At an elite baseball camp, 60% of players can bat both right-handed and left-handed. If 25% of the players who bat left-handed do not bat right-handed, what is the probability that a player selected at random does not bat left-handed?

The probability that a player selected at random does not bat left-handed is 20%.

Step-by-step explanation:

Assume that there are a total of 100 baseball players at an elite baseball camp.

The information provided is:

That is, the number of players can bat both right-handed and left-handed is,

n (L and R) = 60.

Now, if 25% of the players who bat left-handed do not bat right-handed, then 75% of all left-handed players can also bat right-handed.

⇒ n (L and R) = n (L) × 75%

            60     = n (L) × 0.75

           n (L)    =  60/0.75

           n (L)    = 80

So there are 80 left handed batters.

Compute the number of only left handed batters as follows:

n (Only L) = n (L) × 25%

                = 80 × 0.25

                = 20

So there are 20 only left handed batters.

Compute the number of only right handed batters as follows:

Total = n (Only L) + n (Only R) + n (L and R)

 100 = 20 + n (Only R) + 60

n (Only R) = 20

Thus, the probability that a player selected at random does not bat left-handed is 20%.

x = 5

z = 106

Step-by-step explanation:

a)

(10x + 24) + 106 = 180

10x + 130 = 180

10x = 50

x = 5

b) z = 106 (from the definition of vertical angles)