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labwork [276]
3 years ago
14

The college student senate is sponsoring a spring break Caribbean cruise raffle, with proceeds going to the local homeless shelt

er. A local travel agency donated the cruise, worth an estimated $3000. The students sold 2900 tickets at $5 per ticket. Find the mathematical expectation of a student who purchases 10 tickets.
Mathematics
1 answer:
drek231 [11]3 years ago
8 0

Answer:

The mathematical expectation of a student who purchases 10 tickets is -$39.65.

Step-by-step explanation:

A student that purchases 10 tickets out of 2900 has a probability of winning the cruise that can be calculated as:

p=10/2900\approx0.00345

Each ticket cost $5, so he has spent $50 for the 10 tickets.

Then, the expected value of this operation is equal to the expected value of the earnings (probability of winning the prize multiplied by the value of the prize), minus the costs:

E(X)=p\cdot R-C\\\\E(X)=0.00345\cdot3000-50=10.35-50=-39.65

The mathematical expectation of a student who purchases 10 tickets is -$39.65.

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Step-by-step explanation:

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Yves bought 420 tropical fish for a museum display. He bought 6 times as many parrotfish as angelfish. How many of each type of
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The answer would be d , because you have to add how many of each fish to equal up to the sum of 420 . which is x y . so the first equation is x + y = 420 . then it says he bought 6 times as many parrotfish as he did angelfish so you would have to multiply x which represents angelfish by 6 . to get the second equation of y = 6x .
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The table displays the number of students in each grade at Richmond High School who will or will not be attending the amusement
Pani-rosa [81]

Answer:

P( eleventh\: \: |\: \: FT)=  0.2453 \\\\P( eleventh\: \: |\: \: FT)=  24.53\% \\\\

Step-by-step explanation:

We are given a joint probability table.

There are four different graders in a school

1. Grade Ninth

2. Grade Tenth

3. Grade Eleventh

4. Grade Twelfth

Field trip refers to the students who will attending the amusement park field trip.

No field trip refers to the students who will not be attending the amusement park field trip.

We want to find out the probability that the selected student is an eleventh grader given that the student is going on a field trip.

P( eleventh\: \: |\: \: FT)= \frac{P( eleventh\: \: and \: \: FT)}{P(FT)}

Where P(eleventh and FT) is the probability of students who are in eleventh grade and will be going to field trip

P(eleventh\:  and \: FT) = \frac{13}{92}  \\\\P(eleventh \:  and \: FT) = 0.1413

Where P(FT) is the probability of students who will be going to field trip

P(FT) = \frac{12}{92} + \frac{9}{92} + \frac{13}{92} + \frac{19}{92}\\\\P(FT) = 0.1304 + 0.0978 + 0.1413 + 0.2065\\\\P(FT) = 0.576 \\\\

So the required probability is

P( eleventh\: \: |\: \: FT)= \frac{P( eleventh\: \: and \: \: FT)}{P(FT)} \\\\P( eleventh\: \: |\: \: FT)= \frac{0.1413}{0.576} \\\\P( eleventh\: \: |\: \: FT)=  0.2453 \\\\P( eleventh\: \: |\: \: FT)=  24.53\% \\\\

6 0
3 years ago
The population of a small town in central Florida has shown a linear decline in the years 1992-2000. In 1992 the population was
wlad13 [49]

Answer:

  • P = -390t +48,000
  • P = 43,710 in 2003

Step-by-step explanation:

A) We are given two pairs of (t, P): (0, 48000), (8, 44880). We can use these with the 2-point form of the equation of a line:

  y = (y2 -y1)/(x2 -x1)(x -x1) +y1

  P = (44880 -48000)/(8 -0)(t -0) +48000

  P = -3120/8t +48000

  P = -390t +48000

__

B) In 2003, t = 11, so we have ...

  P = -390(11) +48000 = 43710

The population in 2003 is predicted to be 43,710.

7 0
3 years ago
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