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slega [8]
3 years ago
15

You are out with friends. Half of you want to go bowling and the other half want to go to a movie. How will you make a fair deci

sion about whether to go to a movie or go bowling using a fair coin and assuming that all of you want to go to either of the places together? (Let H = heads and T = tails)
1) Flip the coin twice. If the outcome is in the sequence HH, go to the movie. If it is not HH, go bowling.

2) Flip the coin twice. If the outcome is in the sequence HT, go to the movie. If it is TH, go bowling or repeat the process.

3) Flip the coin three times. If the outcome is in the sequence HHT, go to the movie. If it is TTT or HHH, go bowling; otherwise, repeat the process.

4) Flip the coin three times. If the outcome is in the sequence HHH or TTT, go to the movie; otherwise, go bowling.
Mathematics
1 answer:
artcher [175]3 years ago
7 0

Answer:

Option B

Step-by-step explanation:

Given that  Half of you want to go bowling and the other half want to go to a movie

So we must get probability for bowling = prob for movie

Option 1:  Here P(HH) = 1/4 and P(no HH) = 3/4

SInce the two are not equal this is not fair

2) Here P(HT ) =1/4 and P(TH)=1/4 and other outcomes repeat the process

This is fine and hence this option is right

3) P(HHT) = 1/8 and P(TTT or HHH) =1/8+1/8=1/4

This is biased and hence not fair

4) P(HHH or TTT) = 1/4 and P(Other options)=3/4

Biased and hence not fair

Only option B is right.

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3 years ago
the axis of symmetry for the graph of the function is f(x) = 1/4x2 bx 10 is x = 6. what is the value of b?
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we have

f(x)=\frac{1}{4} x^{2} +bx+10

This is a vertical parabola open upward

The axis of symmetry is equal to the x-coordinate of the vertex

The vertex is the point (h,k)

the equation of the axis of symmetry is x=h

In this problem

x=6

so the x-coordinate of the vertex is h=6

<u>Convert the quadratic equation in vertex form</u>

Group terms that contain the same variable, and move the constant to the opposite side of the equation

f(x)-10=\frac{1}{4} x^{2} +bx

Factor the leading coefficient

f(x)-10=\frac{1}{4}(x^{2} +4bx)

Complete the square. Remember to balance the equation by adding the same constants to each side

f(x)-10+b^{2}=\frac{1}{4}(x^{2} +4bx+4b^{2})

Rewrite as perfect squares

f(x)+(b^{2}-10)=\frac{1}{4}(x+2b)^{2}

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remember that

h=6

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substitute in the equation

f(x)=\frac{1}{4}(x-6)^{2}-((-3)^{2}-10)

f(x)=\frac{1}{4}(x-6)^{2}+1

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b=-3


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