Cooper just started a running plan where he runs 8 miles the first week and then

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

jenyasd209 [6]

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

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

3 0

3 years ago

A sample of 300 urban adult residents of a particular state revealed 65 who favored increasing the highway speed limit from 55 t

forsale [732]

Answer and explanation:

Null hypothesis(H0) says sentiment for increasing speed limit in the two populations is the same

Alternative hypothesis(Ha) says sentiment for increasing speed limit in the two populations is different

Where p1 is the first population proportion =65/297= 0.22

And p2 is the second population proportion = 78/189= 0.41

P= p1+p2/n1+n2= 65+78/297+189= 0.29

Hence H0= p1-p2=0

Ha=p1-p2≠0

Test statistic= p1-p2/√p(1-p) (1/n1+1/n2)

= 0.22-0.41/√0.29(1-0.29)(1/297+1/189)

= -4.49

Critical value at 95% significance level= 1.96( from tables)

We therefore reject null hypothesis as critical value is greater than test statistic

Therefore the sentiment for increasing speed limit in the two populations is different

b. at proportions of 0.24 and 0.40 for p1 and p2 respectively

Test statistic = 0.24-0.40/0.29(1-0.29)(1/297+1/189)

= -3.78

P value is 0.0002 at 0.05 significance level

Hence probability =0.4998

7 0

3 years ago

See attached picture

lions [1.4K]

Answer:

ok i see it

Step-by-step explanation:

4 0

3 years ago

Please help!!! will mark brainliest!! Hurryyyy

brilliants [131]

Answer:

cjtsitdtiugxjfzursitdpyzot dig totxitzitzitzktzkfzifzifzkfxofzktzkfz

Step-by-step explanation:

7 0

3 years ago

HELP WILL GIVE BRAINLIEST

Ganezh [65]

Answer:

2 boom

Step-by-step explanation:

2

4 0

2 years ago

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