<h3>3 answers: Choice A, Choice C, Choice D</h3>
Basically, everything but choice B
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Explanation:
Lets go through the answer choices to see which are true or false.
For each, we'll be considering the graph of y = 1/x as shown below.
- A. True. We have two branches that are completely separate and never meet up, and those two branches are in quadrants Q1 and Q3. Both curves are always decreasing (going downhill as you move from left to right). This is because if x is positive, then so is y = 1/x. If x is negative, then so is y = 1/x. Any point (x,y) is either composed of two positive coordinates or two negative coordinates.
- B. False. A parabola is one single curve and we have two separate curves here.
- C. True. A hyperbola is composed of two curves as the graph of y = 1/x shows.
- D. True. The graph shows y = 1/x does not go through the origin. This is because we cannot have x = 0 in y = 1/x, or otherwise we have a division by zero error. A similar situation happens with y = 0 as well.
Answer:
156
Step-by-step explanation:
156
156
156
156
156
156
15
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:
True
aoehfampecfajoifa (sorry had to have at least 20 characters for it to post)
Answer:
Danny will watch all of them on the same day again on August 31
Step-by-step explanation:
Here, we have a word problem.
The shows and frequency of watching are;
Show A Every 2 days
Show B Every 3rd day
Show C Every 5th day
Show D Every 10th day
Danny watched all four shows on August 1, therefore, the day he will watch all four shows can be found by finding the Lowest Common Multiple, LCM of the frequency of his watching each of the shows. That is, the LCM of 2, 3, 5 and 10
Therefore we have, since 10 is the highest frequency and 3 is not a factor of 10, then the LCM that applies to 3 and 10 is 3×10 = 30.
Since the other frequencies are also a factor of 30, then the LCM of 2, 3, 5, and 10 is 30
Therefore, Danny will watch all four shows again on the same day again in 30 days after August 1. That is August (1 + 30) = August 31.
