Let us take number of $5 bills = x and
number of $10 bills = y.
Give that "number of $10 bills is twice the number of $5 bills".
So, y is twice of x,
We can setup an equation.
y= 2x ............................... equation(1)
Total value of all bills = $125.
We can setup another equation,
5*(number of $5 bills) + 10*(number of $10 bills) =125.
5(x) +10(y) = 125 ................................... equation(2)
Plugging y=2x in equation(2), we get
5(x) +10(2x) = 125 .
5x+20x = 125.
Adding like terms
25x = 125
Dividing both sides by 25.
25x /25 = 125/25
x= 5.
Plugging x=5 in first equation, we get
y= 2(5) = 10.
Therefore, number of number of $5 bills=5 bills and number of $10 bills = 10 bills.
Here we have a situation where the probability of a driver wearing seat belts is known and remains constant throughout the experiment of stopping 20 drivers.
The drivers stopped are assumed to be random and independent.
These conditions are suitable for modelling using he binomial distribution, where
where n=number of drivers stopped (sample size = 20)
x=number of drivers wearing seatbelts (4)
p=probability that a driver wears seatbelts (0.35), and
C(n,x)=binomial coefficient of x objects chosen from n = n!/(x!(n-x)!)
So the probability of finding 4 drivers wearing seatbelts out of a sample of 20
P(4;20;0.35)
=C(20,4)*(0.35)^4*(0.65)^16
= 4845*0.0150061*0.0010153
= 0.07382
It has one solution. when discriminant=0, tangent to the curve( means touches at one point ) hence one solution.
Answer:
4y is your answer........
The answer is A. 4(n 3) - 6n