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
(10 1/2)/(3/8)
= (21/2)/(3/8) Convert to improper
= (21/2)*(8/3) Multiply by reciprocal of second
=(7/1)*(4/1) Cross reduce
=28 Multiply
Answer:
2/5
Step-by-step explanation:
First you want to find the least common denominator, which in this case would be 15. If you multiply 2/5 by 3, you get 6/15 which is less than 7/15
