Use the remainder term to find the minimum order of the Taylor polynomial, centered at 0, that is required to approximate the following quantity with an absolute error no greater than 10^-2.

√1.06.
n>= __________

Answer:

V = 32/3 pi cm^3

Step-by-step explanation:

The volume of a sphere is given by

V = 4/3 pi r^3

The diameter is 4 so the radius is d/2 = 4/2 = 2

V = 4/3  pi (2)^3

V = 32/3 pi cm^3

Answer:

12024.7

Step-by-step explanation:

Searched it up.

then rounded

Answer:

a) 0.2

b) 0.0009

c) 0.0298

d) 0.0465 = 4.65% probability that the sample proportion is less than 0.15.

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal Probability Distribution

Problems of normal distributions can be solved using the z-score formula.

In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the z-score of a measure X is given by:

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

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 p-value, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem establishes that, for a normally distributed random variable X, with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]s = \frac{\sigma}{\sqrt{n}}[/tex].

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

For a proportion p in a sample of size n, the sampling distribution of the sample proportion will be approximately normal with mean [tex]\mu = p[/tex] and standard deviation [tex]s = \sqrt{\frac{p(1-p)}{n}}[/tex]

20% of customers entering her store make a purchase.

This means that [tex]p = 0.2[/tex]

180 people

This means that [tex]n = 180[/tex]

a. What is the mean of the distribution of the sample proportion of customers making a purchase?

By the Central Limit Theorem, [tex]\mu = p = 0.2[/tex].

b) What is the variance of the sample proportion?

The standard deviation is:

[tex]s = \sqrt{\frac{0.2*0.8}{180}} = 0.0298[/tex]

Variance is the square of the standard deviation, so:

[tex]s^2 = (0.0298)^2 = 0.0009[/tex]

c) What is the standard error of the sample proportion?

As found in the previous item, 0.0298.

d) What is the probability that the sample proportion is less than 0.15?

This is the p-value of Z when X = 0.15. So

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

By the Central Limit Theorem

[tex]Z = \frac{X - \mu}{s}[/tex]

[tex]Z = \frac{0.15 - 0.20}{0.0298}[/tex]

[tex]Z = -1.68[/tex]

[tex]Z = -1.68[/tex] has a p-value of 0.0465.

0.0465 = 4.65% probability that the sample proportion is less than 0.15.

Answer:

2

Step-by-step explanation:

8 ^ (5/3) ^ 1/5

We know a^b^c = a^(b*c)

8^ (5/3*1/5)

8^ 1/3

Rewriting 8 as 2^3

2^3 ^1/3

2 ^(3*1/3)

2^1

2

Answer:

2

Step-by-step explanation:

((2^3)^5/3)^1/5

= (2^5)^1/5

= 2

Answered by Gauthmath

Answer:

(a)The probability that both marbles are blue=5/18

The probability that both marbles are yellow=1/6

(b)The probability that exactly one marble is blue​=5/9

Step-by-step explanation:

Blue marbles=5

Yellow marbles=4

Total marbles=5+4=9

(a)

Probability of drawing first  blue marble=5/9

Probability of drawing second blue marble without replacement=4/8

The probability that both marbles are blue

[tex]=\frac{5}{9}\times \frac{4}{8}=\frac{5}{18}[/tex]

Probability of drawing first  yellow marble=4/9

Probability of drawing second yellow marble without replacement=3/8

The probability that both marbles are yellow

[tex]=\frac{4}{9}\times \frac{3}{8}=\frac{1}{6}[/tex]

(b)

The probability that exactly one marble is blue​

=Probability of first blue marble (Probability of second yellow marble)+Probability of first yellow marble (Probability of second blue marble)

The probability that exactly one marble is blue​

=[tex]\frac{5}{9}\times \frac{4}{8}+\frac{4}{9}\times \frac{5}{8}[/tex]

=[tex]\frac{5}{18}+\frac{5}{18}[/tex]

=[tex]\frac{10}{18}=\frac{5}{9}[/tex]

Step-by-step explanation:

X= - 1

X=0

X=4/3

X=3

We solve for x by simplifying both sides of the equation, then isolate the variable.

Answer :

C (x=4/3)

[tex]\boxed{ \sf{Answer}} [/tex]

[tex] \sf \: 3(3x - 4) - 5(x + 3) \\ \sf =( 3 \times 3x) - (3 \times 4) + ( - 5 \times x) +( - 5 \times 3 ) \\ \sf = 9x - 12 - 5x - 15 \\ \sf = 9x - 5x - 12 - 15 \\ = \underline{ \bf 4x - 27}[/tex]

ʰᵒᵖᵉ ⁱᵗ ʰᵉˡᵖˢ

꧁❣ ʀᴀɪɴʙᴏᴡˢᵃˡᵗ2²2² ࿐

[tex]\\ \sf\longmapsto 3(3x-4)-5(x+3)[/tex]

[tex]\\ \sf\longmapsto 9x-12-5x-15[/tex]

[tex]\\ \sf\longmapsto 9x-5x-15-12[/tex]

[tex]\\ \sf\longmapsto (9-5)x-27[/tex]

[tex]\\ \sf\longmapsto 4x-27[/tex]

9514 1404 393

Answer:

Step-by-step explanation:

Let s represent the number of orders Scott served. Then we have Joe served 3s, and Laura served (s+9). The total of orders served is ...

  3s +s +(s +9) = 104

  5s = 95 . . . . . . . . . . . subtract 9 and collect terms

  s = 19 . . . . . . . . . divide by 5

  3s = 3×19 = 57

  s+9 = 19+9 = 28

Joe served 57 orders, Scott served 19, and Laura served 28 orders.

Answer:

the first option because I took the test

Answer:

i believe it'll cost 200 dollars

Answer:

was assigned with this problem (the reference text is attached):

Which of the following, if included in a student's paper, would NOT be an example of plagiarism?

1. In the game of baseball, which is rather boring, the batter stands on home base (Hughes 1).

2. Baseball is rather surprisingly known as "America's Favorite Pastime."

3. Baseball is "a rather boring sport played between two teams of nine players" (Hughes 1).

4. All of these are plagiarism.

The answer tells that only the third choice is NOT a plagiarism. My question is, why is the first choice a plagiarismwas assigned with this problem (the reference text is attached):

Which of the following, if included in a student's paper, would NOT be an example of plagiarism?

1. In the game of baseball, which is rather boring, the batter stands on home base (Hughes 1).

2. Baseball is rather surprisingly known as "America's Favorite Pastime."

3. Baseball is "a rather boring sport played between two teams of nine players" (Hughes 1).

4. All of these are plagiarism.

The answer tells that only the third choice is NOT a plagiarism. My question is, why is the first choice a plagiarism

Step-by-step explanation:

was assigned with this problem (the reference text is attached):

Which of the following, if included in a student's paper, would NOT be an example of plagiarism?

1. In the game of baseball, which is rather boring, the batter stands on home base (Hughes 1).

2. Baseball is rather surprisingly known as "America's Favorite Pastime."

3. Baseball is "a rather boring sport played between two teams of nine players" (Hughes 1).

4. All of these are plagiarism.

The answer tells that only the third choice is NOT a plagiarism. My question is, why is the first choice a plagiarism

Answer:

x = 38.4

Step-by-step explanation:

tan(38) = 30/x

x = 30/tan(38)

x = 38.4

Answered by GAUTHMATH

It looks like the integral in polar coordinates is given to be

[tex]\displaystyle\int_{\pi/2}^\pi \int_0^2 r^3\sin(\theta)\cos(\theta)\,\mathrm dr\,\mathrm d\theta[/tex]

Converting back to Cartesian, we take

x = r cos(θ)

y = r sin(θ)

dx dy = r dr dθ

so we can easily recover the integrand in Cartesian:

[tex]r^3\sin(\theta)\cos(\theta)\,\mathrm dr\,\mathrm d\theta = (r\sin(\theta))(r\cos(\theta))(r\,\mathrm dr\,\mathrm d\theta) = xy\,\mathrm dx\,\mathrm dy[/tex]

This leaves us with the limits:

• π/2 ≤ θ ≤ π corresponds to the second quadrant of the (x, y)-plane (that is, where x < 0 and y > 0)

• 0 ≤ r ≤ 2 correspond to the disk of radius 2 centered at the origin

Taken together, we see the region of integration is a quarter-disk of radius 2 in the second quadrant, which we can capture by the set

{(x, y) : -√2 ≤ x ≤ 0 and 0 ≤ y ≤ √(2 - x ²)}

So, in Cartesian coordinates, the integral would be

[tex]\displaystyle \boxed{\int_{-\sqrt2}^0 \int_0^{\sqrt{2-x^2}} xy\,\mathrm dy\,\mathrm dx}[/tex]

Answer:

$199.29

Step-by-step explanation:

Total payments = $7,174.24

Total interest = $1,174.24

From the data given, we estimate the population mean and population standard deviation. Then, we use this estimate to find a 95% confidence interval for the population variance and the population standard deviation.

Sample:

Salaries in millions of dollars: 2.2, 1.5, 0.5, 1.3, 2.4, 1.5, 2.7, 0.3, 2.0, 0.3

Question a:

The mean is the sum of all values divided by the number of values. So

[tex]\overline{x} = \frac{2.2 + 1.5 + 0.5 + 1.3 + 2.4 + 1.5 + 2.7 + 0.3 + 2.0 + 0.3}{10} = 1.42[/tex]

The sample mean salary is of 1.42 million.

Question b:

The standard deviation is the square root of the difference squared between each value and the mean, divided by one less than the number of values.

So

[tex]s = \sqrt{\frac{(2.2-1.42)^2 + (1.5-1.42)^2 + (0.5-1.42)^2 + (1.3-1.42)^2 + (2.4-1.42)^2 + (1.5-1.42)^2 + (2.7-1.42)^2 + ...}{9}} = 0.8772[/tex]

Thus, the estimate for the population standard deviation is of 0.8772 million.

Question c:

The sample size is [tex]n = 10[/tex]

The significance level is [tex]\alpha = 1 - 0.05 = 0.95[/tex]

The estimate, which is the sample standard deviation, is of [tex]s = 0.8772[/tex].

Now, we have to find the critical values for the Pearson distribution. They are:

[tex]\chi^2_{\frac{\alpha}{2},n-1} = \chi^2_{0.025,9} = 19.0228[/tex]

[tex]\chi^2_{1-\frac{\alpha}{2},n-1} = \chi^2_{0.975,9} = 2.7004[/tex]

The confidence interval for the population variance is:

[tex]\frac{(n-1)s^2}{\chi^2_{\frac{\alpha}{2},n-1}} < \sigma^2 < \frac{(n-1)s^2}{\chi^2_{1-\frac{\alpha}{2},n-1}}[/tex]

[tex]\frac{9*0.8772^2}{19.0228} < \sigma^2 < \frac{9*0.8772^2}{2.7004}[/tex]

[tex]0.3641 < \sigma^2 < 2.5646[/tex]

Thus, the 95% confidence interval for the population variance is (0.3641, 2.5646)

Question d:

Standard deviation is the square root of variance, so:

[tex]\sqrt{0.3641} = 0.6034[/tex]

[tex]\sqrt{2.5646} = 1.6014[/tex]

The 95% confidence interval for the population standard deviation is (0.6034, 1.6014).

For more on confidence intervals for population mean/standard deviation, you can check https://brainly.com/question/13807706

Answer:

540 miles/hr and 50 miles/hr respectively

Step-by-step explanation:

Let the speed of plane in still air be x and the speed of wind be y.

ATQ, (x+y)*7.5=4425 and (x-y)*7.5=3675. Solving it, we get x=540 and y=50

Answer:

option D is the answer

Step-by-step explanation:

using the HH,ll,ha and la,

where h is the hypotenuse and l is the leg

Answer:

A. 5

General Formulas and Concepts:

Pre-Algebra

Order of Operations: BPEMDAS

Algebra I

Algebra II

Step-by-step explanation:

Step 1: Define

Identify points from graph.

Point (8, 5)

Point (4, 2)

Step 2: Find distance d

Simply plug in the 2 coordinates into the distance formula to find distance d

Answer:

the write answer to your question is tan 25 degree

With some rewriting, you get

[tex]\displaystyle \sum_{k=0}^\infty (-1)^k\frac{x^{k+2}}{4^k} = x^2 \sum_{k=0}^\infty \left(-\frac x4\right)^k[/tex]

Recall that for |x| < 1, you have

[tex]\displaystyle \frac1{1-x} = \sum_{k=0}^\infty x^k[/tex]

So as long as |-x/4| = |x/4| < 1, or |x| < 4, your series converges to

[tex]\displaystyle x^2 \sum_{k=0}^\infty \left(-\frac x4\right)^k = \frac{x^2}{1-\left(-\frac x4\right)} = \frac{x^2}{1+\frac x4} = \boxed{\frac{4x^2}{4+x}}[/tex]

Based on known expressions from Taylor series, the power series [tex]\sum \limits_{k = 0}^{\infty} (-1)^{k}\cdot \frac{x^{k+2}}{4^{k}}[/tex]Taylor series-derived formula of the rational function [tex]\frac{4\cdot x^{2}}{4+x}[/tex].

Taylor series are polynomic approximations used to estimate values both from trascendental and non-trascendental functions. It is commonly used in trigonometric, potential, logarithmic and even rational functions.

In this question we must use series properties and common Taylor series-derived formulas to infer the expression behind the given series. Now we proceed to find the expression:

[tex]\sum \limits_{k = 0}^{\infty} (-1)^{k}\cdot \frac{x^{k+2}}{4^{k}}[/tex]

[tex]x^{2}\cdot \sum\limits_{k = 0}^{\infty} \left(-\frac{x}{4} \right)^{k}[/tex]

[tex]x^{2}\cdot \left(\frac{1}{1+\frac{x}{4} } \right)[/tex]

[tex]\frac{4\cdot x^{2}}{4+x}[/tex]

Based on power and series properties and most common Taylor series- derived formulas, the power series [tex]\sum \limits_{k = 0}^{\infty} (-1)^{k}\cdot \frac{x^{k+2}}{4^{k}}[/tex] represents a Taylor series-derived formula of the rational function [tex]\frac{4\cdot x^{2}}{4+x}[/tex]. [tex]\blacksquare[/tex]

To learn more on Taylor series, we kindly invite to check this verified question: https://brainly.com/question/12800011

Answer:

4^15

Step-by-step explanation:

We know a^b^c = a^(b*c)

4^3^5

4^(3*5)

4^15

(1) D

[tex]L_s\left\{t\right\} = \displaystyle\int_0^\infty te^{-st}\,\mathrm dt[/tex]

Integrate by parts, taking

[tex]u = t \implies \mathrm du=\mathrm dt[/tex]

[tex]\mathrm dv = e^{-st}\,\mathrm dt \implies v=-\dfrac1se^{-st}[/tex]

Then

[tex]L_s\left\{t\right\} = \displaystyle \left[-\frac1ste^{-st}\right]\bigg|_{t=0}^{t\to\infty}+\frac1s\int_0^\infty e^{-st}\,\mathrm dt[/tex]

[tex]L_s\left\{t\right\} = \displaystyle \frac1s\int_0^\infty e^{-st}\,\mathrm dt[/tex]

[tex]L_s\left\{t\right\} = \displaystyle -\frac1{s^2}e^{-st}\bigg|_{t=0}^{t\to\infty}[/tex]

[tex]L_s\left\{t\right\} = \displaystyle \boxed{\frac1{s^2}}[/tex]

(2) A

[tex]L_s\left\{1\right\} = \displaystyle\int_0^\infty e^{-st}\,\mathrm dt[/tex]

[tex]L_s\left\{1\right\} = \displaystyle\left[-\frac1se^{-st}\right]\bigg|_{t=0}^{t\to\infty}[/tex]

[tex]L_s\left\{1\right\} = \displaystyle\boxed{\frac1s}[/tex]

========================================================

Explanation:

You can use the AAS (angle angle side) theorem to prove that triangle ABD is congruent to triangle CBD.

From there, we can then say that AD and DC are the same length

AD = DC

3y+6 = 5y-18

3y-5y = -18-6

-2y = -24

y = (-24)/(-2)

y = 12

9514 1404 393

Answer:

  A. 7√6/2

Step-by-step explanation:

The side ratios of the 30-60-90 triangle are 1 : √3 : 2. This means the horizontal line segment is 7√3.

The side ratios of the 45-45-90 triangle are 1 : 1 : √2. This means ...

  x = (horizontal segment)/√2 = (√2)/2 × 7√3 = (7/2)√(2·3)

  x = 7√6/2

Answer:

set notation _ A set is denoted or represented by a capital letter and enclosed in a curly bracket For example {A,B,P,Q}.

Answer:

[tex]Pr= \frac{1}{21}[/tex]

Step-by-step explanation:

Given

[tex]n(S) = 7[/tex] --- number of games

Required

Probability of bike and movie in consecutive order

This probability is represented as:

[tex]Pr = P(Bike\ and\ Movie) \ or\ P(Movie\ or\ Bike)[/tex]

So, we have:

[tex]Pr = P(Bike\ and\ Movie) \ +\ P(Movie\ or\ Bike)[/tex]

The probability is an illustration of selection without replacement;

So, we have:

[tex]P(Bike\ and\ Movie) = P(Bike) * P(Movie)[/tex]

[tex]P(Bike\ and\ Movie) = \frac{n(Bike)}{n(S)} * \frac{n(Movie)}{n(S) - 1}[/tex] ---- without replacement

Bike and Movie appear in the game list 1 time.

So, the equation becomes

[tex]P(Bike\ and\ Movie) = \frac{1}{7} * \frac{1}{7 - 1}[/tex]

[tex]P(Bike\ and\ Movie) = \frac{1}{7} * \frac{1}{6}[/tex]

[tex]P(Bike\ and\ Movie) = \frac{1}{42}[/tex]

Similarly,

[tex]P(Movie\ and\ Bike) = \frac{1}{42}[/tex]

So, we have:

[tex]Pr = P(Bike\ and\ Movie) \ +\ P(Movie\ or\ Bike)[/tex]

[tex]Pr= \frac{1}{42}+\frac{1}{42}[/tex]

Take LCM

[tex]Pr= \frac{1+1}{42}[/tex]

[tex]Pr= \frac{2}{42}[/tex]

[tex]Pr= \frac{1}{21}[/tex]

Answer:

x = 8/3

Step-by-step explanation:

Log₂(9x) – Log₂3 = 3

The value of x can be obtained as follow:

Log₂(9x) – Log₂3 = 3

Recall

Log M – Log N = Log (M/N)

Thus,

Log₂(9x) – Log₂3 = 3

Log₂(9x/3) = 3

Log₂3x = 3

3x = 2³

3x = 8

Divide both side by 3

x = 8/3

Answer:

a. 168000 for Eyeconnect, 132000 for Topconnect

b. 100,000

Step-by-step explanation:

a.

Because the change in customers are only due to leaving companies, we can say that, after one year, Eyeconnect loses 10% of its customers to Topconnect and Topconnect loses 5% of its customers to Eyeconnect. This represents all changes in customers.

First, we can calculate how much Eyeconnect loses, which is 10% of 180,000 = 0.1 * 180,000 = 18,000 . They then have 180,000 - 18,000 = 162,000 employees

Next, Topconnect loses 120,000 * 5% = 120,000 * 0.05 = 6,000. They then have 120,000-6,000 = 114,000 employees

We can then add the customer amounts. Note that we are subtracting both sides before adding as both companies gain and lose customers simultaneously.

We can then add how much one company lost to the other company's customers.

Eyeconnect gains 6,000 customers, so they then have 162,000 + 6,000 = 168000 employees. Topconnect gains 18,000 customers so they then have 114,000 + 18,000 = 132,000 employees

b.

After many years, the number of customers Eyeconnect has will be less than the number of customers that Topconnect has. One way to find the end amount of customers that Eyeconnect has is to figure out when the customer bases even out, or when Eyeconnect loses the same amount of customers as Topconnect so the customer base stays the exact same. We know that no customers leave or join the companies except to leave/join the other, so the total amount of customers between the two companies stays the exact same. The amount of customers is 180,000 + 120,000 = 300,000. Therefore, at the end amount,

Eyeconnect customers (E) + Topconnect customers (T) =300,000

Furthermore, if the amount of customers that leave Eyeconnect is the same that leaves Topconnect, we can say

E * 0.1 = T * 0.05

divide both sides by 0.05 to isolate the T

E * 0.1 / 0.05 = T

2 * E = T

plug that into the first equation

E + 2 * E = 300,000

3 * E = 300,000

divide both sides by 3 to isolate E

E = 100,000 after many years

[tex]Hello[/tex] [tex]There[/tex]

The answer is...

[tex]5.[/tex]

[tex]HopeThisHelps!![/tex]

[tex]AnimeVines[/tex]