Antawn jamison 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 73%, percent of the time. what is the probability of antawn jamison making all of his next 9 free throw attempts?

Probability of Antawn Jamison making all of his next 9 free throw attempts is 0.058 i.e. 5.8%

Step-by-step explanation:

Since we have given that

Probability that he will make his next one and he makes his free throws = 73%

Number of throws = 9

We will use "Binomial distribution":

Hence, probability of Antawn Jamison making all of his next 9 free throw attempts is 0.058 i.e. 5.8%

(0.73)^9

Step-by-step explanation:

Khan Academy

(x3 + 6x2 + 2x - 6) • (x - 1)

step-by-step explanation:

polynomial long division :

3.2     polynomial long division

dividing :   x4+5x3-4x2-8x+6

                              ("dividend")

by         :     x-1     ("divisor")

dividend     x4   +   5x3   -   4x2   -   8x   +   6

- divisor   * x3     x4   -   x3            

remainder         6x3   -   4x2   -   8x   +   6

- divisor   * 6x2         6x3   -   6x2        

remainder             2x2   -   8x   +   6

- divisor   * 2x1             2x2   -   2x    

remainder               -   6x   +   6

- divisor   * -6x0               -   6x   +   6

remainder                     0

quotient :   x3+6x2+2x-6   remainder:   0

think about it deeply and passionately about this property

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