Suppose an NBA basketball player shooting an uncontested 2-point shot will make the basket with probability 0.6. How- ever, if you foul t he shooter, t he shot 'vill be missed, but t'vo free thro, vs will be a'varded. Each free thro'v is an independent Bernoulli trial 'vith success probability p. Based on the expected number of points the shooter 'vill score, for what values of p may it be desirable to foul the shooter?

0.06

Step-by-step explanation:

The computation of the value of p is shown below;

P in the case (2 points if not fouled) = 0.6

Now Expected points if not fouled is

= 0.6 × 2

= 1.2

Now Expected points from free throw is

0 × (1-p)^2 + 1 × 2p(1-p) + 2 × p^2 = 2p

As we can see that 2p would be lower tha  1.2 in order to make a foul on shooter that represent all values of p<0.06 would be considered to foul in the shooter

1     3    

step-by-step explanation:

hopefully im right

b. 6

step-by-step explanation:

we are given the table representing an exponential function.

we can see from the table that, as x is increasing, the value of y is increasing in the power of 6.

i.e. the following relation is obtained,

x = 1, y = 6

x = 2, y = 36 =

x = 3, y = 216 =

x = 4, y = 1296 =

x = 5, y = 7776 =

so, the geometric series obtained is .

thus, the common ratio is given by = = 6.

hence, the common ratio is 6.