All categories

3 years ago

How do you do this question with the given power series?

Mathematics

1 answer:

Sunny_sXe [5.5K]3 years ago

8 0

Step-by-step explanation:

tan⁻¹(x) = ∑ₙ₌₀°° (-1)ⁿ x²ⁿ⁺¹ / (2n+1)

tan⁻¹(1/√3) = ∑ₙ₌₀°° (-1)ⁿ (1/√3)²ⁿ⁺¹ / (2n+1)

tan⁻¹(1/√3) = ∑ₙ₌₀°° (-1)ⁿ (1/√3) (1/√3)²ⁿ / (2n+1)

tan⁻¹(1/√3) = (1/√3) ∑ₙ₌₀°° (-1)ⁿ (1/3)ⁿ / (2n+1)

π/6 = (1/√3) ∑ₙ₌₀°° (-1)ⁿ (1/3)ⁿ / (2n+1)

π = (6/√3) ∑ₙ₌₀°° (-1)ⁿ (1/3)ⁿ / (2n+1)

π = 2√3 ∑ₙ₌₀°° (-1)ⁿ / (3ⁿ (2n+1))

You might be interested in

Annual starting salaries in a certain region of the U. S. for college graduates with an engineering major are normally distribut

defon

Answer:

0.8665 = 86.65% probability that the sample mean would be at least $39000

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 $\mu$ and standard deviation $\sigma$ , the z-score 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 p-value, 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 normally distributed random variable X, with mean $\mu$ and standard deviation $\sigma$ , the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$ .

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

Mean $39725 and standard deviation $7320.

This means that $\mu = 39725, \sigma = 7320$

Sample of 125:

This means that $n = 125, s = \frac{7320}{\sqrt{125}} = 654.72$

The probability that the sample mean would be at least $39000 is about?

This is 1 subtracted by the pvalue of Z when X = 39000. So

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

By the Central Limit Theorem

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

$Z = \frac{39000 - 39725}{654.72}$

$Z = -1.11$

$Z = -1.11$ has a pvalue of 0.1335

1 - 0.1335 = 0.8665

0.8665 = 86.65% probability that the sample mean would be at least $39000

4 0

3 years ago

How can you use addition and subtraction to put together and separate measures of an angle and its parts?

Oksanka [162]

We can use addition and subtraction to put together and separate measures of an angle and its parts by looking for the complementary or the supplementary angles.

For example, What is the other angle in a complementary angle in which one of the angles is 45 degrees. This is how should be answered. Ccomplementary angles are two angles that forms a total of 90 degrees.
90 = A + B
90 = 45 + B
B = 90 - 45
B = 45

7 0

3 years ago

Choose two values, a and b, each between 8 and 15. Show how to use the identity a^3+b^3=(a+b)(a^2-ab+b^2) to calculate the sum o

qaws [65]

Answer:

Step-by-step explanation:

a = 8

b = 10

a^3 =8^3 = 512

b^3= 10^3 =1000

a^3 + b^3 = 1512

a^2 = 64

-ab = - 80

b^2 = 100

a + b=18

(a + b) (64 - 80+ 100)

18 * (84)

1512'

5 0

3 years ago

Hello i'll attached them Okay

Phantasy [73]

So megan is four years younger than her aunt

m+4=a
or
m=a-4
the sum of their ages is at least 40
m+a>40
subsitute a-4 for m
a-4+a>40
combine like terms
2a-4>40
the answer is the third from the top

3 0

3 years ago

Which expression is equivalent to StartFraction (5 a b) cubed Over 30 a Superscript negative 6 Baseline b Superscript negative 7

m_a_m_a [10]

Answer:

Option D.

Step-by-step explanation:

The given expression is

$\dfrac{(5ab)^3}{30a^{-6}b^{-7}}$