Create random data for boys and girls ensuring that the data on each column in equal to 10 . Using that data you can answer the questions .
1. Type of favourite fruits 
The rest you need the data for .
        
             
        
        
        
Answer:
The answer is "
The volume is greater than 50 cubic units
"
Step-by-step explanation:
In this question, the value of the units and figure is missing that's why its solution can be defined as follows:
Let's the unit value =50 and please find the figure in the attachment file.
The given figure shows its box is not filled with 50 cubes. These were ten empty places where cubes could be placed.
This figure indicates that only the box size will be four units per five units per three units, so they realize it will be in...

That's why the volume value exceeds 50 cubic units.
 
        
             
        
        
        
To find the answer you do 19×19= 361 and do 361×3.14= 1,133.54 The area of the stage is 1,133.54
        
                    
             
        
        
        
Answer:
               a₁ = -5, d = 7,  a₂ = 2, a₃ = 9, a₄ = 16
equation of sequence:    
Step-by-step explanation:
a₁ + a₂ + a₃ = 6    
a₁ + a₁ + d + a₁ + 2d = 6 
3a₁ + 3d = 6
a₁ + d= 2     ⇒ a₁ = 2 - d
a₄ = 16
a₁ + 3d = 16
2 - r + 3d = 16
2d = 14
d = 7
a₁ = 2-7 = -5
a₁ = -5, d = 7   ⇒  a₂ = -5+7 = 2, a₃ = 2+7 = 9, a₄ = 9+7 = 16
equation of arithmetic sequence:

 
        
             
        
        
        
In order to do this, you must first find the "cross product" of these vectors. To do that, we can use several methods. To simplify this first, I suggest you compute: 
 ‹1, -1, 1› × ‹0, 1, 1› 
 You are interested in vectors orthogonal to the originals, which don't change when you scale them. Using 0,-1,1 is much easier than 6s and 7s. 
 So what methods are there to compute this? You can review them here (or presumably in your class notes or textbook): 
http://en.wikipedia.org/wiki/Cross_produ... 
 In addition to these methods, sometimes I like to set up: 
 ‹1, -1, 1› • ‹a, b, c› = 0 
 ‹0, 1, 1› • ‹a, b, c› = 0 
 That is the dot product, and having these dot products equal zero guarantees orthogonality. You can convert that to: 
 a - b + c = 0 
 b + c = 0 
 This is two equations, three unknowns, so you can solve it with one free parameter: 
 b = -c 
 a = c - b = -2c 
 The computation, regardless of method, yields: 
 ‹1, -1, 1› × ‹0, 1, 1› = ‹-2, -1, 1› 
 The above method, solving equations, works because you'd just plug in c=1 to obtain this solution. However, it is not a unit vector. There will always be two unit vectors (if you find one, then its negative will be the other of course). To find the unit vector, we need to find the magnitude of our vector: 
 || ‹-2, -1, 1› || = √( (-2)² + (-1)² + (1)² ) = √( 4 + 1 + 1 ) = √6 
 Then we divide that vector by its magnitude to yield one solution: 
 ‹ -2/√6 , -1/√6 , 1/√6 › 
 And take the negative for the other: 
 ‹ 2/√6 , 1/√6 , -1/√6 ›