Answer:
a) 0.125
b) 7
c) 0.875 hr
d) 1 hr
e) 0.875
Step-by-step explanation:l
Given:
Arrival rate, λ = 7
Service rate, μ = 8
a) probability that no requests for assistance are in the system (system is idle).
Let's first find p.
a) ρ = λ/μ
Probability that the system is idle =
1 - p
= 1 - 0.875
=0.125
probability that no requests for assistance are in the system is 0.125
b) average number of requests that will be waiting for service will be given as:
λ/(μ - λ)
= 7
(c) Average time in minutes before service
= λ/[μ(μ - λ)]
= 0.875 hour
(d) average time at the reference desk in minutes.
Average time in the system js given as: 1/(μ - λ)
= 1 hour
(e) Probability that a new arrival has to wait for service will be:
λ/μ =
= 0.875
Answer:
Step-by-step explanation:
It can be convenient to compute the length of the hypotenuse of this triangle (AC). The Pythagorean theorem tells you ...
AC^2 = AB^2 + CB^2
AC^2 = 4^2 + 3^2 = 16 + 9 = 25
AC = √25 = 5
The altitude divides ∆ABC into similar triangles ∆AHB and ∆BHC. The scale factor for ∆AHB is ...
scale factor ∆ABC to ∆AHB = AB/AC = 4/5 = 0.8
And the scale factor to ∆BHC is ...
scale factor ∆ABC to ∆BHC = BC/AC = 3/5 = 0.6
Then the side AH is 0.8·AB = 0.8·4 = 3.2
And the side CH is 0.6·BC = 0.6·3 = 1.8
These two side lengths should add to the length AC = 5, and they do.
The remaining side BH can be found from either scale factor:
BH = AB·0.6 = BC·0.8 = 4·0.6 = 3·0.8 = 2.4
_____
The sides of interest are ...
AH = 3.2
CH = 1.8
BH = 2.4
I think that it would be C, but I am debating on C and D. I'm pretty confident that it's C, though. Pretty sure.
Answer:
126 groups
Step-by-step explanation:
we know that
This is a combination problem, because the order doesn't matter. He has 9 toys and wants to know the number of different combinations of 5 he can take.
The formula for the number of possible combinations of r objects from a set of n objects is given by
In this problem we have
substitute
Answer:
The two triangles are related by A - S - A test of similarity , so the triangles are congruent.
