A = p(1 + r/n)^nt
p = principal
r = rate, change to a decimal
n = number of times it is compounded per year
t = time in years
A = 10,000(1 + .055/12)^(12*(44/12))
Since the time is in months and the formulas is in years, I put the (44/12) to represent the time. This will actually simplify to 44.
A = 10,000(1 + .055/12)^(44)
A = $ 12,228.77
Answer:
The answer is in the attachment.
Step-by-step explanation:
Look at the picture.
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
