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Serga [27]
3 years ago
9

Consider a multiple-choice question exam consisting of 20 questions. Assume that each question has four possible choices, and on

ly one of them is correct. If a student is going to guess solutions at the exam, what is the probability that he answers at most three of them are correct
Mathematics
1 answer:
elena-s [515]3 years ago
8 0

Answer:

0.22514

Step-by-step explanation:

Given :

1 answer from 4 options

p = 1/4 = 0.25 ; q = 1 - 0.25 = 0.75

n = 20

P(x ≤ 3) = p(x =0) +p(x=1) + p(x=2) +p(x=3)

P(x = x) = nCr * p^x * q^(n-r)

p(x =0) = 20C0 * 0.25^0 * 0.75^20 = 0.00317

p(x =1) = 20C1 * 0.25^1 * 0.75^19 = 0.02114

p(x =2) = 20C2 * 0.25^2 * 0.75^18 = 0.06694

p(x =3) = 20C3 * 0.25^3 * 0.75^17 = 0.13389

P(x ≤ 3) = 0.00317+0.02114+0.06694+0.13389 = 0.22514

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Answer: An adult’s ticket costs $9 while a children’s ticket costs $5.

Step-by-step explanation:

1). 2x + 3y = 33

2). 5x + 2y = 55

3). I’m going to use substitution.

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2x = -3y + 33

X = (-3y + 33)/2

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Substitute and solve for y:

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Y = 5 <— children’s ticket.

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Check:

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The graph below shows the cost of pens based on the number of pens in a pack. What would be the cost of one pen?
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