Answer:
a) All of them are out of charge = 9.31x10⁻¹⁰
b) 20% of them are out of charge = 5.529x10⁻⁴
Step-by-step explanation:
This problem can be modeled as a binomial distribution since
There are n repeated trials and all of them are independent of each other.
There are only two possibilities: battery is out of charge and battery is not out of charge.
The probability of success does not change with trial to trial.
Since it is given that it is equally likely for the battery to be out of charge or not out of charge so probability of success is 50% or 0.50
P = 0.50
1 - P = 0.50
a) All of them are out of charge?
Probability = nCx * P^x * (1 - P)^n-x
Probability = ₃₀C₃₀(0.50)³⁰(0.50)⁰
Probability = 9.31x10⁻¹⁰
b) 20% of them are out of charge?
0.20*30 = 6 batteries are out of charge
Probability =₃₀C₆(0.50)²⁴(0.50)⁶
Probability = 5.529x10⁻⁴
Answer:
39 or 1/39
Step-by-step explanation:
You need to divide these but it depends on how you divide them
The equation for this is 100,200(2)^n where n is the number of years
so in 3 years the population will be 100,200(2)^3 = 801,600
9, because one square megameter is equal to one million square kilometers.
