Answer:
No, because it does not have a constant rate of change.
Step-by-step explanation:
On the <em>x</em><em> </em>side of the table, there is a constant rate of change (+1). However, on the y side, it is not. The first change is +7, the next +5, and the last +6. The rate of change has to be constant on both sides for the table to be considered linear.
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
Using the properties of arcs and inscribed angles, B is 110/2=55.
Answer:
to need interdependence in
the country
Step-by-step explanation:
