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3 years ago

Given: A circle with inscribed quadrilateral ABCD

Mathematics

2 answers:

hammer [34]3 years ago

8 0

Step-by-step explanation:

$m\angle BAD = \frac{1}{2} m(\overset {\frown} {BCD}) ... (1)\\$

(By inscribed angle theorem)

$m\angle BCD = \frac{1}{2} m(\overset {\frown} {BAD}) ... (2)\\$

(By inscribed angle theorem)

Adding equations (1) & (2)

$m\angle BAD+m\angle BCD \\= \frac{1}{2} m(\overset {\frown} {BCD}) +\frac{1}{2} m(\overset {\frown} {BAD}) \\\\= \frac{1}{2} (m\overset {\frown} {BCD} +m\overset {\frown} {BAD}) \\\\= \frac{1}{2}\times 360°\\\\= 180°\\\\\purple {\boxed {\bold {\therefore m\angle BAD+m\angle BCD =180°}}} \\\\\therefore \angle A \: and \: \angle C \: are \: supplementary\\$

nadezda [96]3 years ago

8 0

Answer: a/2

Step-by-step explanation:

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3 years ago

A binomial distribution

Alina [70]

Answer:

0.324

Step-by-step explanation:

Given that :

Success rate = 30%

p = 30% = 0.3

q = 1 - p = 1 - 0.3 = 0.7

Number of trials, n = 6

Probability of having exactly 2 successes ; x = 2

P(x = 2)

Usibgbtge binomial probability relation :

P(x = x) = nCx * p^x * q^(n-x)

P(x = 2) = 6C2 * 0.3^2 * 0.7^4

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P(x = 2). = 0.324135

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