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
Answer:
Washing cars= 4 hours
Walking dogs= 10 hours
Step-by-step explanation:
You want to start by creating equations. So one thing we know is that he makes $9 an hour washing cars(x) and $8 walking dogs(y).
$9x+$8y=$116
The second Equation is based off of the hours worked. We know that he worked 6 hours more walking the dogs than he did washing cars, so we can take x(being the washing hours) and add 6 to it to equal y (the number of dog hours).
y=x+6
Now You plug what y equals into the first equation to solve for x.
9x+8(x+6)=116 Next distribute the 8 to each term.
9x+8(x)+8(6)=116
9x+8x+48=116 Add the like terms together (9x+8x)
17x+48=116 Subtract the 48 from both sides
-48 -48
17x=68 Now divide by 17 on both sides.
______
17 17
x=4 Finally we can take x and plug it back in to one of the equations in order to solve for y. I'm going to choose the second equation.
y=(4)+6
y=10
Answer:
Yes mark me brainliest
Step-by-step explanation:
Could it be an Imaginary number?
