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Solnce55 [7]
2 years ago
9

A deck of playing cards has four suits, with thirteen cards in each suit consisting of the numbers 2 through 10, a jack, a queen

, a king, and an ace. The four suits are hearts, diamonds, spades, and clubs. A hand of five cards will be chosen at random. Which statements are true? Check all that apply.
The total possible outcomes can be found using 52C5.

The total possible outcomes can be found using 52P5.

The probability of choosing two diamonds and three hearts is 0.089.

The probability of choosing five spades is roughly 0.05

The probability of choosing five clubs is roughly 0.0005.
Mathematics
2 answers:
Rudik [331]2 years ago
6 0
A. True. There are 52 cards total and you only pick 5. Order doesn't matter.

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

B. False. Order doesn't matter so you use a combination (instead of a permutation) as choice A shows.

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

C. False. The answer is roughly 0.000858 as shown by the calculation below

(13/52)*(12/51)*(13/50)*(12/49)*(11/48) = 0.000858

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

D. False. See calculation below

(13/52)*(12/51)*(11/50)*(10/49)*(9/48) = 0.000495

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

E. True. See calculation below

(13/52)*(12/51)*(11/50)*(10/49)*(9/48) = 0.000495
this value rounds to roughly 0.0005
adoni [48]2 years ago
3 0

Answer:A & E

Step-by-step explanation:

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Option (B)

Step-by-step explanation:

Given question is not complete; find the complete question in the attachment.

In the graph attached,

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From Statistics and Data Analysis from Elementary to Intermediate by Tamhane and Dunlop, pg 265. A thermostat used in an electri
Leviafan [203]

Answer:

t=\frac{201.77-200}{\frac{2.41}{\sqrt{10}}}=2.32    

The degrees of freedom are given by:

df=n-1=10-1=9  

The p value for this case is given by:

p_v =2*P(t_{(9)}>2.32)=0.0455    

For this case since the p value is lower than the significance level we have enough evidence to reject the null hypothesis and we can conclude that the true mean is significantly different from 200 F.

Step-by-step explanation:

Information given

data: 202.2 203.4 200.5 202.5 206.3 198.0 203.7 200.8 201.3 199.0

We can calculate the sample mean and deviation with the following formulas:

\bar X= \frac{\sum_{i=1}^n X_i}{n}

\sigma=\sqrt{\frac{\sum_{i=1}^n (X_i -\bar X)^2}{n-1}}

\bar X=201.77 represent the sample mean    

s=2.41 represent the sample standard deviation    

n=10 sample size    

\mu_o =200 represent the value that we want to test    

\alpha=0.05 represent the significance level for the hypothesis test.    

t would represent the statistic

p_v represent the p value for the test

Hypothesis to test

We want to determine if the true mean is equal to 200, the system of hypothesis are :    

Null hypothesis:\mu = 200    

Alternative hypothesis:\mu = 200    

The statistic for this case is given by:

t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}} (1)    

The statistic is given by:

t=\frac{201.77-200}{\frac{2.41}{\sqrt{10}}}=2.32    

The degrees of freedom are given by:

df=n-1=10-1=9  

The p value for this case is given by:

p_v =2*P(t_{(9)}>2.32)=0.0455    

For this case since the p value is lower than the significance level we have enough evidence to reject the null hypothesis and we can conclude that the true mean is significantly different from 200 F.

4 0
2 years ago
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