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Tcecarenko [31]
3 years ago
15

The blades of a helicopter's propeller are perpendicular and congruent. The distance between consecutive blade tips is 24 feet.

Find the length (b) of each blade.

Mathematics
1 answer:
Leno4ka [110]3 years ago
7 0

Check the picture below.

now, recall, both blades are congruent, thus a = b, so c² = a² + b² -> c² = b² + b².

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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
Please answer fast please
juin [17]

#1:

A,B,C,E

#2:

12 edges, 12 vertices

7 0
3 years ago
Read 2 more answers
14. Please help. What is the length of PQ⎯⎯⎯⎯⎯⎯⎯⎯?<br> Enter the correct value.
Pepsi [2]

Answer:

The correct value is 5.

5 0
4 years ago
A rectangle is three centimeters.longer than it is wide. if its length were to be decreased by two centimeters, its area would d
Scorpion4ik [409]
Given/equations formed from text:
I: l=w+3
II: A=l*w
III: new area B=(l-2)*w
IV: B=A-30

substitute l in II with I to remove l:
II': A=l*w=
(w+3)*w=
w^2+3w

substitute l in III with I to remove l:
III': B=(l-2)*w=
(w+3-2)*w=
(w+1)*w=
w^2+w

substitute A and B from II' and III' into IV:
B=A-30
w^2+w=w^2+3w-30
30=2w
w=15

insert w=15 into I:
l=15+3=18
->
A) dimensions of full size rectangle are width 15 and length 18
dimensions of reduced size rectangle are width 15 and length 16

B) full size: 15*18=10*18+5*18=180+90=270cm^2
reduced size: 15*16=10*16+5*16=160+80=240cm^2


7 0
4 years ago
Which of the following theorems verifies that DEF=XYZ? a)AA b) LL c) HA d)HL​
Snowcat [4.5K]

Answer:

is it AA..

I have no idea what to do how...

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4 years ago
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