & 0\\ 0 & 0 & 2\end{pmatrix} =\begin{pmatrix} 5 & 0 & - 3\\ 6 & 5 & 0 \\ 9 & 6 & 5 \end{pmatrix} \\\\3X  =\begin{pmatrix} 5 & 0 & -3 \\ 6 & 5 & 0 \\ 9 & 6 & 5 \end{pmatrix} - \begin{pmatrix} 2 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 2 \end{pmatrix} \\\\3X  =\begin{pmatrix} 5-2 & 0-0 & -3-0 \\ 6-0 & 5-2 & 0-0 \\ 9-0 & 6-0 & 5-2 \end{pmatrix} \\\\3X  =\begin{pmatrix} 3 & 0 & - 3 \\ 6 & 3 & 0 \\ 9 & 6 & 3 \end{pmatrix} \\\\X  =\frac{1}{3} \begin{pmatrix} 3 & 0 & - 3 \\ 6 & 3 & 0 \\ 9 & 6 & 3 \end{pmatrix} \\\\X  =\begin{pmatrix} \frac{3}{3}  & \frac{0}{3}  & \frac{-3}{3} \\\\ \frac{6}{3}  & \frac{3}{3}  & \frac{0}{3} \\\\ \frac{9}{3}  & \frac{6}{3}  & \frac{3}{3} \end{pmatrix} \\\\\huge\purple {X} =\orange{\begin{pmatrix} 1  & 0 & - 1\\ 2  & 1 & 0 \\ 3  & 2  & 1 \end{pmatrix}}\\

8 0
3 years ago
Hint: You won't flip the sign in every problem. ONLY FUP it when you or by a 9 - 2x < 6x - 15 Solutions You solve and graph t
svetoff [14.1K]

You only flip the inequality sign when you multiply or divide both sides by a negative number.

================================================

Problem 1

3 - 8x \ge -29\\\\-8x \ge -29-3\\\\-8x \ge -32\\\\x \le \frac{-32}{-8} \ \text{ inequality sign flip}\\\\x \le 4

The inequality sign flip happens because we divided both sides by -8.

The graph will have a closed circle at 4 with shading to the left.

Three solutions are x = 0, x = 1, x = 2. You can pick any three numbers you want as long as they are 4 or smaller.

================================================

Problem 2

1x+2x-7 < 6\\\\3x-7 < 6\\\\3x < 6+7\\\\3x < 13\\\\x < \frac{13}{3}\\\\x < 4 \frac{1}{3}

The graph will have an open circle at 13/3 = 4&1/3 = 4.333 approx. The shading is to the left. No inequality sign flip happens because we divided both sides by a positive number.

Your choice of three solutions is correct. You can pick anything smaller than 4.3333

================================================

Problem 3

9-2x \le 6x-15\\\\-2x-6x \le -15-9\\\\-8x \le -24\\\\x \ge \frac{-24}{-8} \ \text{ inequality sign flip}\\\\x \ge 3

The solution set is any value 3 or larger. Three solutions are x = 5, x = 6 and x = 7.

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