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Tomtit [17]
3 years ago
8

The mean weight of five complete computer stations is 167.2 pounds. The weights of four of the computer stations are 158.4 pound

s, 162.8 pounds, 165 pounds, and 178.2 pounds respectively. What is the fifth computer station?
Mathematics
2 answers:
devlian [24]3 years ago
8 0
The fifth computer station's weight is 171.6 pounds.
Lynna [10]3 years ago
3 0

The mean weight of five complete computer stations is 167.2 pounds.

The weights of four of the computer stations are 158.4 pounds, 162.8 pounds, 165 pounds, and 178.2 pounds.

We have to determine the weight of fifth computer station.

Let the weight of fifth computer station be 'x' pounds.

Mean is calculated by sum of all the observations divided by the total number of observations.

So, \frac{158.4+162.8+165+178.2+x}{5}=167.2

\frac{664.4+x}{5}=167.2

{664.4+x}=167.2 \times 5

{664.4+x}=836

x= 171.6 pounds

The weight of fifth computer station is 171.6 pounds.

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Please hurry attachment below will Brainly if answer correctly 30 POINTS
Rzqust [24]
Put everything into a calculator to change to decimal form to make things easier:
27/5=5.4
5.36=5.36
5 3/5=5.6
33/6=5.5
So now in order to least to greatest it would be:
5.36, 27/5, 33/6, 5 3/5
6 0
3 years ago
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 .
___________________________________________________
4 0
3 years ago
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Vanyuwa [196]

Let's note the information.

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