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3 years ago

Point C is the center of the circle. The measure of angle ACB is 12x – 7. Arc ADB measures 10x – 7. Find the value of x.

1 answer:

6 0

We are asked to solve for the value of x given that the center is the point C and the ∠ACB is 12x-7 while the given arc is equal to 10x-7. To solve this problem, we need to remember that the summation of angle ACB and angle ADB is equal to 360°. Hence, the solution is shown below:
∠ACB + ∠ADB = 360°
12X - 7 + 10X -7 = 360 , combine same term such as:
22x - 14 = 360 , transpose -14 to the right side such as:
22x = 360 + 14
22x = 374 , divide both sides bu 22
22x/ 22 = 374 / 22
x = 17

The answer to this problem is 17 and this is the value of x.

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Step-by-step explanation:

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(a)

Then; we can Determine the conditional mean E[W1|X(t)=2] as follows;

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$= 1- P (X(s) \leq 0|X(t) = 2) \\ \\ = 1 - \dfrac{P(X(s) \leq 0 , X(t) =2) }{P(X(t) =2)}$

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$= 1 - \dfrac{P(X(s) \leq 0 ,P((3 \eq X(t)) - X(s) \leq 5 ) }{P(X(t) = 2)}$

Now $P(X(s) \leq 0) = P(X(s) = 0)$

(b) We can Determine the conditional mean E[W3|X(t)=5] as follows;

$E(W_1|X(t) =2 ) = \int\limits^t_0 X (\dfrac{d}{dx}P(X(s) \geq 3 |X(t) =5 )) \\ \\ = 1- P (X(s) \leq 2 | X (t) = 5 ) \\ \\ = 1 - \dfrac{P (X(s) \leq 2, X(t) = 5 }{P(X(t) = 5)} \\ \\ = 1 - \dfrac{P (X(s) \LEQ 2, 3 (t) - X(s) \leq 5 )}{P(X(t) = 2)}$

Now; $P (X(s) \leq 2 ) = P(X(s) = 0 ) + P(X(s) = 1) + P(X(s) = 2)$

So ; the conditional probability density function of $W_2$ given that X(t)=5 is:

$f_{W_2|X(t)=5}}= (W_2|X(t) = 5) \\ \\ =\dfrac{d}{ds}P(W_2 \leq s | X(t) =5 ) \\ \\ = \dfrac{d}{ds}P(X(s) \geq 2 | X(t) = 5)$

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