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denis-greek [22]
3 years ago
10

someone plz help me. with promblems above extrax points. i will mark brainliest the first answer and the one that is expalined

Mathematics
1 answer:
madam [21]3 years ago
8 0
14= A.39°
You solved this one so I'm not going to show the work for it.

15=B,C,D
a.-30(-3x+2)=90x-60- incorrect
b.30(-3x+2)=-90x+60
c.10(-9x+6)=-90x+60
d.-10(-9x-6)=90x+60
e.-10(9x+6)=-90x-60- incorrect

16=D
-18x-12-(-13x+17)
-18x-12+13x+17
-5x+5

17=C
9x+3-5x-7
4x-4

18=C
You solved this one...

19=B
8.4x+2.9+(-3.7x+5)
8.4x+2.9-3.7x+5
4.7x+7.9

20=C

21=D
D is the only one that ends with +32 and not -32, which would be incorrect.

22=Cant see >.<

23=A
4x+12+15-3x
x+27

24=B
2x+3

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vredina [299]

Answer:

x = 10

y = 6

Step-by-step explanation:

<u>Vertical Angle Theorem</u>:  When two straight lines intersect, the opposite vertical angles are always equal to each other.

⇒ m∠1 = m∠3   and   m∠2 = m∠4

⇒ m∠1 = m∠3

⇒ 10x = 100

⇒ x = 10

<u>Linear pair:</u>  Two adjacent angles which sum to 180°.

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7 0
2 years ago
The number of texts per day by students in a class is normally distributed with a 
kobusy [5.1K]

Answer:

1, 2, 6

Step-by-step explanation:

The z score shows by how many standard deviations the raw score is above or below the mean. The z score is given by:

z=\frac{x-\mu}{\sigma} \\\\where\ x=raw\ score,\mu=mean, \ \sigma=standard\ deviation

Given that mean (μ) = 130 texts, standard deviation (σ) = 20 texts

1) For x < 90:

z=\frac{x-\mu}{\sigma} \\\\z=\frac{90-130}{20} =-2

From the normal distribution table, P(x < 90) = P(z < -2) = 0.0228 = 2.28%

Option 1 is correct

2) For x > 130:

z=\frac{x-\mu}{\sigma} \\\\z=\frac{130-130}{20} =0

From the normal distribution table, P(x > 130) = P(z > 0) = 1 - P(z < 0) = 1 - 0.5 = 50%

Option 2 is correct

3) For x > 190:

z=\frac{x-\mu}{\sigma} \\\\z=\frac{190-130}{20} =3

From the normal distribution table, P(x > 3) = P(z > 3) = 1 - P(z < 3) = 1 - 0.9987 = 0.0013 = 0.13%

Option 3 is incorrect

4)  For x < 130:

z=\frac{x-\mu}{\sigma} \\\\z=\frac{130-130}{20} =0

For x > 100:

z=\frac{x-\mu}{\sigma} \\\\z=\frac{100-130}{20} =-1.5

From the normal table, P(100 < x < 130) = P(-1.5 < z < 0) = P(z < 0) - P(z < 1.5) = 0.5 - 0.0668 = 0.9332 = 93.32%

Option 4 is incorrect

5)  For x = 130:

z=\frac{x-\mu}{\sigma} \\\\z=\frac{130-130}{20} =0

Option 5 is incorrect

6)  For x = 130:

z=\frac{x-\mu}{\sigma} \\\\z=\frac{160-130}{20} =1.5

Since 1.5 is between 1 and 2, option 6 is correct

5 0
2 years ago
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