Kobe bryant is shooting free throws. making or missing free throws doesn't change the probability that he will make his next one, and he makes his free throws 83\%83% of the time. what is the probability of kobe bryant making all of his next 6 free throw attempts? choose 1 a)(1−0.83)^6 b)0.83^6 c)6⋅0.83 d)6⋅(1−0.83)

b)0.83^6

probability of Kobe Bryant making all of his next 6 free throw attempts

=probability of Kobe Bryant making first free throw attempts x

probability of Kobe Bryant making second free throw attempts x

probability of Kobe Bryant making third free throw attempts x

probability of Kobe Bryant making fourth free throw attempts x

probability of Kobe Bryant making fifth free throw attempts x

probability of Kobe Bryant making sixth free throw attempts

= P(make 1) x P(make 2nd) x P(make 3rd) x P(make 4th) x P(make 5th)x P(make 6th) = 0.83^6

0.83^6

Step-by-step explanation:

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step-by-step explanation:

answer: option a.

step-by-step explanation:

you know that the quadrilateral opqr ( a four sided figure) is inscribed in circle. this means that all its vertices are on the circle.

then, by definition, the opposite angles of the quadrilateral are supplementary, this means that the sum of their measures is 180 degrees.

you can observe in the figure that the angles ∠opq   and the angle ∠orq   are opposite, therefore:

then, can be used to d to calculate the measure of ∠opq