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Sedaia [141]
3 years ago
8

Suppose that a coin is tossed three times and the side showing face up on each toss is noted. Suppose also that on each toss hea

ds and tails are equally likely. Let HHT indicate the outcome heads on the first two tosses and tails on the third, THT the outcome tails on the first and third tosses and heads on the second, and so forth.
(a) Using set-roster notation, list the eight elements in the sample space whose outcomes are all the possible head-tail sequences obtained in the three tosses.
(b) Write each of the following events as a set, in set-roster notation, and find its probability.
(i) The event that exactly one toss results in a head
(ii) The event that at least two tosses result in a head
Mathematics
1 answer:
sleet_krkn [62]3 years ago
7 0

Answer:

1a

Step-by-step explanation:

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Read 2 more answers
8х + 5у = -7<br> — 7х + бу = — 4
Levart [38]

Answer: x = -0.2651 and y = -0.9759

Step-by-step explanation:

Solution by substitution method

8x+5y= -7

-7x+6y= -4 = 7x-6y=4

Suppose,

8x+5y=-7→(1)

7x-6y=4→(2)

Taking equation (1), we have

8x+5y= -7

⇒8x= -5y -7

⇒x= -5y-7 / 8 →(3)

Putting x= -5y-7 / 8

in equation (2), we get 7x-6y=4

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⇒-35y-49-48y=32

⇒-83y-49=32

⇒-83y=32+49

⇒-83y=81

⇒y= 81 / -83

⇒y= -0.9759→(4)

Now, Putting y= -0.9759 in equation (3), we get

x= -5y-7

x= -5(-0.9759)-7/8

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⇒x= -2.1205/8

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Step-by-step explanation:

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