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Pepsi [2]
3 years ago
15

At the city museum, child admission is $ 5.20 and adult admission is $ 9.40 . On Monday, 148 tickets were sold for a total sales

of $ 1051.00 . How many child tickets were sold that day?
Mathematics
1 answer:
Olin [163]3 years ago
7 0
5.20x+9.40(148-x)=1051
Solve for x
X=81 child tickets

148-81=67 adult tickets

Check
5.20*81+9.40*67=1051
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Simplify the expression 6(2x+1)+3(6x+2)
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3 years ago
A small regional carrier accepted 19 reservations for a particular flight with 17 seats. 14 reservations went to regular custome
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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)

                                        =P(X>3)\\\\={5\choose 4}(0.52)^{4}(1-0.52)^{5-4}+{5\choose 5}(0.52)^{5}(1-0.52)^{5-5}\\\\=0.175478784+0.0380204032\\\\=0.2134991872\\\\\approx 0.2135

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)

=P(X

Thus, the probability that the flight has empty seats is 0.4625.

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