Aschool newspaper reporter decides to randomly survey 18 students to see if they will attend tet (vietnamese new year) festivities this year. based on past years, she knows that 24% of students attend tet festivities. we are interested in the number of students who will attend the festivities. part (a) in words, define the random variable x. the types of festivities offered at the school the school newspaper the number of students that will attend tet the number of students in the school part (b) list the values that x may take on. x = 0, 1, 2, 24 x = 1, 2, 3, 24 x = 0, 1, 2, 18 x = 1, 2, 3, 18 part (c) give the distribution of x. x ~ , part (d) how many of the 18 students do we expect to attend the festivities? (round your answer to the nearest whole number.) student(s) part (e) find the probability that at most 4 students will attend. (round your answer to four decimal places.) part (f) find the probability that more than 2 students will attend. (round your answer to four decimal places.)

drop out of school or go to quizlet, quizlet is better

step-by-step explanation:

false

step-by-step explanation:

false : it can not go to 9 the highest that part can go is 7

here is your proof

tell you what. i'll find the length of the medians and you can fill in the blanks where they belong.

br

b = (2a,0)

r = (a/2,b/2) remember these things are medians. they go to the 1/2 way point of the line opposite the vertex they face.

br^2 =   (2a - a/2)^2 + (0 - b/2)^2

br^2 = (3/2 a) ^2 + b^2 / 4

br^2 = 9/4 a^2 + b^2 / 4   we need to find some relationship between a and b.

let's try ab = bc  

ab = 2a

bc = (2a - a)^2 + (b - 0)^2

bc = sqrt(a^2 + b^2)

ab = bc

2a = sqrt(a^2 + b^2) square both sides.

4a^2 = a^2 + b^2 subtract a^2 from both sides.

sqrt(3a^2) = sqrt(b^2)

sqrt(3)a = b

let's put b into br^2

br^2 = 9/4 a^2 + 3b^2 / 4

br^2 = 12 a^2 / 4

br^2 = 3a^2 

br = sqrt(3) * a

cp

c = (a,b) ; p = (a,0) this is another application of the distance formula, and it is a good one.

cp^2 = (a - a)^2 + (b - 0)^2

cp^2 = 0 + b^2

cp^2 = b^2 which you can see from the diagram. 

cp = sqrt(b^2)

cp = b but b = sqrt(3) * a 

cp = sqrt(3)*a

aq

a = (0,0)

q = (3/2) a, b/2

aq^2 = (3a/2 -0)^2 + (b/2 - 0)^2

aq^2 = 9a^2/4 + b^2/4

but b = sqrt(3) * a 

aq^2 = 9a^2 / 4 + (sqrt(3)*a)^2 /4

aq^2 = 9 a^2/ 4 + 3 a^2 / 4

aq^2 = 12 a^2/4

aq^2 = 3 a^2

aq = (sqrt  3) * a which agrees with the other two.