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:
-1
Step-by-step explanation:
1-2k=-3k
1-2k+2k=-3k+2k
1=-1k
1/-1=-1k/-1
-1=k
Answer:
5
Step-by-step explanation:
95+8=103
95+3=98
103-98=5
Answer:
Add 4 positive unit tiles to each side.
Step-by-step explanation:
The original equation is:
3x - 4 = 2x + 5
After adding 4 positive unit tiles to each side, we get:
3x - 4 + 4 = 2x + 5 + 4
3x = 2x + 9
which simplifies the original equation, because at the beginning there were independent terms at both sides of the equal sign and now all independent terms are on the right side (the +9 term). Notice that the other options don't give this result.
