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mario62 [17]
3 years ago
13

1 + 10 square theta (1 - sin square theta is equals to 1​

Mathematics
1 answer:
KengaRu [80]3 years ago
3 0

Let's prove

\boxed{\sf 1+tan^2\theta(1-sin^2\theta)=1}

LHS

\\ \sf\longmapsto 1+tan^2\theta(1-sin^2\theta)

\\ \sf\longmapsto 1+\dfrac{sin^2\theta}{cos^2\theta}(1-sin^2\theta)

\\ \sf\longmapsto \dfrac{cos^2\theta+sin^2\theta}{cos^2\theta}(1-sin^2\theta)

\\ \sf\longmapsto \dfrac{1}{cos^2\theta}(1-sin^2\theta)

\\ \sf\longmapsto \dfrac{1-sin^2\theta}{cos^2\theta}

\\ \sf\longmapsto \dfrac{1-sin^2\theta}{1-sin^2\theta}

\\ \sf\longmapsto 1

Hence provee

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Based on probability and if it is appropriate for a decision to be left up to chance, choose every situation that is both statis
svet-max [94.6K]

The situations that are both statistically fair and morally fair are Situation 2, Situation 3 and Situation 5

<h3>How to categorize the situations?</h3>

For a situation to be fair, the probability of every outcome must be equal.

We can now analyze each option using the above highlight

<u>Situation 1</u>

  • Alphabets = 26
  • Probability of vowel = 5/26
  • Probability of consonant = 21/26

The probabilities of vowels and consonant are not equal.

Hence, this situation is not fair

<u>Situation 2</u>

People = 4

P(Each) = 1/4

The probability of each participant is equal in the above scenario i.e. 1/4

Hence, this situation is fair

<u>Situation 3</u>

  • People = 5
  • P(Each) = 1/5

The probability of each cousin is equal in the above scenario i.e. 1/5

Hence, this situation is fair

<u></u>

<u>Situation 4</u>

  • Numbers = 5 i.e. 2 to 6
  • P(Prime) = 3/5
  • P(Composite) = 4/6

The probabilities of prime and composite numbers are not equal.

Hence, this situation is not fair

<u>Situation 5</u>

  • Roommates = 5
  • Cards = 5
  • Aces = 4
  • P(Each) = 1/5

All probabilities are equal in the above scenario i.e. 1/5

Hence, this situation is fair

Read more about probability at:

brainly.com/question/251701

#SPJ1

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