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
Without the instructions, I can only assume that you are to "expand the given expression."
Following order of operations rules requires that (3x-2)^2 and -6(2x-1.5) be evaluated first.
(3x-2)^2 = 9x^2 - 12x + 4 and -6(2x-1.5) = -12x + 9
Then we have 9x^2 - (9x^2 - 12x + 4) -12x + 9
the 9x^2 terms cancel, leaving us with 12x - 4 - 12x + 9 = 5 (answer)
Answer:
it's a repeating decimal and is that all the question says?
42 is 21% of 200. This is because 200*0.21 = 42.
