James is playing an arcade game with tokens. In the first
2 answers:
You might be interested in
The probability that a randomly selected car with no 4-wheel drive has third-row seats is given by: Option B: 0.4
<h3>How to calculate the probability of an event?</h3>Suppose that there are finite elementary events in the sample space of the considered experiment, and all are equally likely.
Then, suppose we want to find the probability of an event E.
Then, its probability is given as
where favorable cases are those elementary events who belong to E, and total cases are the size of the sample space.
<h3>What is chain rule in probability?</h3>For two events A and B, by chain rule, we have:
where P(A|B) is probability of occurrence of A given that B already occurred.
For this case, the table given is:
Entries 4-wheel drive No 4-wheel drive Total
Third Row Seats 18 12 30
No Third Row Seats 7 28 35
Total 25 40 65
Let we take
A = event that a randomly selected car has no-4 wheel drive
B = event that a randomly selected car has third row seats
The total ways a car can be selected = 65 = n(S)
Total ways A can happen = n(A) = 40 (from the table).
Similarly, n(B) = 30
P(A) = n(A)/n(S) = 40/65
P(B) = n(B)/n(S) = 30/65
P(A∩B) = n(A∩B)/n(S) = 12/65
As by chain rule, we have:
P(A∩B) = P(A)P(B|A) = P(B)P(A|B)
We need P( A randomly selected car has three seats given that the selected car is with no 4-wheel drive)
which is symbolically P( B | A)
Thus, we use: P(A∩B) = P(A)P(B|A)
or
Thus, the probability that a randomly selected car with no 4-wheel drive has third-row seats is given by: Option B: 0.4
Learn more about probability here:
Answer:
C(x) = $850 - $350x
Monthly Profit = -500x² + 1550x - 850
x = $1.55
the largest possible monthly profit = $351.25
Step-by-step explanation:
Given:
Demand equation:
q = -500x + 1200
where, q is the number of users who log on per month
x is the log-on fee you charge
Site maintenance fee = $10 per month
High-volume access fee = $0.70 per log-on
Now,
C(x) = $10 + ( $0.70 × q )
or
C(x) = $10 + ( $0.70 × (-500x + 1200) )
or
C(x) = $10 + ( - $350x + 840 )
or
C(x) = $850 - $350x
Total monthly income = qx
or
Total monthly income = ( -500x + 1200 ) × x
or
Total monthly income = -500x² + 1200x
now,
Profit = Income - cost
or
Monthly Profit = -500x² + 1200x - ( 850 - 350x )
or
Monthly Profit, P(x) = -500x² + 1550x - 850
For the largest possible monthly profit
= 0
or
=0
or
-2 × 500x + 1550 = 0
or
-1000x + 850 = 0
or
x = $1.55
Now,
the largest possible monthly profit will be at x = $1.55
substitute in the function of profit
P(1.55) = ( -500× 1.55² ) + ( 1550 × 1.55 ) - 850
or
the largest possible monthly profit = - 1201.25 + 2402.5 - 850
or
the largest possible monthly profit = $351.25