Probability distribution for a family who has four children. Let X represent the number of boys. Find the possible outcome of the random variable X, and find: a. The probability of having two or three boys in the family. (1 pt. ) b. The probability of having at least 2 boys in the family. (1 pt. ) c. The probability of having at most 3 boys in the family. (1 pt. )

Since opening night, attendance at play a has increased steadily, while attendance at play b first rose and then fell. equations modeling the daily attendance y at each play are shown below, where x is the number of days since opening night. on what day(s) was the attendance the same at both plays? what was the attendance? play a: y = 8x + 191 play b: y = -x^2 + 26x + 126 a. the attendance was never the same at both plays. b. the attendance was the same on day 5. the attendance was 231 at both plays on that day. c. the attendance was the same on day 13. the attendance was 295 at both plays on that day. d. the attendance was the same on days 5 and 13. the attendance at both plays on those days was 231 and 295 respectively.

3questions show how you solved the problem 1. subtract & simplify (–y2 – 4y – 8) – (–4y2 – 6y + 3) 2.multiply and simplify 2x2y3z2 · 4xy4x2 3.multiply and simplify (x – 4) (x2 – 5x – 6)

Im so bad at fractions they are not my best math thing to work on