Answer: 210
Explanation:
Each box has 12 packets of fruit snacks and 18 packets of cashews. That gives 12+18 = 30 packets overall per box. Since we have 7 boxes, that tells us the total count is 7*30 = 210
Answer:
4.8%
Step-by-step explanation:
28+12=40
40÷100=0.4
1% - 0.4 teachers
0.4×12=4.8
<h3>
Answer: C. 3/10</h3>
A terminating decimal only happens when the fraction has a denominator with factors of 2 and/or 5. No other factors are allowed (other than 1).
Something like 1/12 has the denominator 12 with factors 2*2*3. That '3' means we won't have a terminating decimal. Similarly, the fractions 2/9 and 2/3 won't convert to a terminating decimal either.
3/10 on the other hand does convert to a terminating decimal and 3/10 = 0.3
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:
