Mathematics: What is the end behavior of the function f(x) = −3x4−x3 + 2x2 + 4x + 5? select one: a. up on the left, up on the right b. up on the left, down on the right c. down on the left, up on the right d. down on the left, down on the right

What is the end behavior of the function f(x)

Ответ:

evan1v1123

26.06.2021

Mathematics

a) P=0.353

b) P=0.950

Step-by-step explanation:

If the time between calls is exponentially distributed with a mean rate of t minutes, we can say that the expected number of calls per unit of time follows a Poisson distribution with parameter 1/t.

In this case, the parameter for the Poisson distribution is:

$r=\frac{1}{10}=0.1\, min^{-1}$

The Poisson distribution for k amount of calls in a period ot t minutes is described as:

$P(k,t)=\frac{(rt)^ke^{-rt}}{k!}$

The probability of having more than 3 calls in on-half hour (30 min) is:

$P(k3;t=30)=1-(P(0)+P(1)+P(2)+P(3))\\\\\\ \lambda=rt=0.1*30=3\\\\P(0)=\frac{(3)^0e^{-3}}{0!} =\frac{1*0.0498}{1}=0.050\\\\P(1)=\frac{(3)^1e^{-3}}{1!}=\frac{3*0.0498}{1} = 0.149\\\\ P(2)=\frac{(3)^2e^{-3}}{2!}=\frac{9*0.0498}{2} = 0.224\\\\P(3)=\frac{(3)^3e^{-3}}{3!}=\frac{27*0.0498}{6} =0.224\\\\\\ P(k3)=1-(0.050+0.149+0.224+0.224)\\\\P(k3)=1-0.647=0.353$

b) These distribution are memory-less, so they are independent of the past results.

We can calculate then the probability of having at least on call in the next half hour as:

$P(k1;t=30)=1-P(0)\\\\\\ \lambda=rt=0.1*30=3\\\\P(0)=\frac{(3)^0e^{-3}}{0!} =\frac{1*0.0498}{1}=0.050\\\\\\ P(k1)=1-0.050=0.950$

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What is the end behavior of the function f(x) = −3x4−x3 + 2x2 + 4x + 5? select one: a. up on the left, up on the right b. up on the left, down on the right c. down on the left, up on the right d. down on the left, down on the right

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