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Nitella [24]
3 years ago
11

Who has pad let plz let me know bc I really wanna kill my self bc no does and they won't talk to me

Mathematics
1 answer:
Ymorist [56]3 years ago
8 0

I have it please like me or whatever

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Find the reference angle for 105 degree angle.
m_a_m_a [10]

Answer:

75° degree angle

Step-by-step explanation:

Since the angle 105° is in the second quadrant, subtract 105° from 180° .

180-105=75

6 0
3 years ago
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A survey of pet owners found that on average, they spend $1,225 annually per pet, with a standard deviation of $275. Between whi
earnstyle [38]

Answer:

B.$675 and $1,775

Step-by-step explanation:

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3 years ago
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A rain gauge had 0.6 centimeter of rain after 2 hours, 0.9 centimeter after 3 hours, 1.2 centimeters after 4 hours, and 1.5 cent
Ulleksa [173]

Answer: 0.3 cm per hour

Step-by-step explanation:

got it off of quizizzes

3 0
3 years ago
Mathematically verify the outlier(s) in the data set using the 1.5 rule.
statuscvo [17]

Given:

The data values are:

7, 8, 11, 13, 14, 14, 14, 15, 16, 16, 18, 19

To find:

The outliers of the given data set.

Solution:

We have,

7, 8, 11, 13, 14, 14, 14, 15, 16, 16, 18, 19

Divide the data set in two equal parts.

(7, 8, 11, 13, 14, 14), (14, 15, 16, 16, 18, 19)

Divide each parenthesis in two equal parts.

(7, 8, 11), (13, 14, 14), (14, 15, 16), (16, 18, 19)

Now,

Q_1=\dfrac{11+13}{2}

Q_1=\dfrac{24}{2}

Q_1=12

And

Q_3=\dfrac{16+16}{2}

Q_3=\dfrac{32}{2}

Q_3=16

The interquartile range is:

IQR=Q_3-Q_1

IQR=16-12

IQR=4

The data values lies outside the interval [Q_1-1.5IQR,Q_3+1.5IQR] are known as outliers.

[Q_1-1.5IQR,Q_3+1.5IQR]=[12-1.5(4),16+1.5(4)]

[Q_1-1.5IQR,Q_3+1.5IQR]=[12-6,16+6]

[Q_1-1.5IQR,Q_3+1.5IQR]=[6,22]

All the data values lie in the interval [6,22]. So, there are no outliers.

Hence, the correct option is 4.

3 0
3 years ago
What is 320 times 10^2 equal
SIZIF [17.4K]

Answer:

Step-by-step explanation:

10^2=100

320×100=32000

8 0
2 years ago
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