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forsale [732]
3 years ago
5

Graphing rational functions. I need help asap!!!!

Mathematics
1 answer:
Anastaziya [24]3 years ago
6 0

Answer:

1.  

Vert. asymptote: x = {-3, 2}

Horiz. asymptote: y = 0

x-int: None

Question 3.

a. There is no hole

b. Vert. asymptote: x = {-2, 2}

c. f(x) = 0: x = {0, -1/2}

d. The graph has no hole at (-2, 4)

Question 4.

a. Vert. asymptote: x = {-2, 2}

b. f(x) = 0: x = {0, -1/2}

c. Horizontal asymptote: y = 2

d. The graph has no hole

I'm a bit confused.  Some of the things stated in the question aren't true like how there are holes in places where there aren't.

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Customers arrive at a restaurant according to a Poisson distribution at a rate of 20 customers per hour. The restaurant opens fo
jenyasd209 [6]

Answer:

0.087

Step-by-step explanation:

Given that there were 17 customers at 11:07, probability of having 20 customers in the restaurant at 11:12 am could be computed as:

= Probability of having 3 customers in that 5 minute period. For every minute period, the number of customers coming can be modeled as:

X₅ ~ Poisson (20 (5/60))

X₅ ~ Poisson (1.6667)

Formula for computing probabilities for Poisson is as follows:

P (X=ₓ) = ((<em>e</em>^(-λ)) λˣ)/ₓ!

P(X₅= 3) = ((<em>e</em>^(-λ)) λˣ)/ₓ! =  (e^-1.6667)((1.6667²)/3!)

P(X₅= 3) = (2.718^(-1.6667))((2.78)/6)

P(X₅= 3) = (2.718^(-1.6667))0.46

P(X₅= 3) = 0.1889×0.46

P(X₅= 3) = 0.086894

P(X₅= 3) = 0.087

Therefore, the probability of having 20 customers in the restaurant at 11:12 am given that there were 17 customers at 11:07 am is 0.087.

8 0
4 years ago
fter production, a computer component is given a quality score of A, B, and C. On the average U of the components were given a q
RoseWind [281]

Answer:

Follows are the solution to this question:

Step-by-step explanation:

In the given question some of the data is missing so, its correct question is defined in the attached file please find it.

Let

A is quality score of A

B is quality score of B

C is quality score of C

\to P[A] =0.55\\\\\to P[B] =0.28\\\\\to P[C] =0.17\\

Let F is a value of the content so, the value is:

\to P[\frac{F}{A}] =0.15\\\\\to P[\frac{F}{B}] =0.12\\\\\to P[\frac{F}{C}] =0.14\\

Now, we calculate the tooling value:

\to p[\frac{C}{F}]

using the baues therom:

\to p[\frac{C}{F}] =  \frac{p[C] \times p[\frac{F}{C} ]}{p[A] \times p[\frac{F}{A}] + p[B] \times p[\frac{F}{B}]+p[C] \times p[\frac{F}{C}] }  

            =  \frac{ 0.17  \times 0.14 }{0.55 \times 0.15 + 0.28 \times 0.12 + 0.17 \times 0.14  } \\\\=  \frac{0.0238}{0.0825 + 0.0336 + 0.0238} \\\\=  \frac{0.0238}{0.1399} \\\\=0.1701

7 0
3 years ago
2 (h-8)-h = h-16 what is the solution
Svetach [21]
Let's solve your equation step-by-step.<span><span><span>2<span>(<span>h−8</span>)</span></span>−h</span>=<span>h−16</span></span>Step 1: Simplify both sides of the equation.<span><span><span>2<span>(<span>h−8</span>)</span></span>−h</span>=<span>h−16</span></span><span>Simplify: (Show steps)</span><span><span>h−16</span>=<span>h−16</span></span>Step 2: Subtract h from both sides.<span><span><span>h−16</span>−h</span>=<span><span>h−16</span>−h</span></span><span><span>−16</span>=<span>−<span>16
</span></span></span>Step 3: Add 16 to both sides.<span><span><span>−16</span>+16</span>=<span><span>−16</span>+16</span></span><span>0=0</span>Answer:<span>All real numbers are solutions.</span>
6 0
3 years ago
V (t)=32t <br> If a rock dropped off a bridge, how fast will it be falling after 3 seconds
kondor19780726 [428]
V(t) = 32t

when t = 3 seconds,

V = 32*3

V = 96

So velocity will be 96 units/s     after 3 seconds.
6 0
3 years ago
Please help me on this question
viva [34]

I am stuck on same thing as u

Step-by-step explanation:

3 0
3 years ago
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