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
First Equation:

Second Equation can be written as:

Slope of first equation is -3/4 and slope of second equation is 3/4.
Slope of parallel lines must be equal, and slope of perpendicular lines are the negative reciprocal of each other. None of these conditions can be seen for given two equations.
So, the two lines are neither parallel nor perpendicular.
So correct option is C
Answer:
we have the expression as;
1/sin u cos u
Step-by-step explanation:
tan u = sin u/cos u
cot u = cos u/sin u
Thus;
sin u/cos u + cos u/sin u
The lcm is sin u cos u
Thus, we have that;
(sin^2 u + cos^2 u)/sin u cos u
But ; sin^2 u + cos^2 u = 1
so we have ;
1/sin u cos u
Answer:
It’s 5 mph I would be tired if I had to walk 4 miles
5
Step-by-step explanation: