The box plot show the weights, in pounds, of the dogs in two different animal shelters.
1 answer:
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Answer:
The graph crosses the x-axis 2 times
The solutions are x = -8 & x = 4
Step-by-step explanation:
Qaudratics are in the form
Where a, b, c are constants
Now, let's arrange this equation in this form:
Where
a = 1
b = 4
c = -32
We need to know the discriminant to know nature of roots. The discriminant is:
If
- D = 0 , we have 2 similar root and there is 2 solutions and that touches the x-axis
- D > 0, we have 2 distinct roots/solutions and both cut the x-axis
- D < 0, we have imaginary roots and it never cuts the x-axis
Let's find value of Discriminant:
Certainly D > 0, so there are 2 distinct roots and cuts the x-axis twice.
We get the roots/solutions by factoring:
Thus,
The graph crosses the x-axis 2 times
The solutions are x = -8 & x = 4
Answer:
(a) The probability of overbooking is 0.2135.
(b) The probability that the flight has empty seats is 0.4625.
Step-by-step explanation:
Let the random variable <em>X</em> represent the number of passengers showing up for the flight.
It is provided that a small regional carrier accepted 19 reservations for a particular flight with 17 seats.
Of the 17 seats, 14 reservations went to regular customers who will arrive for the flight.
Number of reservations = 19
Regular customers = 14
Seats available = 17 - 14 = 3
Remaining reservations, n = 19 - 14 = 5
P (A remaining passenger will arrive), <em>p</em> = 0.52
The random variable <em>X</em> thus follows a Binomial distribution with parameters <em>n</em> = 5 and <em>p</em> = 0.52.
(1)
Compute the probability of overbooking as follows:
P (Overbooking occurs) = P(More than 3 shows up for the flight)
Thus, the probability of overbooking is 0.2135.
(2)
Compute the probability that the flight has empty seats as follows:
P (The flight has empty seats) = P (Less than 3 shows up for the flight)
Thus, the probability that the flight has empty seats is 0.4625.