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Irina18 [472]
3 years ago
10

A person who claims to be psychic says that the​ probability, p, that he can correctly predict the outcome of the value of of a

card drawn from a deck of cards in another room is greater than 1 divided by 13​, the value that applies with random guessing. If we want to test this​ claim, we could use the data from an experiment in which he predicts the outcomes for n trials. State hypotheses for a significance​ test, letting the alternative hypothesis reflect the​ psychic's claim.
Mathematics
1 answer:
Yuki888 [10]3 years ago
5 0

Answer:

The required null and alternative hypothesis are H_0=\frac{1}{13} and H_a>\frac{1}{13}.

Step-by-step explanation:

Consider the provided information.

A person who claims to be psychic says that the​ probability, p, that he can correctly predict the outcome of the value of a card drawn from a deck of cards in another room is greater than 1/13​, the value that applies with random guessing.

To test this claim we need to use the data from an experiment in which he predicts the outcomes for n trials.

Since, alternative hypothesis represents the effect and null hypothesis represents no effect,

Therefore null hypothesis will be: The person can predict outcome of the value of a card drawn in another room 1/13.  H_0=\frac{1}{13}

The alternative hypothesis will be the person can predict outcome of the value of card is greater than 1/3.  H_a>\frac{1}{13}

Hence, the required null and alternative hypothesis are H_0=\frac{1}{13} and H_a>\frac{1}{13}.

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How many different 5​-letter radio station call letters can be made a. if the first letter must be Upper C comma Upper X comma U
Lady_Fox [76]

Answer:

a) 1,518,000

b) 2,284,880

c) 60,720

Step-by-step explanation:

a) a. if the first letter must be Upper C comma Upper X comma Upper T comma or Upper M and no letter may be​ repeated?

We draw 5 boxes, and based on that we will see the total possible cases. There are 26 alphabets

The first box should have C or X or T or M .No letter may be repeated.

Any                    Any             Any                 Any                    Any

5 alphabets    of the            of the              of the                 of the

C,X , T , M      remaining     remaining       remaining          remaining

                   25 alphabets   24 alphabets   23 alphabets   22 alphabets

Therefore; total possible call letters = 5 × 25 × 24 × 23 × 22 = 1,518,000

b)

The first box should have C or X or T or M Repeats as allowed

Any                    Any             Any                 Any                    Any

5 alphabets    of the            of the              of the                 of the

C,X , T , M      remaining     remaining       remaining          remaining

                   26 alphabets   26 alphabets   26 alphabets   26 alphabets

Therefore Total possible call letters = 5 × 26 × 26 × 26 × 26 = 2,284,880

c)   The first box should have   C,X , T , M  and end with S

So the last place if fixed, and we now have 25 alphabets. The first box can go in 5 ways. The next box then will have only 24 letters to choose from, as the first box has taken a letter and the last box already has S in it. Repetition not allowed

Any                    Any             Any                 Any                       S

5 alphabets    of the            of the              of the                  is fixed

C,X , T , M      remaining     remaining       remaining           here

                   24 alphabets   23 alphabets   22 alphabets  

Therefore Total possible call letters = 5 × 24 × 23 × 22  × 1 =  60,720

3 0
3 years ago
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4 0
2 years ago
Which statement compares the two numbers correctly?
Pie

Hey there! We are given two decimal numbers to compare.

52.311 and 52.31

These two numbers are similar except that 52.311 has another 1 in while 51.31 doesn't.

Keep in mind that 52.31 can also refer to 52.310

So we are comparing 52.311 and 52.310

Since 1 is greater than 0.

52.311 is greater than 52.31 (52.311 > 52.31)

Let me know if you have any questions!

Topic: Decimal Numbers

6 0
2 years ago
A pet store has 5 cats. They need to split 7 cups of cat food evenly between each cat. How many cups of cat food will each cat g
kumpel [21]

Answer

1.4 cups

when you divide 7/5 you get 1.4 cups

1.4 lies between 1 and 2

so:

1-------1.4------2

3 0
3 years ago
Chad wants to buy some books over the Internet. Each book costs $10.01 and has a shipping cost of $9.96 per order. If Chad wants
inysia [295]

Answer:

9.96 + 10.01p ≤ 50, so p ≤ 4

Step-by-step explanation:

4 0
2 years ago
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