D, E and F will lie on the line
This problem is a combination of the Poisson distribution and binomial distribution.
First, we need to find the probability of a single student sending less than 6 messages in a day, i.e.
P(X<6)=P(X=0)+P(X=1)+P(X=2)+P(X=3)+P(X=4)+P(X=5)
=0.006738+0.033690+0.084224+0.140374+0.175467+0.175467
= 0.615961
For ALL 20 students to send less than 6 messages, the probability is
P=C(20,20)*0.615961^20*(1-0.615961)^0
=6.18101*10^(-5) or approximately
=0.00006181
Answer:
85/1
Step-by-step explanation:
because both numbers are divisible by 16
237.3f= 1b. Explanation: is 712/3(237.3) so 1b is 237.3f
