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zubka84 [21]
1 year ago
12

João claims he can read minds. His friend Pedro asked him to guess a number he was thinking of between 1 and 7 (inclusive). João

guessed Pedro's number correctly 4 out of 9 times.
Let's test the hypothesis that João cannot read minds and therefore has a chance of
1/7 of guessing the correct number each time versus the alternative that his chance is somehow greater.
The table below sums up the results of 1000 simulations, each simulating 9 guesses with a chance of 1/7 start fraction, 1, divided by, 7, end fraction of being correct.
According to the simulations, what is the probability of having 4 or more correct guesses out of ?




Let's agree that if the observed outcome has a probability less than 1\%1%1, percent under the tested hypothesis, we will reject the hypothesis.
What should we conclude regarding the hypothesis?
Choose 1 answer:
Choose 1 answer:

(Choice A)

We cannot reject the hypothesis.

(Choice B)

We should reject the hypothesis.

correct guesses out of 9 Frequency
0 250
1 375
2 250
3 97
4 24
5 4
6 0
7 0
8 0
9 0
Mathematics
1 answer:
Ksju [112]1 year ago
5 0

It can be deduced that the probability of having 4 or more correct guesses will be 0.0296.

<h3>How to calculate the probability</h3>

From the given information, the null hypothesis is p = 1/7 and the alternate hypothesis is p>1/7. In this case, the mind reader gets a 4 out of 9 chances.

In order to test the hypothesis, simulations for 10000 cases were run with a probability of the correct answer being 1/7.

In this case, the p-value is the probability of correct answers that are greater than 4, assuming the null hypothesis is true. Here the p-value is 0.0296.

Also, the significance level to reject the hypothesis is less than 0.01. Since p > 0.01, we then fail to reject the hypothesis. Therefore the null hypothesis stands and Joao is not a mind reader.

Learn more about probability on:

brainly.com/question/24756209

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Jen Butler has been pricing​ Speed-Pass train fares for a group trip to New York. Three adults and four children must pay $ 99. 
8090 [49]

Answer:

An adult ticket is 17

A childs ticket costs 12 dollars

Step-by-step explanation:

Let a = adult price

c = child price

3a + 4c = 99

2a + 3c = 70

Multiply the first equation by 2

2(3a + 4c) = 99*2

6a + 8c = 198

Multiply the second equation by -3

-3 (2a + 3c) = 70*-3

-6a -9c = -210

Add these new equations together

6a + 8c = 198

-6a -9c = -210

----------------------

     -c = -12

Multiply by -1

        c = 12

A childs ticket costs 12 dollars

Now we need to find the price of an adult ticket

2a + 3c = 70

2a + 3(12) = 70

2a +36 = 70

Subtract 36 from each side

2a +36-36 =70-36

2a = 34

Divide by 2

2a/2 = 34/2

a = 17

An adult ticket is 17

7 0
3 years ago
What is the value of x if 2x - 7 = 103 x=__________
Greeley [361]

Answer:

x=55

Step-by-step explanation:

2x-7=103

2x=103+7

2x=110

x=55

5 0
2 years ago
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0.51 + 0.63 = ?<br> and<br> 1.93 + 1.73 = ?
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For the first one: 1.14
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3 years ago
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How many seats are in the library?
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The rectangle below has an area of 70y⁸ 30y⁶. The width of the rectangle is equal to the greatest common monomial factor of 70y⁸
Lorico [155]

Answer:

The Width of the Rectangle =10y^6

\text{Length of the Rectangle}=210y^8

Step-by-step explanation:

The area of the rectangle =70y^8 30y^6.

We are told that the width of the rectangle is equal to the greatest common monomial factor of 70y^8 \: and\: 30y^6.

Let us determine the greatest common monomial factor of 70y^8 \: and\: 30y^6.

Express each term as a product to pick out the common factors:

70y^8 =7X10Xy^6Xy^2\\30y^6=3X10Xy^6

In the two terms, the common terms are 10 and y^6. Therefore their greatest monomial factor =10y^6

The Width of the Rectangle =10y^6

Recall: Area of a Rectangle =Length X Width

70y^8 30y^6=Length X 10y^6\\Length =70y^8 30y^6 \div 10y^6 \\=\dfrac{70X30Xy^8Xy^6}{10y^6} =210y^8\\\text{Length of the Rectangle}=210y^8

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3 years ago
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