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KonstantinChe [14]
3 years ago
9

Suppose you just flipped a fair coin 8 times in a row and you got heads each time! What is the probability that the next coin fl

ip will result in a heads
Mathematics
1 answer:
Svetllana [295]3 years ago
8 0

Answer:

The probability is 1

Step-by-step explanation:

Given

Number of flips = 8

Outcomes = 8 heads

Required

Probability of getting a head in the next row

This problem can be attributed to experimental probability and it'll be solved using experimental probability formula, which goes as follows;

Probability = \frac{Number\ of\ Occurence}{Total\ Trials}

Let P(Head) represents the probability of getting a head in the next row;

P(Head)= \frac{Outcome\ of\ head}{Total\ Flips}

P(Head)= \frac{8}{8}

P(Head)= 1

Hence, the probability of obtaining a head in the next flip is 1

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Step-by-step explanation: 10 times ten = 100

4 times 6 = 24

24 divided by 2= 12

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Solve for y<br> -5x-y&lt;3
valentinak56 [21]
-5x - y < 3.....add y to both sides
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3 years ago
What is the unit rate of 88 students for 4 classes
Nat2105 [25]
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5 0
3 years ago
Find all solutions of the equation: 2cos^2x-cosx=1
Art [367]
If you're using the app, try seeing this answer through your browser:  brainly.com/question/3166243

——————————

Solve the trigonometric equation:

     \mathsf{2\,cos^2\,x-cos\,x=1}\\\\ \mathsf{2\,cos^2\,x-cos\,x-1=0}


Make a substitution:

     \mathsf{cos\,x=t\qquad (-1\le t\le 1)}

and the equation becomes

     \mathsf{2t^2-t-1=0}


Rewrite conveniently  – t  as  + t – 2t,  and then factor the left-hand side by grouping:

      \mathsf{2t^2+t-2t-1=0}\\\\ \mathsf{t\cdot (2t+1)-1\cdot (2t+1)=0}


Factor out  2t + 1:

     \mathsf{(2t+1)\cdot (t-1)=0}\\\\ \begin{array}{rcl} \mathsf{2t+1=0}&~\textsf{ or }~&\mathsf{t-1=0}\\\\ \mathsf{2t=1}&~\textsf{ or }~&\mathsf{t=1}\\\\ \mathsf{t=\dfrac{\,1\,}{2}}&~\textsf{ or }~&\mathsf{t=1} \end{array}


Substitute back for  t = cos x:

     \begin{array}{rcl}\mathsf{cos\,x=\dfrac{\,1\,}{2}}&~\textsf{ or }~&\mathsf{cos\,x=1}\\\\ \mathsf{cos\,x=cos\,60^\circ}&~\textsf{ or }~&\mathsf{cos\,x=cos\,0} \end{array}


Therefore,

     \begin{array}{rcl} \mathsf{x=\pm\,60^\circ+k\cdot 360^\circ}&~\textsf{ or }~&\mathsf{cos\,x=0+k\cdot 360^\circ} \end{array}

where  k  is an integer.


Solution set:   

\mathsf{S=\left\{x\in\mathbb{R}:~~x=-\,60^\circ+k\cdot 360^\circ~~or~~x=60^\circ+k\cdot 360^\circ~~or~~x=k\cdot 360^\circ,~~k\in\mathbb{Z}\right\}}


I hope this helps. =)

3 0
3 years ago
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