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butalik [34]
3 years ago
8

On the circle below, tangent line BC¯¯¯¯¯ is constructed by striking an arc from point D that intersects circle A at point B. Th

e measure of EC¯¯¯¯¯ is 8 units and other measures are shown on the diagram below. Enter the distance from point D to point B.

Mathematics
1 answer:
weqwewe [10]3 years ago
5 0

Answer:

\huge\boxed{BD = 12\ units}

Step-by-step explanation:

If AB = 5 , then AE = 5 [Radii of the same circle]

So,

AC = AE + EC

AC = 8+5

AC = 13 units

Now, Using Pythagorean theorem to find the missing side i.e. BD because tangent strikes the circle at 90 degrees making the triangle a right angled triangle

c^2=a^2+b^2

Where c = AC , a = BD and b = AB

13^2 = BD^2+5^2

169 = BD² + 25

Subtracting 25 to both sides

169 - 25 = BD²

BD² = 144

Taking square root on both sides

BD = 12 units

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Answer:

the minimum records to be retrieved by using Chebysher - one sided inequality is 17.

Step-by-step explanation:

Let assume that n should represent the number of the students

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∴  Var (x)  = E(x^2) - E^2(x) \\ \\  = .3348711

Now; P(\bar \geq 2.3) = P( \bar x - 2.991013 \geq 2.3 - 2.991013) \\ \\ = P( \omega  \geq .691013)  \ \ \ \  \ \ \ \ \ \ (x = E(\bar x ) - \mu)

Using Chebysher one sided inequality ; we have:

P(\omega \geq -.691013) \geq \dfrac{(.691013)^2}{Var ( \omega) +(.691013)^2}

So; (\omega = \bar x - \mu)

⇒ E(\omega ) = 0 \\ \\ Var (\omega ) = \dfrac{Var (x_i)}{n}

∴ P(\omega \geq .691013) \geq \dfrac{(.691013)^2}{\frac{.3348711}{n}+(691013)^2}

To determine n; such that ;

\dfrac{(.691013)^2}{\frac{.3348711}{n}+(691013)^2} \geq 0.96 \\ \\ \\ (.691013)^2(1-.96) \geq \dfrac{-3348711*.96}{n}

⇒ n \geq \dfrac{.3348711*.96}{.04*(.691013)^2}

n \geq 16.83125

Thus; we can conclude that; the minimum records to be retrieved by using Chebysher - one sided inequality is 17.

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