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3 years ago

Answer to math class

Mathematics

1 answer:

Fantom [35]3 years ago

5 0

1/3 of 12,000 ballots were thrown out.

1/3 * 12,000 = 12,000/3 = 4,000

4,000 ballots were thrown out.

12,000 - 4,000 = 8,000

There were 8,000 ballots left.

1/4 of 8,000 ballots are for candidate A.

1/4 * 8,000 = 8,000/4 = 2,000

Candidate A received 2,000 votes.

8,000 - 2,000 = 6,000

Candidate B received 6,000 votes.

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The world's largest marathon, The New York City Marathon, is held on the first Sunday in November each year. Between 2 million a

e-lub [12.9K]

Answer:

25% is the percent error.

Step-by-step explanation:

Given:

The world's largest marathon, The New York City Marathon, is held on the first Sunday in November each year. Between 2 million and 2.5 million spectators will line the streets to cheer on the marathon runners.

Now, to find the percent error.

As given, between 2 million and 2.5 million spectators will line the streets to cheer on the marathon runners.

The actual measure = 2 million.

The estimated measure = 2.5 million.

So, the error = $2.5-2=0.5\ million.$

Now, to get the percent error:

$\frac{0.5}{2} \times 100.$

$=0.25\times 100$

$=25\%.$

Therefore, 25% is the percent error.

7 0

3 years ago

This problem asks for Taylor polynomials forf(x) = ln(1 +x) centered at= 0. Show Your work in an organized way.(a) Find the 4th,

stich3 [128]

Answer:

a) The 4th degree , 5th degree and sixth degree polynomials

$f^{lV} (x) = \frac{(2(-3))}{(1+x)^4} (1)= \frac{((-1)^3(3!))}{(1+x)^4}$

$f^{V} (x) = \frac{(2(-3)(-4))}{(1+x)^5} =\frac{(-1)^4 (4!)}{(1+x)^5}$

$f^{V1} (x) = \frac{(-120))}{(1+x)^6} (1) = \frac{(-1)^5 5!}{(1+x)^6}$

b)The nth degree Taylor polynomial for f(x) centered at x = 0, in expanded form.

$log(1+x) = x - \frac{x^2}{2} +\frac{x^3}{3} - \frac{x^4}{4} + \frac{x^5}{5} - \frac{x^6}{6}+\\.. (-1)^{n-1}\frac{x^n}{n} +..$

Step-by-step explanation:

Given the polynomial function f(x) = log(1+x) …...(1) centered at x=0

f(x) = log(1+x) ……(1)

using formula $\frac{d}{dx} logx =\frac{1}{x}$

Differentiating Equation(1) with respective to 'x' we get

$f^{l} (x) = \frac{1}{1+x} (\frac{d}{dx}(1+x)$

$f^{l} (x) = \frac{1}{1+x} (1)$ ….(2)

At x= 0

$f^{l} (0) = \frac{1}{1+0} (1)= 1$

using formula $\frac{d}{dx} x^{n-1} =nx^{n-1}$

Again Differentiating Equation(2) with respective to 'x' we get

$f^{l} (x) = \frac{-1}{(1+x)^2} (\frac{d}{dx}((1+x))$

$f^{ll} (x) = \frac{-1}{(1+x)^2} (1)$ ….(3)

At x=0

$f^{ll} (0) = \frac{-1}{(1+0)^2} (1)= -1$

Again Differentiating Equation(3) with respective to 'x' we get

$f^{lll} (x) = \frac{(-1)(-2)}{(1+x)^3} (\frac{d}{dx}((1+x))$

$f^{lll} (x) = \frac{(-1)(-2)}{(1+x)^3} (1)= \frac{(-1)^2 (2)!}{(1+x)^3}$ ….(4)

At x=0

$f^{lll} (0) = \frac{(-1)(-2)}{(1+0)^3} (1)$

$f^{lll} (0) = 2$

Again Differentiating Equation(4) with respective to 'x' we get

$f^{lV} (x) = \frac{(2(-3))}{(1+x)^4} (\frac{d}{dx}((1+x))$

$f^{lV} (x) = \frac{(2(-3))}{(1+x)^4} (1)= \frac{((-1)^3(3!))}{(1+x)^4}$ ....(5)

$f^{lV} (0) = \frac{(2(-3))}{(1+0)^4}$

$f^{lV} (0) = -6$

Again Differentiating Equation(5) with respective to 'x' we get

$f^{V} (x) = \frac{(2(-3)(-4))}{(1+x)^5} (\frac{d}{dx}((1+x))$

$f^{V} (x) = \frac{(2(-3)(-4))}{(1+x)^5} =\frac{(-1)^4 (4!)}{(1+x)^5}$ .....(6)

At x=0

$f^{V} (x) = 24$

Again Differentiating Equation(6) with respective to 'x' we get

$f^{V1} (x) = \frac{(2(-3)(-4)(-5))}{(1+x)^6} (\frac{d}{dx}((1+x))$

$f^{V1} (x) = \frac{(-120))}{(1+x)^6} (1) = \frac{(-1)^5 5!}{(1+x)^6}$

and so on...

The nth term is

$f^{n} (x) = = \frac{(-1)^{n-1} (n-1)!}{(1+x)^n}$

Step :-2

Taylors theorem expansion of f(x) is

$f(x) = f(a) + \frac{x}{1!} f^{l}(x) +\frac{(x-a)^2}{2!}f^{ll}(x)+\frac{(x-a)^3}{3!}f^{lll}(x)+\frac{(x-a)^4}{4!}f^{lV}(x)+\frac{(x-a)^5}{5!}f^{V}(x)+\frac{(x-a)^6}{6!}f^{VI}(x)+...….. \frac{(x-a)^n}{n!}f^{n}(x)$

At x=a =0

$f(x) = f(0) + \frac{x}{1!} f^{l}(0) +\frac{(x)^2}{2!}f^{ll}(0)+\frac{(x)^3}{3!}f^{lll}(0)+\frac{(x)^4}{4!}f^{lV}(0)+\frac{(x)^5}{5!}f^{V}(0)+\frac{(x)^6}{6!}f^{VI}(0)+...….. \frac{(x-0)^n}{n!}f^{n}(0)$

Substitute all values , we get

$f(x) = f(0) + \frac{x}{1!} (1) +\frac{(x)^2}{2!}(-1)+\frac{(x)^3}{3!}(2)+\frac{(x)^4}{4!}(-6)+\frac{(x)^5}{5!}(24)+\frac{(x)^6}{6!}(-120)+...….. \frac{(x-0)^n}{n!}f^{n}(0)$

On simplification we get

$log(1+x) = x - \frac{x^2}{2} +\frac{x^3}{3} - \frac{x^4}{4} + \frac{x^5}{5} - \frac{x^6}{6}+\\.. (-1)^{n-1}\frac{x^n}{n} +..$

4 0

3 years ago

The breaking strength of a rivet has a mean value of 10,000 psi and a standard deviation of 500 psi. (a) What is the probability

velikii [3]

Answer:

a) 89.05% probability that the sample mean breaking strength for a random sample of 40 rivets is between 9900 and 10,200

b) No, because one of the requirements of the central limit theorem is a sample size of at least 30.

Step-by-step explanation:

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

Normal probability distribution:

Problems of normally distributed samples are solved using 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.

Central limit theorem:

The Central Limit Theorem estabilishes that, for a random variable X, with mean $\mu$ and standard deviation $\sigma$ , the sample means with size n of at least 30 can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$

In this problem, we have that:

$\mu = 10000, \sigma = 500$

(a) What is the probability that the sample mean breaking strength for a random sample of 40 rivets is between 9900 and 10,200?

Here we have $n = 40, s = \frac{500}{\sqrt{40}} = 79.06$

This probability is the pvalue of Z when X = 10200 subtracted by the pvalue of Z when X = 9900. So

X = 10200

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

By the Central Limit Theorem

$Z = \frac{X - \mu}{s}$

$Z = \frac{10200 - 10000}{79.06}$

$Z = 2.53$

$Z = 2.53$ has a pvalue of 0.9943.

X = 9900

$Z = \frac{X - \mu}{s}$

$Z = \frac{9900 - 10000}{79.06}$

$Z = -1.26$

$Z = -1.26$ has a pvalue of 0.1038.

0.9943 - 0.1038 = 0.8905

89.05% probability that the sample mean breaking strength for a random sample of 40 rivets is between 9900 and 10,200

(b) If the sample size had been 15 rather than 40, could the probability requested in part (a) be calculated from the given information?

No, because one of the requirements of the central limit theorem is a sample size of at least 30.

3 0

3 years ago

The short sides of a rectangle are 2 inches. The long sides of the same rectangle are three less than a certain number of inches

Georgia [21]

So the “certain number” will be S.
Since 2width + 2length, the first part would be 2 times 2 which is 4
The length is S - 3
We then have to multiply this by 2
So it is 2(S-3) which is 2S-6
So the answer is 4+2S-6 which is 2S-2

8 0

4 years ago

A recipe used 2/3 cup of sugar for every 2 teaspoon of vanilla. How much sugar was used per teaspoon of vanilla

Brums [2.3K]

The correct answer would be 1/3 cup of sugar for every 1 teaspoon of vanilla, so A.

Hope this helps :)

3 0

3 years ago

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