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just olya [345]
3 years ago
12

Prove the following: 1 by sin 2A + cos4a by sin 4A equals to COT A minus Cosec4A​

Mathematics
2 answers:
n200080 [17]3 years ago
4 0

Answer:

cotA-cosec4A

Step-by-step explanation:

LHS=1/sin2A + cos4A/sin4A

=1/sin2A +cos4A/2sin2.Acos2A

=1/sin2A (1+cos4A/2cos2A)

=1/sin2A(2cos2A+cos^2A-sin^3A)/2Cos2A

=1/sin2A(2cos2A+cos^2Ac-(1-cos^2A)/2cos2A

=1/sin2A(2cosA(1+cos2A)-1)/2cos2A

=1/sin2A(1+cos2A-1/2cos2A)

=1+cos2A/sin2A-1/sin2A.cos2A

=1+2cos^A-1/2sinA.cosA-1/sin4A

=2cos^A/2sinA.cosA-1/sin4A

=cosA/sinA-1/sin4A

=cotA-cosec4A

=LHS=RHS

Solnce55 [7]3 years ago
3 0

Answer:

nice thanks for answer

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How to solve 5a + c = -8a
sergey [27]

Solve for a:

5a+c=-8a\ \ \ |-c\\\\5a=-8a-c\ \ \ |+8a\\\\13a=-c\ \ \ |:13\\\\\boxed{a=-\dfrac{c}{13}}

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6 0
4 years ago
According to the Knot, 22% of couples meet online. Assume the sampling distribution of p follows a normal distribution and answe
Ann [662]

Using the <em>normal distribution and the central limit theorem</em>, we have that:

a) The sampling distribution is approximately normal, with mean 0.22 and standard error 0.0338.

b) There is a 0.1867 = 18.67% probability that in a random sample of 150 couples more than 25% met online.

c) There is a 0.2584 = 25.84% probability that in a random sample of 150 couples between 15% and 20% met online.

<h3>Normal Probability Distribution</h3>

In a normal distribution with mean \mu and standard deviation \sigma, the z-score of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

  • It measures how many standard deviations the measure is from the mean.
  • After finding the z-score, we look at the z-score table and find the p-value associated with this z-score, which is the percentile of X.
  • By the Central Limit Theorem, for a proportion p in a sample of size n, the sampling distribution of sample proportion is approximately normal with mean \mu = p and standard deviation s = \sqrt{\frac{p(1 - p)}{n}}, as long as np \geq 10 and n(1 - p) \geq 10.

In this problem:

  • 22% of couples meet online, hence p = 0.22.
  • A sample of 150 couples is taken, hence n = 150.

Item a:

The mean and the standard error are given by:

\mu = p = 0.22

s = \sqrt{\frac{p(1 - p)}{n}} = \sqrt{\frac{0.22(0.78)}{150}} = 0.0338

The sampling distribution is approximately normal, with mean 0.22 and standard error 0.0338.

Item b:

The probability is <u>one subtracted by the p-value of Z when X = 0.25</u>, hence:

Z = \frac{X - \mu}{\sigma}

By the Central Limit Theorem:

Z = \frac{X - \mu}{s}

Z = \frac{0.25 - 0.22}{0.0338}

Z = 0.89

Z = 0.89 has a p-value of 0.8133.

1 - 0.8133 = 0.1867.

There is a 0.1867 = 18.67% probability that in a random sample of 150 couples more than 25% met online.

Item c:

The probability is the <u>p-value of Z when X = 0.2 subtracted by the p-value of Z when X = 0.15</u>, hence:

X = 0.2:

Z = \frac{X - \mu}{s}

Z = \frac{0.2 - 0.22}{0.0338}

Z = -0.59

Z = -0.59 has a p-value of 0.2776.

X = 0.15:

Z = \frac{X - \mu}{s}

Z = \frac{0.15 - 0.22}{0.0338}

Z = -2.07

Z = -2.07 has a p-value of 0.0192.

0.2776 - 0.0192 = 0.2584.

There is a 0.2584 = 25.84% probability that in a random sample of 150 couples between 15% and 20% met online.

To learn more about the <em>normal distribution and the central limit theorem</em>, you can check brainly.com/question/24663213

4 0
2 years ago
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