All categories

3 years ago

What is the limit of f (×) as x approaches - infinty

Mathematics

1 answer:

tatyana61 [14]3 years ago

5 0

If the degree of numerator and denominator are equal, then limit will be leading coefficient of numerator divided by the leading coefficient of denominator.

So then the limit would be 3/1 = 3.

Alternatively,

$\displaystyle \lim_{x\to\infty}\dfrac{3x^2+6}{x^2-4}=\displaystyle \lim_{x\to\infty}\dfrac{3x^2+6}{x^2-4}\cdot\dfrac{1/x^2}{1/x^2}=\lim_{x\to\infty}\dfrac{3+\frac6{x^2}}{1-\frac4{x^2}} = \dfrac{3+0}{1-0}=\boxed{3}$

Hope this helps.

You might be interested in

A rectangular park is 100 yards long and 65 yards wide. Give the length and width of another rectangular park that has the same

Akimi4 [234]

100 by 65 = 330 perimeter = 6,500 area
90 by 75 = 330 perimeter = 6,750 area

5 0

3 years ago

Which is bigger 5/9 or 11/20

Lina20 [59]

5/9 = 0.56
11/20 = 55/100 = 55%
So that means 5/9ths is bigger than 11/20ths
Hope this helped, trust me i know best:)

7 0

3 years ago

The table below shows the price in dollars for the number of roses indicated is the price proportional to the number of roses ex

makvit [3.9K]

Answer:

Price of roses is proportional to the number of roses.

Step-by-step explanation:

Let the price of roses are proportional to the number of roses.

Equation representing this phenomenon will be,

P = k(R) ⇒ k = $\frac{P}{R}$

where 'P' = price of the roses

R = Number of roses

k = Proportionality constant

If we substitute the values of P and R and we get the same value of constant 'k', then the equation will be true.

For R = 3 and P = $9

k = $\frac{9}{3}$

k = 3

For R = 6 and R = 18

k = $\frac{18}{6}$

k = 3

K is same in both the conditions, therefore, Price of roses are proportional to the number of roses.

6 0

3 years ago

A sanitation department is interested in estimating the mean amount of garbage per bin for all bins in the city. In a random sam

soldi70 [24.7K]

Answer:

a) $52.8-2.02\frac{3.9}{\sqrt{40}}=51.554$

b) $52.8+2.02\frac{3.9}{\sqrt{40}}=54.046$

c) $52.8-2.71\frac{3.9}{\sqrt{40}}=51.129$

d) $52.8+2.71\frac{3.9}{\sqrt{40}}=54.471$

e) Yes, depends of the confidence level.

At 95 % of confidence the value 54.1985 pounds is not included on the interval. At 5% of significance the statement is FALSE.

At 99 % of confidence the value 54.1985 pounds is included on the interval. So at 1% of significance the statement is TRUE.

Step-by-step explanation:

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

$\bar X =52.8$ represent the sample mean for the sample

$\mu$ population mean (variable of interest)

s=3.9 represent the sample standard deviation

n=40 represent the sample size

a) What is the lower limit of the 95% interval? Give your answer to three decimal places

The confidence interval for the mean is given by the following formula:

$\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}$ (1)

In order to calculate the mean and the sample deviation we can use the following formulas:

In order to calculate the critical value $t_{\alpha/2}$ we need to find first the degrees of freedom, given by:

$df=n-1=40-1=39$

Since the Confidence is 0.95 or 95%, the value of $\alpha=0.05$ and $\alpha/2 =0.025$ , and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.025,39)".And we see that $t_{\alpha/2}=2.02$

Now we have everything in order to replace into formula (1):

$52.8-2.02\frac{3.9}{\sqrt{40}}=51.554$

b) What is the upper limit of the 95% interval? Give your answer to three decimal places

$52.8+2.02\frac{3.9}{\sqrt{40}}=54.046$

So on this case the 95% confidence interval would be given by (51.554;54.046)

c) What is the lower limit of the 99% interval? Give your answer to three decimal places

Since the Confidence is 0.99 or 99%, the value of $\alpha=0.01$ and $\alpha/2 =0.005$ , and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.005,39)".And we see that $t_{\alpha/2}=2.71$

Now we have everything in order to replace into formula (1):

$52.8-2.71\frac{3.9}{\sqrt{40}}=51.129$

d) What is the upper limit of the 99% interval? Give your answer to three decimal places

$52.8+2.71\frac{3.9}{\sqrt{40}}=54.471$

So on this case the 99% confidence interval would be given by (51.129;54.471)

e) Consider the claim that the mean amount of garbage per bin is 54.1985 pounds. Is the following statement true or false? The decision about the claim would depend on whether we use a 95% or 99% confidence interval: True/False

Yes, depends of the confidence level.

At 95 % of confidence the value 54.1985 pounds is not included on the interval. At 5% of significance the statement is FALSE.

At 99 % of confidence the value 54.1985 pounds is included on the interval. So at 1% of significance the statement is TRUE.

3 0

3 years ago

What is the slope of the line y=2x+5?

sweet [91]

Answer:

The slope is 2

Explanation:

The equation is written in y=mx+b
M would always be the slope and b the y intercept

Hope this helps!!

4 0

2 years ago

Read 2 more answers