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11111nata11111 [884]
3 years ago
13

A fair coin is tossed 3 times in a row. What is the probability that heads appears on only the last toss?

Business
2 answers:
Sveta_85 [38]3 years ago
5 0
Head and tails is a game that only gives a 50/50 chance so if it lands on tails it could land on head and repeat but it always very so nothing is so sure my answer is 50/50 chance
olga_2 [115]3 years ago
3 0

Answer:

12.5%

Explanation:

The answer to this is 12.5% and you can calculate by multiplying the possibility of getting heads or tails, which would be 1/2, and then multiply it by the next one:

(1/2)(1/2)(1/2)=1/8

1/8=.125

The answer in percentage for the probability that heads only appears on the last one would be 12.5%

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The answer to your question is B - Changes of economic structure.

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Duerr Company makes a $69,000, 30-day, 10% cash loan to Ryan Company. The note and interest to be collected at maturity is: (Use
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Answer:

the journal entry to record the loan:

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2 years ago
A small market orders copies of a certain magazine for its magazine rack each week. Let X 5 demand for the magazine, with pmf Su
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Answer:

See explanation below.

Explanation:

Let X the random variable that represent the demand for the magazine, the pmf for X is given by:

X       1            2           3          4        5        6      

P(X)  1/15      2/15       3/15     4/15   3/15     2/15

3 magazines

For this case the total spent is 2*3 = $ 6

And the net revenue for this case would be:

$4-$6 = -$2 , X=1 (demand 1)

$4*2-$6 = $2 , X=2 (demand 2)

$4*3-$6 = $6 , X=3 (demand 3)

For the values of X=4,5,6 the net revenue will be $6 since the number of magazines is 3

And the expected value for the net revenue would be:

E(R) = \frac{1}{15} *(-2) +\frac{2}{15} *(2) +\frac{3}{15}*(6) + \frac{4}{15}*(6) +\frac{3}{15}*(6) +\frac{2}{15}*(6) = \frac{74}{15}=4.93

4 magazines

For this case the total spent is 2*4 = $ 8

And the net revenue for this case would be:

$4-$8 = -$4 , X=1 (demand 1)

$4*2-$8 = $0 , X=2 (demand 2)

$4*3-$8 = $4 , X=3 (demand 3)

$4*4-$8 = $8 , X=4 (demand 4)

For the values of X=5,6 the net revenue will be $8 since the number of magazines is 4

And the expected value for the net revenue would be:

E(R) = \frac{1}{15} *(-4) +\frac{2}{15} *(0) +\frac{3}{15}*(4) + \frac{4}{15}*(8) +\frac{3}{15}*(8) +\frac{2}{15}*(8) = \frac{80}{15}=5.33

As as we can see we have a higher expected value for the case with 4 magazines.

5 0
3 years ago
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Answer:

d

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price - what consumer pays for the good

place - this is where the good is being marketed

promotion - this are the various forms of advertising carried out for the good

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