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Alex Ar [27]
3 years ago
14

Each question on a multiple-choice exam has four choices. One of the choices is the correct answer, worth 5 points, another choi

ce is wrong but still carries partial credit of 1 point, and the other two choices are worth 0 points.
If a student picks answers at random, what is the expected value of his or her score for a problem?
_____points

If the exam has 30 questions, what is his or her expected score?
_____points
Mathematics
1 answer:
Harlamova29_29 [7]3 years ago
3 0

Answer:

1.5

45

Step-by-step explanation:

The probability for each answer is 1/4

  • (5+1+0+0)/4 = 6/4= 1.5

Each question has an expected value of <u>1.5</u> points.

The expected score of the test is <u>30*1.5</u>= 45 points

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Read 2 more answers
The projected rate of increase in enrollment at a new branch of the UT-system is estimated by E ′ (t) = 12000(t + 9)−3/2 where E
nexus9112 [7]

Answer:

The projected enrollment is \lim_{t \to \infty} E(t)=10,000

Step-by-step explanation:

Consider the provided projected rate.

E'(t) = 12000(t + 9)^{\frac{-3}{2}}

Integrate the above function.

E(t) =\int 12000(t + 9)^{\frac{-3}{2}}dt

E(t) =-\frac{24000}{\left(t+9\right)^{\frac{1}{2}}}+c

The initial enrollment is 2000, that means at t=0 the value of E(t)=2000.

2000=-\frac{24000}{\left(0+9\right)^{\frac{1}{2}}}+c

2000=-\frac{24000}{3}+c

2000=-8000+c

c=10,000

Therefore, E(t) =-\frac{24000}{\left(t+9\right)^{\frac{1}{2}}}+10,000

Now we need to find \lim_{t \to \infty} E(t)

\lim_{t \to \infty} E(t)=-\frac{24000}{\left(t+9\right)^{\frac{1}{2}}}+10,000

\lim_{t \to \infty} E(t)=10,000

Hence, the projected enrollment is \lim_{t \to \infty} E(t)=10,000

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