The available balance online may not reflect all debit transactions or all checks that were written. Some recent debit transactions and check written may not have been posted to his account yet. The available balance would need to subtract any outstanding debit transactions and any written checks that have not cleared his account. This would yield the correct available balance.
Step-by-step explanation:
first you have to see the triangle BCD
then hypotheses and perpendicular are given so you have to find base
after finding base. In rectangle ABCD DC is length and BC is breadth so now you can find area by using the formula A = l×b
Answer:
what?///////
Step-by-step explanation:
A x = 0
using the law of exponents 
= 1
for (6² )^ x = 1 then x = 0
B note that 
= 1 ⇒ x = 1
2 → 2^8 × 3^(-5) × 1^(-2) × 3^(-8) × 2^(-12) × 2^(28)
= 2^(8 -12 + 28) × 1 × 3^(- 5 - 8)
= 2^24 × 3^(- 13) = 2^(24)/3^(13) = 10.523 ( 3 dec. places)
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
