Coram is at basketball practice preparing to shoot free throws. the probability of coram making a free throw is 77%. what is the probability that he makes his first free throw on the second shot? a. 79% b. 95% c. 88% d. 83% i think the answer is c

 The answer is 17.71%.

Step-by-step explanation:  Given that Coram is preparing to shoot free throws at basketball practice.

The probability that Coram will make a free throw = 77% = 0.77.

So, the probability of Coram miss the throw = 1 - 0.77 = 0.23.

Therefore, the probability that he makes his first throw on the second shot

= probability of missing the throw in first shot × probability of making a throw

= 0.23 × 0.77

=0.1771

17.71%.

Thus, the required probability is 17.71%.

no association

step-by-step explanation:

if you have eyes and you can use them efficiently you can see that it just looks like random dots like do a line of best fit and like they're so spread out it's such a weak association you might as well consider it nonexistent

80x^8−68x^6+16^5−20x^4−4x^3+8x^2

hope i'm right lol