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blagie [28]
2 years ago
11

You and a friend are playing a game of chance. Every time you roll a 1 or 2 you are successful, and your friend will pay you $1.

Every time you roll (using a fair die) a 3, 4, 5 or 6, you must pay your friend $2. If 71% of your rolls are successful and 29% of your rolls are unsuccessful, how much money do you expect to have earned/owed by the end of the game
Mathematics
1 answer:
Viktor [21]2 years ago
8 0

Considering the mean of a discrete distribution, you are expected to earn $0.13 by the end of the game.

<h3>What is the mean of a discrete distribution?</h3>

The expected value of a discrete distribution is given by the <u>sum of each outcome multiplied by it's respective probability</u>.

Since 71% of your rolls are successful and 29% of your rolls are unsuccessful, the distribution of your earnings is:

  • P(X = 1) = 0.71.
  • P(X = -2) = 0.29

Hence the expected value is:

E(X) = 1 x 0.71 - 2 x 0.29 = 0.71 - 0.58 = 0.13

More can be learned about the mean of a discrete distribution at brainly.com/question/24245882

#SPJ1

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Michael thought he could read 23 pages of his book in class. He was only able to read 20. what is his percent error
Cerrena [4.2K]
Percent error is calculated using

((Actual - predicted) / actual) * 100

In this case, the actual value of the number of pages he read is 20. The expected value was 23. So simply plug in:

((20 - 23) / 20) * 100 = -15%

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2 years ago
Which inequality is NOT true when x= -2?
Levart [38]

Answer: D

Step-by-step explanation:

-7 isn't greater than -4

3 0
2 years ago
A laptop producing company also produces laptop batteries, and claims that the batteries
gregori [183]

Answer:

There is enough evidence to support the claim that the batteries power the laptops for significantly less than 4 hours. (P-value = 0).

The null and alternative hypothesis are:

H_0: \mu=4\\\\H_a:\mu< 4

Step-by-step explanation:

<em>The question is incomplete: To test this claim a sample or population standard deviation is needed.</em>

<em>We will estimate that the sample standard deviation is 2 hours, and use a t-test to test that claim.</em>

<em> NOTE (after solving): The difference between the sample mean and the mean of the null hypothesis is big enough to reject the null hypothesis, even when we have a sample standard deviation of 3.5 hours, which can be considered bigger than the maximum standard deviation for the sample.</em>

This is a hypothesis test for the population mean.

The claim is that the batteries power the laptops for significantly less than 4 hours.

Then, the null and alternative hypothesis are:

H_0: \mu=4\\\\H_a:\mu< 4

The significance level is 0.05.

The sample has a size n=500.

The sample mean is M=3.5.

As the standard deviation of the population is not known, we estimate it with the sample standard deviation, that has a value of s=2.

The estimated standard error of the mean is computed using the formula:

s_M=\dfrac{s}{\sqrt{n}}=\dfrac{2}{\sqrt{500}}=0.0894

Then, we can calculate the t-statistic as:

t=\dfrac{M-\mu}{s/\sqrt{n}}=\dfrac{3.5-4}{0.0894}=\dfrac{-0.5}{0.0894}=-5.5902

The degrees of freedom for this sample size are:

df=n-1=500-1=499

This test is a left-tailed test, with 499 degrees of freedom and t=-5.5902, so the P-value for this test is calculated as (using a t-table):

P-value=P(t

As the P-value (0) is smaller than the significance level (0.05), the effect is  significant.

The null hypothesis is rejected.

There is enough evidence to support the claim that the batteries power the laptops for significantly less than 4 hours.

5 0
3 years ago
Can someone help me with the ones that aren’t done
garri49 [273]

5:

C = 5/9(F - 32)

C = 5/9(F) - 17.777

-5/9(F) + C = -17.777

-5/9(F) = -C - 17.777

F = 5/9(C) + 32.0306

7 0
3 years ago
Read 2 more answers
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