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Luda [366]
3 years ago
12

Find the solution set for this inequality. SHOW YOUR WORK OR NO CREDIT -2x+4>-4x+8

Mathematics
2 answers:
Digiron [165]3 years ago
7 0

Answer:

x<2 is your answer

Step-by-step explanation:

-2x+4>-4x+8

subtract each term by 8

-2x+4-8>-4x+8-8

add both side by 2x

-2x+2x-4>-4x+2x

-4>-2x

dividing both term by -2

-2x/-2<-4/-2

x<2

grigory [225]3 years ago
7 0
<h2>Answer:</h2>

= - 2x + 4 > - 4x + 8

[ Transforming like terms in same side ]

= - 2x + 4x > 8 - 4

= 2x > 4

= x > 2.

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Solve each of the following equations. Show its solution set on the number line. |3x-4(x+1)|=2
Colt1911 [192]

Answer:

x= -6, -2

Step-by-step explanation:

4 0
2 years ago
I need help. Part A: Create a five-number summary and calculate the interquartile range for the two sets of data. (5 points)
Dahasolnce [82]

Five number summary and IQR of both the data sets are different.

For school A: Minimum=6, Q₁=6.5, Median= 14, Q₃=16, Maximum=17, IQR=9.5

For school B: Minimum=5, Q₁=8, Median= 12, Q₃=15.5, Maximum=19, IQR=7.5

No, the box plots are not symmetric.

Part A:

The given data sets are

School A : 9,14,15,17,17,7,15,6,6

School B : 12,8,13,11,19,15,16,5,8

Arrange the data in ascending order.

School A : 6,6,7,9,14,15,15,17,17

School B : 5,8,8,11,12,13,15,16,19

Divide each data set into four equal parts.

School A : (6,6),(7,9),14,(15,15),(17,17)

School B : (5,8),(8,11),12,(13,15),(16,19)

For school A:

Minimum=6, Q₁=6.5, Median= 14, Q₃=16, Maximum=17

The interquartile range of the data is

For school B:

Minimum=5, Q₁=8, Median= 12, Q₃=15.5, Maximum=19

IQR=Q_3-Q_1\\     =Q_3-Q_1\\     =16-6.2\\     =9.5

The interquartile range of the data is

IQR=Q_3-Q_1\\     =Q_3-Q_1\\     =15.5-8\\     =7.5

Part B:

The box plots are not symmetric because the data values are different. Five number summary and IQR of both the data sets are different.

To learn more about the symmetric visit:

brainly.com/question/1002723

#SPJ1

6 0
1 year ago
Can someone explain to me how this has multiple answers? I really don't understand! When I did it I only got the first answer. P
Vesna [10]

in \: firt \: quadrant \\ sin \frac{\pi}{4}  =  \frac{ \sqrt{2} }{2}  \\ in \: second \: quadrant \\ sin \frac{3\pi}{4} =  \frac{ \sqrt{2} }{2}  \\ in \: third \: quadrant \:  \\ sin \frac{5\pi}{4}  = \frac{ -  \sqrt{2} }{2}  \\ in \: forth \: quadrant \\  sin \frac{7\pi}{4}  =  - \frac { \sqrt{ 2} }{2}  \\ in \: this \: way \: it \: has \: multiple \: soution

3 0
2 years ago
263/4 options 65r2 64r7 65 65r3
Reika [66]

Answer:

65R3

Step-by-step explanation:

260/4 is a whole number, so then there would be a remainder 3. The only one with a r3 is D.

Hope this helps :D

4 0
2 years ago
Jose baked a cake that was partially eaten. The eaten cake is represented by the shaded
garik1379 [7]
The ratio of that question is 7:12











7 0
2 years ago
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