Pleass answer the following

I will mark you

Nana76 [90]

Answer:

Yes, because the x's do not have more than one y.

Step-by-step explanation:

x y

0 0

5 10

6 12

10 20

(That's meant to be a table). Looking at the x and y table values, you can see that none of the x's have two y's. So, this relation is a function.

7 0

3 years ago

Find the equation of the line passing through points (-1,-6) and (3,5)

Phoenix [80]

Answer:

y=11/4x-13/4

Step-by-step explanation:

Use the slope formula and slope-intercept form y=mx+b to find the equation.

7 0

2 years ago

A parabola passes through the points (–5, –10) and (5, –10), and the vertex is located at (0, 0). Find the equation of the parab

vitfil [10]

The equation of the parabola is -0.4x^2

8 0

3 years ago

(09.06 HC) The function H(t) = −16t2 + 112t + 24 shows the height H(t), in feet, of a cannon ball after t seconds. A second cann

erma4kov [3.2K]

PART A:

The table of h(t) and g(t) for the value of x between 4 and 7 inclusive are shown below.

The solution for h(x)=g(x) lies between x = 4 and x = 5, this is becuase the range of value given by h(4) and h(5) and g(4) and g(5).

The graph is shown in picture 2 to confirm this. The two functions intersect each other between x = 4 and x = 5

PART B:

The point of intersection is the point where the cannon balls will collide.

3 0

3 years ago

Read 2 more answers

A new shopping mall is considering setting up an information desk manned by one employee. Based upon information obtained from s

quester [9]

Answer:

a) $P=1-\frac{\lambda}{\mu}=1-\frac{20}{30}=0.33$ and that represent the 33%

b) $p_x =\frac{\lambda}{\mu}=\frac{20}{30}=0.66$

c) $L_s =\frac{20}{30-20}=\frac{20}{10}=2 people$

d) $L_q =\frac{20^2}{30(30-20)}=1.333 people$

e) $W_s =\frac{1}{\lambda -\mu}=\frac{1}{30-20}=0.1hours$

f) $W_q =\frac{\lambda}{\mu(\mu -\lambda)}=\frac{20}{30(30-20)}=0.0667 hours$

Step-by-step explanation:

Notation

P represent the probability that the employee is idle

$p_x$ represent the probability that the employee is busy

$L_s$ represent the average number of people receiving and waiting to receive some information

$L_q$ represent the average number of people waiting in line to get some information

$W_s$ represent the average time a person seeking information spends in the system

$W_q$ represent the expected time a person spends just waiting in line to have a question answered

This an special case of Single channel model

Single Channel Queuing Model. "That division of service channels happen in regards to number of servers that are present at each of the queues that are formed. Poisson distribution determines the number of arrivals on a per unit time basis, where mean arrival rate is denoted by λ".

Part a

Find the probability that the employee is idle

The probability on this case is given by:

In order to find the mean we can do this:

$\mu = \frac{1question}{2minutes}\frac{60minutes}{1hr}=\frac{30 question}{hr}$

And in order to find the probability we can do this:

$P=1-\frac{\lambda}{\mu}=1-\frac{20}{30}=0.33$ and that represent the 33%

Part b

Find the proportion of the time that the employee is busy

This proportion is given by:

$p_x =\frac{\lambda}{\mu}=\frac{20}{30}=0.66$

Part c

Find the average number of people receiving and waiting to receive some information

In order to find this average we can use this formula:

$L_s= \frac{\lambda}{\lambda -\mu}$

And replacing we got:

$L_s =\frac{20}{30-20}=\frac{20}{10}=2 people$

Part d

Find the average number of people waiting in line to get some information.

For the number of people wiating we can us ethe following formula"

$L_q =\frac{\lambda^2}{\mu(\mu-\lambda)}$

And replacing we got this:

$L_q =\frac{20^2}{30(30-20)}=1.333 people$

Part e

Find the average time a person seeking information spends in the system

For this average we can use the following formula:

$W_s =\frac{1}{\lambda -\mu}=\frac{1}{30-20}=0.1hours$

Part f

Find the expected time a person spends just waiting in line to have a question answered (time in the queue).

For this case the waiting time to answer a question we can use this formula:

$W_q =\frac{\lambda}{\mu(\mu -\lambda)}=\frac{20}{30(30-20)}=0.0667 hours$

6 0

2 years ago

Read 2 more answers

Pleass answer the following​

Pleass answer the following