O2, or Oxygen, and C6H12O6, or glucose.
        
             
        
        
        
Answer:
1/2
Step-by-step explanation:
 
        
                    
             
        
        
        
Answer:
Mean and Variance of the number of defective bulbs are 0.5 and 0.475 respectively.
Step-by-step explanation:
Consider the provided information,
Let X is the number of defective bulbs.
Ten light bulbs are randomly selected.
The likelihood that a light bulb is defective is 5%.
Therefore sample size is = n = 10
Probability of a defective bulb = p = 0.05.
Therefore, q = 1 - p = 1 - 0.05 = 0.95
Mean of binomial random variable: 
Therefore, 
Variance of binomial random variable: 
Therefore, 
Mean and Variance of the number of defective bulbs are 0.5 and 0.475 respectively.
 
        
             
        
        
        
<span>A container holds 15 pennies, 8 nickels, and 10 dimes. 
You will randomly select two coins without replacement. 
-->Fill in the probabilities on a tree diagram.</span>
        
             
        
        
        
To find the specification limit such that only 0.5% of the bulbs will not exceed this limit we proceed as follows;
From the z-table, a z-score of -2.57 cuts off 0.005 in the left tail; given the formula for z-score
(x-μ)/σ
we shall have:
(x-5000)/50=-2.57
solving for x we get:
x-5000=-128.5
x=-128.5+5000
x=4871.50