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:
AG = 16
Step-by-step explanation:
If you have a parallelogram, then the diagonals are bisected, meaning GS and SA are congruent, same with OS and ST. So, because 8 + 8 = 16, the length of AG = 16.
9514 1404 393
Answer:
(x -t -250)/2
Step-by-step explanation:
The amount left from x after paying t and 250 is ...
x -t -250
If this is split into two equal parts, each share is ...
share = (x -t -250)/2
Answer:
He owned Cat for 5 years.
Step-by-step explanation:
Given: Owned dog for 9 years.
Dog owned year is one year less than twice as long as he owned a cat.
Lets assume the years that cat has been owned be "x".
Now, finding the number years that Cat hes been owned.
As given, Dog owned year is one year less than twice as long as he owned a cat.
∴ using equation to find the Cat´s owned years.
⇒ 
adding 1 on both side.
⇒ 
dividing both side by 2
⇒
∴x= 
Hence, Cat has been owned for 5 years.
Answer:
17 yards and 31 yards.
Step-by-step explanation:
Since length is 14 more than width, lets call width: w and length: w+14
So perimeter is 96 so: 2w+2(w+14)=96
2w+2w+28=96 so 4w+28=96
4w=68 and finally w=17
So 17 yards and 31 yards.
