1:2 heads 4:6 less than 5 so 4:12
<h3>
Answer:</h3>
B) (x + 7)^2 = 20
<h3>
Step-by-step explanation:</h3>
You can subtract 34 from the original equation to get
... x^2 +14x = -29
Now, you can add the square of half the x-coefficient, and you will have ...
... x^2 +14x +49 = -29 +49
... (x +7)^2 = 20 . . . . . . matches selection B
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<em>Comment on answer choice A</em>
The question asks for the form (x -p)^2 = q, but that does not mean that p must be positive. Here, p=-7 and q=20.
Answer:
slope
Step-by-step explanation:
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Answer:
nu stiu nimic scuze data vitoare
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:
