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3 years ago

How do you do this question?

Mathematics

1 answer:

gregori [183]3 years ago

8 0

Answer:

½√2

Step-by-step explanation:

∑ₙ₌₀°° (-1)ⁿ π²ⁿ / (4²ⁿ (2n)!)

∑ₙ₌₀°° (-1)ⁿ (π/4)²ⁿ / (2n)!

Using the Taylor expansion for cos x:

cos (π/4)

½√2

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Pls help... Prove (1+tanx)/(1+cotx)=tanx by algebra

katen-ka-za [31]

$\text{L.H.S}\\\\=\dfrac{1+\tan x}{ 1+ \cot x}\\\\\\=\dfrac{1+ \tan x}{ 1+ \tfrac 1{\tan x}}\\\\\\=\dfrac{1+\tan x}{\tfrac{\tan x+1}{\tan x}}\\\\\\=(1+\tan x) \times \dfrac{\tan x}{1+ \tan x}\\\\=\tan x\\\\=\text{R.H.S}\\\\\text{Proved.}$

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2 years ago

3. (02.01)

sashaice [31]

Answer:

A. 1

Step-by-step explanation:

3(x+1)=-2(x-1)+6

3x+3=2x-2+6

x+3=4

x=1

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3 years ago

Please help!!!! attached picture!!!

butalik [34]

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3 years ago

A population of 1,750 cheetahs decreases by 10% per year. How many cheetahs will there be in the population after 11 years? Roun

Triss [41]

Answer:

The answer is C. 546.

If a population decreases by 11%, that means that 89% (100% - 11% = 89%) of cheetahs remains each number. 89% can be expressed as 0.89, so to calculate the change of the population, we must each year multiply the number of cheetahs by 0.89.

After 1 year: 1750 * 0.89 ≈ 1558

After 2 years: 1558 * 0.89 ≈ 1387

After 3 years: 1387 * 0.89 ≈ 1234

After 4 years: 1234 * 0.89 ≈ 1098

After 5 years: 1098 * 0.89 ≈ 977

After 6 years: 977 * 0.89 ≈ 870

After 7 years: 870 * 0.89 ≈ 774

After 8 years: 774 * 0.89 ≈ 689

After 9 years: 689 * 0.89 ≈ 613

After 10 years: 613 * 0.89 ≈ 546

Step-by-step explanation:

4 0

3 years ago

Location is known to affect the number, of a particular item, sold by an auto parts facility. Two different locations, A and B,

Mama L [17]

We have two samples, A and B, so we need to construct a 2 Samp T Int using this formula:

$\displaystyle \overline {x}_1 - \overline {x}_2 \ \pm \ t^{*} \sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2} }$

In order to use t*, we need to check conditions for using a t-distribution first.

Random for both samples -- NOT STATED in the problem ∴ proceed with caution!
Independence for both samples: 130 < all items sold at Location A; 180 < all items sold at Location B -- we can reasonably assume this is true
Normality: CLT is not met; n < 30 for both locations A and B ∴ proceed with caution!

Since 2/3 conditions aren't met, we can still proceed with the problem but keep in mind that the results will not be as accurate until more data is collected or more information is given in the problem.

Solve for t*:

We need the tail area first.

$\displaystyle \frac{1-.9}{2}= .05$

Next we need the degree of freedom.

The degree of freedom can be found by subtracting the degree of freedom for A and B.

The general formula is df = n - 1.

df for A: 13 - 1 = 12
df for B: 18 - 1 = 17
df for A - B: |12 - 17| = 5

Use a calculator or a t-table to find the corresponding t-score for df = 5 and tail area = .05.

t* = -2.015

Now we can use the formula at the very top to construct a confidence interval for two sample means.

$\overline {x}_A=39$
$s_A=8$
$n_A=13$
$\overline {x}_B = 55$
$s_B=2$
$n_B=18$
$t^{*}=-2.015$

Substitute the variables into the formula: $\displaystyle \overline {x}_1 - \overline {x}_2 \ \pm \ t^{*} \sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2} }$ .

$39-55 \ \pm \ -2.015 \big{(}\sqrt{\frac{(8)^2}{13} +\frac{(2)^2}{18} } } \ \big{)}$

Simplify this expression.

$-16 \ \pm \ -2.015 (\sqrt{5.1453} \ )$
$-16 \ \pm \ 3.73139$

Adding and subtracting 3.73139 to and from -16 gives us a confidence interval of:

$(-20.5707,-11.4293)$

If we want to interpret the confidence interval of (-20.5707, -11.4293), we can say...

We are 90% confident that the interval from -20.5707 to -11.4293 holds the true mean of items sold at locations A and B.

5 0

1 year ago