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Ratling [72]
3 years ago
8

The estimated probability of a bowler getting a strike during a particular frame is 63%. If several simulations of the bowler bo

wling two frames were performed, in what percentage of the simulations would the bowler be most likely to get a strike in each of the two frames?
Mathematics
1 answer:
Vilka [71]3 years ago
3 0

Answer: Our required probability is 39.69%.

Step-by-step explanation:

Since we have given that

Probability of a bowler getting a strike during a particular frame = 63%

Number of frames are performed = 2

We need to find the probability that the bowler would be most likely to get a strike in each of the two frames.

So, it becomes,

\dfrac{63}{100}\times \dfrac{63}{100}=(0.63)^2\\\\=0.3969\\\\=39.69\%

Hence, Our required probability is 39.69%.

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Please hurry! This is due soon.
kotykmax [81]

Hi!

We can see here that this is a composition question.

And since the composition of g of f of x is x, we can conclude that g(x) is the inverse of f(x) (if you're confused, search up the definition of an inverse function).

To find an inverse function, we can take the f(x) function and change the positions of the x and y variables.

f(x)=\frac{e^7^x+\sqrt{3}}{2}

y=\frac{e^7^x+\sqrt{3}}{2}

x=\frac{e^7^y+\sqrt{3}}{2}

2x=e^7^y+\sqrt{3}

e^7^y=2x-\sqrt{3}

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y=\frac{ln(2x-\sqrt{3})}{7}

Which is answer choice A, to check your work, you can solve the composition of g(f(x)), which will get you x.

g(f(x))

g(\frac{e^7^x+\sqrt{3}}{2})

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2s cancel.

\frac{ln(e^7^x+\sqrt{3})-\sqrt{3}}{7}

The natural log and e cancel.

\frac{7x+\sqrt{3}-\sqrt{3}}{7}

\sqrt{3}s cancel.

\frac{7x}{7}

7s cancel.

x

Hope this helps!

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Answer:

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Hope this helps ∝

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