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Elena L [17]
3 years ago
11

Any of you guys know this answer?

Mathematics
1 answer:
Fantom [35]3 years ago
5 0
The answer would be 10
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Which of the following are solution to the inequality below -4w - 2 < -38
FromTheMoon [43]

Answer:

w > 9

Step-by-step explanation:

Solve the inequality by using inverse operations.

-4w - 2 < -38

-4w < -36

w > 9

Like solving equations, inequalities should be solved using inverse operations. However, unlike equations, when you divide or multiply by a negative the sign changes.

8 0
4 years ago
Can someone plz help me on this thank you sooo much
Stells [14]

Answer:

the last answer

Step-by-step explanation:

6 0
3 years ago
I NEED THIS ASAP, PLEASE HELP!!!
WINSTONCH [101]

Answer: 22.5 ; 5 ; 14

Step-by-step explanation:

Given the dataset:

{20,22,23,24,26,26,28,29,30}

The lower quartile (Q1) = 1/4(n + 1)th term

Where n = number of observations, n = 9

Q1 = 1/4 (9 + 1)th term

Q1 = 1/4(10) = 2.5

We average the 2nd and 3rd term:

(22 + 23) / 2

45 / 2 = 22.5

B) The interquartile range(IQR) of the dataset :

{62,63,64,65,67,68,68,68,69,74}

IQR = Q3 - Q1

The lower quartile (Q1) = 1/4(n + 1)th term

Where n = number of observations, n = 10

Q1 = 1/4 (10 + 1)th term

Q1 = 1/4(11) = 2.75 term

We take the average of the 2nd and 3rd term:

(63 + 64) / 2

45 / 2 = 63.5

The upper quartile (Q3) = 3/4(n + 1)th term

Where n = number of observations, n = 10

Q3 = 3/4 (10 + 1)th term

Q3 = 3/4(11) = 8.25 term

We take the average of the 8th and 9th term:

(68 + 69) / 2

137 / 2 = 68.5

IQR = Q3 - Q1

IQR = 68.5 - 63.5

IQR = 5

C) give the dataset :

{7,8,8,9,10,12,13,15,16}

The upper quartile (Q3) = 3/4(n + 1)th term

Where n = number of observations, n = 9

Q3 = 3/4 (9 + 1)th term

Q3 = 3/4(10) = 7.5 term

We take the average of the 7th and 8th term:

(13 + 15) / 2

28 / 2 = 14

7 0
4 years ago
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
3 - 5y = -37 for math 8th
hram777 [196]

Answer:

y = 8

Step-by-step explanation:

Step 1: Write equation

3 - 5y = -37

Step 2: Subtract 3 on both sides

-5y = -40

Step 3: Divide both sides by -5

y = 8

5 0
3 years ago
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