Can someone help and show the work for this

On a quiz there are four multiple-choice questions worth 3 points each and two true/false questions worth 1 point each. Each mul

Pani-rosa [81]

Answer: Expected value would be $\dfrac{17}{5}$

Step-by-step explanation:

Since we have given that

Number of multiple choice question = 4

Number of true false questions = 2

Number of points for each multiple choice question = 3

Number of points for each true false question = 1

Number of choices in multiple choice = 5

Probability of getting correct answer in multiple choice question = $\dfrac{1}{5}$

Probability of getting correct answer in true false question = $\dfrac{1}{2}$

Expected value of the student's score on the test would be

$E(x)=\sum xp(x)\\\\E(x)=4\times 3\times \dfrac{1}{5}+2\times 1\times \dfrac{1}{2}\\\\E(x)=\dfrac{12}{5}+1\\\\E(x)=\dfrac{12+5}{5}=\dfrac{17}{5}$

Hence, expected value would be $\dfrac{17}{5}$

3 0

4 years ago

Read 2 more answers

When network cards are communicating, bits can occasionally be corrupted in transmission. Engineers have de- termined that the n

ycow [4]

Answer:

a) 11.40% probability of 5 bits being in error during the transmission of 1 kb

b) 11.60% probability of 8 bits being in error during the transmission of 2 kb

c) 0.01% probability of no error bits in 3kb

Step-by-step explanation:

In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by the following formula:

$P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}$

In which

x is the number of sucesses

e = 2.71828 is the Euler number

$\mu$ is the mean in the giveninterval.

Poisson distribution with mean of 3.2 bits/kb (per kilobyte).

This means that $\mu = 3.2kb$ , in which kb is the number of kilobytes.

(a) What is the probability of 5 bits being in error during the transmission of 1 kb?

This is P(X = 5) when $\mu = 3.2$

$P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}$

$P(X = 5) = \frac{e^{-3.2}*3.2^{5}}{(5)!} = 0.1140$

11.40% probability of 5 bits being in error during the transmission of 1 kb

(b) What is the probability of 8 bits being in error during the transmission of 2 kb?

This is P(X = 8) when $\mu = 2*3.2 = 6.4$

$P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}$

$P(X = 8) = \frac{e^{-6.4}*6.4^{8}}{(8)!} = 0.1160$

11.60% probability of 8 bits being in error during the transmission of 2 kb

(c) What is the probability of no error bits in 3kb?

This is P(X = 0) when $\mu = 3*3.2 = 9.6$

Then

$P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}$

$P(X = 0) = \frac{e^{-9.6}*9.6^{0}}{(0)!} = 0.0001$

0.01% probability of no error bits in 3kb

7 0

3 years ago

The sound intensity of rustling leaves is 100 times the reference intensity. Use your graph to determine the sound intensity of

wariber [46]

The sound intensity of rustling leaves, y = 20 decibels

<u>Step-by-step explanation:</u>

y = 10logx

The sound intensity of rustling leaves is 100 times the reference intensity.

x = ratio of intensity of sound to reference to intensity

x = 100:1 or x = 100

y = 10 log(100) = 10 * 2 = 20 decibels

y = 20 decibels

4 0

3 years ago

Read 2 more answers

Which function has a domain of all real numbers?

Blizzard [7]

Answer:

c part ....

please mark brainlest

4 0

3 years ago

a physician orders 62.25 miligrams of a drug. It is available in 20.75 miligrams tablets. how many tablets should the nurses adm

Kruka [31]

Answer:

3 tablets

Step-by-step explanation:

Since the drug is to be administered in 20.75 mg tablets, we divide 62.25 by 20.75 to get the number of tablets required.

62.25 ➗ 20.75 = 3

3 tablets are to be administered

5 0

3 years ago