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Alecsey [184]
2 years ago
11

Kayla has a standard deck of 52 cards and a six-sided die. What is the probability that she will pull a diamond from the deck of

cards and roll a 1?
Mathematics
2 answers:
gayaneshka [121]2 years ago
4 0
Probability( Pulling a diamond) = 1/4
Propability(Rolling a 1)  = 1/6 

The 2 events are independent so the probabilities are mulitplied:-

Required probability  = 1/4 * 1/6 = 1/24 Answer


stiv31 [10]2 years ago
3 0

Answer: 4.2%

Step-by-step explanation:

The deck has 52 cards, and a standard deck has 13 cards of each type, so there are 13 diamond cards.

Then the probability of drawing a random card that is a diamond card is equal to the number of diamonds card divided by the total number of cards:

p1 = 13/52 = 0.25

now, for the dice we have that the dice has 6 sides, but only one side has a 1, so the probability is;

p2 = 1/6 = 0.167

Now, the probability of both events happening at the same time, is equal to the product of both probabilities is:

P = p1*p2 = 0.25*0.167 = 0.042

So the probability is 0.042*100% = 4.2%

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Answer : y>0

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f(x) is an exponential function

f(x) = 9*2^x

When we plug in positive value for x , the value of y is positive

When we plug in negative value for x , the value y is also positive

So for any value of x, the y value is positive always.

Range is the set of y values for which the function is defined

y values are positive , so range is y >0

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2 years ago
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8) Write an absolute value equation that has 5 and 15 as its solutions?
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(I'm going to use brackets as my absolute value bars lol)
 [5 x -3]
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2 years ago
Suppose you are an expert on the fashion industry and wish to gather information to compare the amount earned per month by model
Ann [662]

Answer:

(1) The degrees of freedom for unequal variance test is (14, 11).

(2) The decision rule for the 0.01 significance level is;

  • If the value of our test statistics is less than the critical values of F at 0.01 level of significance, then we have insufficient evidence to reject our null hypothesis.      
  • If the value of our test statistics is more than the critical values of F at 0.01 level of significance, then we have sufficient evidence to reject our null hypothesis.  

(3) The value of the test statistic is 0.3796.

Step-by-step explanation:

We are given that you are an expert on the fashion industry and wish to gather information to compare the amount earned per month by models featuring Liz Claiborne's attire with those of Calvin Klein.

The following is the amount ($000) earned per month by a sample of 15 Claiborne models;

$3.5, $5.1, $5.2, $3.6, $5.0, $3.4, $5.3, $6.5, $4.8, $6.3, $5.8, $4.5, $6.3, $4.9, $4.2 .

The following is the amount ($000) earned by a sample of 12 Klein models;

$4.1, $2.5, $1.2, $3.5, $5.1, $2.3, $6.1, $1.2, $1.5, $1.3, $1.8, $2.1.

(1) As we know that for the unequal variance test, we use F-test. The degrees of freedom for the F-test is given by;

\text{F}_(_n__1-1, n_2-1_)

Here, n_1 = sample of 15 Claiborne models

         n_2 = sample of 12 Klein models

So, the degrees of freedom = (n_1-1, n_2-1) = (15 - 1, 12 - 1) = (14, 11)

(2) The decision rule for 0.01 significance level is given by;

  • If the value of our test statistics is less than the critical values of F at 0.01 level of significance, then we have insufficient evidence to reject our null hypothesis.      
  • If the value of our test statistics is more than the critical values of F at 0.01 level of significance, then we have sufficient evidence to reject our null hypothesis.  

(3) The test statistics that will be used here is F-test which is given by;

                          T.S. = \frac{s_1^{2} }{s_2^{2} } \times \frac{\sigma_2^{2} }{\sigma_1^{2} }  ~ \text{F}_(_n__1-1, n_2-1_)

where, s_1^{2} = sample variance of the Claiborne models data = \frac{\sum (X_i-\bar X)^{2} }{n_1-1} = 1.007

s_2^{2} = sample variance of the Klein models data = \frac{\sum (X_i-\bar X)^{2} }{n_2-1} = 2.653    

So, the test statistics =  \frac{1.007}{2.653 } \times 1  ~ \text{F}_(_1_4,_1_1_)

                                   = 0.3796

Hence, the value of the test statistic is 0.3796.

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