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MariettaO [177]
3 years ago
11

Which are solutions for -7x+14>-3x-6

Mathematics
1 answer:
dybincka [34]3 years ago
5 0
Step1. -7x+14>-3x-6
step2. -7x+3x+14>-6
step3. -7x+3x>-6-14
step4. -4>-6-14
step5. -4x>20
step6.x<5
The answer is x<5
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TR = 6
QR = 2(TR)
QR = 2(6)
QR = 12
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Select all the statements that are true for the graph shown
SVEN [57.7K]

Answer:

there is nothing to prove

Step-by-step explanation:

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What is 2,789 divided by 36
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It's 77.472 , but since you probably need too round it will be 77.5
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The age of some lecturers are 42,54,50,54,50,42,46,46,48 and 48 calculate the mean age and standard deviation
rjkz [21]

Answer:

Mean age: 48

Standard deviation: 4

Step-by-step explanation:

a) Mean

The formula for Mean = Sum of terms/ Number of terms

Number of terms

= 42 + 54 + 50 + 54 + 50 + 42 + 46 + 46 + 48+ 48/ 10

= 480/10

= 48

The mean age is 48

b) Standard deviation

The formula for Standard deviation =

√(x - Mean)²/n

Where n = number of terms

Standard deviation =

√[(42 - 48)² + (54 - 48)² + (50 - 48)² +(54 - 48)² + (50 - 48)² +(42 - 48)² + (46 - 48)² + (46 - 48)² + (48 - 48)² + (48 - 48)² / 10]

= √-6² + 6² + 2² + 6² + 2² + -6² + -2² + -2² + 0² + 0²/10

=√36 + 36 + 4 + 36 + 4 + 36 + 4 + 4 + 0 + 0/ 10

=√160/10

= √16

= 4

The standard deviation of the ages is 4

3 0
3 years ago
Consider the following set of vectors. v1 = 0 0 0 1 , v2 = 0 0 3 1 , v3 = 0 4 3 1 , v4 = 8 4 3 1 Let v1, v2, v3, and v4 be (colu
Alona [7]

Answer:

We have the equation

c_1\left[\begin{array}{c}0\\0\\0\\1\end{array}\right] +c_2\left[\begin{array}{c}0\\0\\3\\1\end{array}\right] +c_3\left[\begin{array}{c}0\\4\\3\\1\end{array}\right] +c_4\left[\begin{array}{c}8\\4\\3\\1\end{array}\right] =\left[\begin{array}{c}0\\0\\0\\0\end{array}\right]

Then, the augmented matrix of the system is

\left[\begin{array}{cccc}0&0&0&8\\0&0&4&4\\0&3&3&3\\1&1&1&1\end{array}\right]

We exchange rows 1 and 4 and rows 2 and 3 and obtain the matrix:

\left[\begin{array}{cccc}1&1&1&1\\0&3&3&3\\0&0&4&4\\0&0&0&8\end{array}\right]

This matrix is in echelon form. Then, now we apply backward substitution:

1.

8c_4=0\\c_4=0

2.

4c_3+4c_4=0\\4c_3+4*0=0\\c_3=0

3.

3c_2+3c_3+3c_4=0\\3c_2+3*0+3*0=0\\c_2=0

4.

c_1+c_2+c_3+c_4=0\\c_1+0+0+0=0\\c_1=0

Then the system has unique solution that is (c_1,c_2c_3,c_4)=(0,0,0,0) and this imply that the vectors v_1,v_2,v_3,v_4 are linear independent.

4 0
3 years ago
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