Consider 4 trials, each having the same probability of success. let x denote the total number of successes in these trials.

if e[x]=0.8, find each of the following.

(a) the largest possible value of p{x=4}

(b) the smallest possible value of p{x=4}.

in this case, give possible values for the remaining probabilities:

p{x=0};
p{x=1};
p{x=2};
p{x=3};

a) 0.2

b) 0

Step-by-step explanation:

E[X] = 0*P(X=0)+ 1*P(X=1)+2*P(X=2)+3*P(X=3)+4*P(X=4) = 0.8

(a) The largest possible value of P(X=4) occurs when

P(X=1) = P(X=2) = P(X=3) = 0.

In that case,  

4*P(X=4) = 0.8 ===> P(X=4) = 0.8/4 = 0.2

(b) The smallest possible value of P(X=4).

Obviously, the smallest possible value is P(X=4) = 0

In that case, possible values for the remaining probabilities

could be, for example

P(X=0) = P(X=2) = P(X=3) = 0

P(X=1) = 0.8

step-by-step explanation:

hunhgunyt7 in u uuuhiuteu

hello from mrbilldoesmath!

answer:

25x^2 + 90x + 84

discussion:

mn(x) = m(n(x)) = m( 5x + 9) =

(5x + 9)^2 + 3 =>

25x^2 + 90x + 9^2 + 3 =

25x^2 + 90x + 84                 (as 9^2 = 9*9 = 81)

which is the second choice

regards,  

mrb

p.s.   i'll be on vacation from friday, dec 22 to jan 2, 2019. have a great new year!