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
(1,2,2,2,2),(2,1,2,2,2)
Observe, the 1 can go in 5 diff places
so 5 ways
Answer:
They do not.
Step-by-step explanation:
a² + b² = c² c is the hypotenuse
9² + 12² = 17²
81 + 144 = 289
225 ≠ 289
The formula for this is A=P(1+r/n)^nt
Where A is the total amount. P is the principal amount. 1 represent 100%. R is the rate. N is the number for annual, quarter, or etc. T is the time given in the question. If you put the numbers that is given in the question and put it in the formula, you will have:
$25,710 = $8192(1 + 0.10/12)^12t
3.14= (1.00833333)^12t
12t= log1.00833333 3.14
t= (log1.00833333 3.14)/12
t is approximately 11.5 yrs
