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inessss [21]
2 years ago
5

Pete entered a lucky draw. The probability of winning a free ticket for a soccer game is 15%. What are the ODDS AGAINST winning

a free ticket to the soccer game?
3 to 17
17 to 3
3 to 20
17 to 20
Mathematics
1 answer:
svet-max [94.6K]2 years ago
6 0

Answer:

3 to 20

Step-by-step explanation:

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x = negative 9/4

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-2x-11=6x+7

-2x-11+11=6x+7+11

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5 and 13

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Solve the porportion Solve the porportion Solve the porportion Solve the porportion Solve the porportion Solve the porportion So
RideAnS [48]
Y=21/7

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3 years ago
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A small regional carrier accepted 19 reservations for a particular flight with 17 seats. 14 reservations went to regular custome
Ludmilka [50]

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.

4 0
3 years ago
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