Savings account, retirement savings, value of second car, balance in checking account i believe those are the only ones not positive though
Answer:
x = 75 and y = -72
Step-by-step explanation:
It is given that,
1/5 x + 1/8 y = 1 ------(1)
1/2 x − 1/3 y = 1 -------(2)
<u>To find the solutions of the system of equations</u>
Step 1: eq(1) * 5 ⇒
x + 5/8y = 5 ----(3)
Step 2: eq(2) * 2 ⇒
x - 2/3y = 2 -----(4)
Step 3: eq(3) - eq(4) ⇒
x + 5/8y = 5 ----(3)
<u>x - 2/3y = 2 -</u>----(4)
0 +(5/8 - 2/3)y = 3
-1/24 y = 3
y = -24*3 = -72
Step 4: Substitute the value of y in eq(1)
1/5 x + 1/8 y = 1 ------(1)
1/5 x + 1/8 (-72) = 1 ------(1)
1/5 x - 24 = 1
1/5 x = 25
x = 5*25 = 75
Therefor x = 75 and y = -72
Answer:
h=2/ab
Step-by-step explanation:
That answer is about 11.1 repeating.
Answer:
Lamda= 4 students/min, µ= 5 students/min
P= Lamda/µ= 4/5= 0.8
a.) Probability that system is empty= P0= 1-P= 1-0.8= 0.2
b.) Probability of more than 2 students in the system= ∑(n=3 to inf) P^n*P0= (1-P)*(1/(1-P) – (1-P) –(1-P)*P –(1-P)*P^2)= (.2)*(5- - .2 - (.8)*.2 – (.2)*.8^2))= 0.848
Probability of more than 3 students in the system= ∑(n=4 to inf) P^n*P0= (1-P)*(1/(1-P) – (1-P) –(1-P)*P –(1-P)*P^2 – (1-P)*P^3)= 0.768
c.) W(q)= Waiting time in Queue= lamda/µ(µ- lamda)= 4/5(1)= 0.8 minutes
d.) L(q)= lamda*W(q)= 4*.8= 3.2 students
e.) L(System)= lamda/(µ-lamda)= 4 students.
f.) If another server with same efficiency as the 1st one is added, then µ= 6 sec/student= 10 students/min.
P= 4/10= 0.4
Probability that system is empty= P0= 1-.4= 0.6
W(q)= 4/10(10-4)= 0.0667 minutes
L(q)= Lamda*W(q)= 4*.0667=0.2668
L(system)= Lamda/(µ-lamda)= 4/6= .667
Step-by-step explanation:
