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Karolina [17]
1 year ago
14

Let P and Q be polynomials with positive coefficients. Consider the limit below. lim x→[infinity] P(x) Q(x) (a) Find the limit i

f the degree of P is less than the degree of Q. 0 (b) Find the limit if the degree of P is greater than the degree of Q.
Mathematics
1 answer:
jenyasd209 [6]1 year ago
3 0

Answer:

If the limit that you want to find is \lim_{x\to \infty}\dfrac{P(x)}{Q(x)} then you can use the following proof.

Step-by-step explanation:

Let P(x)=a_{n}x^{n}+a_{n-1}x^{n-1}+\cdots+a_{1}x+a_{0} and Q(x)=b_{m}x^{m}+b_{m-1}x^{n-1}+\cdots+b_{1}x+b_{0} be the given polinomials. Then

\dfrac{P(x)}{Q(x)}=\dfrac{x^{n}(a_{n}+a_{n-1}x^{-1}+a_{n-2}x^{-2}+\cdots +a_{2}x^{-(n-2)}+a_{1}x^{-(n-1)}+a_{0}x^{-n})}{x^{m}(b_{m}+b_{m-1}x^{-1}+b_{n-2}x^{-2}+\cdots+b_{2}x^{-(m-2)}+b_{1}x^{-(m-1)}+b_{0}x^{-m})}=x^{n-m}\dfrac{a_{n}+a_{n-1}x^{-1}+a_{n-2}x^{-2}+\cdots +a_{2}x^{-(n-2)}+a_{1}x^{-(n-1)})+a_{0}x^{-n}}{b_{m}+b_{m-1}x^{-1}+b_{n-2}x^{-2}+\cdots+b_{2}x^{-(m-2)}+b_{1}x^{-(m-1)}+b_{0}x^{-m}}

Observe that

\lim_{x\to \infty}\dfrac{a_{n}+a_{n-1}x^{-1}+a_{n-2}x^{-2}+\cdots +a_{2}x^{-(n-2)}+a_{1}x^{-(n-1)})+a_{0}x^{-n}}{b_{m}+b_{m-1}x^{-1}+b_{n-2}x^{-2}+\cdots+b_{2}x^{-(m-2)}+b_{1}x^{-(m-1)}+b_{0}x^{-m}}=\dfrac{a_{n}}{b_{m}}

and

\lim_{x\to \infty} x^{n-m}=\begin{cases}0& \text{if}\,\, nm\end{cases}

Then

\lim_{x\to \infty}=\lim_{x\to \infty}x^{n-m}\dfrac{a_{n}+a_{n-1}x^{-1}+a_{n-2}x^{-2}+\cdots +a_{2}x^{-(n-2)}+a_{1}x^{-(n-1)}+a_{0}x^{-n}}{b_{m}+b_{m-1}x^{-1}+b_{n-2}x^{-2}+\cdots+b_{2}x^{-(m-2)}+b_{1}x^{-(m-1)}+b_{0}x^{-m}}=\begin{cases}0 & \text{if}\,\, nm \end{cases}

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The daily dinner bills in a local restaurant are normally distributed with a mean of $28 and a standard deviation of $6. What ar
Degger [83]

Answer:

The minimum value of the bill that is greater than 95% of the bills is $37.87.

Step-by-step explanation:

When the distribution is normal, we use the z-score formula.

In a set with mean \mu and standard deviation \sigma, the zscore of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this question, we have that:

\mu = 28, \sigma = 6

What are the minimum value of the bill that is greater than 95% of the bills?

This is the 95th percentile, which is X when Z has a pvalue of 0.95. So X when Z = 1.645.

Z = \frac{X - \mu}{\sigma}

1.645 = \frac{X - 28}{6}

X - 28 = 6*1.645

X = 37.87

The minimum value of the bill that is greater than 95% of the bills is $37.87.

4 0
2 years ago
A circle with radius of \greenD{1\,\text{cm}}1cmstart color #1fab54, 1, start text, c, m, end text, end color #1fab54 sits insid
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Answer:

8.86cm^2

Step-by-step explanation:

Given the rectangle of 3cm by 4cm and a circle of 1 cm inside it:

The Area of the shaded portion =Area of Rectangle-Area of Circle

Area of a Rectangle=Length X Width=3 X 4 =12cm^2

Area of the circle =\pi r^2=\pi X 1^1=3.142cm^2

Area  of the shaded portion=12-3.142

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6 0
1 year ago
A botanist collected one leaf at random from each of 10 randomly selected mature maple trees of the same species. The mean and t
Zarrin [17]

Answer:

One sample t-test for population mean would be the most appropriate method.

Step-by-step explanation:

Following is the data which botanist collected and can use:

  • Sample mean
  • Sample Standard Deviation
  • Sample size (Which is 10)
  • Distribution is normal

We have to find the best approach to construct the confidence interval for one-sample population mean. Two tests are used for constructing the confidence interval for one-sample population mean. These are:

  • One-sample z test for population mean
  • One-sample t test for population mean

One sample z test is used when the distribution is normal and the population standard deviation is known to us. One sample t test is used when the distribution is normal, population standard deviation is unknown and sample standard deviation is known.

Considering the data botanist collected, One-sample t test would be the most appropriate method as we have all the required data for this test. Using any other test will result in flawed intervals and hence flawed conclusions.

Therefore, One-sample t-test for population mean would be the most appropriate method.

8 0
2 years ago
Ivan and Tanya share £150 in the ratio 4:1 how much more Ivan has then tanya
Alexeev081 [22]

Hey!

--------------------------------------------------------

Solution:

Ratio is 4/1

Add.

4 + 1 = 5

Divide.

150 / 5 = 30

Multiply for Ivan.

30 x 4 = 120

30 x 1 = 30

Check.

120 + 30 = 150

Subtract.

120 - 30 = 90

--------------------------------------------------------

Answer:

Ivan has £90 more than Tanya has!

--------------------------------------------------------

Hope This Helped! Good Luck!

3 0
1 year ago
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Answer:

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Step-by-step explanation:

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