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Paladinen [302]
3 years ago
8

How is it going to tell me the answer

Mathematics
1 answer:
Anon25 [30]3 years ago
7 0
I dont know man you gotta ask a question
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The following data represent the weights in pounds of a sample of 25 police officers:
DochEvi [55]

Given:

The data values are:

164, 148, 137, 157, 173, 156, 177, 172, 169, 165, 145, 168, 163, 162, 174, 152, 156, 168, 154, 151, 174, 146, 134, 140, and 171.

To find;

a. Lower quartile.

b. Upper quartile.

c. Interquartile range.

Solution:

We have,

164, 148, 137, 157, 173, 156, 177, 172, 169, 165, 145, 168, 163, 162, 174, 152, 156, 168, 154, 151, 174, 146, 134, 140, 171.

Arrange the data values in ascending order.

134, 137, 140, 145, 146, 148, 151, 152, 154, 156, 156, 157, 162, 163, 164, 165, 168, 168, 169, 171, 172, 173, 174, 174, 177.

Divide the data values in two equal parts.

(134, 137, 140, 145, 146, 148, 151, 152, 154, 156, 156, 157), 162, (163, 164, 165, 168, 168, 169, 171, 172, 173, 174, 174, 177)

Divide each parentheses in two equal parts.

(134, 137, 140, 145, 146, 148), (151, 152, 154, 156, 156, 157), 162, (163, 164, 165, 168, 168, 169), (171, 172, 173, 174, 174, 177)

a. Location of lower quartile is:

Q_1=\dfrac{1}{4}(n+1)\text{th term}

Q_1=\dfrac{1}{4}(25+1)\text{th term}

Q_1=\dfrac{26}{4}\text{th term}

Q_1=6.5\text{th term}

The lower quartile of the weights is:

Q_1=\dfrac{148+151}{2}

Q_1=\dfrac{299}{2}

Q_1=149.5

Therefore, the location of the lower quartile of the weights is between 6th term and the 7th term. The value of the lower quartile is 149.5.

b. Location of upper quartile is:

Q_3=\dfrac{3}{4}(n+1)\text{th term}

Q_3=\dfrac{3}{4}(25+1)\text{th term}

Q_3=\dfrac{3\cdot 26}{4}\text{th term}

Q_3=19.5\text{th term}

The upper quartile of the weights is:

Q_3=\dfrac{169+171}{2}

Q_3=\dfrac{340}{2}

Q_3=170

Therefore, the location of the upper quartile of the weights is between 19th term and the 20th term. The value of the upper quartile is 170.

c. The interquartile range of the given data set is:

IQR=Q_3-Q_1

IQR=170-149.5

IQR=20.5

Therefore, the interquartile range of the weights is 20.5.

7 0
3 years ago
I need help can s1 help meee
IrinaVladis [17]
6,042.74m cubed

Cylinder volume is 282.74
Rectangle Volume is 5,760
282.74+5,760= 6,042.74
4 0
4 years ago
Read 2 more answers
Fill in the blanks. <br><br> sedrftgyuhihkgfndt7y
kvv77 [185]

HMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMM

5 0
3 years ago
Read 2 more answers
please help Can you solve each system and type the correct code? please remember yo type in all caps with no spaces.
Stels [109]

Answer:

C?EB

Step-by-step explanation:

1) the answer is- ( 3, - 1 ) so letter will be - C

2) not sure?

3) the answer is - ( 2, 3 ) so letter will be - E

4) the answer is - ( 3, 6 ) so letter will be - B

I'm sorry, maybe someone can get number 2 for you, I got the rest.

Goodluck!

6 0
2 years ago
A.
Pepsi [2]

Answer:

(a)Attached

(b)69 degrees

(b)The ladder is safe

Step-by-step explanation:

(a)See attached diagram

The wall is length |AB| and the ladder is length |AC|

(b)

sin \theta =\dfrac{|AB|}{|AC|} \\sin \theta =\dfrac{14}{15}\\\theta=arcsin (\dfrac{14}{15})\\\theta=69^\circ

(c)Jamie wants the angle of the ladder makes with the ground to be no greater than 75 degrees for safety reasons.

Since the angle formed between the ground and the ladder (69 degrees) is less than 75 degrees, the ladder is safe the way it is positioned.

4 0
3 years ago
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