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bazaltina [42]
3 years ago
8

pls pls pls, I beg you to answer this question only if you know the correct answer please please please I beg you

Mathematics
1 answer:
Likurg_2 [28]3 years ago
3 0

Answer:

Median is the middle line of the box, Lower quartile is the number of the lower end of the box, Upper quartile is the line at the larger end of the box, Maximum is the largest number in the box plot (end of right whisker), Minimum is the lowest number (end of left whisker).

Step-by-step explanation:

32.

Median: 88

Maximum (largest number): 102

Minimum (lowest number): 38

Interquartile range:

96 - 72 = 24

b.

25 % higher

75 % lower?

c.

You can see the range is more towards the right of the box meaning they have higher scores, and that the median or middle is a higher number so it means that most of the students understood and did well on the test.

???.

First quartile: 16

Second: 19

Third: 20

Range: 4

34.  Since they are already in least to greatest order, all we have to do is graph it!

85 - 99 is 2

100 - 114 is 1

115 - 129 is 5

130 - 144 is 6

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A cylinder has a volume of 384 cubic inches and a height of 8 inches, what is the radius
Triss [41]
The answer is:  3.91 inches .
___________________________________________

Note:  Volume of cylinder: V = (base area) * (height);

in which: V = volume = 384 in.³ ;
              h = height = 8 in. ; 
              Base area = area of the base (that is; "circle") = π r² ;
                                         in which; "r" = radius; 
___________________________________________
Solve for "r" :
___________________________________________
 V = π r² * (8 in.) ; 

384 in.³ = (8 in.) * (π r²) ;
___________________________________________
Divide EACH SIDE of the equation by "8" ; 
___________________________________________
 (384 in.³) / 8 = [ (8 in.) * (π r²) in.] / 8 ; 
___________________________________________
 to get: 
___________________________________________
   48 in.³ = (π r²) in.² * in.   ;
___________________________________________
 ↔  (π r²) in.² * in. =  48 in.³  ;  
___________________________________________
Rewrite this equation; using "3.14" as an approximation for: π ;
__________________________________________________
 (3.14 * r²) in.² * in. =  48 in.³ 
_______________________________________
Divide EACH SIDE of the equation by:

"[(3.14)*(in.²)*(in.)]" ;  to isolate "r² " on one side of the equation; 
                                 (since we want to solve for "r") ;
_____________________________________________________
→ [(3.14 * r²) in.² * in.] / [(3.14)*(in.²)*(in.)]  = 48 in.³ / [(3.14)*(in.²)*(in.)] ; 
__________________________________________________
→ to get:   r² = 48/3.14 ;
________________________
      → r² = 15.2866242038216561 ;
_______________________________________
To solve for "r" (the radius; take the "positive square root" of EACH side of the equation:
__________________________________________________
     → +√(r²) = +√(15.2866242038216561)
__________________________________________________
     →  r = 3.9098112747064475286  ; round to 3.91 inches .
___________________________________________________
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3 years ago
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