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4vir4ik [10]
2 years ago
11

HELP ASAP!!! ILL GIVE BRAINLY TO THE ONE THAT ACTUALLY HELPS ME!!!! PLEASE ASAP HELP NO JOKING! PLUS THERE IS 50 POINTS TO THE B

RAINLIST
Question 4 (Essay Worth 10 points)
(06.05 MC)
Gina spins a penny 100 times and it lands head side up 38 times. Explain why Gina's experimental probability may be different from the theoretical probability of
spinning a coin

Mathematics
2 answers:
RUDIKE [14]2 years ago
8 0

Answer:

The theoretical probability of spinning a head = 1/2 = 0.5

Therefore, if Gina spins a penny 100 times, the theoretical number of times it will land head side up = 0.5 x 100 = 50 times

Gina's experimental probability is 38/100 = 0.38

Therefore, the experimental probability is less than the theoretical probability.

The main difference between experimental and theoretical probability is that experimental probability is based on the actual results of an experiment (gathered by experimenting repeatedly), whereas theoretical probability is based on possible outcomes based on assumptions.

If Gina continued to spin the penny, as the number of times she spun it increased, you would expect for the experimental probability to get nearer to the theoretical probability of 0.5.

ella [17]2 years ago
7 0

The random nature of the process is why Gina doesn't get the theoretical probability. If she were to repeat this experiment say 1000 or perhaps 10,000 times, then her experimental probability value should get closer to 1/2. It likely won't land *exactly* on 1/2 because again of the random nature of the outcomes.

For more information, check out the Law of Large Numbers.

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It is proved that \frac{\partial P}{\partial y}=\frac{\partial Q}{\partial x}, \frac{\partial P}{\partial z}=\frac{\partial R}{\partial x}, \frac{\partial Q}{\partial z}=\frac{\partial R}{\partial y}

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Given vector field,

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\frac{\partial P}{\partial y}=\frac{\partial Q}{\partial x}, \frac{\partial P}{\partial z}=\frac{\partial R}{\partial x}, \frac{\partial Q}{\partial z}=\frac{\partial R}{\partial y}

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\frac{\partial P}{\partial y}=\frac{\partial}{\partial y}(\frac{\partial f}{\partial x})=\frac{\partial^2 f}{\partial y\partial x}=\frac{\partial^2 f}{\partial x\partial y}=\frac{\partial }{\partial x}(\frac{\partial f}{\partial y})=\frac{\partial Q}{\partial x}

\frac{\partial P}{\partial z}=\frac{\partial}{\partial z}(\frac{\partial f}{\partial x})=\frac{\partial^2 f}{\partial z\partial x}=\frac{\partial^2 f}{\partial x\partial z}=\frac{\partial }{\partial x}(\frac{\partial f}{\partial z})=\frac{\partial R}{\partial x}

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Hence proved.

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