16 cm -10 days
1 cm --- x days
x=1*10/16=5/8=0.625 days
I believe 10c-4d would be your answer
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Answer:
0.5 ; 0.475 ; 0.689 ; 0.4013
Step-by-step explanation:
Given that:
Rate of production of defective batteries p = 0.05
Number of batteries produced (n) = 10
The expected number of defective batteries = mean = n * p = 10 * 0.05 = 0.5 batteries
Variance of defective batteries :
Var(X) = n * p * q ; q = 1 - p
Hence,
Var(X) = 10 * 0.05 * 0.95 = 0.475
Standard deviation (X) = sqrt(variance) = sqrt(0.475) = 0.689
Probability that atleast 1 battery is defective :
Using the binomial probability function
P(x ≥ 1) = 1 - p(x = 0)
= 1 - q^n
= 1 - 0.95^10
= 1 - 0.59873693923837890625
= 0.40126306076162109375
= 0.4013
Answer:
(4,3)
Step-by-step explanation:
Midpoint formula
