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
3x=18 because you add the like terms 6 and 12
Answer:
4951 students
Step-by-step explanation:
If there was 99 students from each state, that would make the total to be:
99 * 50 = 4950
We can use the pigeon hole principle to make this the minimum number required.
The pigeon hole principle basically tells us that if you have more "pigeons" than "holes" then there must be one hole with multiple objects in it.
So, using this idea, we see that:
we need at least 1 more to ensure this
So, min number required would be
4950 + 1 = 4951
The computer will finish in 2 seconds
2/3 = 4/6
1 = 6/6
6/6 - 4/6 = 2/6
Answer:
3.2×10⁷ liters/week
Step-by-step explanation:
