10 different names are put into a computer, and one of them is Colin, the computer chooses 2 names at random. The computer is set so that the same name can be chosen twice. What is the probability that Colin is chosen at least once?

0.19

Step-by-step explanation:

Given that :

Total possible outcomes = 10

If the same name can be chosen twice, then probability is with replacement ;

Probability = required outcome / Total possible outcomes

Required outcome = number of Collins in name = 1

Probability of choosing Collins atleast once ;

2[P(Collins) * P(not Collins)] + P(Collins) * P(Collins)

P(colins) = 1/10

P(not Collins) = 9/10

2(1/10 * 9/10) + (1/10 * 1/10)

2(9/100) + (1/100)

18/100 + 1/100

19/100

= 0.19

hope it you

step-by-step explanation:

sin(7pi/6 - pi/3)  

= sin(7pi/6 - 2pi/6)  

= sin(5pi/6)  

= sin(pi - pi/6)  

= sin(pi/6)  

= 1/2  

sin 7pi/6 - sin pi/3  

= sin (pi + pi/6) - sin pi/3  

= -sin pi/6 - sin pi/3  

= -1/2 - sqrt(3)/2  

= -(1+sqrt(3))2

answer: my brother can figure it out, imma go get him so i can answer : ) edit: my brother is sleeping, sorry man

step-by-step explanation: