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2 years ago

6

Harlene tosses two number cubes. If a sum of 8 or 12 comes up, she gets 9 points. If not, she loses 2 points. What is the expect

ed value of the number of points for one roll?

2 answers:

Alisiya [41]2 years ago

5 0

<span>It's easy enough. Solving looks like that: p(roll of 8)+p(roll of 12) =</span> $\frac{5}{36} + \frac{1}{36} = \frac{1}{6} ; \frac{1}{6} *9 +\frac{5}{6}*2 = \frac{3}{2} - \frac{5}{3} = -\frac{1}{6}$ Hope everything is clear.

konstantin123 [22]2 years ago

3 0

Answer: $-\frac{1}{6}$

Step-by-step explanation:

Since, the total outcomes when two dices are thrown = 36

Also, the total outcomes of getting the sum of 8 or 12 in throwing = {(2,6)(6,2)(3,5),(5,3),(4,4)}= 6

Hence, the probability of getting the sum of 8 or 12 in throwing two dices

$= \frac{6}{36}=\frac{1}{6}$

Also, the probability of getting the sum of 8 or 12 in throwing two dices $= 1-\frac{1}{6}=\frac{5}{6}$

Thus, the expected point he can earn in one throw $= 9\times \frac{1}{6}-2\times {5}{6} = \frac{9}{6}-\frac{10}{6}=-\frac{1}{6}$

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2 years ago

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An aquarium with a square base has no top. There is a metal frame. Glass costs 5 dollars/m2 and the frame costs 2 dollars/m. The

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Answer:

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Step-by-step explanation:

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1 year ago

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In high-school 135 freshmen were interviewed.

timama [110]

Answer:

a) n(none) = 25

b) n(PE but not Bio) = 25

c) n(ENG but not both BIO and PE) = 55

d) n(students that did not take Eng or Bio) = 40

e) P( Students did not take exactly two subjects) = 0.65

Step-by-step explanation:

From the Venn diagram drawn:

a) Number of students that took none

n(Freshmen) = 135

n(all three) = 5

n (PE and Bio) = 10

n(PE and Eng) = 15

n(Bio and Eng) = 7

n (PE and Bio only) = 10 - 5 = 5

n(PE and Eng only) = 15 - 5 = 10

n(Bio and Eng only) = 7 - 5 = 2

n(PE only) = 35 - 5 - 5 - 10 = 15

n(Bio only) = 42 - 5 - 5 - 2 = 30

n(Eng only) = 60 - 10 - 5 -2 = 43

n(Freshmen) = n(PE only) + n(Bio only) + n(Eng only) + n(PE and Bio only) + n(PE and Eng only) + n(Bio and Eng only) + n(all three) + n(none)

135 = 15 + 30 + 43 + 5 + 10 + 2 + 5 + n(none)

135 = 110 + n(none)

n(none) = 135 - 110

n(none) = 25

b)Number of students that too PE but not Bio

n(PE but not bio)= n(PE only) + n(PE and Eng only)

n(PE but not Bio) = 15 + 10

n(PE but not Bio) = 25

c) Number of students that took ENG but not both BIO and PE

n(ENG but not both BIO and PE) = n(Eng only) + n(Eng and Bio only) + n(Eng and PE only) = 43 + 2 + 10

n(ENG but not both BIO and PE) = 55

d) Number of students that did not take ENG or BIO

n( students that did not take Eng or Bio) = n(PE only) + n(none)

n(students that did not take Eng or Bio) = 15 + 25

n(students that did not take Eng or Bio) = 40

e) Probability that a randomly-chosen student from this group did not take exactly two subjects

n( Students that did not take exactly two subjects) = n(PE only) + n(Bio only) + n(Eng only)

n( Students that did not take exactly two subjects) = 15 + 30 + 43

n( Students that did not take exactly two subjects) = 88

P( Students did not take exactly two subjects) = 88/135

P( Students did not take exactly two subjects) = 0.65

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2 years ago

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