Answer:
1. *2/multiplied by 2, since each number gets higher by *2. 
2. *3/multiplied by 3, since each number gets higher by *3
3. ÷2/divided by 2, since each number gets lower by ÷2.
 
        
             
        
        
        
Answer:
The probability that 2 or more of the original 5,100 components may fail during the useful life of the product is:
= 0.001
Step-by-step explanation:
Probability of operating without failure = 0.999
The probability of failed component = 0.001 (1 - 0.999)
The number of components of the electronic office product = 5,100
The number of components that may fail, given the above successful operation = 5.1 (0.001 * 5,100).
Therefore, the probability that 2 or more of the original 5,100 components may fail during the useful life of the product = 0.001
Probability of component failure is the likelihood that a component fails during the useful life of the product.  It is expressed as the number of likely failed components divided by the total number of components.  This result can be left in decimal form or expressed as a percentage.
 
        
             
        
        
        
Answer:

Step-by-step explanation:
(a/5) - (b/3) = (a/2) - (b/6)
 +(B/3) + (b/3)
NOTE: use property of equality to isolate <em>b</em> from <em>a</em>
(a/5) = (b/3) + (a/2) - (b/6)
 -(a/2) - (a/2)
NOTE: continue to use property of equality to isolate <em>a</em><em> </em>from <em>b</em>
(a/5) - (a/2) = (b/3) - (b/6)
60((a/5) - (a/2)) = ((b/3) - (b/6))60
NOTE: 60 is a common multiple of 5, 2, 3, and 6
12a - 30a = 20b - 10b
NOTE: combine alike terms
-17a = 10b
-17a/-17 = 10b/-17
NOTE: decide by -16 to find what a is equal to

Final answer
 
        
             
        
        
        
Answer: you subtract the angles by the one you are trying to solve 
Step-by-step explanation:  learned about it in 6th grade
 
        
                    
             
        
        
        
Answer:
X=5/4 or 1.25
Solution:
Make algebraic equation to solve
4x-3=2
Solve 
4x=2+3=5
X=5/4
Hope this helps. Please mark brainliest.